(2x-3)^2-13=36
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a) 42 - x = 5
x = 42 - 5
x = 37
b) 15.(x + 4) = 75
x + 4 = 75 : 15
x + 4 = 5
x = 5 - 4
x = 1
d) x + 18 = 12
x = 12 - 18
x = -6
e) 2x - 15 = -19
2x = (-19) + 15
2x = -4
x = -4 : 2
x = -2
f) 36 : (x^3 - 12) = -3
x^3 - 12 = 36 : (-3)
x^3 - 12 = -12
x^3 = 0
=> x = 0
g)x + 15 = 35
x = 35 - 15
x = 20
h) 2x - 13 = 3^2
2x - 13 = 9
2x = 9 + 13
2x = 22
x = 22 : 2
x = 11
a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252
\(\frac{2x}{4}:-5=\frac{1}{7}\)
\(\frac{x}{2}=-\frac{5}{7}\)
\(x=-\frac{10}{7}\)
a) Ta có: \(2-\left(15+2x\right)=-5\)
\(\Leftrightarrow2-15-2x+5=0\)
\(\Leftrightarrow-2x-8=0\)
\(\Leftrightarrow-2x=8\)
\(\Leftrightarrow x=-4\)
Vậy: x=-4
b) Ta có: \(\left(x-3\right)^3=-27\)
\(\Leftrightarrow x-3=-3\)
hay x=0
Vậy: x=0
c) Ta có: \(\left(x+1\right)^2=36\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=6\\x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-7\end{matrix}\right.\)
Vậy: x∈{5;-7}
d) Ta có: \(2x-\left(x+5\right)=1-x\)
\(\Leftrightarrow2x-x-5-1+x=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow2x=6\)
hay x=3
Vậy: x=3
e) Ta có: \(15-\left(-3x+13\right)=2\left(x+1\right)\)
\(\Leftrightarrow15+3x-13-2x-2=0\)
\(\Leftrightarrow x=0\)
Vậy: x=0
b: \(\Leftrightarrow\sqrt{3x^2+1}\left(\sqrt{2}-1\right)=\sqrt{13}\left(\sqrt{2}-1\right)\)
=>3x^2+1=13
=>3x^2=12
=>x=2 hoặc x=-2
b: \(=\dfrac{4x\left(x-1\right)\left(x+1\right)}{6x\left(x-1\right)}=\dfrac{2\left(x+1\right)}{3}\)
c: \(=\dfrac{\left(5-x-1\right)\left(5+x+1\right)}{\left(x+6\right)^2}=\dfrac{\left(4-x\right)\left(x+6\right)}{\left(x+6\right)^2}=\dfrac{4-x}{x+6}\)
d: \(=\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\dfrac{x+3}{x+2}\)
\(\left(2x-3\right)^2-13=36.\)
\(\left(2x-3\right)^2=36+13\)
\(\left(2x-3\right)^2=49\)
\(\left(2x-3\right)^2=7^2\)hoặc \(\left(-7\right)^2\)
\(\Rightarrow\orbr{\begin{cases}2x-3=7\\2x-3=-7\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}2x=10\\2x=-4\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=-2\end{cases}}\)