Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
a.\(\left(x+y\right)^2-\left(x-y\right)^2=\left(x+y-x+y\right)\left(x+y+x-y\right)=2\left(x+y\right)\)
b.\(2\left(x+y\right)\left(x-y\right)+\left(x+y\right)^2+\left(x-y\right)^2=\left(x+y+x-y\right)^2=4x^2\)
a) Ta có: \(\dfrac{3x^2-12x+12}{x^2-4}\)
\(=\dfrac{3\left(x^2-4x+4\right)}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\)
\(=\dfrac{3\left(x-2\right)}{x+2}\)
\(=\dfrac{3\cdot\left(\dfrac{-1}{4}-2\right)}{\dfrac{-1}{4}+2}=-\dfrac{27}{7}\)
b) Ta có: \(\dfrac{x^2-5x-6}{x^2-9}\)
\(=\dfrac{\left(x-6\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)
\(=\dfrac{\left(-1-6\right)\left(-1+1\right)}{\left(-1-3\right)\left(-1+3\right)}\)
=0
Bài 1:
a) Ta có: \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow x^3+2x^2+x^2+2x+x+2=0\)
\(\Leftrightarrow x^2\left(x+2\right)+x\left(x+2\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
mà \(x^2+x+1>0\forall x\)
nên x+2=0
hay x=-2
Vậy: x=-2
b) Ta có: \(x^3-12x^2+48x-72=0\)
\(\Leftrightarrow x^3-6x^2-6x^2+36x+12x-72=0\)
\(\Leftrightarrow x^2\left(x-6\right)-6x\left(x-6\right)+12\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x^2-6x+12\right)=0\)
mà \(x^2-6x+12>0\forall x\)
nên x-6=0
hay x=6
Vậy: x=6