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...
c) \(C=25x^2-2xy+\frac{1}{25}y^2\)
\(C=\left(5x-\frac{1}{5}y\right)^2\)
\(C=\left(\frac{-1}{5}\cdot5-\frac{1}{5}\cdot\left(-5\right)\right)^2\)
\(C=0\)
Vậy...
a) A = (3x + 5)(2x - 1) + (4x - 1)(3x + 2)
Ta có: |x| = 2
x = 2; -2
+) x = 2
A = (3.2 + 5)(2.2 - 1) + (4.2 - 1)(3.2 + 2)
A = 89
+) x = -2
A = (3.-2 + 5)(2.-2 - 1) + (4.-2 - 1)(3.-2 + 2)
A = 41
b) B = 9x2 + 42x + 49
B = 9.12 + 42.1 + 49
B = 100
c) C = 25x2 - 2xy + 1/25y2
C = 25.(-1/5)2 - 2.(-1/5)(-5) + 1/25.(-5)2
C = 0
