\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

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

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

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

a) \(\sqrt {x + 2}  = x\)

Điều kiện: \(x \ge 0\)

Bình phương 2 vế của phương trình ta được:

\(x + 2 = {x^2} \Leftrightarrow {x^2} - x - 2 = 0\)\( \Leftrightarrow \left[ \begin{array}{l}x =  - 1\\x = 2\end{array} \right.\)

b) \(\sqrt {2{x^2} + 3x - 2}  = \sqrt {{x^2} + x + 6} \)

Bình phương 2 vế của phương trình ta được:

\(\begin{array}{l}2{x^2} + 3x - 2 = {x^2} + x + 6\\ \Leftrightarrow {x^2} + 2x - 8 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x = 2\\x =  - 4\end{array} \right.\end{array}\)

Thay vào bất phương trình \(2{x^2} + 3x - 2 \ge 0\) ta thấy cả 2 nghiệm đều thỏa mãn.

Vậy tập nghiệm là \(S = \left\{ { - 4;2} \right\}\)

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

Điều kiện: \(x + 3 \ge 0 \Leftrightarrow x \ge  - 3\)

Bình phương 2 vế của phương trình ta được:

\(\begin{array}{l}2{x^2} + 3x - 1 = {\left( {x + 3} \right)^2}\\ \Leftrightarrow {x^2} - 3x - 10 = 0\\ \Leftrightarrow \left[ \begin{array}{l}x =  - 2\left( {tm} \right)\\x = 5\left( {tm} \right)\end{array} \right.\end{array}\)

Vậy tập nghiệm là \(S = \left\{ { - 2;5} \right\}\)

1/a/\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-5\\x=-6\end{cases}}}\)

Vậy ...................

b/ ĐKXĐ:\(x\ne2;x\ne5\)

.....\(\Rightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x^2-10x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\left(nhận\right)\\x=5\left(loại\right)\end{cases}}}\)

Vậy ..............

`Answer:`

`1.`

a. \(\left(x+5\right)\left(2x+1\right)-x^2+25=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x^2-25\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1\right)-\left(x+5\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(2x+1-x+5\right)=0\)

\(\Leftrightarrow\left(x+5\right)\left(x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+5=0\\x+6=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-6\\x=-5\end{cases}}}\)

b. \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\left(ĐKXĐ:x\ne2;x\ne5\right)\)

\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow\frac{3x\left(x-5\right)-x\left(x-2\right)+3x}{\left(x-2\right)\left(x-5\right)}=0\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)+3x=0\)

\(\Leftrightarrow3x^2-15x-x^2+2x+3x=0\)

\(\Leftrightarrow2x\left(x-5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x-5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\text{(Không thoả mãn)}\end{cases}}}\)

`2.`

\(ĐKXĐ:x\ne-m-2;x\ne m-2\)

Ta có: \(\frac{x+1}{x+2+m}=\frac{x+1}{x+2-m}\left(1\right)\)

a. Khi `m=-3` phương trình `(1)` sẽ trở thành: \(\frac{x+1}{x-1}=\frac{x+1}{x+5}\left(x\ne1;x\ne-5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\\frac{1}{x-1}=\frac{1}{x+5}\end{cases}\Leftrightarrow\orbr{\begin{cases}x+1=0\\x-1=x+5\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\-1=5\text{(Vô nghiệm)}\end{cases}}}\)

b. Để phương trình `(1)` nhận `x=3` làm nghiệm thì

\(\Leftrightarrow\hept{\begin{cases}\frac{3+1}{3+2-m}=\frac{3+1}{3+2-m}\\3\ne-m-2\\3\ne m-2\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{4}{5+m}=\frac{4}{5-m}\\m\ne\pm5\end{cases}}\Leftrightarrow\hept{\begin{cases}5+m=5-m\\m\ne\pm5\end{cases}}\Leftrightarrow m=0\)

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: \(S=\left\{-3\right\}\)

[---]

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

1) Ta có: \(x^3-3x^2+2x=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=2\end{matrix}\right.\)

Vậy: S={0;1;2}

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

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

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

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

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

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

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

Vậy: S={0;-2}

       3x²+2x=0

<=>x(3x+2)=0

<=>x=0 hoặc 3x+2=0

từ đó bạn giải ra x thuộc{0;-2/3}

chúc bạn học tốt và nhớ tích đúng cho mình

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):}`

\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

\(\Leftrightarrow x-1=3x-2\)

\(\Leftrightarrow2x=1\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

 \(\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right).\)

\(\Leftrightarrow\left(3x+2\right)\left(x-1\right)\left(x+1\right)-\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0.\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(x-1-3x+2\right)=0.\)

\(\Leftrightarrow\left(3x+2\right)\left(x+1\right)\left(-2x+1\right)=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+2=0.\\x+1=0.\\-2x+1=0.\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}.\\x=-1.\\x=\dfrac{1}{2}.\end{matrix}\right.\)

c: =>(x-3)(x²+3x+5)=0

=>x-3=0

hay x=3

d: =>(3x-1)(x²+2-7x+10)=0

=>(3x-1)(x-3)(x-4)=0

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)

a: =>4x-2x-2-3x-2=0

=>-x-4=0

=>x=-4

b: =>x+2-2x-2+x=0

=>0x=0(luôn đúng)

d: =>3x=3

hay x=1

e: =>2x=1

hay x=1/2

f: =>4x=-4

hay x=-1

g: =>3x=-3

hay x=-1

c: =>2x+3-5-4+x=0

=>3x-6=0

=>x=2

d: =>3x=3

hay x=1

e: =>2x=1

hay x=1/2

f: =>4x=-4

hay x=-1

g: =>3x=-3

hay x=-1

\(a,4x-2\left(x+1\right)=3x+2\\ \Leftrightarrow4x-2x-2-3x-2=0\\ \Leftrightarrow-x-4=0\\ \Leftrightarrow x+4=0\\ \Leftrightarrow x=-4\)

Vậy pt có tập nghiệm \(S=\left\{-4\right\}\)

\(b,x+2-2\left(x+1\right)=-x\\ \Leftrightarrow x+2-2x-2+x=0\\ \Leftrightarrow0=0\)

Vậy pt có tập nghiệm \(S=R\)

\(c,2\left(x+3\right)-5=4-x\\ \Leftrightarrow2x+6-5-4+x=0\\ \Leftrightarrow3x-3=0\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)

Vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(d,3x-2=1\\ \Leftrightarrow3x=3\\ \Leftrightarrow x=1\)

Vậy pt có tập nghiệm \(S=\left\{1\right\}\)

\(e,2x-1=0\\ \Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

Vậy pt có tập nghiệm \(S=\left\{\dfrac{1}{2}\right\}\)

\(f,4x+3=-1\\ \Leftrightarrow4x=-4\\ \Leftrightarrow x=-1\)

Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)

\(g,3x+2=-1\\ \Leftrightarrow3x=-3\\ \Leftrightarrow x=-1\)

Vậy pt có tập nghiệm \(S=\left\{-1\right\}\)

a,4x-2(x+1)=3x+2

⇔ 4x - 2x -2 = 3x + 2

⇔ x = -4

b,x+2-2(x+1)=-x

⇔ x + 2 - 2x - 2 = -x

⇔ 0 = 0

c,2(x+3)-5=4-x

⇔ 2x + 6 - 5 = 4 - x

⇔ 3x = 3

⇔ x = 1

d,3x-2=1

⇔ 3x = 3

⇔ x = 1

e,2x-1=0

⇔ x = \(\dfrac{1}{2}\)

f,4x+3=-1

⇔ x = -1

g,3x+2=-1

⇔ x = -1