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a, 3.(2\(x\) + 4) + 198 = (-3)2.10
3.(2\(x\) + 4) + 198 = 90
3.(2\(x\) + 4) = 90 - 198
3.(2\(x\) + 4) = - 108
2\(x\) + 4 = -108 : 3
2\(x\) + 4 = -36
2\(x\) = - 36 - 4
2\(x\) = - 40
\(x\) = -40 : 2
\(x\) = - 20
b, 2.(\(x\) + 7) - 6 = 18
2.(\(x\) + 7) = 18 + 6
2.(\(x\) + 7) =24
\(x\) + 7 = 24 : 2
\(x\) + 7 = 12
\(x\) = 12 - 7
\(x\) = 5
a, 12 - (2\(x^2\) - 3) = 7
2\(x^2\) - 3 = 12 - 7
2\(x^2\) - 3 = 5
2\(x^2\) = 8
\(x^2\) = 4
\(\left[{}\begin{matrix}x=-2\\x=2\end{matrix}\right.\)
a) -2(2x - 8) + 3(4 - 2x) = -72 - 5(3x - 7)
=> -4x + 18 + 12 - 6x = -72 - 15x + 35
=> -10x + 15x = -37 - 30
=> 5x = -37
=> x = -7,4
b) 3|2x2 - 7| = 33
=> |2x2 - 7| = 11
=> \(\orbr{\begin{cases}2x^2-7=11\\2x^2-7=-11\end{cases}}\)
=> \(\orbr{\begin{cases}2x^2=18\\2x^2=-4\left(loại\right)\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)
\(2\left(x-3\right)-5\left(x-4\right)=-7+14.\left(-3\right)\)
\(2x-6-5x+20=-7+\left(-42\right)\)
\(\left(2x-5x\right)+\left(-6+20\right)=-49\)
\(-3x+14=-49\)
\(-3x=-63\)
\(x=21\)
Bài 1:
Ta có: \(4-2\left(x+1\right)=2\)
\(\Leftrightarrow2\left(x+1\right)=2\)
\(\Leftrightarrow x+1=1\)
hay x=0
Bài 2:
Ta có: \(\left|2x-3\right|-1=2\)
\(\Leftrightarrow\left|2x-3\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=3\\2x-3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=6\\2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
\(\frac{2}{5}-x+2x=-\frac{3}{4}\)
\(x+2x=\frac{2}{5}+\frac{3}{4}=\frac{23}{20}\)
\(1x+2x=x\left(1+2\right)=3x=\frac{23}{20}\)
\(x=\frac{23}{20}:3=\frac{23}{60}\)