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\(4x^3-13x^2+9x-18\)
\(=4x^3-12x^2-x^2+3x+6x-18\)
\(=4x^2\left(x-3\right)-x\left(x-3\right)+6\left(x-3\right)\)
\(=\left(x-3\right)\left(4x^2-x+6\right)\)
\(x^8+3x^4+4\)
\(=\left(x^8-x^6+2x^4\right)+\left(x^6-x^4+2x^2\right)+\left(2x^4-2x^2+4\right)\)
\(=x^4\left(x^4-x^2+2\right)+x^2\left(x^4-x^2+2\right)+2\left(x^4-x^2+2\right)\)
\(=\left(x^4+x^2+2\right)\left(x^4-x^2+2\right)\)
\(4x^4+4x^3+5x^2+2x+1\)
\(=\left(4x^4+2x^3+2x^2\right)+\left(2x^3+x^2+x\right)+\left(2x^2+x+1\right)\)
\(=2x^2\left(2x^2+x+1\right)+x\left(2x^2+x+1\right)+\left(2x^2+x+1\right)\)
\(=\left(2x^2+x+1\right)^2\)
= (3x + 1 - x - 1)(3x + 1 + x + 1)
= 2x(4x + 2)
Em áp dụng hđt số 3 trong sgk nhé.
1) x2- 3x - 6x +18
= (x2- 3x )-(6x -18 )
= x(x-3)- 6(x-3)
= (x-6)(x-3)
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
\(A=3x^2-14x^2+4x+3\)
Giả sử:
\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)
\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)
\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)
Ta có:
\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)
Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)
x3 - x2 - 3x2 + 6x - 3
= x3 - x2 - 3x2 + 3x + 3x - 3
= x2 ( x - 1 ) - 3x ( x - 1 ) + 3 ( x - 1 )
= ( x - 1 ) ( x2 - 3x + 3 )
Ta có:
\(x^2+3x-18\)
\(\Leftrightarrow x^2-3x+6x-18\)
\(\Leftrightarrow\left(x^2-3x\right)+\left(6x-18\right)\)
\(\Leftrightarrow x\left(x-3\right)+6\left(x-3\right)\)
\(\Leftrightarrow\left(x-3\right)\left(x+6\right)\)
\(x^2+3x-18\)
\(\Leftrightarrow x\left(x+3\right)=18\)
\(\Leftrightarrow\orbr{\begin{cases}x=18\\x+3=18\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=18\\x=15\end{cases}}\)
vay \(\orbr{\begin{cases}x=18\\x=15\end{cases}}\)