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a) \(\left|x\right|=2\Leftrightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)
+) TH1: \(x=2\)
\(A=\left(3\cdot2+5\right)\left(2\cdot2-1\right)+\left(4\cdot2-1\right)\left(3\cdot2+2\right)\)
\(A=89\)
+) TH2: \(x=-2\)
\(A=\left(-2\cdot3+5\right)\left(-2\cdot2-1\right)+\left(-2\cdot4-1\right)\left(-2\cdot3+2\right)\)
\(A=-27\)
Vậy...
b) \(B=9x^2+42x+49\)
\(B=\left(3x+7\right)^2\)
\(B=\left(3\cdot1+7\right)^2\)
\(B=100\)
Vậy...

Bài 1:
- \(\dfrac{11}{2}x\) + 1 = \(\dfrac{1}{3}x-\dfrac{1}{4}\)
- \(\dfrac{11}{2}\)\(x\) - \(\dfrac{1}{3}\)\(x\) = - \(\dfrac{1}{4}\) - 1
-(\(\dfrac{33}{6}\) + \(\dfrac{2}{6}\))\(x\) = - \(\dfrac{5}{4}\)
- \(\dfrac{35}{6}\)\(x\) = - \(\dfrac{5}{4}\)
\(x=-\dfrac{5}{4}\) : (- \(\dfrac{35}{6}\))
\(x\) = \(\dfrac{3}{14}\)
Vậy \(x=\dfrac{3}{14}\)
Bài 2: 2\(x\) - \(\dfrac{2}{3}\) - 7\(x\) = \(\dfrac{3}{2}\) - 1
2\(x\) - 7\(x\) = \(\dfrac{3}{2}\) - 1 + \(\dfrac{2}{3}\)
- 5\(x\) = \(\dfrac{9}{6}\) - \(\dfrac{6}{6}\) + \(\dfrac{4}{6}\)
- 5\(x\) = \(\dfrac{7}{6}\)
\(x\) = \(\dfrac{7}{6}\) : (- 5)
\(x\) = - \(\dfrac{7}{30}\)
Vậy \(x=-\dfrac{7}{30}\)