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Câu 12:
\(\left(2x^3+4x^2-16x\right):2x\)
\(=2x\cdot\left(x^2+2x-8\right):2x\)
\(=x^2+2x-8\)
⇒ A
Tự luận:
Câu 1:
a) \(5x+3x-12x\)
\(=8x-12x\)
\(=-4x\)
b) \(2x\left(x-5\right)\)
\(=2x^2-10x\)
c) \(\left(x-3\right)\left(x+4\right)\)
\(=x^2+4x-3x-12\)
\(=x^2+x-12\)
d) \(\left(4x^2-5x-6\right):\left(x-2\right)\)
\(=\left(4x^2+3x-8x-6\right):\left(x-2\right)\)
\(=\left[x\left(4x+3\right)-2\left(4x+3\right)\right]:\left(x-2\right)\)
\(=\left(x-2\right)\left(4x+3\right):\left(x-2\right)\)
\(=4x+3\)
Câu 2:
\(A=\left(x-3\right)\left(x+2\right)+\left(x-4\right)\left(x+4\right)-\left(2x-1\right)x\)
\(A=x^2+2x-3x-6+x^2+4x-4x-16-2x^2+x\)
\(A=\left(x^2+x^2-2x^2\right)+\left(2x-3x+4x-4x+x\right)-\left(6+16\right)\)
\(A=-22\)
Vậy giác trị của A không phụ thuộc vào biến
Câu 12:
\(\dfrac{2x^3+4x^2-16x}{2x}\\ =\dfrac{2x^3}{2x}+\dfrac{4x^2}{2x}-\dfrac{16x}{2x}\\ =x^2+2x-8\)
Vậy chọn A
Tự luận:
Câu 1:
a)
\(5x+3x-12x\\ =\left(5+3-12\right)x=-4x\)
b)
\(2x\left(x-5\right)\\ =2x.x-2x.5\\ =4x^2-10x\)
c)
\(\left(x-3\right)\left(x+4\right)\\ =x.x+x.4-3.x-3.4\\ =x^2+4x-3x-12\\ =x^2+x-12\)
d)
\(\dfrac{4x^2-5x-6}{x-2}\\ =\dfrac{4x^2+3x-8x-6}{x-2}\\ =\dfrac{x\left(4x+3\right)-2\left(4x+3\right)}{\left(x-2\right)}\\ =\dfrac{\left(x-2\right)\left(4x+3\right)}{x-2}\\ =4x+3\)
