a) Nếu \(x-1\ge0\Rightarrow x\ge1\) thì \(x-1=2x-5\Rightarrow-x=-4\Rightarrow x=4\) (nhận)

   Nếu \(x-1< 0\Rightarrow x< 1\) thì \(x-1=-\left(2x-5\right)\Rightarrow3x=6\Rightarrow x=2\) (loại)

Vậy x = 4

b) Nếu \(9-7x\ge0\Rightarrow x\le\frac{9}{7}\) thì \(9-7x=5x-3\Rightarrow-12x=-12\Rightarrow x=1\) (nhận)

    Nếu \(9-7x< 0\Rightarrow x>\frac{9}{7}\) thì \(9-7x=-\left(5x-3\right)\Rightarrow-2x=-6\Rightarrow x=3\) (nhận)

Vậy x = 1 hoặc x = 3

c) \(8x-\left|4x+1\right|=x+2\)

\(\Rightarrow7x-2-\left|4x+1\right|=0\Rightarrow7x-2=\left|4x+1\right|\)

Cách giải tương tự hai câu a;b

mk cảm ơn bạn nhiều nhé MMS_Hồ Khánh Châu

\(a,\Rightarrow\left(2x-1\right)\left(2x+1\right)-x\left(2x+1\right)=0\\ \Rightarrow\left(2x+1\right)\left(2x-1-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=1\end{matrix}\right.\\ b,\Rightarrow\left(x-3\right)\left(x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow\left(x^2-8x+16\right)-10=0\\ \Rightarrow\left(x-4\right)^2-10=0\\ \Rightarrow\left(x-4-\sqrt{10}\right)\left(x-4+\sqrt{10}\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4+\sqrt{10}\\x=4-\sqrt{10}\end{matrix}\right.\)

`@` `\text {Ans}`

`\downarrow`

`a)`

`3x(4x-1) - 2x(6x-3) = 30`

`=> 12x^2 - 3x - 12x^2 + 6x = 30`

`=> 3x = 30`

`=> x = 30 \div 3`

`=> x=10`

Vậy, `x=10`

`b)`

`2x(3-2x) + 2x(2x-1) = 15`

`=> 6x- 4x^2 + 4x^2 - 2x = 15`

`=> 4x = 15`

`=> x = 15/4`

Vậy, `x=15/4`

`c)`

`(5x-2)(4x-1) + (10x+3)(2x-1) = 1`

`=> 5x(4x-1) - 2(4x-1) + 10x(2x-1) + 3(2x-1)=1`

`=> 20x^2-5x - 8x + 2 + 20x^2 - 10x +6x - 3 =1`

`=> 40x^2 -17x - 1 = 1`

`d)`

`(x+2)(x+2)-(x-3)(x+1)=9`

`=> x^2 + 2x + 2x + 4 - x^2 - x + 3x + 3=9`

`=> 6x + 7 =9`

`=> 6x = 2`

`=> x=2/6 =1/3`

Vậy, `x=1/3`

`e)`

`(4x+1)(6x-3) = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + (3x-2)(8x+9)`

`=> 24x^2 - 12x + 6x - 3 = 7 + 24x^2 +11x - 18`

`=> 24x^2 - 6x - 3 = 24x^2 + 18x -11`

`=> 24x^2 - 6x - 3 - 24x^2 + 18x + 11 = 0`

`=> 12x +8 = 0`

`=> 12x = -8`

`=> x= -8/12 = -2/3`

Vậy, `x=-2/3`

`g)`

`(10x+2)(4x- 1)- (8x -3)(5x+2) =14`

`=> 40x^2 - 10x + 8x - 2 - 40x^2 - 16x + 15x + 6 = 14`

`=> -3x + 4 =14`

`=> -3x = 10`

`=> x= - 10/3`

Vậy, `x=-10/3`

Hello các bạn còn đó ko?

b: \(\Leftrightarrow x^4-4x^2+2x^2-8=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2

\(P=\frac{2x^5-x^4-2x+1}{4x^2-1}+\frac{8x^2-4x+2}{8x^3+1}\)

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

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

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

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

\(=\frac{x^4+1}{2x+1}\)

1: \(\Leftrightarrow\left(x-4\right)^2+14=-9\left(x-4\right)\)

\(\Leftrightarrow x^2-8x+16+14+9x-36=0\)

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

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

=>x=-3(nhận) hoặc x=2(nhận)

2: \(\Leftrightarrow\left(8x+1\right)\left(2x-1\right)-2x\left(2x+1\right)-12x^2+9=0\)

\(\Leftrightarrow16x^2-8x+2x-1-4x^2-2x-12x^2+9=0\)

=>-8x+8=0

hay x=1(nhận)

c: \(\dfrac{1}{2\left(x-3\right)}-\dfrac{3x-5}{\left(x-3\right)\left(x-1\right)}=\dfrac{1}{2}\)

\(\Leftrightarrow x-1-2\left(3x-5\right)=\left(x-3\right)\left(x-1\right)\)

\(\Leftrightarrow x^2-4x+3=x-1-6x+10=-5x+9\)

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

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

=>x=-3(nhận) hoặc x=2(nhận)

a) (x²-6xy+9y²):(3y-x)

= (x-3y)²:(3y-x)

=(3y-x)²:(3y-x)

= 3y-x

b) (8x³-1):(4x²+2x+1)

=[(2x)³-1]:(4x²+2x+1)

= (2x-1)(4x²+2x+1):(4x²+2x+1)

= 2x-1

c) (4x⁴-9):(2x²-3)

=(2x²-3)(2x²+3):(2x²-3)

=2x²+3

d) (8x³-27):(4x²+6x+9)

=(2x-3)(4x²+6x+9):(4x²+6x+9)

=2x-3

a) \(\left(8x+5\right)^2\left(4x+3\right)\left(2x+1\right)=9\)

\(\Leftrightarrow\left(64x^2+8x+25\right)\left(8x^2+10x+3\right)-9=0\)

Đặt a = \(8x^2+10x+3\)

\(\left(8a+1\right)a-9=0\)

\(\Leftrightarrow8a^2+a-9=0\)

\(\Leftrightarrow\left(a-1\right)\left(8a+9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a=1\\a=-\frac{9}{8}\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}8x^2+10x+3=1\\8x^2+10x+3=-\frac{9}{8}\end{cases}}\)

mà \(8x^2+10x+3=1\Rightarrow8x^2+10x+2=0\)

\(\Rightarrow2\left(x+1\right)\left(4x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=-1\\x=-0,25\end{cases}}\)

cảm ơn bạn còn mấy phần còn lại ạ