\(\Leftrightarrow\frac{6x+5}{12x+9}-\frac{3x-7}{12x-9}=\frac{4x^2+10x-7}{16x^2-9}.\)

\(\Leftrightarrow\frac{\left(6x+5\right)\left(12x-9\right)-\left(3x-7\right)\left(12x+9\right)}{\left(3.4.x\right)^2-\left(3.3\right)^2}=\frac{4x^2+10x-7}{16x^2-9}\)

\(\Leftrightarrow\frac{72x^2+6x-45-\left(36x^2-57x-63\right)}{3^2\left(16x^2-9\right)}=\frac{4x^2+10x-7}{16x^2-9}\)

ĐK: \(16x^2-9\ne0\Leftrightarrow x^2\ne\left(\frac{3}{4}\right)^2\Rightarrow x\ne\pm\frac{3}{4}\)

\(\Leftrightarrow72x^2+6x-45-36x^2+57x+63=36x^2+90x-63\)

\(\Leftrightarrow27x=81\Leftrightarrow x=3\)

a:

ĐKXĐ: \(x^2+3x>=0\)

=>x(x+3)>=0

=>\(\left[{}\begin{matrix}x>=0\\x< =-3\end{matrix}\right.\)

 \(\sqrt{16}-\sqrt{x^2+3x}=0\)

=>\(\sqrt{x^2+3x}=\sqrt{16}\)

=>x^2+3x=16

=>x^2+3x-16=0

\(\text{Δ}=3^2-4\cdot1\cdot\left(-16\right)=9+64=73>0\)

Do đó: Phương trình có 2 nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{-3-\sqrt{73}}{2}\\x_2=\dfrac{-3+\sqrt{73}}{2}\end{matrix}\right.\)

b:

ĐKXĐ: \(x\in R\)

 \(3x-1-\sqrt{4x^2-12x+9}=0\)

=>\(\sqrt{\left(2x-3\right)^2}=3x-1\)

=>\(\left\{{}\begin{matrix}3x-1>=0\\\left(3x-1\right)^2=\left(2x-3\right)^2\end{matrix}\right.\)

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

=>\(\left\{{}\begin{matrix}x>=\dfrac{1}{3}\\\left(x+2\right)\left(5x-4\right)=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(loại\right)\\x=\dfrac{4}{5}\left(nhận\right)\end{matrix}\right.\)

c:

ĐKXĐ: \(\left\{{}\begin{matrix}x^2-6x+8>=0\\2x^2-10x+11>=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left[{}\begin{matrix}x>=4\\x< =2\end{matrix}\right.\\\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=\dfrac{5+\sqrt{3}}{2}\end{matrix}\right.\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x< =\dfrac{5-\sqrt{3}}{2}\\x>=4\end{matrix}\right.\)

 \(\sqrt{2x^2-10x+11}=\sqrt{x^2-6x+8}\)

\(\Leftrightarrow2x^2-10x+11=x^2-6x+8\)

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

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

=>x=3(loại) hoặc x=1(nhận)

bài 1 :

\(\Rightarrow x=-\frac{1}{4}\) hoặc \(x=\frac{1}{2}\)

bài 2 :

\(\Leftrightarrow\left(2x+1\right)\left(3x+2\right)\left(12x+7\right)^2-3=\left(3x+1\right)\left(6x+5\right)\left(48x^2+56x+19\right)\)

\(\Rightarrow3x+1=0\)

\(\Rightarrow3x=-1\)

\(\Rightarrow6x+5=0\)

\(\Rightarrow6x=-5\)

Áp dụng Delta ta có :

\(\Rightarrow48x^2+56x+19=0\)

\(\Rightarrow56^2-4\left(48.19\right)=-512\)

=>D<0 ko có nghiệm thực ( ko có hình tam giác nên thay tạm )

\(\Rightarrow x=-\frac{5}{6}\) hoặc \(x=-\frac{1}{3}\)

tôi nhớ có 1 lần tôi làm mà ông ko tik nhé

a/ 2x(8x - 1)²(4x - 1) = 9

=> (64x² - 16x + 1) (8x² - 2x) = 9

- Nhân 2 vế cho 8 ta đc:

   (64x² - 16x + 1) (64x² - 16x) = 72

- Đặt a = 64x² - 16x  ta đc:

   (a + 1).a = 72

   => a² + a - 72 = 0 

   => (a - 8)(a + 9) = 0

   => a = 8 hoặc a = -9

- Với a = 8 => 64x² - 16x = 8 => 64x² - 16x - 8 = 0 => (2x - 1)(4x + 1) = 0 => x = 1/2 hoặc x = -1/4

- Với a = -9 => 64x² - 16x = -9 => 64x² - 16x + 9 = 0 , mà 64x² - 16x + 9 > 0 => pt vô nghiệm

Vậy x = 1/2 , x = -1/4

\(\dfrac{6x+5}{12x+9}+\dfrac{3x-7}{9-12x}=\dfrac{4x^2+10x-7}{16x^2-9}\)

\(\Leftrightarrow\dfrac{6x+5}{3\left(4x+3\right)}-\dfrac{3x-7}{3\left(4x-3\right)}=\dfrac{12x^2+30x-21}{3\left(4x-3\right)\left(4x+3\right)}\)

\(\Leftrightarrow\left(6x+5\right)\left(4x-3\right)-\left(3x-7\right)\left(4x+3\right)=12x^2+30x-21\)

\(\Leftrightarrow24x^2-18x+20x-15-\left(12x^2+9x-28x-21\right)=12x^2+30x-21\)

\(\Leftrightarrow24x^2+2x-15-12x^2+19x+21=12x^2+30x-21\)

=>31x+6=30x-21

=>x=-27

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4

\(a.\)  \(\left(x^2+1\right)^2+3x\left(x^2+1\right)+2x^2=0\)  \(\left(1\right)\)

Đặt  \(t=x^2+1\)   , khi đó phương trình \(\left(1\right)\)  trở thành:

\(t^2+3xt+2x^2=0\)

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

\(\Leftrightarrow\)  \(^{t+x=0}_{t+2x=0}\)

\(\text{*}\)  \(t+x=0\)

\(\Leftrightarrow\)  \(x^2+x+1=0\)

Vì  \(x^2+x+1=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ne0\)  với mọi  \(x\)  nên phương trình vô nghiệm

\(\text{*}\)  \(t+2x=0\)

\(\Leftrightarrow\)  \(x^2+2x+1=0\)

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

\(\Leftrightarrow\)  \(x+1=0\)

\(\Leftrightarrow\)  \(x=-1\)

Vậy, tập nghiệm của pt là  \(S=\left\{-1\right\}\)

\(b.\)  \(\left(x^2-9\right)^2=12x+1\)

\(\Leftrightarrow\)  \(x^4-18x^2+81-12x-1=0\)

\(\Leftrightarrow\)  \(x^4-18x^2-12x+80=0\)

\(\Leftrightarrow\)  \(x^4-2x^3+2x^3-4x^2-14x^2+28x-40x+80=0\)

\(\Leftrightarrow\)  \(x^3\left(x-2\right)+2x^2\left(x-2\right)-14x\left(x-2\right)-40\left(x-2\right)=0\)

\(\Leftrightarrow\)  \(\left(x-2\right)\left(x^3+2x^2-14x-40\right)=0\)

\(\Leftrightarrow\)  \(\left(x-2\right)\left(x-4\right)\left(x^2+6x+10\right)=0\)

  Vì  \(x^2+6x+10=\left(x+3\right)^2+1\ne0\)  với mọi  \(x\)

\(\Rightarrow\)  \(\left(x-2\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)  \(^{x_1=2}_{x_2=4}\)

Vậy,  phương trình đã cho có các nghiệm  \(x_1=2;\)  \(x_2=4\)

\(a,\Leftrightarrow\left(x-2\right)\left(3x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{1}{3}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-2\right)^3=0\Leftrightarrow x-2=0\Leftrightarrow x=2\\ c,\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\\ \Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{3}{7}\end{matrix}\right.\\ d,\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\\ \Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

`a)16x^2-24x+9=25`

`<=>(4x-3)^2=25`

`+)4x-3=5`

`<=>4x=8<=>x=2`

`+)4x-3=-5`

`<=>4x=-2`

`<=>x=-1/2`

`b)x^2+10x+9=0`

`<=>x^2+x+9x+9=0`

`<=>x(x+1)+9(x+1)=0`

`<=>(x+1)(x+9)=0`

`<=>` \(\left[ \begin{array}{l}x=-9\\x=-1\end{array} \right.\) 

`c)x^2-4x-12=0`

`<=>x^2+2x-6x-12=0`

`<=>x(x+2)-6(x+2)=0`

`<=>(x+2)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=-2\\x=6\end{array} \right.\) 

`d)x^2-5x-6=0`

`<=>x^2+x-6x-6=0`

`<=>x(x+1)-6(x+1)=0`

`<=>(x+1)(x-6)=0`

`<=>` \(\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.\) 

`e)4x^2-3x-1=0`

`<=>4x^2-4x+x-1=0`

`<=>4x(x-1)+(x-1)=0`

`<=>` \(\left[ \begin{array}{l}x=1\\x=-\dfrac14\end{array} \right.\) 

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

`<=>x^4-x^2+5x^2-5=0`

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

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

Vì `x^2+5>=5>0`

`=>x^2-1=0<=>x^2=1`

`<=>` \(\left[ \begin{array}{l}x=1\\x=-1\end{array} \right.\)