a, ĐK: \(x\ge1\)

Đặt \(\sqrt{5x-1}=a;\sqrt{x-1}=b\left(a,b\ge0\right)\)

\(pt\Leftrightarrow\left(a+b\right)\left(\dfrac{a^2+b^2}{2}-ab\right)=a^2-b^2\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2=2\left(a-b\right)\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)\left(a-b-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=b\\a=b+2\end{matrix}\right.\)

TH1: \(a=b\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}\Leftrightarrow x=0\left(l\right)\)

TH2: \(a=b+2\Leftrightarrow\sqrt{5x-1}=\sqrt{x-1}+2\)

\(\Leftrightarrow5x-1=x-1+4+4\sqrt{x-1}\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}=0\)

\(\Leftrightarrow4x-4-4\sqrt{x-1}+1=1\)

\(\Leftrightarrow\left(2\sqrt{x-1}-1\right)^2=1\)

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

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

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

pt phải có dấu bằng giữa hai vế chứ bạn hik như thiếu đề ạk bn ktra lại xem

=>(x-1)(x^2-2x+1-x^2-5x+2)=0

=>(x-1)(-7x+3)=0

=>x=3/7 hoặc x=1

a) \(\left(3x+2\right)^2-\left(3x-2\right)^2=5x+8\)

\(\Rightarrow\left(3x+2+3x-2\right)\left(3x+2-3x+2\right)=5x+8\)

\(\Rightarrow4.6x=5x+8\Rightarrow24x=5x+8\)

\(\Rightarrow19x=8\Rightarrow x=\frac{8}{19}\)

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

\(\Rightarrow3\left(x^2-4x+4\right)+9x-9=3x^2+3x-9\)

\(\Rightarrow3x^2-12x+12+9x-9=3x^2+3x-9\)

\(\Rightarrow-12x+12+9x-9=3x-9\)

\(\Rightarrow-3x+3=3x-9\)

\(\Rightarrow6x=12\Rightarrow x=2\)

c) \(\left(x+3\right)^2-\left(x-3\right)^2=6x+18\)

\(\Rightarrow\left(x+3+x-3\right)\left(x+3-x+3\right)=6x+18\)

\(\Rightarrow2x.6=6x+18\Rightarrow6x=18\Rightarrow x=3\)

a: =>(x^2+4x-5)(x^2+4x-21)=297

=>(x^2+4x)^2-26(x^2+4x)+105-297=0

=>x^2+4x=32 hoặc x^2+4x=-6(loại)

=>x^2+4x-32=0

=>(x+8)(x-4)=0

=>x=4 hoặc x=-8

b: =>(x^2-x-3)(x^2+x-4)=0

hay \(x\in\left\{\dfrac{1+\sqrt{13}}{2};\dfrac{1-\sqrt{13}}{2};\dfrac{-1+\sqrt{17}}{2};\dfrac{-1-\sqrt{17}}{2}\right\}\)

c: =>(x-1)(x+2)(x^2-6x-2)=0

hay \(x\in\left\{1;-2;3+\sqrt{11};3-\sqrt{11}\right\}\)

i) (x - 1)(5x + 3) = (3x - 8)(x - 1)

<=> 5x² + 3x - 5x - 3 = 3x² - 3x - 8x + 8

<=> 5x² - 2x - 3 = 3x² - 11x + 8

<=> 5x² - 2x - 3 - 3x² + 11x - 8 = 0

<=> 2x² + 9x - 11 = 0

<=> 2x² + 11x - 2x - 11 = 0

<=> x(2x + 11) - (2x + 11) = 0

<=> (x - 1)(2x + 11) = 0

<=> x - 1 = 0 hoặc 2x + 11 = 0

<=> x = 0 hoặc x = -11/2

m) 2x(x - 1) = x² - 1

<=> 2x² - 2x = x² - 1

<=> 2x² - 2x - x² + 1 = 0

<=> x² - 2x + 1 = 0

<=> (x - 1)² = 0

<=> x - 1 = 0

<=> x = 1

n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)

<=> 2x + 22 - 3x² - 33x = 6x - 15x² - 4 + 10x

<=> -31x + 22 - 3x² = 16x - 15x² - 4

<=> 31x - 22 + 3x² + 16x - 15x² - 4 = 0

<=> 47x - 18 - 12x² = 0

<=> -12x² + 47x - 26 = 0

<=> 12x² - 47x + 26 = 0

<=> 12x² - 8x - 39x + 26 = 0

<=> 4x(3x - 2) - 13(3x - 2) = 0

<=> (4x - 13)(3x - 2) = 0

<=> 4x - 13 = 0 hoặc 3x - 2 = 0

<=> x = 13/4 hoặc x = 2/3

i) (x - 1)(5x + 3) = (3x - 8)(x - 1)

<=> 5x² + 3x - 5x - 3 = 3x² - 3x - 8x + 8

<=> 5x² - 2x - 3 = 3x² - 11x + 8

<=> 5x² - 2x - 3 - 3x² + 11x - 8 = 0

<=> 2x² + 9x - 11 = 0

<=> 2x² + 11x - 2x - 11 = 0

<=> x(2x + 11) - (2x + 11) = 0

<=> (x - 1)(2x + 11) = 0

<=> x - 1 = 0 hoặc 2x + 11 = 0

<=> x = 0 hoặc x = -11/2

m) 2x(x - 1) = x² - 1

<=> 2x² - 2x = x² - 1

<=> 2x² - 2x - x² + 1 = 0

<=> x² - 2x + 1 = 0

<=> (x - 1)² = 0

<=> x - 1 = 0

<=> x = 1

n) (2 - 3x)(x + 11) = (3x - 2)(2 - 5x)

<=> 2x + 22 - 3x² - 33x = 6x - 15x² - 4 + 10x

<=> -31x + 22 - 3x² = 16x - 15x² - 4

<=> 31x - 22 + 3x² + 16x - 15x² - 4 = 0

<=> 47x - 18 - 12x² = 0

<=> -12x² + 47x - 26 = 0

<=> 12x² - 47x + 26 = 0

<=> 12x² - 8x - 39x + 26 = 0

<=> 4x(3x - 2) - 13(3x - 2) = 0

<=> (4x - 13)(3x - 2) = 0

<=> 4x - 13 = 0 hoặc 3x - 2 = 0

<=> x = 13/4 hoặc x = 2/3

wifi chán thế?

a) \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right).\left(x-2\right)}\) Đk : x \(\ne-1\) ; x \(\ne2\)

\(\Leftrightarrow\frac{2.\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}-\frac{1.\left(x+1\right)}{\left(x+1\right).\left(x-2\right)}=3x-11\)

\(\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow2x-3x-x=-11+4+1\)

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

\(\Leftrightarrow x=3\)

Vậy S = \(\left\{3\right\}\)

\(8,1-\left(x-6\right)=4\left(2-2x\right)\)

\(\Leftrightarrow1-x+6=8-8x\)

\(\Leftrightarrow-x+8x=8-1-6\)

\(\Leftrightarrow7x=1\)

\(\Leftrightarrow x=\dfrac{1}{7}\)

\(9,\left(3x-2\right)\left(x+5\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-5\end{matrix}\right.\)

\(10,\left(x+3\right)\left(x^2+2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\varnothing\end{matrix}\right.\)

`8)1-(x-5)=4(2-2x)`

`<=>1-x+5=8-6x`

`<=>5x=2<=>x=2/5`

`9)(3x-2)(x+5)=0`

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

`10)(x+3)(x^2+2)=0`

  Mà `x^2+2 > 0 AA x`

 `=>x+3=0`

`<=>x=-3`

`11)(5x-1)(x^2-9)=0`

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

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

`12)x(x-3)+3(x-3)=0`

`<=>(x-3)(x+3)=0`

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

`13)x(x-5)-4x+20=0`

`<=>x(x-5)-4(x-5)=0`

`<=>(x-5)(x-4)=0`

`<=>[(x=5),(x=4):}`

`14)x^2+4x-5=0`

`<=>x^2+5x-x-5=0`

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

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

\(11,=>\left[{}\begin{matrix}5x-1=0\\x^2-9=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=3\\x=-3\end{matrix}\right.\\ 12,=>\left(x+3\right)\left(x-3\right)=0\\ =>\left[{}\begin{matrix}x+3=0\\x-3=0\end{matrix}\right.=>\left[{}\begin{matrix}x=-3\\x=3\end{matrix}\right.\\ 13,=>x\left(x-5\right)-4\left(x-5\right)=0\\ =>\left(x-4\right)\left(x-5\right)=0\\ =>\left[{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

\(14,=>x^2+5x-x-5=0\\ =>x\left(x+5\right)-\left(x+5\right)=0\\ =>\left(x-1\right)\left(x+5\right)=0\\ =>\left[{}\begin{matrix}x-1=0\\x+5=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)