Phân tích thành nhân tử (với a, b, x, y là các số không âm)
x 3 - y 3 + x 2 y - x y 2
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a, \(ab+b\sqrt{a}+\sqrt{a}+1=\sqrt{a}b\left(\sqrt{a}+1\right)+\sqrt{a}+1\)
\(=\left(b\sqrt{a}+1\right)\left(\sqrt{a}+1\right)\)
b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\sqrt{x^2}\left(\sqrt{x}+\sqrt{y}\right)-\sqrt{y^2}\left(\sqrt{y}+\sqrt{x}\right)=\left(\left|x\right|-\left|y\right|\right)\left(\sqrt{x}+\sqrt{y}\right)\)
với a,b,x,y không âm ta có
a,\(ab+b\sqrt{a}+\sqrt{a}+1\)
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
b, \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\)
\(=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
a. \(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
b. \(\sqrt{x^3}-\sqrt{y^3}+\sqrt{x^2y}-\sqrt{xy^2}=\left(\sqrt{x}-\sqrt{y}\right)\left(x+\sqrt{xy}+y\right)+\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(x+2\sqrt{xy}+y\right)=\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)^2\)
12 - √x - x
= 16 - x - 4 - √x (tách 12 = 16 - 4 và đổi vị trí)
= [42 - (√x)2] - (4 + √x)
= (4 - √x)(4 + √x) - (4 + √x)
= (4 + √x)(4 - √x - 1)
= (4 + √x)(3 - √x)
ab + b√a + √a + 1 = [(√a)2b + b√a] + (√a + 1)
= b√a(√a + 1) + (√a + 1) = (√a + 1)(b√a + 1)
xy - y√x + √x - 1
= (√x)2.y - y√x + √x - 1
= y√x(√x - 1) + √x - 1
= (√x - 1)(y√x + 1) với x ≥ 1
\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)
a) ab + b√a + √a + 1 = [(√a)2b + b√a] + (√a + 1)
= b√a(√a + 1) + (√a + 1) = (√a + 1)(b√a + 1)
= (√x - √y)(√x + √y)2
= (√x - √y)(√x + √y)(√x + √y)
= (x - y)(√x + √y)
= (√x - √y)(√x + √y)2
= (√x - √y)(√x + √y)(√x + √y)
= (x - y)(√x + √y)