Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có: 56 . 64 = ( 60 - 4 ) ( 60 + 4 ) = 60 2 - 4 2 = 3600 - 16 = 3584
Ta có: 56 . 64 = ( 60 - 4 ) ( 60 + 4 ) = 60 2 - 4 2 = 3600 - 16 = 3584 .
a/(x+1)(x-1)=x² - 1²
b/(x-2y) (x+2y) =x²-(2y)²=x² - 4y
c/ 56.64=(60-4).(60+4)
=60² - 4²=3600-16=3584
*Áp dụng hằng đẳng thức nha bn!
Câu a : \(\left(x+1\right)\left(x-1\right)=x^2-1^2\)
Câu b : \(\left(x-2y\right)\left(x+2y\right)=x^2-4y^2\)
Câu c : \(56.64=\left(60-4\right)\left(60+4\right)=60^2-4^2=3600-16=3584\)
\(56.64=\left(60-4\right).\left(60+4\right)=60^2-4^2=3600-16=3584\)
B1:
\(a,A=\left(\frac{3-x}{x+3}.\frac{x^2+6x+9}{x^2-9}+\frac{x}{x+3}\right):\frac{3x^2}{x+3}\)
\(=\left(\frac{\left(3-x\right)\left(x+3\right)^2}{\left(x+3\right)\left(x^2-9\right)}+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)
\(=\left(\frac{3-x}{x-3}+\frac{x}{x+3}\right).\frac{x+3}{3x^2}\)
\(=\left(\frac{\left(3-x\right)\left(x+3\right)}{x^2-9}+\frac{x\left(x-3\right)}{x^2-9}\right).\frac{x+3}{3x^2}\)
\(=\frac{3x+9-x^2-3x+x^2-3x}{x^2-9}.\frac{x+3}{3x^2}\)
\(=\frac{9-3x}{x^2-9}.\frac{x+3}{3x^2}\)
\(=\frac{3\left(3-x\right)\left(x+3\right)}{\left(x+3\right)\left(x-3\right)3x^2}\)
\(=\frac{3-x}{x^3-3x^2}\)
B2:
\(a,B=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{x^2-4}-\frac{2}{x-2}+\frac{1}{x+2}\right):\left(\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right)\)
\(=\left(\frac{x}{x^2-4}-\frac{2\left(x+2\right)}{x^2-4}+\frac{x+2}{x^2-4}\right):\left(\frac{x^2-4+10-x^2}{x+2}\right)\)
\(=\left(\frac{x-2x-4+x-2}{x^2-4}\right):\frac{6}{x+2}\)
\(=-\frac{6}{x^2-4}.\frac{x+2}{6}\)
\(=\frac{-6\left(x+2\right)}{\left(x+2\right)\left(x-2\right)6}=-\frac{1}{x-2}\)
a) Ta có: ( x - 2 )( x + 2 ) = ( x )2 - 22 = x2 - 4.
b) Ta có: 56.64 = ( 60 - 4 )( 60 + 4 ) = 602 - 42 = 3600 - 16 = 3584.