Phân tích đa thức thành nhân tử
\(a-3\sqrt{ab}+5y\) với \(a\ge0;b\ge0\)
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a) \(3a-2\sqrt{ab}-b=3a-3\sqrt{ab}+\sqrt{ab}-b\)
\(=3\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)=\left(3\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
b) \(5a+3\sqrt{ab}-8b=5a-5\sqrt{ab}+8\sqrt{ab}-8b\)
\(=5\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)+8\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(5\sqrt{a}+8\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\)
a) (\(\sqrt{a}-\sqrt{b}\))(3\(\sqrt{a}+b\))
b) \(\left(\sqrt{a}-\sqrt{b}\right)\left(5\sqrt{a}+8\sqrt{b}\right)\)
a) xy+3x-7y-21
=x(y+3)-7(x+3)
=(x-7)(y+3)
b)2xy-15-6x-5y
=2x(y-3)-5(-3+y)
=(2x-5)(y-3)
c)2x^2y+2xy^2-2x-2y
=2x(xy-1)+2y(xy-1)
=(2x+2y)(xy-1)
x(x+3)-5x(x-5)-5(x+3)
=(x-5)(x+3)-5x(x-5)
=(x-5)(x+3-5x)
Câu cuối mình bị nhầm dòng cuối phải là (x-5)(x+3+x-5)=(x-5)(2x-2)nha bạn
a: \(A=x\sqrt{x}-y\sqrt{y}+x\sqrt{y}-y\sqrt{x}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
b: \(B=5x^2-7x\sqrt{y}+2y\)
\(=5x^2-5x\sqrt{y}-2x\sqrt{y}+2y\)
\(=5x\left(x-\sqrt{y}\right)-2\sqrt{y}\left(x-\sqrt{y}\right)\)
\(=\left(x-\sqrt{y}\right)\left(5x-2\sqrt{y}\right)\)
a: \(=3\left[\left(m-5\right)^3+8\right]\)
\(=3\left(m-5+2\right)\left[\left(m-5\right)^2-2\left(m-5\right)+4\right]\)
\(=3\left(m-3\right)\left(m^2-10m+25-2m+10+4\right)\)
\(=3\left(m-3\right)\left(m^2-12m+39\right)\)
b: \(=5\left(1-6k\right)^3-5\cdot64\)
\(=5\left(1-6k-4\right)\left[\left(1-6k\right)^2+4\left(1-6k\right)+16\right]\)
\(=5\left(-6k-3\right)\left(36k^2-12k+1+4-24k+16\right)\)
\(=-15\left(2k+1\right)\left(36k^2-36k+21\right)\)
\(=-45\left(2k+1\right)\left(12k^2-12k+7\right)\)
c: \(=\left(a+b-2b+a\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(2b-a\right)+\left(2b-a\right)^2\right]\)
\(=\left(2a-b\right)\left(a^2+2ab+b^2+2ab-a^2+2b^2-ab+4b^2-4ab+b^2\right)\)
\(=\left(2a-b\right)\cdot b\cdot\left(6b+a\right)\)