\(\sqrt{x+1}+2=0\)
\(\sqrt{x+1}=-2\)
\(\Rightarrow x\in\varnothing\) vì \(x\ge0\)
\(3\sqrt{x+1}=40\)
\(\sqrt{x+1}=\frac{40}{3}\)
\(x+1=\frac{1600}{9}\)
\(x=\frac{1591}{9}\)
Tìm GTNN A = \(\dfrac{3\sqrt{x}+1}{2\sqrt{x}+2}\) (x ≥ 0)
Tìm x biết:
a)\(\left|\sqrt{2}-x\right|=\sqrt{2}\)
b)\(\left|x-1\right|=\sqrt{3}+2\)
c)\(\left|1-2x\right|=\sqrt{5}-1\)
d)\(\left|1-x\right|=\sqrt{2}-0,\left(1\right)\)
e)\(\left|x-\sqrt{3}\right|=\sqrt{3}-1\)
f)\(\left|x-\sqrt{2}\right|=1,\left(4\right)\)
Tìm x, biết:
a. 2\(\sqrt{x}\)-10= 20
b. 2x-\(\sqrt{x}\)= 0
c. \(x+3\sqrt{x}\)= 0
d. ( x-1).(x2+1) = 0
Tìm x, biết:
5/x+y/4=1/8
A=\(\frac{\sqrt{2}+1}{\sqrt{x}-3}\)(x>=0)
2.|5x-3|-2x=14
x-\(2\sqrt{2}\)=0(x>=0)
tìm x biết :
a)\(2\sqrt{x}+3=0\)
b)\(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
c)\(\sqrt{x+3}+3=0\)
Tìm x
1/ \(\sqrt{x+1+2}=0\)
2/ \(\sqrt{x+3-9}=0\)
3/\(7-\sqrt{x}=0\)
có ai giúp em giải bài toán này với ạ :)
1/Tính A=\(0,5^3+1^3+1,5^3+...+5^3\)
biết \(1^3+2^3+3^3+...+10^3=3025\)
2/Tính:
\(\left(\frac{16}{3}-\sqrt{\frac{40}{9}}^{^0}\right)\left(\frac{17}{3}-\sqrt{\frac{40}{9}}^0\right)...\left(\frac{30}{3}-\sqrt{\frac{40}{9}}^0\right)\)
tìm x biết :
a) \(2\sqrt{x}+3=0\)
b)\(\frac{5}{12}\sqrt{x}-\frac{1}{6}=\frac{1}{3}\)
c) \(\sqrt{x+3}+3=0\)
Tìm số thực x biết:
a)\(\sqrt{x+2}=3\) ; b)\(\sqrt{x^2+3}=\sqrt{7}\); c) \(\sqrt{x}=0\) ; d)\(\sqrt{x}=-3\) ; e)\(3\sqrt{x}=1\)
g)\(4-5\sqrt{x}=-1\)
