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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(3-\sqrt{x}=\)0
\(\sqrt{x}=0+3\)
\(\sqrt{x}=3\)
mà :\(\sqrt{9}=3\)
=> x = 9
![](https://rs.olm.vn/images/avt/0.png?1311)
a, ta có:
\(\sqrt{24}=4,89\\ \sqrt{3}=1,73\)
\(\Rightarrow\sqrt{24}+\sqrt{3}=4,89+1,73=6,62\)
vì 7>6,62 nên 7>\(\sqrt{24}+\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1. a)\(2\&\sqrt{5}\)
\(2=\sqrt{4}\)
=> \(2< \sqrt{5}\)
b)\(5\&\sqrt{23}\)
\(5=\sqrt{25}\)
=> \(5>\sqrt{23}\)
c) \(\sqrt{23}+\sqrt{13}\&\sqrt{83}\)
\(\left(\sqrt{23}+\sqrt{13}\right)^2=36+2\sqrt{229}\)
\(\left(\sqrt{83}\right)^2=83\)
\(\Rightarrow36+2\sqrt{299}< 83\)
=> \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)
2. a) \(\sqrt{x}=5;x\ge0\)
=> x = 25
b) \(3\sqrt{x}=6;x\ge0\)
=> x = 4
c) trùng
d) \(3-\sqrt{3+1}=1\)
\(3-\sqrt{3+1}=3-2=1\)
1)
a)\(2=\sqrt{4}< \sqrt{5}\)
b) \(5=\sqrt{25}>\sqrt{23}\)
c) \(\sqrt{83}>\sqrt{81}=9\)
\(\left\{{}\begin{matrix}\sqrt{23}< \sqrt{25}=5\\\sqrt{13}< \sqrt{16}=4\end{matrix}\right.\)
\(\sqrt{23}+\sqrt{13}< 4+5=9\)
Vậy \(\sqrt{23}+\sqrt{13}< \sqrt{83}\)
2) Ta có:
\(\sqrt{x}=5\Rightarrow x=25\)
\(3\sqrt{x}=6\Rightarrow\sqrt{x}=2\Rightarrow x=4\)
\(3-\sqrt{3+1}=1\)
Nên:
\(3-2=1\)(luôn đúng)
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\left(x+1\right)^3=-27\)
\(\left(x+1\right)^3=\left(-3\right)^3\)
x+1=-3
x=(-3)-1
x=-4
b)6-3x=8
3x=6-8
3x=(-2)
x=\(-\frac{2}{3}\)
a) \(\left(x+1\right)^3=-27\)
\(\Rightarrow\left(x+1\right)^3=\left(-3\right)^3\)
\(\Rightarrow x-1=-3\)
\(\Rightarrow x=-4\)
Vậy \(x=-4\)
b) \(\sqrt{36}-\sqrt{9}.x=\sqrt{64}\)
\(\Rightarrow6-3.x=8\)
\(\Rightarrow3x=-2\)
\(\Rightarrow x=\frac{-2}{3}\)
Vậy \(x=\frac{-2}{3}\)
Ý bạn có phải là:
\(\sqrt{x+27}=\sqrt[3]{3}\)
chắc thế đó bn