a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

1) Ta có: \(4x+8=3x-1\)

\(\Leftrightarrow4x-3x=-1-8\)

\(\Leftrightarrow x=-9\)

2) Ta có: \(10-5\left(x+3\right)>3\left(x-1\right)\)

\(\Leftrightarrow10-5x-15-3x+3>0\)

\(\Leftrightarrow-8x>2\)

hay \(x< \dfrac{-1}{4}\)

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

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

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

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

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

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

Vậy \(S=\left\{3\right\}\)

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

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

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

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

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

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):}`

Ta có: |x – 1| = x – 1 khi x – 1 ≥ 0

⇔ x ≥  1

|x – 1| = 1 – x khi x – 1 < 0

⇔x < 1

Ta có: x – 1 = 3x + 2

⇔ x – 3x = 2 + 1

⇔ x = -1,5

Giá trị x = -1,5 không thỏa mãn điều kiện x  ≥  1 nên loại.

1 – x = 3x + 2

⇔ -x – 3x = 2 – 1

⇔ -4x = 1

⇔ x = -0,25

Giá trị x = -0,25 thỏa mãn điều kiện x < 1 nên -0,25 là nghiệm của phương trình.

Vậy tập nghiệm của phương trình là S = {-0,25}.

Khi đó: |3x – 1| = 3x – 1.

(1) trở thành: 3x – 1 – x = 2 ⇔ 2x = 3 ⇔  (t/m).

+ TH2 :

Khi đó |3x – 1| = -(3x – 1) = 1 – 3x.

(1) trở thành: 1 – 3x – x = 2 ⇔ 1 – 4x = 2 ⇔ 4x = -1 ⇔  (t/m).

Vậy phương trình (1) có tập nghiệm 

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

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

\(a,2x\left(x-5\right)+4\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\2x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Vậy \(x\in\left\{5;-2\right\}\)

\(b,3x-15=2x\left(x-5\right)\\ \Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\\ \Leftrightarrow\left(x-5\right)\left(-2x+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-5=0\\-2x+3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\2x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy \(x\in\left\{5;\dfrac{3}{2}\right\}\)

\(c,\left(2x+1\right)\left(3x-2\right)=\left(5x-8\right)\left(2x+1\right)\\ \Leftrightarrow\left(2x+1\right)\left(3x-2\right)-\left(5x-8\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(3x-2-5x+8\right)=0\\ \Leftrightarrow\left(2x+1\right)\left(-2x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=0\\-2x+6=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=-1\\2x=6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=3\end{matrix}\right.\)

Vậy \(x\in\left\{-\dfrac{1}{2};3\right\}\)

Câu d xem lại đề