Đặt x² + 2x = a ta có

\(\frac{1}{a-3}\)+ \(\frac{18}{a+2}\)= \(\frac{18}{a+1}\)

<=> a² - 15a + 56 = 0

<=> a = (7;8)

Thế vô tìm được nghiệm 

x=-0,384367156686985

x=0,442125301696298

x=2,9422181027264

Thay x = 1 vào phương trình  2(2x+1)+18=3(x+2)(2x+k)2(2x+1)+18=3(x+2)(2x+k), ta có:

2(2.1+1)+18=3(1+2)(2.1+k)

⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)

⇔6+18=18+9k⇔24−18=9k⇔6=9k

⇔k=69=232(2.1+1)+18=3(1+2)(2.1+k)

⇔2(2+1)+18=3.3(2+k)

⇔2.3+18=9(2+k)

⇔6+18=18+9k

⇔24−18=9k⇔6=9k

⇔k=\(\frac{6}{9}\)=\(\frac{2}{3}\)

Vậy khi  thì phương trình  có nghiệm x = 1

\(\dfrac{1}{x^2+2x-3}+\dfrac{18}{x^2+2x+2}=\dfrac{18}{x^2+2x+1}\left(1\right)\)

ĐK: \(x\ne\pm1,x\ne-3\)

Đặt \(y=x^2+2x+1\) (với y > 0,y khác 4) ta được:

\(\left(1\right)\Leftrightarrow\dfrac{1}{y-4}+\dfrac{18}{y+1}=\dfrac{18}{y}\Leftrightarrow\dfrac{y\left(y+1\right)}{y\left(y+1\right)\left(y-4\right)}+\dfrac{18y\left(y-4\right)}{y\left(y+1\right)\left(y-4\right)}=\dfrac{18\left(y+1\right)\left(y-4\right)}{y\left(y+1\right)\left(y-4\right)}\Rightarrow y\left(y+1\right)+18y\left(y-4\right)=18\left(y+1\right)\left(y-4\right)\Leftrightarrow y^2+y+18y^2-72y=18y^2-54y-72\Leftrightarrow y^2-17y+72=0\Leftrightarrow\left(y-8\right)\left(y-9\right)=0\Leftrightarrow\left[{}\begin{matrix}y=8\left(TM\right)\\y=9\left(TM\right)\end{matrix}\right.\)

Với \(y=8\) ta có :

\(x^2+2x+1=8\Leftrightarrow\left(x+1\right)^2=8\Leftrightarrow x+1=\pm\sqrt{8}\Leftrightarrow x=\pm\sqrt{8}-1\)

Với y=9 ta có:

\(x^2+2x+1=9\Leftrightarrow\left(x+1\right)^2=9\Leftrightarrow x+1=\pm\sqrt{9}\Leftrightarrow x=\pm\sqrt{9}-1\)

ĐKXĐ : \(x\ne1;-3\)

Đặt \(x^2+2x+1=a\) , ta có :

\(\frac{1}{a-4}+\frac{18}{a+1}=\frac{18}{a}\)

\(\Leftrightarrow\frac{a+1+18a-72}{\left(a+1\right)\left(a-4\right)}=\frac{18}{a}\)

\(\Leftrightarrow\frac{19a-71}{a^2-3a-4}=\frac{18}{a}\)

\(\Leftrightarrow19a^2-71a-18a^2+54a+72=0\)

\(\Leftrightarrow a^2-17a+72=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}a=8\\a=9\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2=8\\\left(x+1\right)^2=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm\sqrt{8}-1\\\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\end{matrix}\right.\)

Vậy ...

giúp mình bài ni với :3x^2(x+1)-5x(x+1)^2+4(x+1)

a. Thay x = 2 vào phương trình (2x + 1)(9x + 2k) – 5(x + 2) = 40, ta có:

(2.2+1)(9.2+2k)−5(2+2)=40⇔(4+1)(18+2k)−5.4=40⇔5(18+2k)−20=40⇔90+10k−20=40⇔10k=40−90+20⇔10k=−30⇔k=−3(2.2+1)(9.2+2k)−5(2+2)=40⇔(4+1)(18+2k)−5.4=40⇔5(18+2k)−20=40⇔90+10k−20=40⇔10k=40−90+20⇔10k=−30⇔k=−3

Vậy khi k = -3 thì phương trình (2x + 1)(9x + 2k) – 5(x + 2) = 40 có nghiệm x = 2

b. Thay x = 1 vào phương trình  2(2x+1)+18=3(x+2)(2x+k)2(2x+1)+18=3(x+2)(2x+k), ta có:

2(2.1+1)+18=3(1+2)(2.1+k)⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)⇔6+18=18+9k⇔24−18=9k⇔6=9k⇔k=69=232(2.1+1)+18=3(1+2)(2.1+k)⇔2(2+1)+18=3.3(2+k)⇔2.3+18=9(2+k)⇔6+18=18+9k⇔24−18=9k⇔6=9k⇔k=\(\frac{6}{9}\)=\(\frac{2}{3}\)

Vậy khi  thì phương trình  có nghiệm x = 1

thế x vào bấm máy tính nhanh nhứt :)))

\(\Leftrightarrow M\cdot\left(4x^2+2x-9\right)=\left(4x^2+2x-18-4x^2-2x\right)\left(4x^2+2x-18+4x^2+2x\right)\)

\(\Leftrightarrow M\cdot\left(4x^2+2x-9\right)=-18\cdot\left(8x^2+4x-18\right)\)

\(\Leftrightarrow M=-18\cdot2=-36\)

Ta có: \(\left(2x+1\right)\left(-2x+1\right)+\left(1-2x\right)^2=18\)

\(\Leftrightarrow1-4x^2+4x^2-4x+1=0\)

\(\Leftrightarrow-4x=-2\)

hay \(x=\dfrac{1}{2}\)