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\(8x^2-2x-3=8x^2+4x-6x-3=4x\left(2x+1\right)-3\left(2x+1\right)=\left(4x-3\right)\left(2x+1\right)\)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
x3 - 2x2 - 8x
= x( x2 - 2x - 8 )
= x( x2 - 4x + 2x - 8 )
= x[ x( x - 4 ) + 2( x - 4 ) ]
= x( x - 4 )( x + 2 )
\(x^3-2x^2-8x=x\left(x^2-2x-8\right)=x\left(x^2-2x+1-9\right)=x\left[\left(x-1\right)^2-3^2\right]=x\left(x-4\right)\left(x+2\right)\)
\(2x^3-2xy^2-8x^2+8xy\)
\(=2x\left(x^2-y^2-4x+4y\right)\)
\(=2x\left[\left(x^2-y^2\right)-4\left(x-y\right)\right]\)
\(=2x\left[\left(x-y\right)\left(x+y\right)-4\left(x-y\right)\right]\)
\(=2x\left(x-y\right)\left(x+y-4\right)\)
2x^3 - 2xy^2 - 8x^2 + 8xy
= 2x^2 ( x - y ) - 8x ( x - y )
= ( x - y ) ( 2x^2 - 8x )
= ( x - y ) 2x ( x - 4 )
\(a,=3\left(x^2-2\right)\\ b,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ c,=9x^2\left(x-y\right)-4\left(x-y\right)=\left(3x-2\right)\left(3x+2\right)\left(x-y\right)\\ d,=x\left(x^2-2x-8\right)=x\left(x^2+2x-4x-8\right)=x\left(x+2\right)\left(x-4\right)\)
Bài 1
2x2 + 8x + 16 = 2(x2 + 4x + 4) = 2(x + 2)2
Bài 2
\(\frac{x}{x-5}\)\(+\)\(\frac{2}{x^2-25}\)\(=\)\(\frac{x\left(x+5\right)}{\left(x-5\right)\left(x+5\right)}\)\(+\)\(\frac{2}{x^2-25}\)\(=\)\(\frac{x^2+5x+2}{x^2-25}\)
\(\frac{x}{x-5}+\frac{2}{x^2-25}=\frac{x\left(x+5\right)+2}{x^2-25}\)
\(=\frac{x^2+5x+2}{x^2-25}\)
Answer:
\(8x^2-2x-1\)
\(=8x^2-4x+2x-1\)
\(=\left(8x^2-4x\right)+\left(2x-1\right)\)
\(=4x\left(2x-1\right)+\left(2x-1\right)\)
\(=\left(4x+1\right)\left(2x-1\right)\)