K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
1 tháng 1 2021
a, \(A=\frac{x+1}{x-2}+\frac{x-1}{x+2}+\frac{x^2+4x}{4-x^2}\)
\(=\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\frac{x^2+4x}{\left(x-2\right)\left(x+2\right)}\)
\(=\frac{x^2+2x+x+2+x^2-2x-x+2-x^2-4}{\left(x-2\right)\left(x+2\right)}=\frac{x^2}{\left(x-2\right)\left(x+2\right)}\)
b, Thay x = 4 ta có :
\(\frac{4^2}{\left(4-2\right)\left(4+2\right)}=\frac{16}{2.8}=\frac{16}{16}=1\)
Vậy \(A=1\)
28 tháng 9 2021
m khác 3 nhá chứ ko phải -3 đâu bạn ạ
Để đths trên song song <=> \(\hept{\begin{cases}m-1=3-m\\2\ne1\end{cases}}\Leftrightarrow2m=4\Leftrightarrow m=2\)( tm )
VE
0
10 tháng 7 2021
Với \(x>0;x\ne4\)
\(\left(\frac{2}{\sqrt{x}-2}+\frac{3}{2\sqrt{x}+1}-\frac{5\sqrt{x}-7}{2x-3\sqrt{x}-2}\right):\frac{2\sqrt{x}+3}{5x-10\sqrt{x}}\)
\(=\left(\frac{2\left(2\sqrt{x}+1\right)+3\left(\sqrt{x}-2\right)-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right):\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right):\frac{2\sqrt{x}+3}{5\sqrt{x}\left(\sqrt{x}-2\right)}\)
\(=\left(\frac{2\sqrt{x}+3}{\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)}\right).\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
10 tháng 7 2021
\(A=\left[\frac{2\left(2\sqrt{x}+1\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}+\frac{3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}-\frac{5\sqrt{x}-7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\right]\times\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{4\sqrt{x}+2+3\sqrt{x}-6-5\sqrt{x}+7}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)}\times\frac{5\sqrt{x}\left(\sqrt{x}-2\right)}{2\sqrt{x}+3}\)
\(=\frac{5\sqrt{x}\left(2\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(2\sqrt{x}+1\right)\left(2\sqrt{x}+3\right)}=\frac{5\sqrt{x}}{2\sqrt{x}+1}\)
27 tháng 6 2021
Ta có: \(\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4+y^4}+\frac{8y^8}{x^8-y^8}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\left[\frac{4y^4}{x^4+y^4}+\frac{8y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}\right]=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4\left(x^4-y^4\right)+8y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}=4\)
\(\Leftrightarrow\frac{x}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4x^4y^4+4y^8}{\left(x^4-y^4\right)\left(x^4+y^4\right)}=4\)
\(\Leftrightarrow\frac{x}{x+y}+\frac{2y^2}{x^2+y^2}+\frac{4y^4}{x^4-y^4}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\left[\frac{2y^2}{x^2+y^2}+\frac{4y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}\right]=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2\left(x^2-y^2\right)+4y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2x^2y^2+2y^4}{\left(x^2-y^2\right)\left(x^2+y^2\right)}=4\)
\(\Leftrightarrow\frac{y}{x+y}+\frac{2y^2}{x^2-y^2}=4\)
\(\Leftrightarrow\frac{y\left(x-y\right)+2y^2}{\left(x-y\right)\left(x+y\right)}=4\)
\(\Leftrightarrow\frac{xy+y^2}{\left(x+y\right)\left(x-y\right)}=4\)
\(\Leftrightarrow\frac{y}{x-y}=4\)
\(\Leftrightarrow y=4x-4y\Rightarrow4x=5y\)
=> đpcm
\(\Leftrightarrow5.5^{x^2}-5.\frac{1}{5^{x^2}}=24\)
Đặt \(t=5^{x^2}\)
\(5t-\frac{5}{t}=24\Leftrightarrow5t^2-24t-5=0\)\(\Leftrightarrow t=5\) hoặc \(t=-\frac{1}{5}=5^{-1}\)
\(\Rightarrow5^{x^2}=5\) hoặc \(5^{x^2}=5^{-1}\)
\(\Rightarrow x^2=1\Rightarrow x=1\) hoặc \(x=-1\)