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1) \(ĐK:x\in R\)
2) \(ĐK:x< 0\)
3) \(ĐK:x\in\varnothing\)
4) \(=\sqrt{\left(x+1\right)^2+2}\)
\(ĐK:x\in R\)
5) \(=\sqrt{-\left(a-4\right)^2}\)
\(ĐK:x\in\varnothing\)
1) ĐKXĐ: \(x^2+2x-3\ge0\Leftrightarrow\left(x+1\right)^2\ge4\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1\ge2\\x+1\le-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-3\end{matrix}\right.\)
2) ĐKXĐ: \(2x^2+5x+3\ge0\Leftrightarrow2\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{8}\Leftrightarrow\left(x+\dfrac{5}{4}\right)^2\ge\dfrac{1}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{5}{4}\ge\dfrac{1}{4}\\x+\dfrac{5}{4}\le-\dfrac{1}{4}\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x\ge-1\\x\le-\dfrac{3}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(x-1>0\Leftrightarrow x>1\)
4) ĐKXĐ: \(x-3< 0\Leftrightarrow x< 3\)
5) ĐKXĐ: \(x+2< 0\Leftrightarrow x< -2\)
6) ĐKXĐ: \(2a-1>0\Leftrightarrow a>\dfrac{1}{2}\)
1) \(ĐK:3-2a>0\Leftrightarrow a< \dfrac{3}{2}\)
2) \(ĐK:2x-5< 0\Leftrightarrow x< \dfrac{5}{2}\)
3) \(ĐK:3-5a< 0\Leftrightarrow a>\dfrac{3}{5}\)
4) \(ĐK:a< 0\)
5) \(ĐK:-a\ge0\Leftrightarrow a\le0\)
1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\x\ne3\end{matrix}\right.\)
2: Ta có: \(P=\dfrac{x-3}{\sqrt{x-1}-2}\)
\(=\dfrac{x-1-2}{\sqrt{x-1}-2}\)
\(=\sqrt{x-1}+2\)
a)√x−2+12√4x−8=√9x−18−2
=>√x−2+12√4(x−2)=√9(x−2)−2
=>√x−2+12√22(x−2)=√32(x−2)−2
=>√x−2+12.2√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)=3√(x−2)−2
=>√x−2+24√(x−2)-3√(x−2)=-2
=>√x−2(1+24-3)=-2
=>22√x−2=-2
=>√x−2=-2/22
=>√x−2=-1/11
=>x−2=1/121
=>x=1/121+2=243/121
b)√(3x−1)2=5
=>|3x−1|=5
=>3x−1=5 hoặc 3x−1=-5
=>3x=6 hoặc 3x=-4
=>x=2 hoặc x=-4/3
a. ĐKXĐ: $x\geq 0$
PT $\Leftrightarrow -5x-5\sqrt{x}+12\sqrt{x}+12=0$
$\Leftrightarrow -5\sqrt{x}(\sqrt{x}+1)+12(\sqrt{x}+1)=0$
$\Leftrightarrow (\sqrt{x}+1)(12-5\sqrt{x})=0$
Dễ thấy $\sqrt{x}+1>1$ với mọi $x\geq 0$ nên $12-5\sqrt{x}=0$
$\Leftrightarrow \sqrt{x}=\frac{12}{5}$
$\Leftrightarrow x=5,76$ (thỏa mãn)
b. ĐKXĐ: $x^2\geq 5$
PT $\Leftrightarrow \frac{1}{3}\sqrt{4}.\sqrt{x^2-5}+2\sqrt{\frac{1}{9}}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow \frac{2}{3}\sqrt{x^2-5}+\frac{2}{3}\sqrt{x^2-5}-3\sqrt{x^2-5}=0$
$\Leftrightarrow -\frac{5}{3}\sqrt{x^2-5}=0$
$\Leftrightarrow \sqrt{x^2-5}=0$
$\Leftrightarrow x=\pm \sqrt{5}$
1: ĐKXĐ: \(a>-2\)
2: ĐKXĐ: \(x\ne2\)
3: ĐKXĐ: \(a\in\varnothing\)
1)
\(-\dfrac{1}{\sqrt{a+2}}\) có nghĩa khi \(\sqrt{a+2}>0\)
=>a+2>0
a>-2
2)
\(\sqrt{\dfrac{3}{\left(x-2\right)^2}}=\dfrac{\sqrt{3}}{\sqrt{\left(x-2\right)^2}}\)
mà \(\left(x-2\right)^2>0=>\sqrt{\left(x-2\right)^2}>0vớimọix\)
3)
\(\sqrt{\dfrac{-3}{a^2-4a+4}}=\sqrt{\dfrac{-3}{\left(a-2\right)^2}}cónghĩakhi\left(a-2\right)^2< 0mà\left(a-2\right)^2>0=>biểuthứckocónghĩavớimọia\)
1)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\sqrt{\left(\sqrt{11}-\sqrt{3}\right)^2}=\sqrt{11}-\sqrt{3}\)
2)
\(=\sqrt{\left(\sqrt{7}\right)^2-2.\sqrt{7}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}=\sqrt{7}-\sqrt{5}\)
3)
\(=\sqrt{\left(\sqrt{11}\right)^2-2.\sqrt{11}\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(\sqrt{11}-\sqrt{5}\right)}=\sqrt{11}-\sqrt{5}\)
4)
\(=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=3-\sqrt{5}\)
5)
\(=\sqrt{3^2-2.3.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(3-2\sqrt{2}\right)^2}=3-2\sqrt{2}\)
1) ĐKXĐ: \(16x^2-25\ge0\)
\(\Leftrightarrow x^2\ge\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{5}{4}\\x\le-\dfrac{5}{4}\end{matrix}\right.\)
2) ĐKXĐ: \(4x^2-49\ge0\Leftrightarrow x^2\ge\dfrac{49}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge\dfrac{7}{2}\\x\le-\dfrac{7}{2}\end{matrix}\right.\)
3) ĐKXĐ: \(8-x^2\ge0\Leftrightarrow x^2\le8\)
\(\Leftrightarrow-2\sqrt{2}\le x\le2\sqrt{2}\)
4) ĐKXĐ: \(x^2-12\ge0\Leftrightarrow x^2\ge12\)
\(\Leftrightarrow\left[{}\begin{matrix}x\ge2\sqrt{3}\\x\le-2\sqrt{3}\end{matrix}\right.\)
5) ĐKXĐ: \(x^2+4\ge0\left(đúng\forall x\right)\)