2019-04-03

【数学１】数と式--実数

数学1 数学1-数と式

実数

実数

今回のテーマ

今回のテーマは，実数です．

有理数と無理数
絶対値
平方根
分数の有理化

を扱っています．

授業動画はこちらをご覧ください

www.youtube.com

演習問題

${\fbox{1} \ 次の分数を循環小数に，また循環小数を分数にせよ。\\ ( 1 )\ \dfrac{1}{3}\\ ( 2 )\ \dfrac{7}{110}\\ ( 3 )\ 0.\dot{2}\dot{5}\\ ( 4 )\ 1.\dot{2}0\dot{4}\\ }$

${\fbox{2} \ 次の計算をせよ。\\ ( 1 )\ \sqrt{18}\times\sqrt{40}\\ ( 2 )\ 5\sqrt{8}-\sqrt{98}+\sqrt{32}\\ ( 3 )\ (2\sqrt{2}-\sqrt{3})(\sqrt{2}+2\sqrt{3})\\ ( 4 )\ (1+\sqrt{2}-\sqrt{3})^2\\ }$

${\fbox{3} \ 次の式を簡単にせよ。\\ ( 1 )\ \dfrac{1}{2\sqrt{2}}\\ ( 2 )\ \dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{2}}\\ ( 3 )\ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\\ ( 4 )\ \dfrac{1}{3-\sqrt{2}+\sqrt{7}}\\ }$

${\fbox{4} \ 次の二重根号をはずせ。\\ ( 1 )\ \sqrt{6+2\sqrt{5}}\\ ( 2 )\ \sqrt{7-2\sqrt{12}}\\ ( 3 )\ \sqrt{6+\sqrt{32}}\\ ( 4 )\ \sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\\ }$

CheatSheet

f:id:brian_tee:20190403222218p:plain — cheat sheet

演習解答

${\fbox{1}\\ ( 1 )\ \dfrac{1}{3} = 0.\dot{3}\\ ( 2 )\ \dfrac{7}{110} = 0.0\dot{6}\dot{3}\\ ( 3 )\ 0.\dot{2}\dot{5}\\ \begin{align} \ x &= 0.2525...\ と置くと，\\ \ 100x &= 25.2525...\ なので，\\ \ 99x &= 25\\ \ x &= \dfrac{25}{99}\\ ( 4 )\ 1.\dot{2}0\dot{4}\\ \ x &= 1.204204...\ と置くと，\\ \ 1000x &= 1204.204204...\ なので，\\ \ 999x &= 1203\\ \ x &= \dfrac{1203}{999}\\ \ &= \dfrac{401}{333}\\ \end{align} \\ }$

${\fbox{2} \ \\ ( 1 )\ \sqrt{18}\times\sqrt{40}\\ \ = 3\sqrt{2}\times2\sqrt{10}\\ \ = 6\sqrt{20}\\ \ = 12\sqrt{5}\\ ( 2 )\ 5\sqrt{8}-\sqrt{98}+\sqrt{32}\\ \ =10\sqrt{2}-7\sqrt{2}+4\sqrt{2}\\ \ =7\sqrt{2}\\ ( 3 )\ (2\sqrt{2}-\sqrt{3})(\sqrt{2}+2\sqrt{3})\\ \ = 4-6+4\sqrt{6}-\sqrt{6}\\ \ = 3\sqrt{6}-2\\ ( 4 )\ (1+\sqrt{2}-\sqrt{3})^2\\ \ = 1+2+3+2\sqrt{2}-2\sqrt{3}-2\sqrt{6}\\ \ = 6+2\sqrt{2}-2\sqrt{3}-2\sqrt{6}\\ }$

${\fbox{3} \\ ( 1 )\ \dfrac{1}{2\sqrt{2}}\\ \ = \dfrac{1}{2\sqrt{2}}\times\dfrac{\sqrt{2}}{\sqrt{2}}\\ \ =\dfrac{\sqrt{2}}{4}\\ ( 2 )\ \dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{2}}\\ \ =\dfrac{\sqrt{5}}{\sqrt{5}-\sqrt{2}}\times\dfrac{\sqrt{5}+\sqrt{2}}{\sqrt{5}+\sqrt{2}}\\ \ =\dfrac{\sqrt{5}(\sqrt{5}-\sqrt{2})}{(\sqrt{5})^2-(\sqrt{2})^2}\\ \ = \dfrac{5-\sqrt{10}}{3}\\ ( 3 )\ \dfrac{2+\sqrt{3}}{2-\sqrt{3}}\\ \ =\dfrac{2+\sqrt{3}}{2-\sqrt{3}}\times\dfrac{2+\sqrt{3}}{2+\sqrt{3}}\\ \ =\dfrac{(2+\sqrt{3})^2}{2^2-(\sqrt{3})^2}\\ \ =4+3+4\sqrt{3}\\ \ =7+4\sqrt{3}\\ ( 4 )\ \dfrac{1}{3-\sqrt{2}+\sqrt{7}}\\ \ =\dfrac{1}{3-(\sqrt{2}-\sqrt{7})}\times\dfrac{3+(\sqrt{2}-\sqrt{7})}{3+(\sqrt{2}-\sqrt{7})}\\ \ =\dfrac{3+(\sqrt{2}-\sqrt{7})}{3^2-(\sqrt{2}-\sqrt{7})^2}\\ \ =\dfrac{3+(\sqrt{2}-\sqrt{7})}{9-(9-2\sqrt{14})}\\ \ =\dfrac{3+(\sqrt{2}-\sqrt{7})}{2\sqrt{14}}\times\dfrac{\sqrt{14}}{\sqrt{14}}\\ \ =\dfrac{3\sqrt{14} +2\sqrt{7} -7\sqrt{2}}{28}\\ }$

${\fbox{4} \ \\ ( 1 )\ \sqrt{6+2\sqrt{5}}\\ \ =\sqrt{5+1+2\sqrt{5}}\\ \ =\sqrt{(\sqrt{5}+1)^2}\\ \ =\sqrt{5}+1\\ ( 2 )\ \sqrt{8-2\sqrt{12}}\\ \ =\sqrt{6+2 -2\sqrt{12}}\\ \ =\sqrt{(\sqrt{6}-\sqrt{2})^2}\\ \ =\sqrt{6}-\sqrt{2}\ (\because \sqrt{6}\gt \sqrt{2})\\ ( 3 )\ \sqrt{6+\sqrt{32}}\\ \ = \sqrt{4+2+2\sqrt{8}}\\ \ =\sqrt{(2+\sqrt{2})^2}\\ \ =2+\sqrt{2}\\ ( 4 )\ \sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}\\ \ =\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}\times\dfrac{2-\sqrt{3}}{2-\sqrt{3}}}\\ \ = \sqrt{\dfrac{(2-\sqrt{3})^2}{2^2-(\sqrt{3})^2}}\\ \ =2-\sqrt{3}\ (\because 2\gt \sqrt{3})\\ }$

2019-04-03

【演習１】因数分解

数学1 数学1-数と式演習

【演習第１回】因数分解

【演習第１回】因数分解

演習問題

$\fbox{1} \ 因数分解しなさい。\\ ( 1 )\ x^2 y -4y^2 z -4 y^3 + x^2 z \\( 2 )\ x^4 + x^2 + 1 - 2xy - y^2 \\( 3 )\ xy + (x+1)(y+1)(xy+1)$

$\fbox{2} \ 次の問いに答えなさい。\\ ( 1 )\ \sqrt{x^2} + \sqrt{x^2 - 2x + 1}\ の根号を外しなさい。 \\( 2 ) \ 6 -2\sqrt{2}\ の整数部を\ a，小数部を\ b\ とするとき，a^3 - (b^3 + \displaystyle \frac{1}{b^3})\ の値を求めよ。$

動画はこちら

www.youtube.com

解説

$\fbox{1}$
$( 1 ) \ x^2 y -4y^2 z -4 y^3 + x^2 z\\\ =y(x^2-4y^2)+z(x^2-4y^2)\\\ =(y+z)(x+2y)(x-2y)\\$
$( 2 )\ x^4 + x^2 + 1 - 2xy - y^2\\\ =x^4 + 2x^2 + 1 -x^2 - 2xy - y^2\\\ =(x^2 + 1)^2 - (x + y)^2\\\ =(x^2+x+y+1)(x^2-x-y+1)\\$
$( 3 )\ xy + (x+1)(y+1)(xy+1)\\\ =xy + (xy + x + y + 1)(xy + 1)\\\ = x^2 y^2+x^2 y + xy^2+3xy+x+y+1\\\ =(y^2 + y)x^2 +(y^2 + 3y + 1)x + y + 1\\\ = y(y+1)x^2 + (y^2 + 3y + 1)x +(y+1)\\\ = (x(y+1) + 1)(xy + y + 1)\\\ = (xy + x + 1)(xy + y + 1)\\$
$\fbox{別解}\\( 3 )\ xy + (x+1)(y+1)(xy+1)\\\ =xy + (xy + x + y + 1)(xy + 1)\\\ =(xy + 1)^2 +(x + y)(xy + 1) + xy\\\ =(xy + x + 1)(xy + y + 1)\\$

$\fbox{2}$
$( 1 )\ \sqrt{x^2} + \sqrt{x^2 - 2x + 1}\ の根号を外しなさい。$

$\sqrt{x^2} = |x| = \begin{cases} x &(x \geqq 0) \\ -x &(x\lt 0) \end{cases}\\ \sqrt{x^2 - 2x +1} = |x-1| = \begin{cases} x - 1 &(x \geqq 1) \\ -x + 1 &(x\lt 1) \end{cases}\\$

$\begin{cases} \ x < 0のとき，\ &(-x) + (1-x) &= 1 - 2x \\ \ 0 \leqq x < 1 のとき，\ &x + (1 - x) &= 1 \\ \ x \geqq 1 のとき，\ &x + x-1 &=2x-1\\ \end{cases}$

$( 2 ) \ 6 -2\sqrt{2}\ の整数部を\ a，小数部を\ b\ とするとき，a^3 - (b^3 + \dfrac{1}{b^3})\ の値を求めよ。$

\begin{array}
\ (2 \sqrt{2}) ^ 2 = 8 より，\\
2^2 \lt ({2 \sqrt{2})^2} \lt3^2 \ なので, \\
2\lt 2{\sqrt{2}} \lt 3 \\
3\lt 6-2{\sqrt{2}} \lt 4
\end{array}
$これより，a = 3，b = 3-2{\sqrt{2}}\ である。$
$また，$
$\ \begin{array} \ \dfrac{1}{b} &= \dfrac{1}{3-2\sqrt{2}} \\ &= \dfrac{1}{3-2\sqrt{2}}\times \dfrac{3+2\sqrt{2}}{3+2\sqrt{2}}\\ &= \dfrac{3+2\sqrt{2}}{3^2-(2\sqrt{2})^2}\\ &= 3+2\sqrt{2}\\ \end{array}$

$であるから，b + \dfrac{1}{b} = 6$
$\ ここで，$
$\begin{array} \ (b + \dfrac{1}{b})^3 &= b^3 + 3b + \dfrac{3}{b} + \dfrac{1}{b^3}\ より，\\ b^3 + \dfrac{1}{b^3} &= (b + \dfrac{1}{b})^3 -3(b + \dfrac{1}{b}) \ なので，\\ \end{array}$

$\begin{array}
\ a^3 - (b^3 + \dfrac{1}{b^3}) &= 3^3 - (6^3 - 3\times 6)\\
&= 27 - 180\\
&= -153
\end{array}$

2019-04-02

【数学１】数と式--複雑な因数分解

数学1 数学1-数と式

複雑な因数分解

複雑な因数分解

今回のテーマ

今回は，複雑な因数分解を扱いました．

たすきがけを必要とする問題
最小次数の文字の式とみなす問題
複二次式

についての解説授業と問題です．

授業動画はこちらをご覧ください

youtu.be

演習問題

${\fbox{1} \ 次の式を展開せよ。\\ ( 1 )\ (x - 3y)^3\\ ( 2 )\ (x +1)(x +2)(x+3)(x+4)\\ ( 3 )\ (xy-z)(x^2y^2 +xyz + z^2)\\ ( 4 )\ (x^3 + y^3)(x-y)(x^2+xy+y^2) }$

${\fbox{2} \ 次の式を因数分解せよ。\\ ( 1 )\ 5x^2 - 17x + 6\\ ( 2 )\ 6x^2 + x - 12\\ ( 3 )\ 4x^2 + 16x + 15\\ ( 4 )\ (a+b)^2 + 6c(a+b) + 9c^2\\ ( 5 )\ 3ax^2 -24a^2x + 36a^3\\ ( 6 )\ (x^2 + x)^2 -8(x^2+x) + 12\\ ( 7 )\ x^2 - y^2 + 10y - 25\\ ( 8 )\ x^4 + 2x^2 + 9\\ ( 9 )\ x^2 + xy -2y^2 +2x +7y -3\\ ( 10 )\ 2x^2 + xy - 6y^2 - 8x -9y + 6\\ }$

演習解答

${\fbox{1} \ \\ ( 1 )\ (x - 3y)^3\\\ =x^3 -9x^2y + 27xy^2 -27y^3\\ ( 2 )\ (x +1)(x +2)(x+3)(x+4)\\ \ = (x^2+5x+4)(x^2+5x+6)\\ \ = (x^2+5x)^2 + 10(x^2+5x) + 24\\ \ = x^4 + 10x^3 +35x^2 + 50x + 24\\ ( 3 )\ (xy-z)(x^2y^2 +xyz + z^2)\\\ =x^3y^3-z^3\\ ( 4 )\ (x^3 + y^3)(x-y)(x^2+xy+y^2)\\ \ =(x^3+y^3)(x^3-y^3)\\ \ =x^6-y^6 }$

${\fbox{2} \ 次の式を因数分解せよ。\\ ( 1 )\ 5x^2 - 17x + 6\\\ =(x-3)(5x-2)\\ ( 2 )\ 6x^2 + x - 12\\\ =(2x+3)(3x-4)\\ ( 3 )\ 4x^2 + 16x + 15\\\ =(2x+3)(2x+5)\\ ( 4 )\ (a+b)^2 + 6c(a+b) + 9c^2\\\ =(a+b+3c)^2\\ ( 5 )\ 3ax^2 -24a^2x + 36a^3\\\ =3a(x-2a)(x-6a)\\ ( 6 )\ (x^2 + x)^2 -8(x^2+x) + 12\\ \ =\{(x^2+x) -2\}\{(x^2+x) -6\}\\ \ =(x-2)(x-1)(x+2)(x+3)\\ ( 7 )\ x^2 - y^2 + 10y - 25\\ \ =x^2 - (y-5)^2\\ \ =(x+y-5)(x-y+5)\\ ( 8 )\ x^4 + 2x^2 + 9\\ \ =x^4 + 6x^2 + 9 -4x^2\\ \ =(x^2 +3)^2 - 4x^2\\ \ =(x^2+2x+3)(x^2-2x+3)\\ ( 9 )\ x^2 + xy -2y^2 +2x +7y -3\\ \ =x^2 + (y+2)x -2y^2 +7y -3\\ \ =x^2 + (y+2)x -(2y -1)(y-3)\\ \ =\{x - (y -3)\}\{x+(2y-1)\}\\ \ =(x-y+3)(x+2y-1)\\ ( 10 )\ 2x^2 + xy - 6y^2 - 8x -9y + 6\\ \ =2x^2+(y-8)x-6y^2-9y+6\\ \ =2x^2+(y-8)x-3(2y-1)(y+2)\\ \ =\{2x -3(y+2)\}\{x+(2y-1)\}\\ \ =(2x-3y-6)(x+2y-1)\\ }$

2019-04-01

【数学１】数と式--展開と因数分解

数学1 数学1-数と式

展開と因数分解

展開と因数分解

今回のテーマ

今回は数と式の導入と展開，因数分解の基本です．
内容は，

降べきの順
指数の計算
展開の基本
展開の工夫
因数分解の基本

です．

授業動画はこちらをご覧ください

youtu.be

演習問題

${\fbox{1} \ 次の式を【\ 】の文字について降べきの順に並べよ。\\ ( 1 )\ 2x^2 + 3xy - y^2 + 4x - 6y +12 【\ x\ 】 \\ ( 2 )\ ab^3 + a^2b - 4b^2 + 3ab^3 -a^3 +4a^3 【\ a\ 】}$

${\fbox{2} \ 次の計算をせよ。\\ ( 1 )\ (-a^2)^3(-2a)^4(-a^3) \\ ( 2 )\ 4x^2y^3 \div (-6x^3y^2) }$

${\fbox{3} \ 次の式を展開せよ。\\ ( 1 )\ (2a - b)(a + 2b)\\ ( 2 )\ (x + 1)(x^2 -2x - 2)\\ ( 3 )\ (x - 6)^2\\ ( 4 )\ (2a + 3b)^2\\ ( 5 )\ (x + y)(x - y)\\ ( 6 )\ (2a + b)(2a - b)\\ ( 7 )\ (a-2b)(a^2 + 2ab + 4b^2)\\ ( 8 )\ (x - y - z)(x + y - z) }$

${\fbox{4} \ 次の式を因数分解せよ。\\ ( 1 )\ 4a^2b - 2ab + 6a\\ ( 2 )\ x^2 - 14x + 49\\ ( 3 )\ 4a^2 - 9b^2\\ ( 4 )\ x^2 + 7x + 12\\ }$

CheatSheet

演習解答

${\fbox{1} \\ ( 1 )\ 2x^2 + 3xy - y^2 + 4x - 6y +12 \\\ = 2x^2 + 3xy + 4x -y^2 -6y +12\\ ( 2 )\ ab^3 + a^2b - 4b^2 + 3ab^3 -a^3 +4a^3\\\ = 3a^3 +a^2b + 4ab^3 - 4b^2 }$

${\fbox{2} \\ ( 1 )\ (-a^2)^3(-2a)^4(-a^3) \\\ = 16a^{13}\\ ( 2 )\ 4x^2y^3 \div (-6x^3y^2)\\\ = -\dfrac{2y}{3x} }$

${\fbox{3} \ 次の式を展開せよ。\\ ( 1 )\ (2a - b)(a + 2b)\\\ =2a^2 +3ab -2b^2\\ ( 2 )\ (x + 1)(x^2 -2x - 2)\\\ = x^3 -x^2 -4x -2\\ ( 3 )\ (x - 6)^2\\\ = x^2 -12x + 36\\ ( 4 )\ (2a + 3b)^2\\\ = 4a^2 +12ab + 9b^2\\ ( 5 )\ (x + y)(x - y)\\\ = x^2-y^2\\ ( 6 )\ (2a + b)(2a - b)\\\ = 4a^2 -b^2\\ ( 7 )\ (a-2b)(a^2 + 2ab + 4b^2)\\\ = a^3-8b^3\\ ( 8 )\ (x - y - z)(x + y - z)\\ \ = \{(x-z)-y\}\{(x-z)+y\}\\ \ = (x-z)^2-y^2\\ \ = x^2 -2xz + z^2 -y^2\\ }$

${\fbox{4} \ 次の式を因数分解せよ。\\ ( 1 )\ 4a^2b - 2ab + 6a\\\ = 2a(2ab-b+3)\\ ( 2 )\ x^2 - 14x + 49\\\ =(x-7)^2\\ ( 3 )\ 4a^2 - 9b^2\\\ =(2a+3b)(2a-3b)\\ ( 4 )\ x^2 + 7x + 12\\\ =(x+3)(x+4) }$