phân tích thành nhân tử: \(2a-\sqrt{ab}-6b\)
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HT
0
AP
3
3 tháng 10 2021
\(a\sqrt{a}+2a+\sqrt{a}+2=\left(a\sqrt{a}+2a\right)+\left(\sqrt{a}+2\right)\)
\(=a\left(\sqrt{a}+2\right)+\left(\sqrt{a}+2\right)=\left(\sqrt{a}+2\right)\left(a+1\right)\)
22 tháng 7 2023
`a^2 + ab + 2a + 2b = a(a+2) + b(a+2) = (a+b)(a+2)`
3 tháng 10 2016
a) \(2a-4b=2\left(a-2b\right)\)
c) \(2ax-2ay+2a=2a\left(x-y+1\right)\)
e) \(3xy\left(x-4\right)-9x\left(4-x\right)=3x\left(x-4\right)\left(y+3\right)\)
b,d xem lại đề
15 tháng 10 2021
Bài 2:
a: \(x^2+5x-6=\left(x+6\right)\left(x-1\right)\)
b: \(5x^2+5xy-x-y\)
\(=5x\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(5x-1\right)\)
c:\(-6x^2+7x-2\)
\(=-6x^2+3x+4x-2\)
\(=-3x\left(2x-1\right)+2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(-3x+2\right)\)
15 tháng 10 2021
1.
a) \(=x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
b) \(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\)
\(=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
c) \(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]=5\left[\left(x-y\right)^2-4z^2\right]\)
\(=5\left(x-y-2z\right)\left(x-y+2z\right)\)
2.
a) \(=x\left(x+2\right)+3\left(x+2\right)=\left(x+2\right)\left(x+3\right)\)
b) \(=5x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(5x-1\right)\)
c) \(=-\left[3x\left(2x-1\right)-2\left(2x-1\right)\right]=-\left(2x-1\right)\left(3x-2\right)\)
3.
b) \(=2x\left(x-1\right)+5\left(x-1\right)=\left(x-1\right)\left(2x+5\right)\)
c) \(=-\left[5x\left(x-3\right)-1\left(x-3\right)\right]=-\left(x-3\right)\left(5x-1\right)\)
4.
a) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
b) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\)
\(\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
NV
0
HC
2
24 tháng 10 2021
\(ab+b\sqrt{a}+\sqrt{a}+1=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
24 tháng 10 2021
\(=b\sqrt{a}\left(\sqrt{a}+1\right)+\left(\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)\left(b\sqrt{a}+1\right)\)
CN
1
KY
15 tháng 9 2019
\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=a^2b-ab^2-a^2c-ac^2+2abc-b^2c+bc^2\)
\(=a^2b-ab^2-a^2c-ac^2+abc+abc-b^2c+bc^2\)
\(=\left(bc^2-ac^2+abc-a^2c\right)-\left(b^2c-abc-ab^2+a^2b\right)\)
\(=c\left(bc-ac+ab-a^2\right)-b\left(bc-ac-ab+a^2\right)\)
\(=\left(c-b\right)\left(bc-ac+ab-a^2\right)\)
\(=\left(c-b\right)\left[c\left(b-a\right)+a\left(b-a\right)\right]\)
\(=\left(c-b\right)\left(c+a\right)\left(b-a\right)\)