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\(4a^2b^2-\left(a^2+b^2-c^2\right)^2\)
\(=4a^2b^2-2ab\left(a^2+b^2-c^2\right)+2ab\left(a^2+b^2-c^2\right)-\left(a^2+b^2-c^2\right)^2\)
\(=2ab\left[2ab-\left(a^2+b^2-c^2\right)\right]+\left(a^2+b^2-c^2\right)\left[2ab-\left(a^2+b^2-c^2\right)\right]\)
\(=\left(2ab+a^2+b^2-c^2\right)\left(2ab-a^2-b^2+c^2\right)\)
\(=\left(a^2+ab+ab+b^2-c^2\right)\left[c^2-\left(a^2-ab-ab+b^2\right)\right]\)
\(=\left[a\left(a+b\right)+b\left(a+b\right)-c^2\right]\left[c^2-\left(a\left(a-b\right)-b\left(a-b\right)\right)\right]\)
\(=\left[\left(a+b\right)^2-c^2\right]\left[c^2-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)^2-c\left(a+b\right)+c\left(a+b\right)-c^2\right]\left[c^2+c\left(a-b\right)-c\left(a-b\right)-\left(a-b\right)^2\right]\)
\(=\left[\left(a+b\right)\left(a+b-c\right)+c\left(a+b-c\right)\right]\left[c\left(c+a-b\right)-\left(a-b\right)\left(c+a-b\right)\right]\)
\(=\left(a+b+c\right)\left(a+b-c\right)\left(c+a-b\right)\left(c-a+b\right)\)
\(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)
\(M=\left(a^2+b^2-c^2\right)^2-\left(2ab\right)^2\)
\(M=\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)\)
\(M=\left(\left(a^2-2ab+b^2\right)-c^2\right)\left(\left(a^2+2ab+b^2\right)-c^2\right)\)
\(M=\left(\left(a-b\right)^2-c^2\right)\left(\left(a+b\right)^2-c^2\right)\)
\(M=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)
1) a^2 + b^2 + 2a - 2b - 2ab = (a^2 - 2ab + b^2) + (2a-2b) = (a-b)^2 + 2(a-b) = (a-b)(a-b+2)
2) 4a^2 - 4b^2 - 4a + 1 = ( 4a^2 - 4a +1) - 4b^2 = (2a-1)^2 - 4b^2 = (2a-1-2b)(2a-1+2b)
3) a^3+6a^2+12a+8= (a^3+8)+(6a^2+12a)= (a+2)(a^2-2a+4)+6a(a+2)=(a+2)(a^2-2a+4+6a)=(a+2)(a^2+4a+4)=(a+2)(a+2)^2=(a+2)^3
giúp mình đang cần gấp
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\(\dfrac{xy}{2}-x+\dfrac{x^2}{4}=x\left(\dfrac{y}{2}-1+\dfrac{x}{4}\right)\)
\(M=\left(a^2+b^2-c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2+b^2-c^2-2ab\right)\left(a^2+b^2-c^2+2ab\right)\)
\(=\left(\left(a^2-2ab+b^2\right)-c^2\right)\left(\left(a^2+2ab+b^2\right)-c^2\right)\)
\(=\left(\left(a-b\right)^2-c^2\right)\left(\left(a+b\right)^2-c^2\right)\)
\(=\left(a-b-c\right)\left(a-b+c\right)\left(a+b-c\right)\left(a+b+c\right)\)