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\(A=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2-4a^2b^2\)
\(=\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\left(a^2+b^2-c^2-a^2+b^2-c^2\right)-4a^2b^2\)
\(=2a^2.2b^2-4a^2b^2=0\)
\(C=\left(2-6x\right)^2+\left(2-5x\right)^2+2\left(6x-2\right)\left(2-5x\right)\)
\(=\left[\left(2-6x\right)+\left(2-5x\right)\right]^2\)
\(=\left[4-11x\right]^2\)
\(=16-88x+121x^2\)
chúc bn học tốt
a) \(3x\left(x-2\right)-5x\left(1-x\right)-8\left(x^2-3\right)\)
\(=3x^2-6x-5x+5x^2-8x^2+24\)
\(=24-11x\)
b) \(\left(4x^2-3y\right)\cdot2y-\left(3x^2-4y\right)\cdot3y\)
\(=8x^2y-6y^2-9x^2y+12y^2\)
\(=6y^2-x^2y\)
c) \(3y^2\left[\left(2x-1\right)+y+1\right]-y\left(1-y-y^2\right)+y\)
\(=3y^2\cdot\left(2x-1+y+1\right)-y\cdot\left(1-y-y^2\right)+y\)
\(=6xy^2-3y^2+3y^3+3y^2-y+y^2+y^3+y\)
\(=4y^3+y^2+6xy^2\)
1, đa thức đã cho \(\Leftrightarrow\left(2x-y\right)^2-2\left(2x-y\right)\left(x-y\right)+\left(x-y\right)^2=\left[\left(2x-y\right)-\left(x-y\right)\right]^2=\left(2x-y-x+y\right)^2=x^2\)
2, đa thức đã cho \(\Leftrightarrow\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2=\left(x-y+z+y-z\right)^2=x^2\)
--- giải chi tiết lắm rồi đó---
a, \(\left(2x-y\right)^2+2\left(2x-y\right)\left(y-x\right)+\left(x-y\right)^2\)
\(=4x^2-4xy+y^2+2\left(2xy-2x^2-y^2+xy\right)+x^2-2xy+y^2\)
\(=4x^2-4xy+y^2+4xy-4x^2-2y^2+2xy+x^2-2xy+y^2\)
\(=x^2\)
b, \(\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z\right)\left[1+2\left(y-z\right)\right]+y^2-2yz+z^2\)
\(=\left(x-y+z\right)\left(1+2y-2z\right)+y^2-2yz+z^2\)
\(=x+2xy-2xz-y-2y^2+2yz+z+2yz-2z^2+y^2-2yz+z^2\)
\(=x-y+z+2xy-2xz+2yz-y^2-z^2\)
Chúc bạn học tốt!!!
a, \(\left(x^2+\dfrac{2}{5}y\right)\left(x^2-\dfrac{2}{5}y\right)=x^4-\dfrac{4}{25}y^2\)
b, \(\left(3x-2y\right)\left(3x+2y\right)\left(9x^2+4y^2\right)\)
\(=\left(9x^2-4y^2\right)\left(9x^2+4y^2\right)\)
\(=81x^4-16y^4\)
\(a,\left(3x+1\right)^2-2\left(3x+1\right)\left(3x-5\right)+\left(3x-5\right)^2=\left(\left(3x+1\right)-\left(3x-5\right)\right)^2=6^2=36\)
\(b,\left(3x^2-y\right)^2-\left(2x^2+y\right)^2=\left(3x^2-y-2x^2-y\right)\left(3x^2-y+2x^2+y\right)=\left(x^2-2y\right).5x^2\)
a. BT= ((3x+1) - (3x-5))2=62=36
b. BT = (3x2-y-2x2-y). (3x2- y + 2x2+ y) = (x2-2y).5x2
a: Ta có: \(A=\left(2x+y\right)^2-\left(2x-y\right)^2\)
\(=\left(2x+y-2x+y\right)\left(2x+y+2x-y\right)\)
\(=4x\cdot2y=8xy\)
b: Ta có: \(B=\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(2y-1\right)^2\)
\(=\left(3x+2+1-2y\right)^2\)
\(=\left(3x-2y+3\right)^2\)
Câu A) là \(\left(2x+y\right)^2-\left(y-2x\right)^2\)
Chứ ko phải là\(\left(2x+y\right)^2-\left(2x-y\right)^2\)
Nhưng dù sao thì cũng cảm ơn