Bài `1:`

`h)(3/4x-1)(5/3x+2)=0`

`=>[(3/4x-1=0),(5/3x+2=0):}=>[(x=4/3),(x=-6/5):}`

______________

Bài `2:`

`b)3x-15=2x(x-5)`

`<=>3(x-5)-2x(x-5)=0`

`<=>(x-5)(3-2x)=0<=>[(x=5),(x=3/2):}`

`d)x(x+6)-7x-42=0`

`<=>x(x+6)-7(x+6)=0`

`<=>(x+6)(x-7)=0<=>[(x=-6),(x=7):}`

`f)x^3-2x^2-(x-2)=0`

`<=>x^2(x-2)-(x-2)=0`

`<=>(x-2)(x^2-1)=0<=>[(x=2),(x^2=1<=>x=+-2):}`

`h)(3x-1)(6x+1)=(x+7)(3x-1)`

`<=>18x^2+3x-6x-1=3x^2-x+21x-7`

`<=>15x^2-23x+6=0<=>15x^2-5x-18x+6=0`

`<=>(3x-1)(5x-1)=0<=>[(x=1/3),(x=1/5):}`

`j)(2x-5)^2-(x+2)^2=0`

`<=>(2x-5-x-2)(2x-5+x+2)=0`

`<=>(x-7)(3x-3)=0<=>[(x=7),(x=1):}`

`w)x^2-x-12=0`

`<=>x^2-4x+3x-12=0`

`<=>(x-4)(x+3)=0<=>[(x=4),(x=-3):}`

`m)(1-x)(5x+3)=(3x-7)(x-1)`

`<=>(1-x)(5x+3)+(1-x)(3x-7)=0`

`<=>(1-x)(5x+3+3x-7)=0`

`<=>(1-x)(8x-4)=0<=>[(x=1),(x=1/2):}`

`p)(2x-1)^2-4=0`

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

`<=>(2x-3)(2x+1)=0<=>[(x=3/2),(x=-1/2):}`

`r)(2x-1)^2=49`

`<=>(2x-1-7)(2x-1+7)=0`

`<=>(2x-8)(2x+6)=0<=>[(x=4),(x=-3):}`

`t)(5x-3)^2-(4x-7)^2=0`

`<=>(5x-3-4x+7)(5x-3+4x-7)=0`

`<=>(x+4)(9x-10)=0<=>[(x=-4),(x=10/9):}`

`u)x^2-10x+16=0`

`<=>x^2-8x-2x+16=0`

`<=>(x-2)(x-8)=0<=>[(x=2),(x=8):}`

\(\dfrac{\left(2x+1\right)^2}{5}-\dfrac{\left(x-1\right)^2}{3}=\dfrac{7x^2-14x-5}{15}\) 

⇔ \(\dfrac{3\left(2x+1\right)^2}{15}-\dfrac{5\left(x-1\right)^2}{15}=\dfrac{7x^2-14x-5}{15}\)

⇔ \(3\left(2x+1\right)^2-5\left(x-1\right)^2=7x^2-14x-5\)

⇔ \(3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)

⇔ \(12x^2+12x+3-5x^2+10x-5=7x^2-14x-5\)

⇔ \(7x^2+22x-2=7x^2-14x-5\) ⇔ \(36x+3=0\) ⇔ x=\(\dfrac{-1}{12}\)

\(\Leftrightarrow3\left(4x^2+4x+1\right)-5\left(x^2-2x+1\right)=7x^2-14x-5\)

\(\Leftrightarrow12x^2+12x+3-5x^2+10x-5-7x^2+14x+5=0\)

\(\Leftrightarrow36x=-3\)

hay x=-1/12

`(-7x^2+4)/(x^3+1)=5/(x^2-x+1)-1/(x+1)(x ne -1)`

`<=>-7x^2+4=5(x+1)-x^2+x-1`

`<=>-7x^2+4=5x+5-x^2+x-1`

`<=>6x^2+6x=0`

`<=>6x(x+1)=0`

Vì `x ne -1=>x+1 ne 0`

`=>x=0`

Vậy `S={0}`

ĐKXĐ: \(x\ne-1\)

Ta có: \(\dfrac{-7x^2+4}{x^3+1}=\dfrac{5}{x^2-x+1}-\dfrac{1}{x+1}\)

\(\Leftrightarrow\dfrac{5\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{-7x^2+4}{\left(x+1\right)\left(x^2-x+1\right)}\)

Suy ra: \(5x+5-x^2+x-1=-7x^2+4\)

\(\Leftrightarrow-x^2+6x+4+7x^2-4=0\)

\(\Leftrightarrow6x^2+6x=0\)

\(\Leftrightarrow6x\left(x+1\right)=0\)

mà 6>0

nên x(x+1)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-1\left(loại\right)\end{matrix}\right.\)

Vậy: S={0}

\(\frac{x+1}{x-2}+\frac{x-1}{x+2}=\frac{2\left(x^2+2\right)}{x^2-4}\left(x\ne\pm2\right)\)

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

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

Vậy phương trình này có vô số nghiệm x thỏa mãn trừ x khác 2 và -2

a) ĐKXĐ: \(x\notin\left\{-1;0\right\}\)

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

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

Suy ra: \(x^2+3x+x^2-3x+2=2x^2+2x\)

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

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

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

hay x=1(nhận)

Vậy: S={1}

b) ĐKXĐ: \(x\notin\left\{-7;\dfrac{3}{2}\right\}\)

Ta có: \(\dfrac{3x-2}{x+7}=\dfrac{6x+1}{2x-3}\)

\(\Leftrightarrow\left(3x-2\right)\left(2x-3\right)=\left(6x+1\right)\left(x+7\right)\)

\(\Leftrightarrow6x^2-9x-4x+6=6x^2+42x+x+7\)

\(\Leftrightarrow6x^2-13x+6-6x^2-43x-7=0\)

\(\Leftrightarrow-56x-1=0\)

\(\Leftrightarrow-56x=1\)

hay \(x=-\dfrac{1}{56}\)(nhận)

Vậy: \(S=\left\{-\dfrac{1}{56}\right\}\)

c) ĐKXĐ: \(x\ne-\dfrac{2}{3}\)

Ta có: \(\dfrac{5}{3x+2}=2x-1\)

\(\Leftrightarrow5=\left(3x+2\right)\left(2x-1\right)\)

\(\Leftrightarrow6x^2-3x+4x-2-5=0\)

\(\Leftrightarrow6x^2+x-7=0\)

\(\Leftrightarrow6x^2-6x+7x-7=0\)

\(\Leftrightarrow6x\left(x-1\right)+7\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(6x+7\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\6x+7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\6x=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-\dfrac{7}{6}\left(nhận\right)\end{matrix}\right.\)

Vậy: \(S=\left\{1;-\dfrac{7}{6}\right\}\)

d) ĐKXĐ: \(x\ne\dfrac{2}{7}\)

Ta có: \(\left(2x+3\right)\cdot\left(\dfrac{3x+8}{2-7x}+1\right)=\left(x-5\right)\left(\dfrac{3x+8}{2-7x}+1\right)\)

\(\Leftrightarrow\left(2x+3\right)\cdot\left(\dfrac{3x+8+2-7x}{2-7x}\right)-\left(x-5\right)\left(\dfrac{3x+8+2-7x}{2-7x}\right)=0\)

\(\Leftrightarrow\left(2x+3-x+5\right)\cdot\dfrac{-4x+6}{2-7x}=0\)

\(\Leftrightarrow\left(x+8\right)\cdot\left(-4x+6\right)=0\)(Vì \(2-7x\ne0\forall x\) thỏa mãn ĐKXĐ)

\(\Leftrightarrow\left[{}\begin{matrix}x+8=0\\-4x+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\\-4x=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-8\left(nhận\right)\\x=\dfrac{3}{2}\left(nhận\right)\end{matrix}\right.\)

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

1/ \(2\left(x-5\right)=\left(-x-5\right)\)

\(\Leftrightarrow2x-10=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy: \(S=\left\{\dfrac{5}{3}\right\}\)

==========

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

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

\(\Leftrightarrow-x=-7\)

\(\Leftrightarrow x=7\)

Vậy: \(S=\left\{7\right\}\)

==========

3/ \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1=x-19\)

\(\Leftrightarrow0x=0\)

Vậy: \(S=\left\{x|x\text{ ∈ }R\right\}\) 

===========

4/ \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2=10-15x\)

\(\Leftrightarrow14x=1\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy: \(S=\left\{\dfrac{1}{14}\right\}\)

==========

5/ \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x=7x+1\)

\(\Leftrightarrow-2x=6\)

\(\Leftrightarrow x=-3\)

Vậy: \(S=\left\{-3\right\}\)

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Chúc bạn học tốt.

1. \(2\left(x-5\right)=-x-5\)

\(\Leftrightarrow3x=5\)

\(\Leftrightarrow x=\dfrac{5}{3}\)

Vậy \(S=\left\{\dfrac{5}{3}\right\}\)

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

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

\(\Leftrightarrow x=7\)

Vậy \(S=\left\{7\right\}\)

3. \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1-x+19=0\)

\(\Leftrightarrow0x=0\)

Vậy \(S=\left\{x\in R\right\}\)

4. \(7-\left(x-2\right)=5\left(2-3x\right)\)

\(\Leftrightarrow7-x+2-10+15x=0\)

\(\Leftrightarrow14x-1=0\)

\(\Leftrightarrow x=\dfrac{1}{14}\)

Vậy \(S=\left\{\dfrac{1}{14}\right\}\)

4. \(2x-\left(5-3x\right)=7x+1\)

\(\Leftrightarrow2x-5+3x-7x-1=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow x=-3\)

Vậy \(S=\left\{-3\right\}\)

1: Ta có: \(2\left(x-5\right)=\left(-x-5\right)\)

\(\Leftrightarrow2x-10+x+5=0\)

\(\Leftrightarrow3x=5\)

hay \(x=\dfrac{5}{3}\)

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

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

\(\Leftrightarrow-x=-7\)

hay x=7

3: Ta có: \(4\left(x-5\right)-\left(3x-1\right)=x-19\)

\(\Leftrightarrow4x-20-3x+1-x+19=0\)

\(\Leftrightarrow0x=0\)(luôn đúng

\(\frac{7x-\frac{x-3}{2}}{5}-x+1nha.Mình,nhầm\)

Anh ko ghi lại đề nha em gái ! 

\(\Leftrightarrow\frac{\left(\frac{10x-4+5x}{5}\right)}{15}=\frac{\left(\frac{14x-x+3}{2}\right).x}{5}+1\)

\(\Leftrightarrow\frac{\left(\frac{15x-4}{5}\right)}{15}=\frac{\left(\frac{13x^2+3x}{2}\right)}{5}+1\)

\(\Leftrightarrow\frac{\left(\frac{15x-4}{5}\right)}{15}=\frac{\left(\frac{39x^2+9x}{2}\right)+15}{15}\)

\(\Leftrightarrow\frac{15x-4}{5}=\frac{39x^2+9x+30}{2}\)

\(\Leftrightarrow2.\left(15x-4\right)=5.\left(39x^2+9x+30\right)\)

\(\Leftrightarrow30x-8=195x^2+45x+150\)

\(\Leftrightarrow-195x^2-15x-158=0\)

\(\left(a=-195;b=-15;c=-158\right)\)

\(\Delta=b^2-4ac\)

\(=\left(-15\right)^2-4.\left(-195\right).\left(-158\right)=-123015< 0\)

Vì \(\Delta< 0\) nên phương trình vô nghiệm. 

Nếu có gì thắc mắc về bài này cứ hỏi anh ! 

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

\(\Leftrightarrow15x=-203\)

\(\Leftrightarrow x=-\frac{203}{15}\)