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a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\1-x^2\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ne0\\-1\le x\le1\end{matrix}\right.\)
b/ ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4>0\\x+1\ge0\end{matrix}\right.\) \(\Rightarrow x>2\)
c/ ĐKXĐ: \(\left\{{}\begin{matrix}1+x\ge0\\x-3\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x\ge-1\\x\ne3\end{matrix}\right.\)
d/ ĐKXĐ: \(\left\{{}\begin{matrix}x^2-4x+3>0\\x\ge-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\\x\ge-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x>3\\-1\le x< 1\end{matrix}\right.\)
a/ ĐKXĐ: ...
\(\Leftrightarrow2\sqrt{\frac{x}{x-1}}-\sqrt{\frac{x-1}{x}}=\frac{2\left(x-1\right)}{x}+3\)
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(\frac{2}{a}-a=2a^2+3\Leftrightarrow2a^3+a^2+3a-2=0\)
\(\Leftrightarrow\left(2a-1\right)\left(a^2+a+2\right)=0\Leftrightarrow a=\frac{1}{2}\)
\(\Rightarrow\sqrt{\frac{x-1}{x}}=\frac{1}{2}\Leftrightarrow4\left(x-1\right)=x\)
b/ ĐKXĐ: ...
\(\Leftrightarrow3\sqrt{\frac{2x}{x-1}}+4\sqrt{\frac{x-1}{2x}}=\frac{3\left(x-1\right)}{2x}+10\)
Đặt \(\sqrt{\frac{x-1}{2x}}=a>0\)
\(\frac{3}{a}+4a=3a^2+10\Leftrightarrow3a^3-4a^2+10a-3=0\)
\(\Leftrightarrow\left(3a-1\right)\left(a^2-a+3\right)=0\Leftrightarrow a=\frac{1}{3}\)
\(\Leftrightarrow\sqrt{\frac{x-1}{2x}}=\frac{1}{3}\Leftrightarrow9\left(x-1\right)=2x\)
c/ ĐKXĐ: ...
\(\Leftrightarrow\sqrt{\frac{x}{3-2x}}+5\sqrt{\frac{3-2x}{x}}=\frac{4\left(3-2x\right)}{x}+5\)
Đặt \(\sqrt{\frac{3-2x}{x}}=a>0\)
\(\frac{1}{a}+5a=4a^2+5\Leftrightarrow4a^3-5a^2+5a-1=0\)
\(\Leftrightarrow\left(4a-1\right)\left(a^2-a+1\right)=0\Leftrightarrow a=\frac{1}{4}\)
\(\Leftrightarrow\sqrt{\frac{3-2x}{x}}=\frac{1}{4}\Leftrightarrow16\left(3-2x\right)=x\)
d/ ĐKXĐ: ...
Đặt \(\sqrt{\frac{x-1}{x}}=a>0\)
\(a^2-2a=3\Leftrightarrow a^2-2a-3=0\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=3\end{matrix}\right.\)
\(\Leftrightarrow\sqrt{\frac{x-1}{x}}=3\Leftrightarrow x-1=9x\)
a)\(ĐK:-3\le x\le6\)
\(PT\Leftrightarrow\sqrt{x+3}+\sqrt{6-x}=3\)
\(\Leftrightarrow x+3+6-x+2\sqrt{\left(x+3\right)\left(6-x\right)}=9\)
\(\Leftrightarrow\sqrt{\left(x+3\right)\left(6-x\right)}=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=6\end{matrix}\right.\left(tm\right)\)
b/ ĐKXĐ: \(x\ge7\)
\(\sqrt{3x-2}=1+\sqrt{x-7}\)
\(\Leftrightarrow3x-2=x-6+2\sqrt{x-7}\)
\(\Leftrightarrow x+2=\sqrt{x-7}\)
\(\Leftrightarrow x^2+4x+4=x-7\)
\(\Leftrightarrow x^2+3x+11=0\) (vô nghiệm)
c/ ĐKXĐ: \(x\ge1;x\ne50\)
\(1-\sqrt{3x+1}=\sqrt{x-1}-7\)
\(\Leftrightarrow\sqrt{x-1}+\sqrt{3x+1}=8\)
\(\Leftrightarrow4x+2\sqrt{3x^2-2x-1}=64\)
\(\Leftrightarrow\sqrt{3x^2-2x-1}=32-2x\) (\(x\le16\))
\(\Leftrightarrow3x^2-2x-1=\left(32-2x\right)^2\)
ĐK: \(x^4-4x^3+14x-11\ge0\) (*)
\(PT\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3+14x-11=x^2-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\x^4-4x^3-x^2+16x-12=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\le1\\\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)(tm)
e/ ĐKXĐ: \(x\ge1\)
\(\Leftrightarrow x+3-\sqrt{x-1}=4\)
\(\Leftrightarrow\sqrt{x-1}=x-1\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-3x+2=0\Rightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
f/ \(\Leftrightarrow\left\{{}\begin{matrix}x+5\ge0\\x^3+x^2+6x+28=\left(x+5\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\x^3-4x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-5\\\left(x-1\right)\left(x^2+x-3\right)=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{-1\pm\sqrt{13}}{2}\\\end{matrix}\right.\)
a) \(Y=\frac{\sqrt{3-2x}}{\sqrt{1-x}}+\frac{\sqrt{2x+1}}{x}\)
\(\Rightarrow\left\{{}\begin{matrix}3-2x\ge0\\1-x>0\\2x+1\ge0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\le\frac{3}{2}\\x< 1\\x\ge\frac{-1}{2}\\x\ne0\end{matrix}\right.\)
TXĐ: \([-\frac{1}{2};\frac{3}{2}]\backslash\left\{0\right\}\)
b) \(Y=\frac{\sqrt{3x+5}}{x-2}+\frac{\sqrt{2x+3}}{\sqrt{4-x}}\)
\(\left\{{}\begin{matrix}3x+5\ge0\\x-2\ne0\\2x+3\ge0\\4-x>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge-\frac{5}{3}\\x\ne2\\x\ge-\frac{3}{2}\\x< 4\end{matrix}\right.\)
TXĐ: \([-\frac{5}{3};4)\backslash\left\{2\right\}\)
Em thử nhá, ko chắc đâu...
Gọi B là tổng các phần tử trong tập hợp A.
Thì \(B=\frac{1}{\sqrt{1}+\sqrt{2}}+\frac{1}{\sqrt{2}+\sqrt{3}}+...+\frac{1}{\sqrt{99}+\sqrt{100}}\)
Xét dạng tổng quát \(\frac{1}{\sqrt{n}+\sqrt{n+1}}\left(n\in Z^+\right)=\frac{\sqrt{n+1}-\sqrt{n}}{n+1-n}=\sqrt{n+1}-\sqrt{n}\)
Suy ra \(B=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+....+\sqrt{100}-\sqrt{99}\)
\(=\sqrt{100}-1=10-1=9\) là một số nguyên (đpcm)