Phân tích thành nhân tử:
B=9x + 6\(\sqrt{xy}\)+y
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\(B=4x^2+4xy-8xy-8y^2\)
\(=4x\left(x+y\right)-8y\left(x+y\right)\)
\(=\left(4x-8y\right)\left(x+y\right)\)
\(=4\left(x-2y\right)\left(x+y\right)\)
\(a,=\sqrt{xy}\left(\sqrt{x}-1\right)+\left(\sqrt{x}-1\right)=\left(\sqrt{xy}+1\right)\left(\sqrt{x}-1\right)\\ b,=\sqrt{xy}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)=\left(\sqrt{x}+1\right)\left(\sqrt{xy}+1\right)\)
\(b,\left(x+2\right)^2-25\)
\(=\left(x+2\right)^2-5^2\)
\(=\left(x-3\right)\left(x+7\right)\)
\(c,36\left(x-y\right)^2\)
\(=36\left(x^2-2xy+y^2\right)\)
\(=36x^2-72xy+36y^2\)
\(d,x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
\(=x^2+2.x.\dfrac{1}{4}+\dfrac{1}{4}^2\)
\(=\left(x+\dfrac{1}{4}\right)^2\)
\(e,2x^4y^3-3x^2y^4+5x^3y^4\)
\(=x^2y^3\left(2x^2-3y+5xy\right)\)
Các câu còn lại làm tương tự, chú ý sd HĐT
a,3x2-6xy+3y2
= 3(x2- 2xy+ y2)
= 3(x- y)2
b,xy-9x+y-9
= (xy+ y)- (9x+ 9)
= y(x+ 1)- 9(x+ 1)
= (x+1)(y- 9)
Chúc bạn học tốt
a,\(3x^2-6xy+3y^2\)
=\(3\left(x^2-2xy+y^2\right)\)
=\(3\left(x-y\right)^2\)
b,xy-9x+y-9
=\(\left(xy+y\right)-\left(9x+9\right)\)
=\(y\left(x+1\right)-9\left(x+1\right)\)
=\(\left(x+1\right)\left(y-9\right)\)
\(x\sqrt{x}+x-y+y\sqrt{x}-xy\sqrt{x}-xy\sqrt{y}=\left(x\sqrt{y}+y\sqrt{x}\right)+\left(x-y\right)-\left(xy\sqrt{x}+xy\sqrt{y}\right)\)
\(=\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)+\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)-xy\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{xy}+\sqrt{x}-\sqrt{y}-xy\right)\)
a) \(=9x-9\sqrt{xy}+4\sqrt{xy}-4y\)
\(=\left(9x-9\sqrt{xy}\right)+\left(4\sqrt{xy}-4y\right)\)
\(=9\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)+4\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(9\sqrt{x}+4\sqrt{y}\right)\)
b)\(=\left(xy+\sqrt{x}.y\right)+\left(\sqrt{x}+1\right)\)
\(=\sqrt{x}y\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}+1\right)\left(\sqrt{x}.y+1\right)\)
\(Sửa:x+y-9-2\sqrt{xy}\\ =\left(\sqrt{x}-\sqrt{y}\right)^2-9=\left(\sqrt{x}-\sqrt{y}-3\right)\left(\sqrt{x}-\sqrt{y}+3\right)\)
a) \(x^6-y^6\)
\(=\left(x^3\right)^2-\left(y^3\right)^2\)
\(=\left(x^3-y^3\right)\left(x^3+y^3\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)\left(x+y\right)\left(x^2-xy+y^2\right)\)
b) \(10ab+0,25a^2+100b^2\)
\(=\left(0,5a\right)^2+2\cdot0,5a\cdot10b+\left(10b\right)^2\)
\(=\left(0,5a+10b\right)^2\)
c) \(9x^2-xy+\frac{1}{36}y^2\)
\(=\left(3x\right)^2-2\cdot3x\cdot\frac{1}{6}y+\left(\frac{1}{6}y\right)^2\)
\(=\left(3x-\frac{1}{6}y\right)^2\)
\(B=\left(3\sqrt{x}\right)^2+2.3\sqrt{x}.\sqrt{y}+\sqrt{y}^2=\left(3\sqrt{x}+\sqrt{y}\right)^2\)
Hằng đẳng thức dễ mà bạn: a^2 + 2ab + b^2 = (a+b)^2
Còn điều kiện s b?