a, \(\dfrac{6-x}{4x-3}=\dfrac{2}{4x-3}\)

ĐKXĐ: \(x\ne\dfrac{3}{4}\)

PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(6-x\right)\left(4x-3\right)}{4x-3}=\dfrac{2\left(4x-3\right)}{4x-3}\)

                  \(\Rightarrow6-x=2\)

                  \(\Leftrightarrow x=4\)(thỏa mãn ĐKXĐ)

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

ĐKXĐ: \(x\ne\dfrac{3}{2}\)

PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(3-x\right)\left(2x-3\right)}{2x-3}+\left(x+1\right)\left(2x-3\right)=\dfrac{-4\left(2x-3\right)}{2x-3}\)

                  \(\Rightarrow3-x+2x-3x+2x-3=-8x+12\)

                  \(\Leftrightarrow8x=12\)

                  \(\Leftrightarrow x=\dfrac{3}{2}\)(không thỏa mãn ĐKXĐ)

Vậy \(x\in\varnothing\).

a) ĐK: \(x\ne\dfrac{3}{4}\)

PT \(\Rightarrow27x-18-4x^2=8x-6\)

\(\Leftrightarrow4x^2-19x+12=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{3}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy phương trình có nghiệm \(x=4\)

b) ĐK: \(x\ne\dfrac{3}{2}\)

PT \(\Rightarrow3-x+2x^2-5x+3=-4\) 

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

  Vậy phương trình vô nghiệm

c) ĐK: \(x\ne3\)

PT \(\Rightarrow2x^2-5x-3=2x-4\)

\(\Leftrightarrow2x^2-7x+1=0\) \(\Leftrightarrow x=\dfrac{7\pm\sqrt{41}}{4}\)

  Vậy phương trình có nghiệm \(x=\dfrac{7\pm\sqrt{41}}{4}\)

d: Ta có: \(\dfrac{2x+1}{3}-\dfrac{1-x}{2}\ge1-\dfrac{x}{4}\)

\(\Leftrightarrow8x+4-6+6x\ge12-3x\)

\(\Leftrightarrow14x+3x\ge12+2=14\)

\(\Leftrightarrow x\ge\dfrac{14}{17}\)

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

\(\Leftrightarrow6x+12+4x-8< 6x-9\)

\(\Leftrightarrow4x< -9+8-12=-13\)

hay \(x< -\dfrac{13}{4}\)

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

Suy ra: \(3x-6-x+1=2x+4\)

\(\Leftrightarrow2x-5=2x+4\left(vôlý\right)\)

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

Suy ra: \(\left(x-5\right)\left(2x+3\right)-x\left(2x-3\right)=1-6x\)

\(\Leftrightarrow2x^2-7x-15-2x^2+6x+6x-1=0\)

\(\Leftrightarrow5x=16\)

hay \(x=\dfrac{16}{5}\)

a: \(\Leftrightarrow7\left(7-3x\right)+12\left(5x+2\right)=84\left(x+13\right)\)

\(\Leftrightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow39x-84x=1092-73\)

=>-45x=1019

hay x=-1019/45

b: \(\Leftrightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

=>21x+63-14=20x+36-49x+63

=>21x+49=-29x+99

=>50x=50

hay x=1

c: \(\Leftrightarrow7\left(2x+1\right)-3\left(5x+2\right)=21x+63\)

=>14x+7-15x-6-21x-63=0

=>-22x-64=0

hay x=-32/11

d: \(\Leftrightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-17\cdot105\)

=>70x-105-30x-45=84x+63-1785

=>40x-150-84x+1722=0

=>-44x+1572=0

hay x=393/11

a, msc 12.7=84 

Chuyển vế về =0 rồi làm

b,msc 28

c,làm tương tự

a, \(\Rightarrow49-21x+60x+24=84x+1092\)

\(\Leftrightarrow-45x=1019\Leftrightarrow x=-\dfrac{1019}{45}\)

b, \(\Rightarrow21\left(x+3\right)-14=4\left(5x+9\right)-7\left(7x-9\right)\)

\(\Leftrightarrow21x+63-14=20x+36-49x+63\)

\(\Leftrightarrow50x=50\Leftrightarrow x=1\)

c, \(\Rightarrow14x+7-15x-6=21x+63\Leftrightarrow-22x=62\Leftrightarrow x=-\dfrac{31}{11}\)

d, \(\Rightarrow35\left(2x-3\right)-15\left(2x+3\right)=21\left(4x+3\right)-105.17\)

\(\Leftrightarrow70x-105-30x-45=84x+63-1785\)

\(\Leftrightarrow-44x=-1572\Leftrightarrow x=\dfrac{393}{11}\)

1: Ta có: \(\dfrac{-3}{x-4}-\dfrac{3-5x}{x^2-16}=\dfrac{1}{x+4}\)

Suy ra: \(-3\left(x+4\right)-3+5x=x-4\)

\(\Leftrightarrow-3x-12-3+5x-x+4=0\)

\(\Leftrightarrow x=11\left(nhận\right)\)

2. ĐKXĐ: $x\neq \pm 2$

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

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

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

\(\Leftrightarrow 2x-5=2x+4\Leftrightarrow 9=0\) (vô lý)

Vậy pt vô nghiệm

3. ĐKXĐ: $x\neq \pm \frac{3}{2}$

PT \(\Leftrightarrow \frac{(x-5)(2x+3)-x(2x-3)}{(2x-3)(2x+3)}=\frac{1-6x}{(2x-3)(2x+3)}\)

\(\Rightarrow (x-5)(2x+3)-x(2x-3)=1-6x\)

\(\Leftrightarrow 2x^2-7x-15-2x^2+3x+6x-1=0\)

\(\Leftrightarrow 2x-16=0\Leftrightarrow x=8\) (thỏa mãn)

1.

\(x^4-6x^2-12x-8=0\)

\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)

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

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

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

\(\Leftrightarrow x=1\pm\sqrt{5}\)

3.

ĐK: \(x\ge-9\)

\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)

\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)

\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)

Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)

\(\Leftrightarrow...\)

2.

ĐK: \(x\ne\dfrac{2\pm\sqrt{2}}{2};x\ne\dfrac{-2\pm\sqrt{2}}{2}\)

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

\(\Leftrightarrow\dfrac{1}{2x+\dfrac{1}{x}+4}+\dfrac{1}{2x+\dfrac{1}{x}-4}=\dfrac{3}{5}\)

Đặt \(2x+\dfrac{1}{x}+4=a;2x+\dfrac{1}{x}-4=b\left(a,b\ne0\right)\)

\(pt\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{3}{5}\left(1\right)\)

Lại có \(a-b=8\Rightarrow a=b+8\), khi đó:

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

\(\Leftrightarrow\dfrac{2b+8}{\left(b+8\right)b}=\dfrac{3}{5}\)

\(\Leftrightarrow10b+40=3\left(b+8\right)b\)

\(\Leftrightarrow\left[{}\begin{matrix}b=2\\b=-\dfrac{20}{3}\end{matrix}\right.\)

TH1: \(b=2\Leftrightarrow...\)

TH2: \(b=-\dfrac{20}{3}\Leftrightarrow...\)

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

\(\Leftrightarrow\dfrac{20\left(2x-1\right)}{60}+\dfrac{15\left(3x-2\right)}{60}=\dfrac{12\left(4x-3\right)}{60}\)

`<=> 20(2x-1) +15(3x-2) =12(4x-3)`

`<=> 40x - 20 + 45x - 30 = 48x - 36`

`<=> 85x -50 = 48x - 36`

`<=> 85x-48x = -36+50`

`<=> 37x =14`

`<=> x= 14/37`

Vậy phương trình có nghiệm `x=14/37`

__

\(\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{x^2-9}\)

\(\Leftrightarrow\dfrac{5}{x-3}+\dfrac{4}{x+3}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{5\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-6}{\left(x-3\right)\left(x+3\right)}\)

`=> 5x + 15 + 4x -12=x-6`

`<=> 9x + 3=x-6`

`<=> 9x-x=-6-3`

`<=> 8x = -9`

`<=>x=-9/8(tm)`

Vậy phương trình có nghiệm `x=-9/8`

` @ yngoc`

\(1,\dfrac{4x-4}{3}=\dfrac{7-x}{5}\\ \Leftrightarrow5\left(4x-4\right)=3\left(7-x\right)\\ \Leftrightarrow20x-20=21-3x\\ \Leftrightarrow17x=41\Leftrightarrow x=\dfrac{41}{17}\)

\(2,\dfrac{3x-9}{5}=\dfrac{3-x}{2}\\ \Leftrightarrow6x-18=15-5x\\ \Leftrightarrow11x=33\\ \Leftrightarrow x=3\)

\(3,\dfrac{2x-1}{5}-\dfrac{3-x}{3}=1\\ \Leftrightarrow\dfrac{6x-3-15+5x}{15}=1\\ \Leftrightarrow11x-18=1\\ \Leftrightarrow x=\dfrac{19}{11}\)

\(4,\dfrac{x-5}{3}+\dfrac{3x+4}{2}=\dfrac{5x+2}{6}\\ \Leftrightarrow2x-10+9x+12=5x+2\\ \Leftrightarrow6x=0\Leftrightarrow x=0\)

\(5,\dfrac{x-3}{2}+\dfrac{2x+3}{5}=\dfrac{2x+5}{10}\\ \Leftrightarrow5x-15+4x+6=2x+5\\ \Leftrightarrow7x=14\\ \Leftrightarrow x=2\)

Tick nha

2: Ta có: \(\dfrac{3x-9}{5}=\dfrac{3-x}{2}\)

\(\Leftrightarrow6x-18=15-5x\)

\(\Leftrightarrow11x=33\)

hay x=3