Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(A=\frac{1}{x+5}+\frac{2}{x-5}-\frac{2x+10}{\left(x+5\right)\left(x-5\right)}\) ĐK đề bài
\(=\frac{x-5+2\left(x+5\right)-2x-10}{\left(x+5\right)\left(x-5\right)}=\frac{-\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}=-\frac{1}{x-5}\)
b/ có A=-3 => \(-\frac{1}{x-5}=-3 \Rightarrow x-5=\frac{1}{3}\Rightarrow x=\frac{16}{3}\)
có \(9x^2-42x+49=\left(3x-7\right)^2=\left(\frac{3.16}{3}-7\right)^2=81\)
a: ĐKXĐ: x<>0; x<>-3
b: \(=\dfrac{x^2+6x+9}{x\left(x+3\right)}\cdot\dfrac{2}{x+3}=\dfrac{2}{x}\)
c: Khi x=1/5 thì A=2:1/5=10
a: \(P=\dfrac{x\left(x+2\right)}{2\left(x+5\right)}+\dfrac{x-5}{x}-\dfrac{5x-50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+2x^2+2x^2-50-5x+50}{2x\left(x+5\right)}\)
\(=\dfrac{x^3+4x^2-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x+5\right)\left(x-1\right)}{2x\left(x+5\right)}=\dfrac{x-1}{2}\)
\(a,A=\dfrac{2x\left(x-3\right)+8\left(x+3\right)-2x-12}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x^2+6}\\ A=\dfrac{2x^2-6x+8x+24-2x-12}{\left(x-3\right)}\cdot\dfrac{1}{x^2+6}\\ A=\dfrac{2x^2+12}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2\left(x^2+6\right)}{\left(x-3\right)\left(x^2+6\right)}=\dfrac{2}{x-3}\)
\(b,A=5\Leftrightarrow\dfrac{2}{x-3}=5\Leftrightarrow5x-15=2\Leftrightarrow x=\dfrac{17}{5}\)
Bài 1:
\(a,ĐK:x\ne\pm5\\ b,P=\dfrac{x-5+2x+10-2x-10}{\left(x-5\right)\left(x+5\right)}=\dfrac{x-5}{\left(x-5\right)\left(x+5\right)}=\dfrac{1}{x+5}\\ c,P=-3\Leftrightarrow x+5=-\dfrac{1}{3}\Leftrightarrow x=-\dfrac{16}{3}\\ d,P\in Z\Leftrightarrow x+5\inƯ\left(1\right)=\left\{-1;1\right\}\\ \Leftrightarrow x\in\left\{-6;-4\right\}\)
Bài 2:
\(a,\Leftrightarrow\dfrac{3\left(x^2+2x+4\right)}{\left(x-2\right)\left(x^2+2x+4\right)}=\dfrac{3}{x-2}=0\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow\dfrac{x\left(2-x\right)}{\left(x-2\right)\left(x+2\right)}=0\Leftrightarrow\dfrac{-x}{x+2}=0\Leftrightarrow x=0\)
a ) Rút gọn : \(A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2x+10}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2\left(x+5\right)}{\left(x+5\right)\left(x-5\right)}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}+\dfrac{2}{x-5}-\dfrac{2}{x-5}\)
\(\Leftrightarrow A=\dfrac{1}{x+5}.\)
Khi \(A=-3\),thì :
\(\dfrac{1}{x+5}=-3\Leftrightarrow x=-\dfrac{16}{3}\)
Ta có : \(9x^2-42x+49\)
\(=\left(3x\right)^2-2.3x.7+49\)
\(=\left(3x-7\right)^2\)
Thay \(x=-\dfrac{16}{3},\) ta có :
\(\left(3.\dfrac{-16}{3}-7\right)^2=\left(-16-7\right)^2=\left(-23\right)^2=529\)