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a) 2.(x-1)=4/3
<=> (X-1)= 4/3 :2 = 2/3
<=> X= 2/3+1=5/3
b) (x-1)^2= 4
<=> X-1=2 hoặc x-1=-2
<=> X= 3 hoặc x= -1
c) 2.|x+1|=3
<=> Ix+1| = 3/2
<=> X+1= 3/2 hoặc x+1= -3/2
<=> X= 1/2 hoặc x= -5/2
a) Ta có: \(2\left(x-1\right)=\dfrac{4}{3}\)
\(\Leftrightarrow x-1=\dfrac{4}{3}:2=\dfrac{4}{6}=\dfrac{2}{3}\)
hay \(x=\dfrac{5}{3}\)
Vậy: \(x=\dfrac{5}{3}\)
b) Ta có: \(\left(x-1\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-1\right\}\)
c)96 - 3.(x + 1) = 42
3.(x + 1)= 96 - 42
3.(x + 1) = 54
x + 1= 54:3
x + 1= 18
x = 18 - 1
x = 17
96-3(x+1)=42
3(x+1)=96-42
3(x+1)=54
x+1 = 54 : 3
x+1 = 18
x = 18 - 1
x = 17
96 - 3(x + 1) = 42
\(\Rightarrow\)3x + 3 = 54
\(\Rightarrow\)3x = 51
\(\Rightarrow\)x = 17
96 – 3(x + 1) = 42
3(x + 1) = 96 - 42
3(x + 1) = 54
x + 1 = 54:3
x + 1 = 18
x = 18 - 1
x = 17
Vậy x = 17
\(95-3\cdot\left(x+1\right)=42\)
\(=>3\cdot\left(x+1\right)=95-42\)
\(=>3\cdot\left(x+1\right)=53\)
\(=>x+1=53:3\)
\(=>x+1=\frac{53}{3}\)
\(=>x=\frac{53}{3}-1\)
\(=>x=\frac{53}{3}-\frac{3}{3}\)
\(=>x=\frac{50}{3}\)
\(96-3.\left(x+1\right)=42\)
\(\Rightarrow3.\left(x+1\right)=54\)
\(\Rightarrow x+1=18\)
\(\Rightarrow x=17\)
96 - 3(x + 1) = 42
<=> 96 - 3x - 3 = 42
<=> 96 - 3 - 42 = 3x
<=> 3x = 51
<=> x = 17