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1
37.(43-51)-43.(37-51)
=37.43-37.51-43.37-43.51
=(37.43-43.57)-(37.51-43.51)
=0-(-306)
=306
2
a.(x-1).(2x+2)=0
=>x-1 hoặc 2x+2=0
=>x=1 hoặc 2x=-2
=>x=1 hoặc x=-1
Vậy x=+1
b.(6x-12).(x-3)=0
=>6x-12 hoặc x-3=0
=>6x=12 hoặc x=3
=>x=2 hoặc x=3
Vậy x=2 hoặc x=3
|x|=20
=> x= +2 hoặc x=-2
37-|x|=12
|x|=37-12
|x|=25
=>x=\(\pm25\)
|x|-12=8
|x|=8+12
|x|=20
=>x=\(\pm20\)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
a) ( 5x - 15) .37 =74
5x - 15= 74 : 37
5x - 15 = 2
5x = 2 + 15
5x = 17
x = 17: 5
x= \(\frac{17}{5}\)
1) +) \(\left|x\right|=20\Rightarrow\orbr{\begin{cases}x=20\\x=-20\end{cases}}\)
+) \(37-\left|x\right|=12\)
\(\Rightarrow\left|x\right|=37-12=25\)
\(\Rightarrow\orbr{\begin{cases}x=25\\x=-25\end{cases}}\)
+) \(\left|x\right|-12=8\)
\(\Rightarrow\left|x\right|=8+12=20\Rightarrow\orbr{\begin{cases}x=20\\x=-20\end{cases}}\)
+) \(\left|x\right|+8=\left|-20\right|-12\)
\(\Rightarrow\left|x\right|+8=20-12=8\)
\(\Rightarrow\left|x\right|=8-8=0\)
\(\Rightarrow x=0\)
2) Ta có: \(\left|x\right|\ge0\)
\(\left|y\right|\ge0\)
Mà \(\left|x\right|+\left|y\right|\le0\)
\(\Rightarrow\left|x\right|=\left|y\right|=0\)
\(\Rightarrow x=y=0\)