Cho biểu thức:
A=(\(\dfrac{x+1}{x-1}\)-\(\dfrac{x-1}{x+1}\)) . \(\dfrac{5x-5}{2x}\)
A)tìm điều kiện xác định của A
b) Rút gọn A
c)tìm giá trị của A tại x=3;x=-1
em cảm ơn nhiều nhiều ạ !!
K
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19 tháng 11 2021
\(a,ĐK:x\ne0;x\ne1;x\ne\pm2\\ b,A=\left[\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}\right]\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(x+2\right)}\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{4x\left(x+1\right)\cdot x}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x^2}{x+2}\)
24 tháng 11 2021
\(a,ĐK:x>0;x\ne9\\ b,A=\dfrac{\sqrt{x}+3+\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}}\\ A=\dfrac{2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+3\right)}=\dfrac{2}{\sqrt{x}+3}\\ c,A>\dfrac{2}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+3}-\dfrac{2}{5}>0\\ \Leftrightarrow\dfrac{1}{\sqrt{x}+3}-\dfrac{1}{5}>0\\ \Leftrightarrow\dfrac{2-\sqrt{x}}{5\left(\sqrt{x}+3\right)}>0\\ \Leftrightarrow2-\sqrt{x}>0\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow\sqrt{x}< 2\Leftrightarrow0< x< 4\)
3 tháng 2 2022
a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)
b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)
\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)
\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)
\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)
c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)
Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.
6 tháng 2 2022
\(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\) (1)
a) ĐKXĐ: \(x\ne\pm3\)
b) \(\left(1\right)=\left[\dfrac{2x^2}{\left(x-3\right)\left(x+3\right)}+\dfrac{3\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{4}{5x+15}\)
\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5x+15}\)
\(=\dfrac{x^2+6x+9}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)^2}{\left(x-3\right)\left(x+3\right)}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{4}{5\left(x-3\right)}\)
c) Thay \(x=19\) vào \(A=\dfrac{4}{5\left(x-3\right)}\) ta có:
\(A=\dfrac{4}{5.\left(19-3\right)}=\dfrac{4}{80}=\dfrac{1}{20}\)
Vậy \(x=19\) thì \(A=\dfrac{1}{20}\)
MH
6 tháng 2 2022
a) ĐK: \(x\)≠\(+-3\)
b) \(A=\left(\dfrac{2x^2}{x^2-9}+\dfrac{3}{x-3}-\dfrac{x}{x+3}\right).\dfrac{4}{5x+15}\)
\(=\dfrac{2x^2+3\left(x+3\right)-x\left(x-3\right)}{x^2-9}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{2x^2+3x+9-x^2+3x}{\left(x+3\right)\left(x-3\right)}.\dfrac{4}{5\left(x+3\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4\left(x+3\right)^2}{5\left(x+3\right)^2\left(x-3\right)}=\dfrac{4}{5\left(x-3\right)}=\dfrac{4}{5x-15}\)
c) Tại x=19
⇒ \(A=\dfrac{4}{5.19-15}=\dfrac{4}{80}=\dfrac{1}{20}\)
Vậy ...
27 tháng 8 2021
a: ĐKXĐ: \(x\notin\left\{-1;3\right\}\)
Ta có: \(A=\dfrac{x^3-3}{x^2-2x-3}+\dfrac{6-2x}{x+1}+\dfrac{x+3}{3-x}\)
\(=\dfrac{x^3-3-2\left(x-3\right)^2-\left(x+3\right)\left(x+1\right)}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^3-3-2x^2+12x-18-x^2-4x-3}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^4-3x^2+8x-24}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^2\left(x-3\right)+8\left(x-3\right)}{\left(x-3\right)\left(x+1\right)}\)
\(=\dfrac{x^2+8}{x+1}\)
b: Ta có: A=x-2
\(\Leftrightarrow x^2+8=x^2-x-2\)
\(\Leftrightarrow8+x+2=0\)
hay x=-10
20 tháng 11 2023
a: ĐKXĐ: \(x\notin\left\{0;1;-1\right\}\)
b: \(A=\dfrac{x^2+x}{x^2-2x+1}:\left(\dfrac{x+1}{x}+\dfrac{1}{x-1}+\dfrac{2-x^2}{x^2-x}\right)\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{\left(x+1\right)\left(x-1\right)+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x^2-1+x+2-x^2}\)
\(=\dfrac{x^2\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}=\dfrac{x^2}{x-1}\)
c: \(A=\dfrac{x^2}{x-1}=\dfrac{x^2-1+1}{x-1}=x+1+\dfrac{1}{x-1}\)
=>\(A=x-1+\dfrac{1}{x-1}+2>=2\cdot\sqrt{\left(x-1\right)\cdot\dfrac{1}{x-1}}+2=2+2=4\)
Dấu '=' xảy ra khi (x-1)2=1
=>x-1=1 hoặc x-1=-1
=>x=0(loại) hoặc x=2(nhận)
Vậy: \(A_{min}=4\) khi x=2