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12 tháng 8 2018

\(H=\left(x^2-x+1\right)\left(x^2+3x+1\right)+4x^2\)

Đặt \(x^2+1=t\), ta được:

\(H=\left(t-x\right)\left(t+3x\right)+4x^2\)

\(H=t^2+2xt+x^2\)

\(H=\left(t+x\right)^2\)

\(H=\left(x^2+x+1\right)^2\)

Vậy.......

24 tháng 9 2021

\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)

30 tháng 9 2021

giup e với

 

6 tháng 11 2021

\(1,=x\left(x^2-2x+1-y^2\right)=x\left[\left(x-1\right)^2-y^2\right]=x\left(x-y-1\right)\left(x+y-1\right)\\ 2,=\left(x+y\right)^3\\ 3,=\left(2y-z\right)\left(4x+7y\right)\\ 4,=\left(x+2\right)^2\\ 5,Sửa:x\left(x-2\right)-x+2=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

1 tháng 11 2021

1D  2C

Câu 1: D

Câu 2: C

30 tháng 9 2021

e) \(=x^2\left(x+1\right)-2x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x^2-2x+3\right)\)

g) \(=x^2\left(3x-1\right)-x\left(3x-1\right)+4\left(3x-1\right)=\left(3x-1\right)\left(x^2-x+4\right)\)

h) \(=3x^2\left(2x+1\right)-x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(3x^2-x+1\right)\)

i) \(=2x^2\left(2x+1\right)+2x\left(2x+1\right)+\left(2x+1\right)=\left(2x+1\right)\left(2x^2+2x+1\right)\) 

30 tháng 9 2021

ảm ơn nha

 

22 tháng 11 2023

a: \(2y\left(x+2\right)-3x-6\)

\(=2y\left(x+2\right)-3\left(x+2\right)\)

\(=\left(x+2\right)\left(2y-3\right)\)

b: \(3\left(x+4\right)-x^2-4x\)

\(=3\left(x+4\right)-\left(x^2+4x\right)\)

\(=3\left(x+4\right)-x\left(x+4\right)\)

\(=\left(x+4\right)\left(3-x\right)\)

c: \(2\left(x+5\right)-x^2-4x\)

\(=2x+10-x^2-4x\)

\(=-x^2-2x+10\)

\(=-x^2-2x-1+11\)

\(=11-\left(x^2+2x+1\right)\)

\(=11-\left(x+1\right)^2\)

\(=\left(\sqrt{11}-x-1\right)\left(\sqrt{11}+x+1\right)\)

d: \(x^2+6x-3x-18\)

\(=\left(x^2+6x\right)-\left(3x+18\right)\)

\(=x\left(x+6\right)-3\left(x+6\right)\)

\(=\left(x+6\right)\left(x-3\right)\)

24 tháng 7 2018

\(x^2+4x+4=\left(x+2\right)^2 \)

\(4x^2-4x+1=\left(2x-1\right)^2\)

\(c\left(x+1\right)-y\left(x+1\right)=\left(x+1\right)\left(c-y\right)\)

\(x^3-3x^2+3x-1+27y^3=\left(x-1\right)^3+27y^3=\left(x-1+3y\right)\left(x^2-2x+1-3xy+3y+9y^2\right)\)