C/M: (ax^2 - bx^2)^4 + (2ab+bx^2)^4 + (2ab+a^2)^4 = 2(a^2+ab+b^2)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\dfrac{ax+ay-bx-by}{ax-ay-bx+by}=\dfrac{a\left(x+y\right)-b\left(x+y\right)}{a\left(x-y\right)-b\left(x-y\right)}=\dfrac{\left(a-b\right)\left(x+y\right)}{\left(a-b\right)\left(x-y\right)}=\dfrac{x+y}{x-y}\)
b) \(\dfrac{a^2+b^2-c^2+2ab}{a^2-b^2+c^2+2ac}=\dfrac{\left(a+b\right)^2-c^2}{\left(a+c\right)^2-b^2}=\dfrac{\left(a+b+c\right)\left(a+b-c\right)}{\left(a+c+b\right)\left(a+c-b\right)}=\dfrac{a+b-c}{a+c-b}\)
\(=x\left(a-b\right)-\left(a^2-2ab+b^2\right)=\)
\(=\left(a-b\right)x-\left(a-b\right)^2=\)
\(=\left(a-b\right)\left[x-\left(a-b\right)\right]=\left(a-b\right)\left(x-a+b\right)\)
\(1)4x^2-25+\left(2x+7\right).\left(5.2x\right)\)
\(=\left(2x\right)^2-5^2-\left(2x+7\right).\left(2x-5\right)\)
\(=\left(2x.5\right)\left(2x+5\right).\left(2x+7\right)\left(2x-5\right)\)
\(=\left(2x-5\right)\left(2x+5-2x+7\right)\)
\(=\left(2x-5\right).12\)
\(2)3x+4-x^2-4x\)
\(=3(x+4)-\left(x+4\right)\)
\(=\left(3-x\right)\left(x+4\right)\)
\(3)5x^2-2y^2-10x+10y\)
\(=5\left(x^2-y^2\right)-10\left(x-4\right)\)
\(=5\left(x-y\right)\left(x+y\right)-10\left(x-y\right)\)
\(=\left(x-y\right)[5(x+y)-10]\)
Còn lại bn lm nốt nha!