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G
10 tháng 8 2023
\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)
\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)
\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
LM
0
TN
2
2 tháng 9 2021
\(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1=\left(x-3\right)^2\left(x^2-1\right)-\left(x^2-1\right)=\left(x^2-1\right)\left[\left(x-3\right)^2-1\right]=\left(x-1\right)\left(x+1\right)\left(x-4\right)\left(x-2\right)\)
2 tháng 9 2021
\(x^2\left(x-3\right)^2-\left(x-3\right)^2-x^2+1\)
\(=\left(x-3\right)^2\cdot\left(x-1\right)\left(x+1\right)-\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\cdot\left(x+1\right)\left(x-2\right)\left(x-4\right)\)
4L
1
18 tháng 10 2021
a) \(2xy-y+6x-3=\left(2xy+6x\right)-\left(y+3\right)=2x\left(y+3\right)-\left(y+3\right)=\left(2x-1\right)\left(y+3\right)\)
b) \(x^2-2xy-x+2y=\left(x^2-2xy\right)-\left(x-2y\right)=x\left(x-2y\right)-\left(x-2y\right)=\left(x-1\right)\left(x-2y\right)\)
AH
Akai Haruma
Giáo viên
15 tháng 10 2021
Lời giải:
$y-x^2y-2xy^2-y^3=y(1-x^2-2xy-y^2)$
$=y[1-(x^2+2xy+y^2)]=y[1-(x+y)^2]=y(1-x-y)(1+x+y)$
19 tháng 8 2021
a) \(x^3+3x^2+3x=0\Rightarrow x\left(x^2+3x+3\right)=0\Rightarrow x\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\Rightarrow x=0\)
(do \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
b) \(x^3+6x^2+12x=0\Rightarrow x\left(x^2+6x+12\right)=0\Rightarrow x\left[\left(x+3\right)^2+4\right]=0\Rightarrow x=0\)
(do (x+3)2+4≥4>0)
19 tháng 8 2021
a: Ta có: \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)
hay x=0
b: Ta có: \(x^3+6x^2+12x=0\)
\(\Leftrightarrow x\left(x^2+6x+12\right)=0\)
hay x=0
20 tháng 8 2021
\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
20 tháng 8 2021
Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)
HK
3
PH
1
24 tháng 11 2021
\(=x^2-\left(y+1\right)^2=\left(x-y-1\right)\left(x+y+1\right)\)
78: \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc\)
\(=\left(a+b-c-a+c\right)\left(a+b-c+a-c\right)-2b\left(a-c\right)\)
\(=b\left(2a-2c\right)-2b\left(a-c\right)\)
=0