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a) \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Rightarrow5x=10\)
\(\Leftrightarrow x=2\)
Vậy x = 2
b) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Rightarrow-3x=-48\)
\(\Leftrightarrow x=16\)
Vậy x = 16
c) \(\dfrac{1}{9}=\dfrac{-2x}{10}\)
\(\Rightarrow-18x=10\)
\(\Leftrightarrow x=-\dfrac{5}{9}\)
Vậy \(x=-\dfrac{5}{9}\)
d) ĐKXĐ: \(x\ne0\)
\(\dfrac{3}{x}-5=\dfrac{-9}{x}+2\)
\(\Leftrightarrow\dfrac{3-5x}{x}=\dfrac{-9+2x}{x}\)
\(\Rightarrow3-5x=-9+2x\)
\(\Leftrightarrow7x=12\)
\(\Leftrightarrow x=\dfrac{12}{7}\)
Vậy \(x=\dfrac{12}{7}\)
e) ĐKXĐ: \(x\ne0\)
\(\dfrac{x}{-2}=\dfrac{-8}{x}\)
\(\Rightarrow x^2=16\)
\(\Leftrightarrow x=\pm4\)
Vậy \(x=\pm4\)
a) Ta có: \(\dfrac{x}{5}=\dfrac{2}{5}\)
\(\Leftrightarrow x=\dfrac{2\cdot5}{5}=2\)
Vậy: x=2
b) Ta có: \(\dfrac{3}{-8}=\dfrac{6}{-x}\)
\(\Leftrightarrow-x=\dfrac{6\cdot\left(-8\right)}{3}=-16\)
hay x=16
Vậy: x=16
A. x = 2
B. \(\dfrac{3}{8}=\dfrac{6}{x}\)\(\Leftrightarrow x=\dfrac{6.8}{3}=16\)
C. x = 3
D. \(x=\dfrac{4.6}{8}=3\)
E. \(x=\dfrac{7}{3}\)
G.\(\dfrac{14}{13}=\dfrac{28}{10-x}\)
<=>\(14\left(10-x\right)=364\)
<=> 10 - x = 26
<=> x = -16
H. \(3\left(x+2\right)=4\left(x-5\right)\)
<=> 3x + 6 = 4x - 20
<=> -x = -26
<=> x = 26
K. \(\dfrac{x}{2}=\dfrac{8}{x}\)
<=> \(x^2=16\)
<=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
M. \(\left(x-2\right)^2=100\)
<=> \(\left[{}\begin{matrix}x-2=10\\x-2=-10\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-8\end{matrix}\right.\)
a=2
b=16
c=3
d=3
mik chỉ biết thế này thôi(ko chắc đúng=3)
a) \(2.\left(x+\frac{2}{5}\right)+1\frac{1}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)+\frac{5}{4}=\frac{11}{20}\)
\(2.\left(x+\frac{2}{5}\right)=\frac{-7}{10}\)
\(x+\frac{2}{5}=\frac{-7}{20}\)
\(x=\frac{-13}{20}\)
Vậy \(x=\frac{-13}{20}\)
b)\(x-1\frac{1}{8}-\frac{2}{3}x-\frac{5}{6}x=75\%\)
\(\left(x-\frac{2}{3}x-\frac{5}{6}x\right)-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x-\frac{9}{8}=\frac{3}{4}\)
\(\frac{-1}{2}x=\frac{15}{8}\)
\(x=\frac{-15}{4}\)
Vậy \(x=\frac{-15}{4}\)
a. \(158-x=-12\)
\(x=158+12\)
\(x=170\)
b. \(37+x=12\)
\(x=12-37\)
\(x=-25\)
c. \(2x-15=-47\)
\(2x=\left(-47\right)+15\)
\(x=\dfrac{\left(-32\right)}{2}\)
\(x=-16\)
d. \(\left(-5\right)^2-\left(5x-3\right)=43\)
\(25-\left(5x-3\right)=43\)
\(\left(5x-3\right)=25-43\)
\(5x=\left(-18\right)+3\)
\(x=\dfrac{\left(-15\right)}{5}=-3\)
e. \(\left|x-1\right|+\left(-5\right)=2\)
\(\left|x-1\right|=2-\left(-5\right)\)
\(\left|x-1\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)
f. \(\left|x+1\right|=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
g. \(\left|x\right|=12\)
\(\Rightarrow x=\pm12\)
Câu h ko có x sao tìm
a) \(158-x=-12.\)
\(\Rightarrow x=158-\left(-12\right).\)
\(\Rightarrow x=158+12=170.\)
Vậy..........
b) \(37+x=12.\)
\(\Rightarrow x=12-37.\)
\(\Rightarrow x=-25.\)
Vậy..........
c) \(2x-15=-47.\)
\(\Rightarrow2x=-47+15.\)
\(\Rightarrow2x=-32.\)
\(\Rightarrow x=-\dfrac{32}{2}=-16.\)
Vậy..........
d) \(\left(-5\right)^2-\left(5x-3\right)=43.\)
\(\Rightarrow25-\left(5x-3\right)=43.\)
\(\Rightarrow5x-3=25-43.\)
\(\Rightarrow5x-3=-18.\)
\(\Rightarrow5x=-18+3.\)
\(\Rightarrow5x=-15.\)
\(\Rightarrow x=-\dfrac{15}{5}=3.\)
Vậy..........
e) \(\left|x-1\right|+\left(-5\right)=2.\)
\(\Rightarrow\left|x-1\right|=2-\left(-5\right).\)
\(\Rightarrow\left|x-1\right|=2+5.\)
\(\Rightarrow\left|x-1\right|=7.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right..\)
Vậy..........
f) \(\left|x+1\right|=4\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right..\)
Vậy..........
g) \(\left|x\right|=12\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right..\)
Vậy..........
a) \(120=x^2-1\)
\(\Leftrightarrow x^2=121\)
\(\Rightarrow x=\pm11\)
b) \(5^3-5\left(4+x\right)=15\)
\(\Leftrightarrow25-4-x=3\)
\(\Rightarrow x=18\)
c) \(4^3-\left(x-2\right)^2=52\)
\(\Leftrightarrow\left(x-2\right)^2=12\)
\(\Leftrightarrow\orbr{\begin{cases}x=2-2\sqrt{3}\\x=2+2\sqrt{3}\end{cases}}\)
d) \(11^{3-x}-1=120\)
\(\Leftrightarrow11^{3-x}=121=11^2\)
\(\Rightarrow3-x=2\Rightarrow x=1\)
e) \(4^{3x+2}+8=24\)
\(\Leftrightarrow4^{3x+2}=4^2\)
\(\Rightarrow3x+2=2\Rightarrow x=0\)
f) sửa đề