Question

[Nguyễn Thùy Chi]

\(3x^2+10x+\sqrt{3x+3}=x^3+26+\sqrt{5-2x}\)

Answer 1

[Lê Thị Thục Hiền]

Bấm máy may mắn ra nghiệm đẹp

Đk: \(-1\le x\le\frac{5}{2}\)

PT <=> \(6x^2+20x+2\sqrt{3x+3}=2x^3+52+2\sqrt{5-2x}\)

<=> \(\left[2\sqrt{3x+3}-\left(4+x\right)\right]+6x^2+23x=2x^3+2\left[\sqrt{5-2x}-\left(3-x\right)\right]+54\)

Xét \(-1\le x\) => \(2\sqrt{3x+3}+4+x\ge0+4-1=3>0\)

Xét \(-1\le x\le\frac{5}{2}\) => \(\frac{1}{2}\le\sqrt{5-2x}+3-x\le\sqrt{7}+4\) => \(\sqrt{5-2x}+3-x\ne0\)

Pt <=> \(\frac{4\left(3x+3\right)-\left(4+x\right)^2}{2\sqrt{3x+3}+4+x}+6x^2+23x=2x^3+2.\frac{5-2x-\left(3-x\right)^2}{\sqrt{5-2x}+3-x}+54\)

<=>\(\frac{-x^2+4x-4}{2\sqrt{3x+3}+4x+}-2.\frac{-x^2+4x-4}{\sqrt{5-2x}+3-x}-\left(2x^3-6x^2-23x+54\right)=0\)

<=> \(\frac{-\left(x-2\right)^2}{2\sqrt{3x+3}+4+x}+\frac{2\left(x-2\right)^2}{\sqrt{5-2x}+3-x}-\left(x-2\right)\left(2x^2-2x-27\right)=0\)

<=>\(\left(x-2\right)\left[\frac{-\left(x-2\right)}{2\sqrt{3x+3}+4+x}+\frac{2\left(x-2\right)}{\sqrt{5-2x}+3-x}-2x^2+2x+27\right]=0\)

<=>\(\left[{}\begin{matrix}x-2=0\left(1\right)\\-\frac{\left(x-2\right)}{2\sqrt{3x+3}+4+x}+\frac{2\left(x-2\right)}{\sqrt{5-2x}+3-x}-2x^2+2x+27=0\left(2\right)\end{matrix}\right.\)

Từ (1)=> x=2(t/m pt)

Chắc chắn (2) vô nghiệm nhưng chưa biết CM

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Đau mắt quá thì chuyển qua liên hợp kiểu này đi(dễ hơn)

pt <=> \(\left(\sqrt{3x+3}-3\right)-\left(\sqrt{5-2x}-1\right)+3x^2+10x-x^3-24=0\)

Luôn có \(\left\{{}\begin{matrix}\sqrt{3x+3}+3>0\\\sqrt{5-2x}+1>0\end{matrix}\right.\) với mọi x

pt <=> \(\frac{3x+3-9}{\sqrt{3x+3}+3}-\frac{5-2x-1}{\sqrt{5-2x}+1}-\left(x-2\right)\left(x+3\right)\left(x-4\right)=0\)

<=>\(\frac{3\left(x-2\right)}{\sqrt{3x+3}+3}+\frac{2\left(x-2\right)}{\sqrt{5-2x}+1}-\left(x-2\right)\left(x+3\right)\left(x-4\right)=0\)

<=>\(\left(x-2\right)\left[\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}-\left(x+3\right)\left(x-4\right)\right]=0\)

<=>\(\left[{}\begin{matrix}x=2\left(tm\right)\\\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}-\left(x+3\right)\left(x-4\right)=0\left(1\right)\end{matrix}\right.\)

(1) <=>\(\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}=\left(x+3\right)\left(x-4\right)\)

Tại \(-1\le x\le\frac{5}{2}\)=> \(-10\le\left(x+3\right)\left(x-4\right)\le-\frac{33}{4}< 0\)

=> Vế phải của (1) luôn âm

Xét vế trái của (1) có: \(\left\{{}\begin{matrix}\sqrt{3x+3}+3>0\\\sqrt{5-2x}+1>0\end{matrix}\right.\)=> \(\left\{{}\begin{matrix}\frac{3}{\sqrt{3x+3}+3}>0\\\frac{2}{\sqrt{5-2x}+1}>0\end{matrix}\right.\)=> \(\frac{3}{\sqrt{3x+3}+3}+\frac{2}{\sqrt{5-2x}+1}>0\)

=> Vế trái của (1) luôn dương hay (1) vô nghiệm

Vậy pt có 1 nghiệm duy nhất x=2