Ta có: \(\left|x+1\right|+3\left|x-1\right|=x+2+\left|x\right|+2\left|x-2\right|\)

=>\(\left|x+1\right|-\left|x\right|+3\left|x-1\right|-2\left|x-2\right|=x+2\) (1)

TH1: x<-1

=>x+1<0; x<0; x-1<0; x-2<0

(1) sẽ trở thành:

-x-1-(-x)+3(-x+1)-2(-x+2)=x+2

=>-x-1+x-3x+3+2x-4=x+2

=>-x-2=x+2

=>-2x=4

=>x=-2(nhận)

TH2: -1<=x<0

=>x+1>=0; x<0; x-1<0; x-2<0

(1) sẽ trở thành:

x+1-(-x)+3(-x+1)-2(-x+2)=x+2

=>x+1+x-3x+3+2x-4=x+2

=>x=x+2

=>0=2(vô lý)

TH3: 0<=x<1

=>x+1>0; x>=0; x-1<0; x-2<0

(1) sẽ trở thành:

x+1-(x)+3(-x+1)-2(-x+2)=x+2

=>1-3x+3+2x-4=x+2

=>-x=x+2

=>-2x=2

=>x=-1(loại)

TH4: 1<=x<2

=>x+1>0; x>0; x-1>=0; x-2<0

(1) sẽ trở thành:

x+1-x+3(x-1)-2(-x+2)=x+2

=>1+3x-3+2x-4=x+2

=>5x-6=x+2

=>4x=8

=>x=2(loại)

TH5: x>=2

=>x+1>0; x>0; x-1>0; x-2>=0

(1) sẽ trở thành:

x+1-x+3(x-1)-2(x-2)=x+2

=>1+3x-3-2x+4=x+2

=>x+2=x+2(luôn đúng)

=>x>=2

Vậy: x>=2 hoặc x=-2

đk tự giải nhé 

với x tjỏa mãn đk ta có 

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

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

đặt \(\sqrt{x^3+3x}=a\)

ta có pt<=> \(a=\frac{a^2+4x}{2\left(x+1\right)}\Leftrightarrow2a\left(x+1\right)=a^2+4x\)

\(\Leftrightarrow2ax+2a=a^2+4x\Leftrightarrow a^2+4ax-2a-2ax=0\)

\(\Leftrightarrow\left(a^2-2ax\right)-\left(2a-4x\right)=0\Leftrightarrow a\left(a-2x\right)-2\left(a-2x\right)=0\)

\(\Leftrightarrow\left(a-2\right)\left(a-2x\right)=0\)

đến đây tự làm nhé

a: ĐKXĐ: \(\begin{cases}x-2\ge0\\ x+1\ge0\\ x^2-x-2\ge0\end{cases}\Rightarrow\begin{cases}x\ge2\\ x\ge-1\\ \left(x-2\right)\left(x+1\right)\ge0\end{cases}\)

=>x>=2

Ta có: \(3x + 14 + 5\sqrt{x-2} = 7\left(\sqrt{x+1} + \sqrt{x^2 - x - 2}\right)\) (1)

Đặt \(a=\sqrt{x-2};b=\sqrt{x+1}\) (ĐIều kiện: a>=0; b>0)

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

\(b^2-a^2=x+1-\left(x-2\right)=3\)

\(b^2-1=x+1-1=x\)

=>\(3x+14=3\left(b^2-1\right)+14=3b^2+11\)

(1) sẽ tương đương: \(3b^2+11+5a=7b+7ab\)

=>\(3b^2-7b+11+a(5-7b)=0\)

=>\((b - 2)(40b^3 + 52b^2 - 133b + 98) = 0\)

=>b-2=0

=>b=2

=>x+1=4

=>x=3

b: ĐKXĐ: 7/3<=x<=7

\(7\sqrt{3x-7}+(4x-7)\sqrt{7-x}=32\left(2\right)\)

Đặt \(a=\sqrt{3x-7};b=\sqrt{7-x}\)

=>\(a^2+b^2=3x-7+7-x=2x\)

\(3a^2+b^2=3\left(3x-7\right)+7-x=9x-21+7-x=8x-14\)

\(3x-7=a^2\)

=>\(3x=a^2+7\)

=>\(x=\frac{a^2+7}{3}\)

=>\(4x-7=4\cdot\frac{a^2+7}{3}-7=\frac{4\left(a^2+7\right)-21}{3}=\frac{4a^2+7}{3}\)

(2) sẽ trở thành: \(7a+\frac{4a^2+7}{3}b=32\iff21a+(4a^2+7)b=96\)

mà \(a^2 + 3b^2 = 14\)

nên \(a=\sqrt5;b=\sqrt3\)

=>3x-7=5

=>3x=12

=>x=4(nhận)

a ) \(\left(2x-1\right)\left(x-3\right)\left(x+7\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}2x+1=0\\x-3=0\\x+7=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=-\frac{1}{2}\\x=3\\x=-7\end{array}\right.\)

Vậy phương trình đã cho các nghiệm \(x=-\frac{1}{2};x=3;x=-7.\)

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

\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-3=0\end{array}\right.\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=3\end{array}\right.\)

Vậy phương trình đã cho các nghiệm \(x=1,x=3\).

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

Điều kiện: \(\hept{\begin{cases}x+3\ge0\\x-1\ge0\\2x+2\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-3\\x\ge1\\x\ge-1\end{cases}\Leftrightarrow x\ge1}\)

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

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

     \(\Leftrightarrow2x+2-2\sqrt{\left(x+3\right)\left(x-1\right)}=2x+2\)

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

     \(\Leftrightarrow\left(x+3\right)\left(x-1\right)=0\)

      \(\Leftrightarrow\orbr{\begin{cases}x+3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\left(l\right)\\x=1\left(n\right)\end{cases}}\)

Vậy \(S=\left\{1\right\}\)