Question

Giai PT.

a.\(x^2+4x+5=2\sqrt[]{2x+3}\)

 

Answer 1

dat \(\sqrt{2x+3}=a\left(a\ge0\right)\)

=> \(a^2=2x+3\)

=> \(x=\frac{a^2-3}{2}\)

pt <=> \(\frac{\left(a^2-3\right)^2}{4}+4\times\frac{a^2-3}{2}+5=2a\)

<=> \(\left(a^4-6a^2+9\right)+8a^2-4=8a\)

<=> \(a^4+2a^2-8a+5=0\)

<=> \(a^4-a^3+a^3-a^2+3a^2-3a-5a+5=0\)

<=.> \(a^3.\left(a-1\right)+a^2.\left(a-1\right)+3a\left(a-1\right)+5\left(a-1\right)=0\)

<=> \(\left(a-1\right)\left(a^3+a^2+3a-5\right)=0\)

<=> \(\orbr{\begin{cases}a=1\\a^3+a^2+3a+5=0\left(1\right)\end{cases}}\)

bạn bấm máy tính cái pt 1 thì a=1 (tm) 

thay a=1 vao \(\sqrt{2x+3}=a\)

tìm ra x