1: \(P=\left(\dfrac{2x}{x^2-9}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x^2-3x}\right)\)

\(=\left(\dfrac{2x}{\left(x-3\right)\left(x+3\right)}-\dfrac{1}{x+3}\right):\left(\dfrac{2}{x}-\dfrac{x-1}{x\cdot\left(x-3\right)}\right)\)

\(=\dfrac{2x-x+3}{\left(x-3\right)\left(x+3\right)}:\dfrac{2\left(x-3\right)-x+1}{x\left(x-3\right)}\)

\(=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x\left(x-3\right)}{2x-6-x+1}\)

\(=\dfrac{x}{x-5}\)

giúp mình giải bài toán trên với. Mình cảm ơn rất nhiều

a: =>1/2x-3/4x=-5/6+7/3

=>-1/4x=14/6-5/6=3/2

=>x=-3/2*4=-6

b: =>4/5x-3/2x=1/2+6/5

=>-7/10x=17/10

=>x=-17/7

c: =>6/5x+6/20=6/5-1/3x

=>6/5x+1/3x=6/5-3/10=12/10-3/10=9/10

=>x=27/46

d: =>6x+3/2+4/5=1/2-2x

=>8x=1/2-3/2-4/5=-1-4/5=-9/5

=>x=-9/40

1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)

\(\Leftrightarrow2x-8+12x=4x-2\)

\(\Leftrightarrow10x=6\)

hay \(x=\dfrac{3}{5}\)

2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)

\(\Leftrightarrow15x-6-30=10-20x\)

\(\Leftrightarrow35x=46\)

hay \(x=\dfrac{46}{35}\)

3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)

\(\Leftrightarrow3x-6-4=6x-6\)

\(\Leftrightarrow-3x=4\)

hay \(x=-\dfrac{4}{3}\)

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

\(\Leftrightarrow\dfrac{\left(x-4\right).2}{3.2}+\dfrac{2x.6}{6}=\dfrac{4x-2}{6}\)

\(\Rightarrow2x-8+12x=4x-2\\ \Leftrightarrow10x=6\\ \Leftrightarrow x=\dfrac{3}{5}\)

4: Ta có: \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)

\(\Leftrightarrow40x-20+45x-30=48x-36\)

\(\Leftrightarrow37x=14\)

hay \(x=\dfrac{14}{37}\)

5: Ta có: \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)

\(\Leftrightarrow2x-6-3x-6=x+4-9\)

\(\Leftrightarrow-x-x=-5-12=-17\)

hay \(x=\dfrac{17}{2}\)

d: ĐKXĐ: \(x\notin\left\{2;-3\right\}\)

\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{5}{6-x^2-x}\)

=>\(\dfrac{1}{x-2}-\dfrac{6}{x+3}=\dfrac{-5}{\left(x+3\right)\left(x-2\right)}\)

=>\(x+3-6\left(x-2\right)=-5\)

=>x+3-6x+12=-5

=>-5x+15=-5

=>-5x=-20

=>x=4(nhận)

e: ĐKXĐ: x<>-2

\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{x^3+8}=\dfrac{5}{x^2-2x+4}\)

=>\(\dfrac{2}{x+2}-\dfrac{2x^2+16}{\left(x+2\right)\left(x^2-2x+4\right)}=\dfrac{5}{x^2-2x+4}\)

=>\(2\left(x^2-2x+4\right)-2x^2-16=5\left(x+2\right)\)

=>\(2x^2-4x+8-2x^2-16=5x+10\)

=>5x+10=-4x-8

=>9x=-18

=>x=-2(loại)

f: ĐKXĐ: \(x\in\left\{1;-1\right\}\)

\(\dfrac{x+1}{x^2+x+1}-\dfrac{x-1}{x^2-x+1}=\dfrac{2\left(x+2\right)^2}{x^6-1}\)

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

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

=>\(\left(x^3+1\right)\left(x^2-1\right)-\left(x^3-1\right)\left(x^2-1\right)=2\left(x^2+4x+4\right)\)

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

=>\(2x^2+8x+8=\left(x^2-1\right)\cdot2=2x^2-2\)

=>8x=-10

=>x=-5/4(nhận)

1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)

\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)

\(\Leftrightarrow5x+1-2x+4=4\)

\(\Leftrightarrow3x=-1\)

hay \(x=-\dfrac{1}{3}\)

2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)

\(\Leftrightarrow9x+27+12-36x=-2x+2\)

\(\Leftrightarrow-27x+2x=2-39\)

hay \(x=\dfrac{37}{25}\)

3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)

\(\Leftrightarrow3x+6-10x=4-4x\)

\(\Leftrightarrow-7x+4x=4-6=-2\)

hay \(x=\dfrac{2}{3}\)

4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)

\(\Leftrightarrow5x-15-x-1=2x-4\)

\(\Leftrightarrow4x-2x=-4+16=12\)

hay x=6

5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)

\(\Leftrightarrow12x+3-9x+5+4x-8=0\)

\(\Leftrightarrow7x=0\)

hay x=0

2) Ta có: \(\dfrac{x+4}{5}-x+4=\dfrac{x}{3}-\dfrac{x-2}{2}\)

\(\Leftrightarrow\dfrac{6\left(x+4\right)}{30}-\dfrac{30\left(x-4\right)}{30}=\dfrac{10x}{30}-\dfrac{15\left(x-2\right)}{30}\)

\(\Leftrightarrow6x+24-30x+120=10x-15x+30\)

\(\Leftrightarrow-24x+144=-5x+30\)

\(\Leftrightarrow-24x+144+5x-30=0\)

\(\Leftrightarrow-19x+114=0\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: x=6

3) Ta có: \(\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\)

\(\Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\)

\(\Leftrightarrow30x+9=36+24+32x\)

\(\Leftrightarrow30x+9-60-32x=0\)

\(\Leftrightarrow-2x-51=0\)

\(\Leftrightarrow-2x=51\)

hay \(x=-\dfrac{51}{2}\)

Vậy: \(x=-\dfrac{51}{2}\)

4) Ta có: \(\dfrac{x+1}{3}-\dfrac{x-2}{6}=\dfrac{2x-1}{2}\)

\(\Leftrightarrow\dfrac{2\left(x+1\right)}{6}-\dfrac{x-2}{6}=\dfrac{3\left(2x-1\right)}{6}\)

\(\Leftrightarrow2x+2-x+2=6x-3\)

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

\(\Leftrightarrow-5x+7=0\)

\(\Leftrightarrow-5x=-7\)

hay \(x=\dfrac{7}{5}\)

Vậy: \(x=\dfrac{7}{5}\)

1) \(\dfrac{5x-2}{3}=\dfrac{5-3x}{2}\)

\(2\left(5x-2\right)=3\left(5-3x\right)\)

\(10x-4=15-9x\)

\(10x+9x=15+4\)

\(19x=19\)

\(x=1\)

Vậy \(x=1\)

2) Ta có: ⇔6(x+4)30−30(x−4)30=10x30−15(x−2)30⇔6(x+4)30−30(x−4)30=10x30−15(x−2)30

hay x=6

Vậy: x=6

3) Ta có: ⇔3(10x+3)36=3636+4(6+8x)36⇔3(10x+3)36=3636+4(6+8x)36

hay x=−512x=−512

4) Ta có: ⇔2(x+1)6−x−26=3(2x−1)6⇔2(x+1)6−x−26=3(2x−1)6

hay x=75

4 câu đầu hìn như sai đề :v

`m)(3/2-2/(-5)):x-1/2=3/2`

`<=>(3/2+2/5):x=3/2+1/2=2`

`<=>19/10:x=2`

`<=>x=19/10:2=19/20`

`n)(3/2-5/11-3/13)(2x-2)=(-3/4+5/22+3/26)`

`<=>(3/2-5/11-3/13)(2x-2)+3/4-5/22-3/26=0`

`<=>(3/2-5/11-3/13)(2x-2)+1/2(3/2-5/11-3/13)=0`

`<=>(3/2-5/11-3/13)(2x-2+1/2)=0`

Mà `3/2-5/11-3/13>0`

`<=>2x-2+1/2=0`

`<=>2x-3/2=0`

`<=>2x=3/2<=>x=3/4`

h, \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\left(x\ne0\right)\)

\(\Leftrightarrow\dfrac{x^2}{2}-1=\dfrac{x}{12}\)

\(\Leftrightarrow x^2-\dfrac{x}{6}-2=0\)

\(\Leftrightarrow x^2-2.x.\dfrac{1}{12}+\dfrac{1}{144}-\dfrac{289}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{1}{12}\right)^2=\dfrac{289}{144}\)

\(\Leftrightarrow x=\dfrac{1}{12}\pm\dfrac{\sqrt{289}}{12}\)

Vậy ...

i, \(\Leftrightarrow x^2-\dfrac{2.x.7}{12}+\dfrac{49}{144}-\dfrac{1}{144}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{2}\right)^2=\dfrac{1}{144}\)

\(\Leftrightarrow x=\dfrac{7}{2}\pm\dfrac{1}{12}\)

Vậy ...

h) Ta có: \(\dfrac{x}{2}-\dfrac{1}{x}=\dfrac{1}{12}\)

\(\Leftrightarrow\dfrac{x^2-2}{2x}=\dfrac{1}{12}\)

\(\Leftrightarrow12x^2-24-2x=0\)

\(\Delta=\left(-2\right)^2-4\cdot12\cdot\left(-24\right)=1156\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{2-34}{24}=\dfrac{-8}{3}\\x_2=\dfrac{2+34}{24}=\dfrac{36}{24}=\dfrac{3}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{-\dfrac{8}{3};\dfrac{3}{2}\right\}\)

m) Ta có: \(\left(\dfrac{3}{2}-\dfrac{2}{-5}\right):x-\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Leftrightarrow\dfrac{19}{10}:x=2\)

hay \(x=\dfrac{19}{20}\)

Vậy: \(S=\left\{\dfrac{19}{20}\right\}\)

a) ĐKXĐ: \(x\ne3\)

Ta có: \(\dfrac{x^2-x-6}{x-3}=0\)

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

Suy ra: x+2=0

hay x=-2(thỏa ĐK)

Vậy: S={-2}

d)

ĐKXĐ: \(x\notin\left\{1;3\right\}\)

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

\(\Leftrightarrow\dfrac{\left(x+5\right)\left(x-3\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-3\right)\left(x-1\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\)

Suy ra: \(x^2-3x+5x-15=x^2-1-8\)

\(\Leftrightarrow2x-15+9=0\)

\(\Leftrightarrow2x-6=0\)

hay x=3(loại)

Vậy: \(S=\varnothing\)