Question

Giải các phương trình sau:

a)\(\sqrt{3x+1}=\sqrt{4x-3}\)

b)\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

Answer 1

Câu a : ĐK : \(x\ge\dfrac{3}{4}\)

\(\sqrt{3x+1}=\sqrt{4x-3}\)

\(\Leftrightarrow3x+1=4x-3\)

\(\Leftrightarrow-x=-4\)

\(\Leftrightarrow x=4\left(TM\right)\)

Vậy \(S=\left\{4\right\}\)

Câu b : ĐK : \(x\ge-2\)

\(\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\)

\(\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6\)

\(\Leftrightarrow2\sqrt{x+2}=6\)

\(\Leftrightarrow\sqrt{x+2}=3\)

\(\Leftrightarrow x+2=9\)

\(\Leftrightarrow x=7\left(TM\right)\)

Vậy \(S=\left\{7\right\}\)

Answer 2

a ,B1 : đặt đk

B2 : tự làm ok

Answer 3

\(a\text{) }\sqrt{3x+1}=\sqrt{4x-3}\left(ĐKXĐ:x\ge\dfrac{3}{4}\right)\\ \Leftrightarrow3x+1=4x-3\\ \Leftrightarrow-x=-4\\ \Leftrightarrow x=4\left(T/m\right)\)

Vậy..........

\(\text{b) }\sqrt{9x+18}-5\sqrt{x+2}+\dfrac{4}{5}\sqrt{25x+50}=6\left(ĐKXĐ:x\ge-2\right)\\\Leftrightarrow3\sqrt{x+2}-5\sqrt{x+2}+4\sqrt{x+2}=6 \\ \Leftrightarrow2\sqrt{x+2}=6\\ \Leftrightarrow\sqrt{x+2}=3\\ \Leftrightarrow x+2=9\\ \Leftrightarrow x=7\left(T/m\right)\)

Vậy...........