Question

Giúp em với ạ em đang cần gấp . Làm hết giúp em nha

Answer 1

1) Ta có: \(\sqrt{4x}=\sqrt{5}\)

nên 4x=5

hay \(x=\dfrac{5}{4}\)

2) Ta có: \(\sqrt{16x}=8\)

nên 16x=64

hay x=4

 

Answer 2

3, \(2\sqrt{x}=\sqrt{9x}-3\left(đk:x\ge0\right)\)

\(< =>2\sqrt{x}-3\sqrt{x}+3=0\)

\(< =>3-\sqrt{x}=0< =>x=9\)(tmđk)

4, \(\sqrt{3x-1}=4\left(đk:x\ge\frac{1}{3}\right)\)

\(< =>3x-1=16< =>3x-17=0\)

\(< =>x=\frac{17}{3}\)(tmđk)

Answer 3

5, \(\sqrt{-3x+4}=12\left(đk:x\le\frac{4}{3}\right)\)

\(< =>-3x+4=144\)

\(< =>-3x-140< =>3x+140=0< =>x=-\frac{140}{3}\)(tmđk)

6, \(\sqrt{2-3x}=10\left(đk:x\le\frac{2}{3}\right)\)

\(< =>2-3x=100< =>3x+98=0< =>x=-\frac{98}{3}\)(tmđk)

Answer 4

7, \(\sqrt{4-5x}=12\left(đk:x\le\frac{4}{5}\right)\)

\(< =>4-5x=144< =>5x+140=0\)

\(< =>x=-\frac{140}{5}=-28\)(tmđk)

8, \(\sqrt{9\left(x-1\right)}=21\left(đk:x\ge1\right)\)

\(< =>3\sqrt{x-1}=21< =>\sqrt{x-1}=7\)

\(< =>x-1=49< =>x=49+1=50\)(tmđk)

Answer 5

9, \(\sqrt{5x+3}=\sqrt{3-\sqrt{2}}\left(đk:x\ge-\frac{3}{5}\right)\)

\(< =>5x+3=3-\sqrt{2}< =>5x+\sqrt{2}=0\)

\(< =>x=-\frac{\sqrt{2}}{5}\left(tmđk\right)\)

10, \(\sqrt{3x^2-5}=2\left(đk:...\right)\)

\(< =>3x^2-5=4< =>3\left(x^2-3\right)=0\)

\(< =>\orbr{\begin{cases}x=\sqrt{3}\\x=-\sqrt{3}\end{cases}\left(...\right)}\)