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a) -21 + (4 - x) = -17 + (-20) + 5
=>-21 + 4 - x = -32
=> -17 - x = -32
=> x = -17 + 32
=> x = 15
b) (15 - x) - (+9) = 34 - (-31)
=> 15 - x - 9 = 34 + 31
=> 6 - x = 65
=> x = 6 - 65
=> x = -59
c) (17 + x) - (-12) = -14 - (-10)
=> 17 + x + 12 = -14 + 10
=> 29 + x = -4
=> x = -4 - 29
=> x = -33
c) (x + 24) + 15 = 8 - 17
=> x + 24 + 15 = -9
=> x + 39 = -9
=> x = -9 - 39
=> x = -48
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) \(\left(x-17\right)\left(x+15\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=17\\x=-15\end{matrix}\right.\)
b) \(\left(6-x\right)\left(x-35\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=6\\x=35\end{matrix}\right.\)
a) \(\left|x\right|+\left|-23\right|=\left|17\right|\)
\(\left|x\right|+23=17\)
\(\left|x\right|=17-23\)
\(\left|x\right|=-6\)
=> x ko có giá trị
b) \(\left|-5\right|.\left|x\right|+20\)
sai đề
\(\text{(−1)+(−3)+...+(−199)+(−201)(−1)+(−3)+...+(−199)+(−201)}\)
=\(\text{−(1+3+...+199+201)=−(1+3+...+199+201)}\)
=\(\dfrac{\left(201+1\right).\left[\left(201-1\right)\right]:2+1}{2}\)
= \(\dfrac{-200.102}{2}=\dfrac{-20400}{2}=-10200\)
\(\text{17 + ( − 20 ) + 23 + ( − 26 ) + . . . + 53 + ( − 56 ) = [ 17 + ( − 20 ) ] + [ 23 + ( − 26 ) ] + . . . + [ 53 + ( − 56 ) ] = ( − 3 ) + ( − 3 ) + . . . + ( − 3 ) = ( − 3 ) . ( 7 ) = − 21}\)
\(\text{=17 + ( − 20 ) + 23 + ( − 26 ) + . . . + 53 + ( − 56 ) = [ 17 + ( − 20 ) ] + [ 23 + ( − 26 ) ] + . . . + [ 53 + ( − 56 ) ] = ( − 3 ) + ( − 3 ) + . . . + ( − 3 ) = ( − 3 ) . ( 7 ) = − 21}\)
\(\text{ = ( − 3 ) + ( − 3 ) + . . . + ( − 3 )}\)
\(\text{= ( − 3 ) . ( 7 ) = − 21}\)
\(a,\left(x+3\right)\left(5-x\right)=0\\ \Rightarrow\left\{{}\begin{matrix}x+3=0\\5-x=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-3\\x=5\end{matrix}\right.\)
\(c,x+17⋮x+3\\ x+3+14⋮x+3\\ 14⋮x+3\\ x+3\inƯ\left(14\right)=\left\{\pm14;\pm7\pm2;\pm1\right\}\)
Từ đó bạn tìm những giá trị của x nha!