(2x+9)/(x+1)(x+8)-(2x+15)/(x+8)(x+7)+(2x+10)/(x+7)(x+3)=4/3

(x+1+x+8)/(x+1)(x+8)-(x+8+x+7)/(x+8)(x+7)+(x+7+x+3)/(x+7)(x+3)=4/3

1/(x+8)+1/(x+1)-1/(x+7)-1/(x+8)+1/(x+7)+1/(x+3)=4/3

1/(x+1)+1/(x+3)=4/3

(x+3+x+1)/(x+3)(x+1)=4/3

(2x+4)/(x+3)(x+1)=4/3

=>(2x+4).3=(x+3)(x+1).4

6(x+2)=4(x+3)(x+1)

3(x+2)=2(x+3)(x+1)

3x+6=2(x^2+4x+3)

3x+6=2x^2+8x+6

2x^2+8x+6-3x-6=0

2x^2+5x=0

x(2x+5)=0

=> x=0 hoac 2x+5=0

=> x=0 hoac x=-5/2 

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

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

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

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

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

Đặt \(a=\frac{x\left(3-x\right)}{x+1}\) 

\(\Leftrightarrow a\left(1+a=2\right)\)

\frac{x\left(3-x\right)}{x+1}\left(x+\frac{3-x}{x+1}\right)=2x+1x(3−x)​(x+x+13−x​)=2 

\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\left(\frac{x^2+x+3-x}{x+1}\right)=2⇔x+1x(3−x)​(x+1x2+x+3−x​)=2 

\Leftrightarrow\frac{x\left(3-x\right)}{x+1}.\frac{x^2+3}{x+1}=2⇔x+1x(3−x)​.x+1x2+3​=2 

\Leftrightarrow\frac{x\left(3-x\right)}{x+1}.\frac{3x+3+x^2-3x}{x+1}=2⇔x+1x(3−x)​.x+13x+3+x2−3x​=2 

\Leftrightarrow\frac{x\left(3-x\right)}{x+1}\left(1+\frac{x^2-3x}{x+1}\right)=2⇔x+1x(3−x)​(1+x+1x2−3x​)=2 

Đặt a=\frac{x\left(3-x\right)}{x+1}a=x+1x(3−x)​ 

\Leftrightarrow a\left(1+a=2\right)⇔a(1+a=2)

a/ -4x(x - 5) - 2x(8 - 2x) = -3

  => -4x² + 20x - 16x + 4x² = -3

  => 4x = -3

   => x = -3/4

b/ \(\frac{x-1}{-15}=-\frac{60}{x-1}\Rightarrow\left(x-1\right)^2=\left(-60\right)\left(-15\right)\)

      \(\Rightarrow\left(x-1\right)^2=900\Rightarrow\orbr{\begin{cases}x-1=30\\x-1=-30\end{cases}\Rightarrow\orbr{\begin{cases}x=31\\x=-29\end{cases}}}\)

                                                        Vậy x = -29 , x = 31

Lớp 5 tụi mk chưa học cái này

a) ĐK: \(x\ne-1\)

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

\(\Leftrightarrow\frac{x^2\left(3-x\right)}{x+1}+\frac{x\left(3-x\right)^2}{\left(x+1\right)^2}-2=0\)

\(\Leftrightarrow\frac{\left(3x^2-x^3\right)\left(x+1\right)+x\left(9-6x+x^2\right)-2\left(x^2+2x+1\right)}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow\left(3x^2-x^3\right)\left(x+1\right)+x\left(9-6x+x^2\right)-2\left(x^2+2x+1\right)=0\)

\(\Leftrightarrow-x^4+3x^3-5x^2+5x-2=0\)

\(\Leftrightarrow\left(x-1\right)\left(-x^3+2x^2-3x+2\right)=0\)

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

Do \(-x^2+x-1\ne0\forall x\) nên \(x-1=0\Leftrightarrow x=1\)

b) Tương tự.

Điều kiện \(x\ne1.\)

Đặt \(y=\frac{x-8}{x-1}\to xy\left(x+y\right)=-15,y\left(x-1\right)=x-8\to xy\left(x+y\right)=-15,xy=x+y-8.\)

Đặt \(a=xy,b=x+y\to ab=-15,a=b-8\to b^2-8b=-15\to b-4=\pm1\to b=5,3.\)
Với \(b=5\to a=-3\to xy=-3,x+y=5\to x,y\) là nghiệm phương trình \(t^2-5t-3=0\), hay \(t=\frac{5\pm\sqrt{37}}{2}\),  suy ra \(x=\frac{5\pm\sqrt{37}}{2}\)

Với \(b=3\to a=-5\to xy=-5,x+y=3\to x,y\) là nghiệm của \(t^2-3t-5=0\to t=\frac{3\pm\sqrt{29}}{2}\) suy ra \(x=\frac{3\pm\sqrt{29}}{2}.\) 
Vậy phương trình có bốn nghiệm \(x=\frac{5\pm\sqrt{37}}{2}\) và \(x=\frac{3\pm\sqrt{29}}{2}.\)

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