\(-16x^2+8xy-y^2+49\)
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\(-16x^2+8xy-y^2+49\)
\(=7^2-\left(4x-y\right)^2\)
\(=\left(7-4x+y\right)\left(7+4x-y\right)\)
\(-16x^2+8xy-y^2+49\)
\(=49-\left(16x^2-8xy+y^2\right)\)
\(=7^2-\left(4x-y\right)^2\)
\(=\left(7-4x+y\right)\left(7+4x-y\right)\)
a. -\(-16x^2+8xy-y^2+49\)
= \(\left(-\left(4x\right)^2+8xy-y^2\right)+49\)
= \(-\left(\left(4x^2\right)-8xy+y^2\right)+49\)
= \(-\left(4x-y\right)^2+49\)
b. \(y^2\left(x^2+y\right)-zx^2-zy\)
= \(y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
= \(\left(x^2+y\right)\left(y^2-z\right)\)
_16x2+8xy_y2+49
=( _(4x)2+2 × 4 × xy _ y2 )+ 72
= _((4x)2_ 2×4×x × xy +y2)+72
= _(4x_y)2+72
=72_(4x_y)2
= (7_(4x_y))×(7+(4x_y))
= (7_4x+y)×(7+4x_y)
2)y2×(x2+y)_zx2_zy
=y×(x2+y)_z(x2+y)
= ( x2+y)×(y_z)
a) Ta có: \(x^2-2xy+y^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
b) Ta có: \(-16x^2+8xy-y^2+49\)
\(=-\left(16x^2-8xy+y^2-49\right)\)
\(=-\left[\left(16x^2-8xy+y^2\right)-49\right]\)
\(=-\left[\left(4x-y\right)^2-7^2\right]\)
\(=-\left(4x-y-7\right)\left(4x-y+7\right)\)
c) Ta có: \(x^6-x^4+2x^3+2x^2\)
\(=x^4\left(x^2-1\right)+2x^2\left(x+1\right)\)
\(=x^4\left(x-1\right)\left(x+1\right)+2x^2\left(x+1\right)\)
\(=\left(x+1\right)\left[x^4\left(x-1\right)+2x^2\right]\)
\(=\left(x+1\right)\left(x^5-x^4+2x^2\right)\)
\(=\left(x+1\right)\left[\left(x^5+x^2\right)-\left(x^4-x^2\right)\right]\)
\(=\left(x+1\right)\left[x^2\left(x^3+1\right)-x^2\left(x^2-1\right)\right]\)
\(=x^2\cdot\left(x+1\right)\cdot\left(x^3+1-x^2+1\right)\)
\(=x^2\cdot\left(x+1\right)\cdot\left(x^3-x^2+2\right)\)
d) Ta có: \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left[\left(x+y\right)-\left(x-y\right)\right]\left[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2\right]\)
\(=\left(x+y-x+y\right)\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\cdot\left(3x^2+y^2\right)\)
\(16x^2+y^2+4y-16y-8xy\)
\(=\left(16x^2-8xy+y^2\right)+4y-16y\)
\(=\left(4x+y\right)^2-12y\)
\(=\left(4x+y-\sqrt{12y}\right)\left(4x+y-\sqrt{12y}\right)\)
P/S : Sai thì thôi nha!
\(a,AB=\left(4x+y\right)\left(16x^2-8xy+y^2\right)=\left(4x+y\right)\left(4x-y\right)^2\\ b,x=1;y=-1\Leftrightarrow AB=\left(4-1\right)\left(4+1\right)^2=3\cdot25=75\\ c,AB=0\Leftrightarrow\left(4x+y\right)\left(4x-y\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}4x=-y\\4x=y\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{y}{4}\\x=\dfrac{y}{4}\end{matrix}\right.\)
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Ta có: \(-16x^2+8xy-y^2+49\)
\(=-\left(16x^2-8xy+y^2-49\right)\)
\(=-\left\{\left[\left(4x\right)^2-2\cdot4x\cdot y+y^2\right]-7^2\right\}\)
\(=-\left[\left(4x-y\right)^2-7^2\right]\)
\(=-\left(4x-y-7\right)\cdot\left(4x-y+7\right)\)