a) `4x(x-5)-(x-1)(x-3)=23`

`<=> 4x^2-20x-x^2+4x-3=23`

`<=>3x^2-16x-26=0`

`<=> x=(8+-\sqrt142)/3

*Nếu đề là: `4x(x-5)-(x-1)(4x-3)=23`

`<=> 4x^2-20x-4x^2+7x-3=23`

`<=>-13x=26`

`<=>x=-2`

b) `(x-5)(x-4)-(x+1)(x-2)=7`

`<=>x^2-9x+20-x^2+x+2=7`

`<=>-8x=-15`

`<=>x=15/8`

\(\text{a) 4x(x-5)-(x-1)(1x-3)=23}\)

\(\Leftrightarrow4x^2-20x-\left(x^2-3x-x+3\right)=23\)

\(\Leftrightarrow4x^2-20x-x^2+3x+x-3=23\)

\(\Leftrightarrow3x^2-16x-26=0\)

Giải phương trình bậc 2 một ẩn ta được :

\(x_1=\dfrac{8-\sqrt{142}}{3};x_2=\dfrac{8+\sqrt{142}}{3}\)

a) \(\left|\frac{3}{2}x+\frac{1}{2}\right|=\left|4x-1\right|\) 

\(\Rightarrow\orbr{\begin{cases}\frac{3}{2}x+\frac{1}{2}=4x-1\\\frac{3}{2}x+\frac{1}{2}=-4x+1\end{cases}}\Rightarrow\orbr{\begin{cases}4x-\frac{3}{2}x-1=\frac{1}{2}\\-4x-\frac{3}{2}x+1=\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}\frac{5}{2}x=\frac{3}{2}\\-\frac{11}{2}x=-\frac{1}{2}\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{5}\\x=\frac{1}{11}\end{cases}}\) 

phần b ở đề bài mình ghi sai, là bằng 0 chứ ko phải bằng 10

a: =>x=3/7+3/5=15/35+21/35=36/35

b: =>x/35=4/5-5/7=28/35-25/35=3/35

=>x=3

c: =>x<3/4+8/4=11/4

=>\(x\in\left\{0;1;2;3\right\}\)

d: =>5/3<x<5/6+24/6=29/6

=>\(x\in\left\{2;3;4\right\}\)

e: =>x<10/12-9/12=1/12

=>x=0

f: =>2/3<x<12/6-5/6=7/6

=>x=1

ko phải "-" đâu mà là nhân đấy

giữa 2 ngoặc đó

https://olm.vn/hoi-dap/detail/55678450227.html

1) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

2) \(\Rightarrow4x^2+4x+1-4x^2+13x-3-15=0\)

\(\Rightarrow17x=17\Rightarrow x=1\)

3) \(\Rightarrow\left(3x-1\right)\left(2x-7+6x-5\right)=0\)

\(\Rightarrow\left(2x-3\right)\left(3x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

2: Ta có: \(\left(2x+1\right)^2-\left(4x-1\right)\left(x-3\right)-15=0\)

\(\Leftrightarrow4x^2+4x+1-4x^2+12x+x-3-15=0\)

\(\Leftrightarrow17x=17\)

hay x=1