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Chỗ dấu bằng thứ hai sai nên bạn làm cũng chưa đúng
x^6 -y^6 = (x^2-y^2)(x^4 +x^2 .y^2 + y^4)
Bạn hiểu ra chỗ sai của mình chưa.Chúc bạn học tốt.
\(5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2=\left(x-y\right)^2\left[5\left(x-y\right)^2-3\left(x-y\right)+4\right]\)
\(\left(y-x\right)^2=\left(x-y\right)^2\)
\(\Rightarrow\left[5\left(x-y\right)^4-3\left(x-y\right)^3+4\left(x-y\right)^2\right]:\left(y-x\right)^2=5\left(x-y\right)^2-3\left(x-y\right)+4\)
\(A=\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)
\(=\frac{x^2}{\left(x-y\right)\left(x-z\right)}-\frac{y^2}{\left(x-y\right)\left(y-z\right)}+\frac{z^2}{\left(x-z\right)\left(y-z\right)}\)
\(=\frac{x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)}{\left(x-y\right)\left(x-z\right)\left(y-z\right)}\)
\(x^2\left(y-z\right)-y^2\left(x-z\right)+z^2\left(x-y\right)\)
\(=x^2y-x^2z-xy^2+y^2z+z^2\left(x-y\right)\)
\(=xy\left(x-y\right)-z\left(x-y\right)\left(x+y\right)+z^2\left(x-y\right)\)
\(=\left(x-y\right)\left[xy-zx-zy+z^2\right]\)
\(=\left(x-y\right)\left[x\left(y-z\right)-z\left(y-z\right)\right]=\left(x-y\right)\left(x-z\right)\left(y-z\right)\)
Vậy A = 1
Dùng hằng đẳng thức đáng nhớ thôi b
Ta có y2 - x2 = (y - x)(y + x)
Mà theo đêc bài thì mẫu có (y + x) rồi nên chỉ cần nhân cho (y - x) nữa là được
b: Ta có: \(\left(4x^4-3x^3\right):\left(-x^3\right)+\left(15x^2+6x\right):3x=0\)
\(\Leftrightarrow-4x+3+5x+2=0\)
\(\Leftrightarrow x=-5\)
5) \(\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)\)
\(=\left(x-y\right)^2-2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2\)
\(=\left[\left(x-y\right)-\left(x+y\right)\right]^2\)
\(=\left(x-y-x-y\right)^2\)
\(=\left(-2y^2\right)\)
\(=4y^2\)
6) \(\left(5-x\right)^2+\left(x+5\right)^2-\left(2x+10\right)\left(x-5\right)\)
\(=\left(x-5\right)^2-2\left(x-5\right)\left(x+5\right)+\left(x+5\right)^2\)
\(=\left[\left(x-5\right)-\left(x+5\right)\right]^2\)
\(=\left(x-5-x-5\right)^2\)
\(=\left(-10\right)^2=100\)
7) \(\left(x-2\right)^2+\left(x+1\right)^2+2\left(x-2\right)\left(-1-x\right)\)
\(=\left(x-2\right)^2-2\left(x-2\right)\left(x+1\right)+\left(x+1\right)^2\)
\(=\left[\left(x-2\right)-\left(x+1\right)\right]^2\)
\(=\left(-3\right)^2=9\)
8) \(-\left(2x+3y\right)^2+\left(2x-3y\right)^2-2\left(4x^2-9y^2\right)\)
\(=\left(2x-3y\right)^2+2\left(2x+3y\right)\left(2x-3y\right)+\left(2x+3y\right)^2\)
\(=\left[\left(2x+3y\right)+\left(2x-3y\right)\right]^2\)
\(=\left(4x\right)^2=16x^2\)
\(\frac{4\left(x-y\right)^5+2\left(x-y\right)^3-\left(x-y\right)^2}{\left(y-x\right)^2}\)
\(=\frac{4\left(x-y\right)^5+2\left(x-y\right)^3-\left(x-y\right)^2}{\left(x-y\right)^2}\)
\(=4\left(x-y\right)^3+2\left(x-y\right)-1\)
Điều kiện
\(x\ne y\)
A=\(-4\left(x-y\right)^3-2\left(x-y\right)+1\)