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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: 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: ĐKXĐ: \(-1< x< 1\)
2: ĐKXĐ: \(\left[{}\begin{matrix}x>2\\x\le-1\end{matrix}\right.\)
3: ĐKXĐ: \(\left[{}\begin{matrix}x< -3\\x\ge2\end{matrix}\right.\)
4: ĐKXĐ: \(2< a\le3\)
Bài 1:
\(a,ĐK:2+8x\ge0\Leftrightarrow x\ge-\dfrac{1}{4}\\ b,ĐK:-\dfrac{1}{5}x+9\ge0\Leftrightarrow-\dfrac{1}{5}x\ge-9\Leftrightarrow x\le45\\ c,ĐK:11-7x\ge0\Leftrightarrow x\le\dfrac{11}{7}\)
Bài 2:
\(a,=\sqrt{144a^2}-2a=12\left|a\right|-2a=12a-2a=10\\ b,=\sqrt{6}-6\sqrt{6}-\sqrt{6}=-6\sqrt{6}\)
Bài 3:
\(a,\Leftrightarrow\left|2x+3\right|=3\Leftrightarrow\left[{}\begin{matrix}2x+3=3\\2x+3=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\\ b,ĐK:x\ge2\\ PT\Leftrightarrow2\sqrt{x-2}-4\sqrt{x-2}+3\sqrt{x-2}=4\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)
1)\(=\left|\sqrt{3}-3\right|+\sqrt{\left(\sqrt{3}-1\right)^2}=3-\sqrt{3}+\left|\sqrt{3}-1\right|=3-\sqrt{3}+\sqrt{3}-1=2\)
1)Để căn có nghĩa \(\Leftrightarrow\dfrac{-a}{3}\ge0\Leftrightarrow a\le0\)
Vậy...
2)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a^2+1}{1-3a}\ge0\\1-3a\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}1-3a>0\left(vìa^2+1>0\right)\\1-3a\ne0\end{matrix}\right.\)
\(\Leftrightarrow1-3a>0\Leftrightarrow3a< 1\Leftrightarrow a< \dfrac{1}{3}\)
Vậy...
3)Để căn có nghĩa
\(\Leftrightarrow a^2-6a+10\ge0\Leftrightarrow\left(a^2-6a+9\right)+1\ge0\Leftrightarrow\left(a-3\right)^2+1\ge0\left(lđ;\forall a\right)\)
Vậy căn luôn có nghĩa với mọi a
4)Để căn có nghĩa \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a-1}{a+2}\ge0\\a+2\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}\left\{{}\begin{matrix}a-1\ge0\\a+2>0\end{matrix}\right.\\\left\{{}\begin{matrix}a-1\le0\\a+2< 0\end{matrix}\right.\end{matrix}\right.\\a+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a\ge1\\a>-2\end{matrix}\right.\\\left\{{}\begin{matrix}a\le1\\a< -2\end{matrix}\right.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}a\ge1\\a< -2\end{matrix}\right.\)
Vậy...
a) Ta có: \(A=\dfrac{a^2-1}{3}\cdot\sqrt{\dfrac{9}{\left(1-a\right)^2}}\)
\(=\dfrac{\left(a+1\right)\cdot\left(a-1\right)}{3}\cdot\dfrac{3}{\left|1-a\right|}\)
\(=\dfrac{\left(a+1\right)\left(a-1\right)}{1-a}\)
=-a-1
b) Ta có: \(B=\sqrt{\left(3a-5\right)^2}-2a+4\)
\(=\left|3a-5\right|-2a+4\)
\(=5-3a-2a+4\)
=9-5a
c) Ta có: \(C=4a-3-\sqrt{\left(2a-1\right)^2}\)
\(=4a-3-\left|2a-1\right|\)
\(=4a-3-2a+1\)
\(=2a-2\)
d) Ta có: \(D=\dfrac{a-2}{4}\cdot\sqrt{\dfrac{16a^4}{\left(a-2\right)^2}}\)
\(=\dfrac{a-2}{4}\cdot\dfrac{4a^2}{\left|a-2\right|}\)
\(=\dfrac{a^2\left(a-2\right)}{-\left(a-2\right)}\)
\(=-a^2\)
`M=sqrt{(3a-1)^2}+2a-3`
`=|3a-1|+2a-3`
`=3a-1+2a-3(do \ a>=1/3)`
`=5a-4`
`N=sqrt{(4-a)^2}-a+5`
`=|4-a|-a+5`
`=a-4-a+5(do \ a>4)`
`=1`
`I=sqrt{(3-2a)^2}+2-7`
`=|3-2a|-5`
`=3-2a-5(do \ a<3/2)`
`=-2-2a`
`K=(a^2-9)/4*sqrt{4/(a-2)^2}`
`=(a^2-9)/4*|2/(a-2)|`
`=(a^2-9)/(2|a-2|)`
Nếu `3>a>2=>|a-2|=a-2`
`=>K=(a^2-9)/(2(a-2))`
Nếu `a<2=>|a-2|=2-a`
`=>K=(a^2-9)/(2(2-a))`
\(M=\left|3a-1\right|+2a-3\)
Mà \(a-\dfrac{1}{3}\ge0\)
\(\Rightarrow M=3a-1+2a-3=5a-4\)
\(N=\left|4-a\right|-a+5\)
Mà \(4-a< 0\)
\(\Rightarrow N=a-4-a+5=1\)
\(I=\left|3-2a\right|-5\)
Mà \(a-\dfrac{3}{2}< 0\)
\(\Rightarrow I=3-2a-5=-2a-2\)
K, Ta có : \(a-3< 0\)
\(\Rightarrow K=\dfrac{2\left(a^2-9\right)}{4\left|a-2\right|}=\dfrac{\left(a-3\right)\left(a+3\right)}{\left|2a-4\right|}\)
a: \(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
b: \(\sqrt{\dfrac{2a}{3}}\cdot\sqrt{\dfrac{3a}{8}}=\sqrt{\dfrac{6a^2}{24}}=\sqrt{\dfrac{a^2}{4}}=\dfrac{a}{2}\)
c: \(\sqrt{5a\cdot45a}-3a=-15a-3a=-18a\)
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) \(ĐK:-a\ge0\Leftrightarrow a\le0\)