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a) 3x - 2 = 0 => 3x = 2 => x = 2/3
b) 2x - 1 = 0 => 2x = 1 => x = 1/2
c) 5 ( 4+2x) = 8+5x
<=> 20 + 10x = 8 + 5x
<=> 10x - 5x = 8 - 20
<=> 5x = -12
x = -12/5
d) \(\frac{1}{2}+\frac{3}{4}x=6-\frac{4}{5}x\)
\(\frac{3}{4}x+\frac{4}{5}x=6-\frac{1}{2}\)
\(\frac{31}{20}x=\frac{11}{2}\)
\(x=\frac{11}{2}:\frac{31}{20}=\frac{110}{31}\)
e) 3 + 2x = 4 - 8x
<=> 2x + 8x = 4 - 3
10 x = 1
x = 1/10
f \(5+\frac{1}{2}\left(x+5\right)=3\)
\(\frac{1}{2}\left(x+5\right)=3-5=-2\)
\(x+5=-2:\frac{1}{2}=-4\)
\(x=-4-5=1\)
Vậy ......

\(\Leftrightarrow45-8x=13\left(because45x^0=45.1=45\right)\)
\(\Leftrightarrow8x=32\)
\(\Leftrightarrow x=4\)
= 45-8x=13(because45x mũ 0=45.1=45
=8x=8.4=32
= x=4
tick mình đi

a)
\(\left|x\right|-2\left|x\right|+3\left|x\right|=16+6\left|x\right|-19\)
\(\left|x\right|-2\left|x\right|+3\left|x\right|-6\left|x\right|=16-19\)
\(\left|x\right|.\left(1-2+3-6\right)=-3\)
\(\left|x\right|.\left(-4\right)=-3\)
\(\left|x\right|=\dfrac{3}{4}\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-\dfrac{3}{4}\\x=\dfrac{3}{4}\end{matrix}\right.\)
b,
2.(|x| - 5) - 15 = 9
\(2.\left(\left|x\right|-5\right)=9+15\)
\(2.\left(\left|x\right|-5\right)=24\)
\(\left|x\right|-5=24:2\)
\(\left|x\right|-5=12\)
\(\left|x\right|=12+5\)
\(\left|x\right|=17\)
\(\Rightarrow\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=-17\\x=17\end{matrix}\right.\)
c,
|8 - 2x| + |4y - 16| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|8-2x\right|=0\\\left|4y-16\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}8-2x=0\\4y-16=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}2x=8\\4y=16\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=4\\y=4\end{matrix}\right.\)
d,
|x - 14| + |2y - x| = 0
\(\Rightarrow\left\{{}\begin{matrix}\left|x-14\right|=0\\\left|2y-x\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-14=0\\2y-x=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=x\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\2y=14\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=14\\y=7\end{matrix}\right.\)
2.Tìm x, y, z biết
a,
2.|3x| + |y + 3| + |z - y| = 0
\(\Rightarrow\left\{{}\begin{matrix}2.\left|3x\right|=0\\\left|y+3\right|=0\\\left|z-y\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left|3x\right|=0\\y+3=0\\z-y=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3x=0\\y=-3\\z=y\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=0\\y=-3\\z=-3\end{matrix}\right.\)
b, (x - 3y)2 + | y + 4|= 0
\(\Rightarrow\left\{{}\begin{matrix}\left(x-3y\right)2=0\\\left|y+4\right|=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-3y=0\\y+4=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3y\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\left(-4\right)\\y=-4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x=-12\\y=-4\end{matrix}\right.\)

\(\left(2x+1\right)^4=0\)
\(\Rightarrow2x+1=0\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\frac{1}{2}\)

a)
(- 2)5 + 7(x - 8) = - 167 + 8x
32 + 7x - 56 + 167 = 8x
8x - 7x = 143
x = 143
b)
(7 - 3x)(4x + 4) = 0
4(x + 1)(7 - 3x) = 0
\(\left[\begin{matrix}x+1=0\\7-3x=0\end{matrix}\right.\)
\(\left[\begin{matrix}x=-1\\3x=7\end{matrix}\right.\)
\(\left[\begin{matrix}x=-1\\x=\frac{7}{3}\end{matrix}\right.\)

a) \(\left|2x-1\right|=5\)
\(2x-1=5\)
\(2x=6\)
\(x=3\)
b) \(x:4\frac{1}{3}=-2,5\)
\(x:\frac{4}{3}=\frac{-5}{2}\)
\(x=\frac{-5}{2}.\frac{4}{3}=\frac{-10}{3}\)
c) \(x^3-\frac{9}{10}.x=0\)
\(xx^2-\frac{9}{10}.x=0\)
\(x\left(x^2-\frac{9}{10}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x^2-\frac{9}{10}=0\Rightarrow x^2=\frac{9}{10}\Rightarrow x=\sqrt{\frac{9}{10}}\end{cases}}\)
Câu c có 2 kết quả là \(\hept{\begin{cases}x=0\\x=\sqrt{\frac{9}{10}}\end{cases}}\)
Câu b nhầm
\(x:4\frac{1}{3}=-2,5\)
\(x:\frac{13}{3}=\frac{-5}{2}\)
\(x=\frac{-5}{2}.\frac{13}{3}=\frac{-65}{6}\)
(8x-16)(x-4)=0
=>\(\left[{}\begin{matrix}8x-16=0\\x-4=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}8x=16\\x=4\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
\(\left(8x-16\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}8x-16=0\\x-4=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}8x=16\\x=4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\)
vậy x ∈ {2; 4}