Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a, dk \(x\ge0.x\ne1\)
\(\left(\frac{1+\sqrt{x}+1-\sqrt{x}}{2\left(1-x\right)}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)=\(\left(\frac{1}{1-x}-\frac{x^2+1}{1-x^2}\right)\left(\frac{x+1}{x}\right)\)
=\(\left(\frac{1+x-x^2-1}{1-x^2}\right)\left(\frac{x+1}{x}\right)=\frac{x\left(1-x\right)\left(x+1\right)}{x\left(1-x\right)\left(1+x\right)}=1\)
phan b,c ban tu lam not nhe dai lam mk ko lam dau mk co vc ban rui
Bài 1:
a) \(\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)
\(=\frac{2\left(\sqrt{3}+1\right)}{2}-\frac{2\left(\sqrt{3}-1\right)}{2}\)
\(=\sqrt{3}+1-\left(\sqrt{3}-1\right)=2\)
b) \(\frac{2}{5-\sqrt{3}}+\frac{3}{\sqrt{6}+\sqrt{3}}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{\left(5-\sqrt{3}\right)\left(5+\sqrt{3}\right)}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{\left(\sqrt{6}+\sqrt{3}\right)\left(\sqrt{6}-\sqrt{3}\right)}\)
\(=\frac{2\left(5+\sqrt{3}\right)}{2}+\frac{3\left(\sqrt{6}-\sqrt{3}\right)}{3}\)
\(=5+\sqrt{3}+\sqrt{6}-\sqrt{3}=5+\sqrt{6}\)
c) ĐK: \(a\ge0;a\ne1\)
\(\left(1+\frac{a+\sqrt{a}}{1+\sqrt{a}}\right).\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{1+\sqrt{a}}\right).\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)+a\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)+a\)
\(=1-a+a=1\)
cho S=1-3+32-33+...+398-399
a. Chứng minh: S chia hêt cho 20
b. Rút gọn S, từ đó suy ra 3100 chia 4 dư 1
chịu
a.\(DK:x\ge0\)
\(A=\frac{x-2\sqrt{x}+1}{x+1}.\frac{\left(x+1\right)\left(\sqrt{x}+1\right)}{x-2\sqrt{x}+1}=\sqrt{x}+1\)
b.Dat \(P=\frac{1}{A}\left(x+3\right)=\frac{x+3}{\sqrt{x}+1}\left(P>0\right)\)
\(\Rightarrow P\sqrt{x}+P=x+3\)
\(\Leftrightarrow x-P\sqrt{x}+3-P=0\)
Dat \(t=\sqrt{x}\left(t\ge0\right)\)
Ta co:
\(\Delta\ge0\)
\(\Leftrightarrow P^2-4\left(3-P\right)\ge0\)
\(\Leftrightarrow P^2+4P-12\ge0\)
\(\Leftrightarrow\left(P-2\right)\left(P+6\right)\ge0\)
TH1:
\(\hept{\begin{cases}P-2\ge0\\P+6\ge0\end{cases}\Leftrightarrow P\ge2}\)
TH2:
\(\hept{\begin{cases}P-2\le0\\P+6\le0\end{cases}\Leftrightarrow P\le2\left(P>0\right)}\)
Vi la de bai tim min nen lay TH1 thoi
Dau '=' xay ra khi \(x=\frac{P}{2}=1\)
Vay \(P_{min}=2\)khi \(x=1\)
a/ đkxđ \(\hept{\begin{cases}\sqrt{1+x}-\sqrt{1-x}\ne0\\\sqrt{1-x^2}-1+x\ne0\\x\ne0\end{cases}}va\hept{\begin{cases}1+x>0\\1-x>0\\1-x^2>0\end{cases}va}\sqrt{\frac{1}{x^2}-1}>0\)
\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne1\\-1< x< 1\end{cases}}vax>0\)
b =/\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x^2}-1+x}\right].\left[\frac{\sqrt{1-x^2}}{x}-\frac{1}{x}\right]\)=
\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{1-x}{\sqrt{1-x}\left[\sqrt{1+x}-\sqrt{1-x}\right]}\right].\frac{\sqrt{1-x^2}-1}{x}\)=\(\left[\frac{\sqrt{1+x}}{\sqrt{1+x}-\sqrt{1-x}}+\frac{\sqrt{1-x}}{\sqrt{1+x}-\sqrt{1-x}}\right].\frac{\sqrt{1-x^2}-1}{x}\)=\(\frac{\left[\sqrt{1+x}+\sqrt{1-x}\right]\left[\sqrt{1-x^2}-1\right]}{\left[\sqrt{1+x}-\sqrt{1-x}\right].x}\)
c/ khi x=1/2 thi A=\(\frac{\left[\sqrt{1+\frac{1}{2}}+\sqrt{1-\frac{1}{2}}\right]\left[\sqrt{1-\frac{1}{4}}-1\right]}{\left[\sqrt{1+\frac{1}{2}}-\sqrt{1-\frac{1}{2}}\right].\frac{1}{2}}=-1\)
a/ đkxđ
va{
va√1x2 −1>0
vax>0
b =/[√1+x√1+x−√1−x +1−x√1−x2−1+x ].[√1−x2x −1x ]=
[√1+x√1+x−√1−x +1−x√1−x[√1+x−√1−x] ].√1−x2−1x =[√1+x√1+x−√1−x +√1−x√1+x−√1−x ].√1−x2−1x =[√1+x+√1−x][√1−x2−1][√1+x−√1−x].x
c/ khi x=1/2 thi A=[√1+12 +√1−12 ][√1−14 −1][√1+12 −√1−12 ].12 =−1