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Phân tích đa thức thành nhân tử:
\(a,x^3-6x^2+11x-6\)
\(=x^3-x^2-5x^2+5x+6x-6\)
\(=x^2\left(x-1\right)-5x\left(x-1\right)+6\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-5x+6\right)\)
\(=\left(x-1\right)\left(x^2-3x-2x+6\right)\)
\(=\left(x-1\right)\left[x\left(x-3\right)-2\left(x-2\right)\right]\)
\(=\left(x-1\right)\left(x-2\right)\left(x-3\right)\)
bạn không biết làm câu b) à ? câu a) dễ hơn mà nên ai mà chả bít làm, trừ bạn "Sana" ở trên thôi !!!!!
a) \(6x^3-6x=0\Leftrightarrow6x\left(x^2-1\right)=0\Leftrightarrow6x\left(x-1\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}6x=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)b) \(2x\left(3x+7\right)-6x^2=28\Leftrightarrow6x^2+14x-6x^2=28\Leftrightarrow14x=28\Leftrightarrow x=2\)
c) \(2\left(4x+4\right)-5\left(x-3\right)=0\Leftrightarrow8x+8-5x+15=0\Leftrightarrow3x=-23\Leftrightarrow x=-\dfrac{23}{3}\)
a. x2 - 6x = -9
<=> x2 - 6x + 9 = 0
<=> (x - 3)2 = 0
<=> x - 3 = 0
<=> x = 3
b. 2(x + 3) - x2 + 3x = 0
<=> 2(x + 3) - x(x + 3) = 0
<=> (2 - x)(x + 3) = 0
<=> \(\left[{}\begin{matrix}2-x=0\\x+3=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
a/ (x-1)2-(4x+3)(2-x)=x2-2x+1-(8x-4x2+6-3x)
=x2-2x+1-8x+4x2-6+3x=5x2-7x-6
b/ (15x3y2 - 6x2y3) : 3x2y2 = 5x - 2y
c/ \(\dfrac{x+7}{x-7}-\dfrac{x-7}{x+7}+\dfrac{4x^2}{x^2-49}\)=\(\dfrac{\left(x+7\right)^2-\left(x-7\right)^2+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{x^2+14x+49-\left(x^2-14x+49\right)+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{28x+4x^2}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x\left(x+7\right)}{\left(x-7\right)\left(x+7\right)}\)=\(\dfrac{4x}{x-7}\)
5(9x-7)-39x=3(7-6x)
45x-35-39x=21-18x
VT=6x-35
pt trở thành 6x-35=21-18x
=>24x=56
=>x=\(\frac{7}{3}\)
a) Ta có: \(x^3+6x-7\)
\(=x^3-x+7x-7\)
\(=x\left(x-1\right)\left(x+1\right)+7\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+7\right)\)
b) Ta có: \(4x^2+8x-5\)
\(=4x^2+10x-2x-5\)
\(=2x\left(2x+5\right)-\left(2x+5\right)\)
\(=\left(2x+5\right)\left(2x-1\right)\)
c) Ta có: \(9x^2-4y^2+6x-4y\)
\(=9x^2+6x+1-\left(4y^2+4y+1\right)\)
\(=\left(3x+1\right)^2-\left(2y+1\right)^2\)
\(=\left(3x+1+2y+1\right)\left(3x+2y\right)\)
\(=\left(3x+2y\right)\left(3x+2y+2\right)\)