Question

[Trần Đình Nam]

Giúp em bài giải phương trình với bài hệ với ạ

Answer 1

[๖ۣۜDũ๖ۣۜN๖ۣۜG]

a) đk \(x\ge\dfrac{-3}{2}\)

PT <=> \(4x^2\left(2x+3\right)=\left(3x^2+6x+1\right)^2\)

<=> \(8x^3+12x^2=9x^4+36x^2+1+36x^3+12x+6x^2\)

<=> \(9x^4+28x^3+30x^2+12x+1=0\)

<=> \(\left(x+1\right)^3\left(9x+1\right)=0\)

<=> \(\left[{}\begin{matrix}x=-1\left(c\right)\\x=\dfrac{-1}{9}\left(l\right)\end{matrix}\right.\)

KL: PT có nghiệm duy nhất x = -1

b) đk: \(x\ge-1;x\ge2y\)

hpt <=> \(\left\{{}\begin{matrix}2x^2-4xy+3y-4x-4=\sqrt{9\left(x-1\right)\left(x+1\right)\left(x-2y\right)}\left(1\right)\\2x-2y+1+2\sqrt{\left(x+1\right)\left(x-2y\right)}=2x-2y+5\left(2\right)\end{matrix}\right.\)

(2) <=> \(\sqrt{\left(x+1\right)\left(x-2y\right)}=2\)

<=> \(\left(x+1\right)\left(x-2y\right)=4\)

(1) <=> 2(x+1)(x-2y) + x - 4 = \(6.\sqrt{x-1}\)

<=> x+4 = \(6\sqrt{x-1}\)

<=> x² + 8x + 16 = 36x - 36

<=> x² -28x + 52 = 0 

<=> (x-26)(x-2) = 0

<=> \(\left[{}\begin{matrix}x=26< =>y=\dfrac{349}{27}\\x=2< =>y=\dfrac{1}{3}\end{matrix}\right.\)