|x|-9=-2+17
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Lời giải:
$\frac{4}{9}\times \frac{-13}{-17}+\frac{-2}{17}\times \frac{4}{9}+\frac{2}{9}\times \frac{4}{17}$
$=\frac{4}{9}\times (\frac{13}{17}+\frac{-2}{17})+\frac{2}{9}\times \frac{4}{17}$
$=\frac{4}{9}\times \frac{11}{17}+\frac{4}{9}\times \frac{2}{17}$
$=\frac{4}{9}\times (\frac{11}{17}+\frac{2}{17})$
$=\frac{4}{9}\times \frac{13}{17}=\frac{52}{153}$
#)Giải :
\(\frac{2}{9}.\frac{5}{17}+\frac{2}{9}.\frac{13}{17}-\frac{2}{9}.\frac{1}{17}+2018\frac{7}{9}\)
\(=\frac{2}{9}\left(\frac{5}{17}+\frac{13}{17}-\frac{1}{17}\right)+2018\frac{7}{9}\)
\(=\frac{2}{9}.1+2018\frac{7}{9}\)
\(=\frac{2}{9}+2018\frac{7}{9}\)
\(=2018\frac{9}{9}=2019\)
#~Will~be~Pens~#
| x | -9 = -2 +17
| x | -9 = 15
=> x = 24 hoặc x = -24
| x-9 | = -2 +17
| x-9 | = 15
xét | x-9 | = { x-9 nếu x-9 >/ 0 => x >/ 9
{ -(x-9) nếu x-9 < 0 => x < 9
Với x >/ 9 thì:
x-9 = 15
x = 15 + 9
x = 24 ( TM)
Với x < 9 thì :
-(x-9) = 15
-x + 9 = 15
x = 15-9
x = 6 (TM)
Vậy x = 24 hoặc x = -6
nhớ tíck cho mk nha ( nếu đúng) . Thanks !!!!!!!!!!!!
\(\dfrac{9}{17}\times\dfrac{21}{13}+\dfrac{9}{17}\times\dfrac{5}{13}-\dfrac{9}{17}\times2\)
\(=\dfrac{9}{17}\times\left(\dfrac{21}{13}+\dfrac{5}{13}-2\right)\)
\(=\dfrac{9}{17}\times\left(\dfrac{26}{13}-2\right)=\dfrac{9}{17}\times\left(2-2\right)\)
\(=\dfrac{9}{17}\times0=0\)
a) x = 4
b) x = -8
c) | x | - 9 = -2 + 17
| x | = 15 + 9
| x | = 24
x = 24 hoặc x = -24
d) |x – 9| = -2 + 17
|x – 9| = 15
x – 9 = 15 hoặc x – 9 = -15
x = 24 hoặc x = -6
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)
\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)
\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)
\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)
\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)
\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)
\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)
\(\Rightarrow x=-\frac{7}{2}\)
\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)
\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)
\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)
\(\Rightarrow x-5=-8\)
\(\Rightarrow x=-3\)
\(\left|x\right|-9=-2+17\)
\(\left|x\right|-9=15\)
\(\left|x\right|=24\)
=> \(\orbr{\begin{cases}x=24\\x=-24\end{cases}}\)
\(\left|x-9\right|=-2+17\)
\(\left|x-9\right|=15\)
=> \(\orbr{\begin{cases}x-9=15\\x-9=-15\end{cases}}\)
=> \(\orbr{\begin{cases}x=24\\x=-6\end{cases}}\)
ĐKXĐ: ...
Đặt \(\left\{{}\begin{matrix}u=v\\v=\sqrt{17-x^2}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}u+v+uv=9\\u^2+v^2=17\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}uv=9-\left(u+v\right)\\\left(u+v\right)^2-2uv=17\end{matrix}\right.\)
\(\Rightarrow\left(u+v\right)^2+2\left(u+v\right)-35=0\)
\(\Rightarrow\left[{}\begin{matrix}u+v=5\Rightarrow uv=4\\u+v=-7\Rightarrow uv=16\end{matrix}\right.\)
\(\Rightarrow...\)
\(\left|x\right|-9=-2+17\)
\(\Leftrightarrow\left|x\right|-9=15\)
\(\Leftrightarrow\left|x\right|=15+9\)
\(\Leftrightarrow\left|x\right|=24\)
\(\Rightarrow x=24\)HOẶC \(x=-24\)
|x|-9= -2+17
|x|-9= 15
|x| = 15 + 9
|x| = 24
x = -24
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