1) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}=\dfrac{4}{x^2-1}\)

\(\Leftrightarrow\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{4}{\left(x-1\right)\left(x+1\right)}\)

Suy ra: \(x^2+2x+1-\left(x^2-2x+1\right)=4\)

\(\Leftrightarrow x^2+2x+1-x^2+2x-1=4\)

\(\Leftrightarrow4x=4\)

hay x=1(loại)

Vậy: \(S=\varnothing\)

2) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

Ta có: \(\dfrac{x+2}{x-2}+\dfrac{x}{x+2}=2\)

\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=\dfrac{2\left(x^2-4\right)}{\left(x-2\right)\left(x+2\right)}\)

Suy ra: \(x^2+4x+4+x^2-2x=2x^2-8\)

\(\Leftrightarrow2x^2+2x+4-2x^2-8=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

hay x=2(loại)

Vậy: \(S=\varnothing\)

a: \(\dfrac{x+5}{x-1}+\dfrac{8}{x^2-4x+3}=\dfrac{x+1}{x-3}\)

=>(x+5)(x-3)+8=x^2-1

=>x^2+2x-15+8=x^2-1

=>2x-7=-1

=>x=3(loại)

b: \(\dfrac{x-4}{x-1}-\dfrac{x^2+3}{1-x^2}+\dfrac{5}{x+1}=0\)

=>(x-4)(x+1)+x^2+3+5(x-1)=0

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

=>2x^2+2x-6=0

=>x^2+x-3=0

=>\(x=\dfrac{-1\pm\sqrt{13}}{2}\)

e: =>x^2-2x+1+2x+2=5x+5

=>x^2+3=5x+5

=>x^2-5x-2=0

=>\(x=\dfrac{5\pm\sqrt{33}}{2}\)

g: (x-3)(x+4)*x=0

=>x=0 hoặc x-3=0 hoặc x+4=0

=>x=0;x=3;x=-4

Bài dài quá, lần sau chia nhỏ câu hỏi nhé!!!!!

đúng vậy

1: 

\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)

\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)

\(\Leftrightarrow x^2+5x=0\)

=>x=0 hoặc x=-5

3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)

=>(x+2)(x-1)=0

=>x=-2 hoặc x=1

1: \(\frac{3x-2}{3}-2=\frac{4x+1}{4}\)

=>\(\frac{3x-2-6}{3}=\frac{4x+1}{4}\)

=>\(\frac{3x-8}{3}=\frac{4x+1}{4}\)

=>3(4x+1)=4(3x-8)

=>12x+3=12x-32

=>3=-32(vô lý)

=>Phương trình vô nghiệm

2: \(\frac{x-3}{4}+\frac{2x-1}{3}=\frac{2-x}{6}\)

=>\(\frac{3\left(x-3\right)+4\left(2x-1\right)}{12}=\frac{2\left(2-x\right)}{12}\)

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

=>3x-9+8x-4=4-2x

=>11x-13=4-2x

=>13x=17

=>\(x=\frac{17}{13}\)

3: \(\frac12\left(x+1\right)+\frac14\left(x+3\right)=3-\frac13\left(x+2\right)\)

=>\(\frac12x+\frac12+\frac14x+\frac34+\frac13x+\frac23=3\)

=>\(x\left(\frac12+\frac14+\frac13\right)+\frac{6}{12}+\frac{9}{12}+\frac{8}{12}=3\)

=>\(x\left(\frac{6}{12}+\frac{3}{12}+\frac{4}{12}\right)=3-\frac{23}{12}=\frac{36}{12}-\frac{23}{12}=\frac{13}{12}\)

=>\(x\cdot\frac{13}{12}=\frac{13}{12}\)

=>x=1

4: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

=>\(\frac{x+4}{5}+\frac{5\left(-x+4\right)}{5}=\frac{2x-3\left(x-2\right)}{6}\)

=>\(\frac{x+4-5x+20}{5}=\frac{2x-3x+6}{6}\)

=>\(\frac{-4x+24}{5}=\frac{-x+6}{6}\)

=>6(-4x+24)=5(-x+6)

=>-24x+144=-5x+30

=>-19x=-114

=>x=6

5: \(\frac{4-5x}{6}=\frac{2\left(-x+1\right)}{2}\)

=>\(\frac{4-5x}{6}=-x+1\)

=>6(-x+1)=-5x+4

=>-6x+6=-5x+4

=>-6x+5x=4-6

=>-x=-2

=>x=2

6: \(-\left(\frac{x-3}{2}-2\right)=\frac{5\left(x+2\right)}{4}\)

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

=>\(\frac{-2\left(x-7\right)}{4}=\frac{5\left(x+2\right)}{4}\)

=>5(x+2)=-2(x-7)

=>5x+10=-2x+14

=>7x=4

=>x=4/7

7: \(\frac{2\left(2x+1\right)}{5}-\frac{6+x}{3}=\frac{5-4x}{15}\)

=>\(\frac{6\left(2x+1\right)-5\left(x+6\right)}{15}=\frac{5-4x}{15}\)

=>6(2x+1)-5(x+6)=-4x+5

=>12x+6-5x-30=-4x+5

=>7x-24=-4x+5

=>7x+4x=5+24

=>11x=29

=>\(x=\frac{29}{11}\)

8: \(\frac{7-3x}{2}-\frac{5+x}{5}=1\)

=>\(\frac{5\left(7-3x\right)-2\left(x+5\right)}{10}=1\)

=>5(7-3x)-2(x+5)=10

=>35-15x-2x-10=10

=>-17x+25=10

=>-17x=-15

=>x=15/17

a, \(-\left(x+3\right)\left(x-4\right)+\left(x+1\right)\left(x-1\right)=10\)

\(\Rightarrow-\left(x^2-4x+3x-12\right)+x^2-1=10\)

\(\Rightarrow-x^2+x+12+x^2-1=10\)

\(\Rightarrow x=10+1-12\Rightarrow x=-1\)

b, \(\left(2x-1\right)\left(x-2\right)-\left(x+3\right)\left(2x-7\right)=3\)

\(\Rightarrow2x^2-4x-x+2-\left(2x^2-7x+6x-21\right)=3\)

\(\Rightarrow2x^2-5x+2-2x^2+x+21=3\)

\(\Rightarrow-4x=3-21-2\Rightarrow-4x=-20\)

\(\Rightarrow x=5\)

Các câu còn lại làm tương tự! Phá ngoặc ra!

Chúc bạn học tốt!!!

\(\frac{1-x}{1+x}+3=\frac{2x+3}{x+1}\left(ĐKXĐ:x\ne-1\right)\)

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

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

\(\Rightarrow1-x+3\left(x+1\right)=2x+3\)

\(\Leftrightarrow1-x+3x+3=2x+3\)

\(\Leftrightarrow2x+4=2x+3\)

\(\Leftrightarrow0x=-1\)(vô nghiệm)

Vậy phương trình vô nghiệm.

\(\frac{\left(x+2\right)^2}{2x-3}-1=\frac{x^2-10}{2x-3}\left(ĐKXĐ:x\ne\frac{3}{2}\right)\)

\(\Leftrightarrow\frac{x^2+4x+4}{2x-3}-\frac{2x-3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Leftrightarrow\frac{x^2+4x+4-2x+3}{2x-3}=\frac{x^2-10}{2x-3}\)

\(\Rightarrow x^2+4x+4-2x+3=x^2-10\)

\(\Leftrightarrow2x+7=-10\)

\(\Leftrightarrow2x=-17\)

\(\Leftrightarrow x=\frac{-17}{2}\)(thỏa mãn ĐKXĐ)

Vậy phương trình có nghiệm duy nhất : \(x=\frac{-17}{2}\)

Bạn cần viết đề bài bằng công thức toán để được hỗ trợ tốt hơn. 

x^2+2x-3/3+2x/4=x^2/3

\(1)\left(x+1\right)^3-\left(x-1\right)^3-6\left(x-1\right)\left(x-1\right)\\ =x^3+3x^2+3x+1-x^3+3x^2-3x+1-6\cdot\left(x-1\right)^2\\ =6x^2+2-6\cdot\left(x^2-2x+1\right)\\ =6x^2+2-6x^2+12x-6\\ =12x-4\)

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

\(3)\left(x-1\right)^3-\left(x+2\right)\left(x^2-2x+4\right)+3\left(x-4\right)\left(x+4\right)\\ =x^3-3x^2+3x-1-(x^3+8)+3\cdot\left(x^2-16\right)\\ =x^3-3x^2+3x-1-x^3-8+3x^2-48\\ =3x-55\)

Thanks bạn