Question

Giải phương trình

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

b,\(\sqrt{2+\sqrt{3x-5}}\)=\(\sqrt{x+1}\)

c,\(\sqrt{\dfrac{5x+7}{x+3}}\)=4

d,\(\dfrac{\sqrt{5x+7}}{\sqrt{x+3}}\)=4

Answer 1

a,\(\sqrt{3x+1}=3x-1\) Đk:\(x\ge\dfrac{-1}{3}\)

\(< =>3x+1=9x^2-6x+1\)

\(< =>9x-9x^2=0\)

\(< =>9x\left(1-x\right)=0\)

\(< =>x=0\) hoặc \(x=1\)
b,\(2+\sqrt{3x-5}=x+1\) Đk:\(x\ge\dfrac{5}{3}\)

\(< =>\sqrt{3x-5}=x-1\)

\(< =>3x-5=x^2-2x+1\)

\(< =>x^2+x+6=0\)(vô lý vì \(x^2\ge\dfrac{25}{9},x\ge\dfrac{5}{3}\))

=>\(x\in\varnothing\)

c,Đk : \(x\ge\dfrac{-7}{5}\)

\(\)\(\dfrac{5x+7}{x+3}=16\)

\(< =>5x+7=16x+48\)

\(< =>-11x=41 \)

\(< =>x=\dfrac{-41}{11}\)(ko tm đk)

\(=>x\in\varnothing\)

d,tương tự câu c bình phương 2 vế cũng ra \(x\in\varnothing\)