ĐK : \(x\ge-2;y\ge-3\)

pt (1) <=> \(x^3+x=\left(y+1\right)^3+\left(y+1\right)\)

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

<=> \(\left(y+1-x\right)\left(\left(y+1\right)^2+\left(y+1\right)x+x^2+1\right)=0\)

<=> \(y+1-x=0\) vì \(\left(y+1\right)^2+\left(y+1\right)x+x^2+1>0\)dễ chứng minh.

<=> \(x=y+1\)(1')

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

<=> \(\left|\sqrt{x+2}-2\right|+\left|\sqrt{y+3}-3\right|=1\)(2')

Thế (1') vào (2') ta có: \(\left|\sqrt{y+3}-2\right|+\left|\sqrt{y+3}-3\right|=1\)

Có: \(\left|\sqrt{y+3}-2\right|+\left|\sqrt{y+3}-3\right|=\left|\sqrt{y+3}-2\right|+\left|3-\sqrt{y+3}\right|\ge1\)

Do đó: \(\left|\sqrt{y+3}-2\right|+\left|\sqrt{y+3}-3\right|=1\)<=> \(\left(\sqrt{y+3}-2\right)\left(3-\sqrt{y+3}\right)\ge0\)

<=> \(2\le\sqrt{y+3}\le3\)

<=> \(4\le y+3\le9\)

<=> \(1\le y\le6\)(tm) 

Khi đó: x = y + 1 với mọi y thỏa mãn \(1\le y\le6\)

Vậy tập nghiệm  \(S=\left\{\left(y+1;y\right):1\le y\le6\right\}\)

Bài làm:

Ta có: \(\sqrt{x}+2>3\)

\(\Leftrightarrow\sqrt{x}>1\)

\(\Rightarrow x>1\)

\(\sqrt{x}>1\) 

\(\orbr{\begin{cases}1>0\left(llđ\right)\\x>1^2\end{cases}}\) 

\(x>1\)

Sửa đề: +6x^2

x^4+4x^3+6x^2-x-10=0

=>x^4-x^3+5x^3-5x^2+11x^2-11x+10x-10=0

=>(x-1)(x^3+5x^2+11x+10)=0

=>(x-1)(x^3+2x^2+3x^2+6x+5x+10)=0

=>(x-1)(x+2)(x^2+3x+5)=0

=>x=1 hoặc x=-2

a: ĐKXĐ: \(\frac{3x+1}{x-1}\ge0\)

=>x>1 hoặc x<=-1/3

Đặt \(t=\sqrt{\frac{3x+1}{x-1}}\)

=>\(\frac{1}{t}=\sqrt{\frac{x-1}{3x+1}}\)

Phương trình ban đầu trở thành: \(t+6\cdot\frac{1}{t}=5\)

=>\(\frac{t^2+6}{t}=5\)

=>\(t^2+6=5t\)

=>\(t^2-5t+6=0\)

=>(t-2)(t-3)=0

=>t=2 hoặc t=3

t=2

=>\(\sqrt{\frac{3x+1}{x-1}}=2\)

=>\(\frac{3x+1}{x-1}=4\)

=>4(x-1)=3x+1

=>4x-4=3x+1

=>x=5(nhận)

t=3

=>\(\sqrt{\frac{3x+1}{x-1}}=3\)

=>\(\frac{3x+1}{x-1}=9\)

=>9(x-1)=3x+1

=>9x-9=3x+1

=>6x=10

=>x=5/3(nhận)

b: ĐKXĐ: x>=2

Đặt \(t = \sqrt{x-2} + \sqrt{x+1}\) (Điều kiện: t>0)

=>\(t^2=x-2+x+1+2\cdot\sqrt{\left(x-2\right)\left(x+1\right)}=2x-1+2\cdot\sqrt{x^2-x-2}\)

=>\(2x-7+2\sqrt{x^2 - x - 2}=t^2-6\)

Phương trình ban đầu trở thành: \(t^2-6=t\)

=>\(t^2-t-6=0\)

=>(t-3)(t+2)=0

=>t-3=0

=>t=3

=>\(\sqrt{x-2} + \sqrt{x+1} = 3\)

=>\(2x - 1 + 2\sqrt{x^2 - x - 2} = 9\)

=>\(2\sqrt{x^2 - x - 2}=10-2x\)

=>\(\sqrt{x^2 - x - 2}=5-x\)

=>\(\begin{cases}5-x\ge0\\ x^2-x-2=\left(5-x\right)^2=x^2-10x+25\end{cases}\Rightarrow\begin{cases}x\le5\\ 9x=25+2=27\end{cases}\)

=>x=3

Đặt căn x=a

=>\(\sqrt{3+2a}+a=6\)

\(\Leftrightarrow\sqrt{2a+3}=6-a\)

\(\Leftrightarrow\left\{{}\begin{matrix}a< =6\\a^2-12a+36=2a+3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a< =6\\a^2-14a+33=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a< =6\\\left(a-3\right)\left(a-11\right)=0\end{matrix}\right.\Leftrightarrow a=3\)

=>x=9

x.(3x - 1) - (3x + 2) . (x - 5) = 0

<=> 3x² - x - 3x² + 15x - 2x + 10 = 0

<=> 12x + 10 = 0

<=> 12x = -10

<=> x = -5/6

Vậy S = {-5/6}

Ta có PT <=> x⁴ + 5x³ - 15x + 9 = 0

<=> (x - 1)(x + 3)(x² + 3x - 3) = 0

Tới đây thì đơn giản rồi

x=-2 hoặc 1

\(a,\left(x-6\right)\left(2x-5\right)\left(3x+9\right)=0\Leftrightarrow\left[{}\begin{matrix}x-6=0\Leftrightarrow x=6\\2x-5=0\Leftrightarrow x=\dfrac{5}{2}\\3x+9=0\Leftrightarrow x=-3\end{matrix}\right.\)

\(b,2x\left(x-3\right)+5\left(x-3\right)=0\Leftrightarrow\left(2x+5\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-3=0\Leftrightarrow x=3\\2x+5=0\Leftrightarrow x=-\dfrac{5}{2}\end{matrix}\right.\)

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

\(x=-7\left(2m-5\right)x-2m^2+8\Leftrightarrow x+7\left(2m-5\right)=8-2m^2\Leftrightarrow x\left(14m-34\right)=8-2m^2\)

\(ycđb\Leftrightarrow14m-34\ne0\Leftrightarrow m\ne\dfrac{34}{14}\)\(\Rightarrow x=\dfrac{8-2m^2}{14m-34}\)

\(3.17\Leftrightarrow4x^2-4x+1-2x-1=0\Leftrightarrow4x^2-6x=0\Leftrightarrow x\left(4x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{3}{2}\end{matrix}\right.\)

3.15:

a, \(\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\2x-5=0\\3x+9=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=\dfrac{5}{2}\\x=-\dfrac{9}{3}=-3\end{matrix}\right.\)

b, \(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)

c, \(\Leftrightarrow\left(x-2\right)\left(x+2\right)-\left(x-2\right)\left(3-2x\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(x+2-3+2x\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x-2=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3.16

\(\Leftrightarrow\left(2m-5\right).-7-2m^2+8=0\)

\(\Leftrightarrow-14m+35-2m^2+8=0\)

\(\Leftrightarrow-14m-2m^2+43=0\)

\(\Leftrightarrow-2\left(7m+m^2\right)=-43\)

\(\Leftrightarrow m\left(7-m\right)=\dfrac{43}{2}\)

\(\Leftrightarrow\dfrac{m\left(7-m\right)}{1}-\dfrac{43}{2}=0\)

\(\Leftrightarrow\dfrac{14m-2m^2}{2}-\dfrac{43}{2}=0\)

pt vô nghiệm