Question

Tìm x, biết:

a)\(\sqrt{4-5x}=12\)

b)\(\sqrt{x^2-14x+49}\)\(-3x=1\)

c)\(\sqrt{9x-9}\)+\(2\sqrt{16x-16}\)\(-\frac{1}{3}\sqrt{25x-25}\)\(=\sqrt{x-1}\)

Answer 1

a, \(\sqrt{4-5x}=12\Leftrightarrow4-5x=144\Leftrightarrow5x=140\Leftrightarrow x=28\)

b,ĐK :  \(x\ge7\)

 \(\sqrt{x^2-14x+49}-3x=1\Leftrightarrow\sqrt{\left(x-7\right)^2}=3x+1\)

\(\Leftrightarrow x-7=3x+1\Leftrightarrow-2x-8=0\Leftrightarrow x=-4\)( vô lí )

c, Bn làm nốt nhé 

Answer 2

a) đk: \(x\le\frac{4}{5}\)

Ta có: \(\sqrt{4-5x}=12\)

\(\Leftrightarrow\left|4-5x\right|=144\)

\(\Rightarrow4-5x=144\)

\(\Leftrightarrow5x=-140\)

\(\Rightarrow x=-28\left(tm\right)\)

b) Ta có: \(\sqrt{x^2-14x+49}-3x=1\)

\(\Leftrightarrow\sqrt{\left(x-7\right)^2}=1+3x\)

\(\Leftrightarrow\left|x-7\right|=3x+1\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=3x+1\\x-7=-3x-1\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-8\\4x=6\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{3}{2}\end{cases}}\)

Answer 3

CTRL là cái gì vậy

Answer 4

c) Ta có: đk: \(x\ge1\)

PT<=> \(3\sqrt{x-1}+8\sqrt{x-1}-\frac{5}{3}\sqrt{x-1}=\sqrt{x-1}\)

\(\Leftrightarrow\frac{25}{3}\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x-1=0\)

\(\Rightarrow x=1\left(tm\right)\)

Ở câu b loại bỏ trường hợp x =-4 giúp mk nhé