Câu 10:
x<0
=>-x>0
=>|-x|=-x
\(B=\frac{x^2+6\left|-x\right|+8}{\left(2x-3\right)^2-\left(1-x\right)^2}\)
\(=\frac{x^2-6x+8}{\left(2x-3+1-x\right)\left(2x-3-1+x\right)}=\frac{\left.\left(x-2\right)\left(x-4\right)\right.}{\left(x-2\right)\left(3x-4\right)}=\frac{x-4}{3x-4}\)
=>Chọn B
Câu 11: \(\frac{4x^3-12x^2}{x\left(x-3\right)^2}=\frac{M}{x-3}\)
=>\(\frac{4x^2\left(x-3\right)}{x\left(x-3\right)^2}=\frac{M}{x-3}\)
=>\(\frac{4x}{x-3}=\frac{M}{x-3}\)
=>M=4x
=>Chọn C
Câu 12: \(\frac{7x^4y^2}{14x^5y}=\frac{7x^4y\cdot y}{7x^4y\cdot2x}=\frac{y}{2x}\)
=>Chọn B
Câu 13: C
Câu 14: C
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