Cho biểu thức A= x2 (2x-1)+8 (x-4) = -15
a, Rút gọn A
b, Tìm x để A = 0
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a, \(A=x^2\left(2x-1\right)+x\left(x+8\right)=2x^3-x^2+x^2+8x=2x^3+8x\)
Thay x = -2, ta có:
\(2\cdot\left(-2\right)^3+8\cdot\left(-2\right)=-32\)
b, \(A=2x^3+8x=0\\ \Leftrightarrow2x\left(x^2+4\right)=0\\ \Leftrightarrow x=0\)
Vậy A=0 khi x=0
a,A = \(x^2\).( 2\(x\) - 1) + \(x\)(\(x+8\))
A = 2\(x^3\) - \(x^2\) + \(x^2\) + 8\(x\)
A = 2\(x^3\) + 8\(x\)
b, \(x=-2\) ⇒ A = 2.(-2)3 + 8.(-2) = - 32
A = 0 ⇔ 2\(x^3\) + 8\(x\) = 0
2\(x\left(x^2+4\right)\) = 0
vì \(x^2\) + 4 > 0 ∀ \(x\) ⇒ \(x\) =0
Bài 1:
a) Ta có: \(P=1+\dfrac{3}{x^2+5x+6}:\left(\dfrac{8x^2}{4x^3-8x^2}-\dfrac{3x}{3x^2-12}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{8x^2}{4x^2\left(x-2\right)}-\dfrac{3x}{3\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\left(\dfrac{4}{x-2}-\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x+2}\right)\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}:\dfrac{4\left(x+2\right)-x-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\dfrac{3}{\left(x+2\right)\left(x+3\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{4x+8-x-x+2}\)
\(=1+3\cdot\dfrac{\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=1+\dfrac{3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{\left(x+3\right)\left(2x+10\right)+3\left(x-2\right)}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+10x+6x+30+3x-6}{\left(x+3\right)\left(2x+10\right)}\)
\(=\dfrac{2x^2+19x-6}{\left(x+3\right)\left(2x+10\right)}\)