\(DK:x\ge-\frac{1}{3}\)

\(\Leftrightarrow\frac{2x-1}{\sqrt{3x+1}+\sqrt{x+2}}\left(\sqrt{3x^2+7x+2}+4\right)-2\left(2x-1\right)=0\)

\(\Leftrightarrow\left(2x-1\right)\left(\frac{\sqrt{3x^2+7x+2}+4}{\sqrt{3x+1}+\sqrt{x+2}}-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{2}\left(1\right)\\\frac{\sqrt{3x^2+7x+2}+4}{\sqrt{3x+1}+\sqrt{x+2}}=2\left(2\right)\end{cases}}\)

Xet PT(2)

Dat \(\hept{\begin{cases}\sqrt{3x+1}=a\\\sqrt{x+2}=b\end{cases}\left(a,b\ge0\right)}\)

PT(2)\(\Leftrightarrow\frac{ab+4}{a+b}=2\)

\(\Leftrightarrow2a+2b-ab-4=0\)

\(\Leftrightarrow\left(a+2\right)\left(2-b\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=-2\left(3\right)\\b=2\left(4\right)\end{cases}}\)

Xet PT(3)

Ta co:\(a\ge0\)

Nen PT vo nghiem

Xet PT (4)

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

\(\Leftrightarrow x+2=4\)

\(\Leftrightarrow x=2\)

Vay PT co 2 nghiem la \(x_1=\frac{1}{2};x_2=2\)

a) \(x^3-4x^2-5x+6=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-7x^2-9x+4+x^3+3x^2+4x+2=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow-\left(7x^2+9x-4\right)+\left(x+1\right)^3+x+1=\sqrt[3]{7x^2+9x-4}\) (*)

Đặt \(\sqrt[3]{7x^2+9x-4}=a;x+1=b\)

Khi đó (*) \(\Leftrightarrow-a^3+b^3+b=a\)

\(\Leftrightarrow\left(b-a\right).\left(b^2+ab+a^2+1\right)=0\)

\(\Leftrightarrow b=a\)

Hay \(x+1=\sqrt[3]{7x^2+9x-4}\)

\(\Leftrightarrow\left(x+1\right)^3=7x^2+9x-4\)

\(\Leftrightarrow x^3-4x^2-6x+5=0\)

\(\Leftrightarrow x^3-4x^2-5x-x+5=0\)

\(\Leftrightarrow\left(x-5\right)\left(x^2+x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{-1\pm\sqrt{5}}{2}\end{matrix}\right.\)

sao đề nhìn bá vậy bạn ...

bài này chắc đặt \(\sqrt{x^3-3x+6}\)cho nó gọn thôi

Điều kiện: 3x² - 6x - 6 \(\ge\) 0 và 2 - x  \(\ge\) 0

pt <=> \(\sqrt{3x^2-6x-6}=3.\left(2-x\right)^2\sqrt{2-x}+\left(7x-19\right)\sqrt{2-x}\)

<=> \(\sqrt{3x^2-6x-6}=\left(3x^2-12x+12+7x-19\right)\sqrt{2-x}\)

<=> \(\sqrt{3x^2-6x-6}=\left(3x^2-5x-7\right)\sqrt{2-x}\) (1)

Đặt \(\sqrt{3x^2-6x-6}=a;\sqrt{2-x}=b;\left(a;b\ge0\right)\)

=> \(3x^2-6x-6=a^2;2-x=b^2\)=> \(a^2-b^2=3x^2-5x-8\) 

=> (1) trở thành: a = (a² - b² + 1).b

<=> a = (a- b)(a+b).b + b

<=> (a - b) - (a- b)(a+b).b = 0

<=> (a - b).(1 - b(a+b)) = 0

<=> a = b  hoặc (a+b).b = 1

+) a = b => ......

+) (a+b).b = 1 <=> ab + b² - 1 = 0

<=> \(\sqrt{3x^2-3x-6}.\sqrt{2-x}+\left(2-x\right)-1=0\)

<=> \(\sqrt{3\left(x^2-x-2\right)\left(2-x\right)}=x-1\)

<=> x \(\ge\) 1; 3(x² - x - 2)(2 - x) = (x-1)²

<=> ........  

1.   3x( x - 2 ) - ( x - 2 ) = 0

<=> ( x-2).(3x-1)  = 0 => x = 2 hoặc x = \(\dfrac{1}{3}\)

2.    x( x-1 ) ( x² + x + 1 ) - 4( x - 1 )

<=> ( x - 1 ).( x (x^2 + x + 1 ) - 4 ) = 0

(phần này tui giải được x = 1 thôi còn bên kia giải ko ra nha )

3 \(\left\{{}\begin{matrix}\sqrt{5}x-2y=7\\\sqrt{5}x-5y=10\end{matrix}\right.\)<=> \(\left\{{}\begin{matrix}y=-1\\x=\sqrt{5}\end{matrix}\right.\)

\(1. 3x^2 - 7x +2=0\)

=>\(Δ=(-7)^2 - 4.3.2\)

        \(= 49-24 = 25\)

Vì 25>0 suy ra phương trình có 2 nghiệm phân biệt:

\(x_1\)=\(\dfrac{-\left(-7\right)+\sqrt{25}}{2.3}=\dfrac{7+5}{6}=2\)

\(x_2\)=\(\dfrac{-\left(-7\right)-\sqrt{25}}{2.3}=\dfrac{7-5}{6}=\dfrac{1}{3}\)

ĐKXĐ: \(x\ge-\frac{1}{3}\)

Do \(\sqrt{3x+1}+\sqrt{x+2}>0;\forall x\ge-\frac{1}{3}\)

Nhân 2 vế của pt với \(\sqrt{3x+1}+\sqrt{x+2}\) và rút gọn ta được:

\(\left(2x-1\right)\left(\sqrt{3x^2+7x+2}+4\right)=2\left(2x-1\right)\left(\sqrt{3x+1}+\sqrt{x+2}\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\\sqrt{3x^2+7x+2}+4=2\left(\sqrt{3x+1}+\sqrt{x+2}\right)\left(1\right)\end{matrix}\right.\)

Xét (1)

\(\Leftrightarrow\sqrt{\left(3x+1\right)\left(x+2\right)}-2\sqrt{3x+1}-2\left(\sqrt{x+2}-2\right)=0\)

\(\Leftrightarrow\sqrt{3x+1}\left(\sqrt{x+2}-2\right)-2\left(\sqrt{x+2}-2\right)=0\)

\(\Leftrightarrow\left(\sqrt{3x+1}-2\right)\left(\sqrt{x+2}-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+1}=2\\\sqrt{x+2}=2\end{matrix}\right.\) \(\Leftrightarrow...\)