=>(9x^2+24x-6x-16)(x^2+2x+1)=-16

=>(9x^2+18x-16)(x^2+2x+1)=-16

=>(9x^2+18x+9-25)(x^2+2x+1)=-16

=>[9(x+1)^2-25](x+1)^2=-16

=>9(x+1)^4-25(x+1)^2+16=0

Đặt (x+1)^2=a

=>9a^2-25a+16=0

=>a=1 hoặc a=16/9

=>(x+1)^2=1 hoặc (x+1)^2=16/9

=>\(x\in\left\{0;-2;\dfrac{1}{3};-\dfrac{7}{3}\right\}\)

CẢM ƠN NHÌU NHA

1.a)|−7x|=3x+16

Vì |-7x| ≥ 0  nên 3x+16 ≥ 0 ⇔ x ≥ \(\dfrac{-16}{3}\)    (*)

Với đk (*), ta có: |-7x|=3x+16

\(\left[\begin{array}{} -7x=3x+16\\ -7x=-3x-16 \end{array} \right.\) ⇔  \(\left[\begin{array}{} -7x-3x=16\\ -7x+3x=-16 \end{array} \right.\)

⇔ \(\left[\begin{array}{} x=-1,6 (t/m)\\ x= 4 (t/m) \end{array} \right.\)

b) \(\dfrac{x-1}{x+2}\) - \(\dfrac{x}{x-2}\) = \(\dfrac{5x-8}{x^2-4}\)

⇔ \(\dfrac{(x-1)(x-2)}{x^2-4}\) - \(\dfrac{x(x+2)}{x^2-4}\) = \(\dfrac{5x-8}{x^2-4}\)

⇒ x² - 2x - x + 2 - x² - 2x = 5x - 8  

⇔ -5x - 5x = -8 - 2

⇔ -10x = -10

⇔ x=1

2.7x+5 < 3x−11

⇔ 7x - 3x < -11 - 5

⇔ 4x < -16

⇔ x < -4

bạn tự biểu diễn trên trục số nha !

\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

\(\Leftrightarrow\left[\left(3x-2\right)\left(3x+8\right)\right]\left[9\left(x+1\right)^2\right]=-16.9\)

\(\Leftrightarrow\left(9x^2+18x-16\right)\left(9x^2+18x+9\right)=-144\)

\(\Leftrightarrow\left(9x^2+18x\right)^2-7\left(9x^2+18x\right)-144=-144\)

\(\Leftrightarrow\left(9x^2+18x\right)^2-7\left(9x^2+18x\right)=0\)

\(\Leftrightarrow\left(9x^2+18x\right)\left(9x^2+18x-7\right)=0\)

\(\Leftrightarrow9x\left(x+2\right)\left(3x-1\right)\left(3x+7\right)=0\)

Tập nghiệm của pt là: \(S=\left\{0;-2;\frac{1}{3};\frac{-7}{3}\right\}\)

\(\left(3x-2\right)\left(x-1\right)^2\left(3x+8\right)=-16\)

\(\Leftrightarrow\left[\left(3x-2\right)\left(3x+8\right)\right]\left[9\left(x+1\right)^2\right]=-16.9=-144\)

\(\Leftrightarrow\left(9x^2+18x\right)^2-7\left(9x^2+18x\right)-144=-144\)

\(\Leftrightarrow\left(9x^2+18x\right)^2-7\left(9x^2+18x\right)=0\)

\(\Leftrightarrow\left(9x^2+18x\right)\left(9x^2+18x-7\right)=0\)

\(\Leftrightarrow9x\left(x+2\right)\left(3x-1\right)\left(3x+7\right)=0\)

Tập nghiệm của phương trình là : \(S=\left\{0;-2;\frac{1}{3};\frac{-7}{3}\right\}\)

Anh Ngáo  coppy trắng trợn là ko tốt :v 

\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

\(9x^4+18x^3+9x^2+24x^3+48x^2+24x-6x^3-12x^2-6x-16x^2-32x-16=-16\)

\(9x^4+36x^3+29x^2-14x-16=-16\)

\(9x^4+36x^3+29x^2-14x=0\)

\(x\left(9x^3+36x^2+29x-14\right)=0\)

\(x=0;-\frac{7}{3};\frac{1}{3};-2\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

\(\left(3x-2\right)\left(x+1\right)^2\left(3x+8\right)=-16\)

<=>   \(\left(3x-2\right)\left(x+1\right)^2.3^2.\left(3x+8\right)+144=0\)

<=>  \(\left(3x-2\right)\left(3x+3\right)^2\left(3x+8\right)+144=0\)   (*)

Đặt  \(3x+3=t\) Khi đó pt (*) trở thành: 

   \(\left(t-5\right)t^2\left(t+5\right)+144=0\)

<=>  \(t^4-25t^2+144=0\)

<=>  \(\left(t-4\right)\left(t-3\right)\left(t+3\right)\left(t+4\right)=0\)

đến đây bn tự giải tiếp nhé

e) ĐK : \(\left\{{}\begin{matrix}1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x\ne-1\\3x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}=\dfrac{\left(1-3x\right)^2-\left(1+3x\right)^2}{\left(1+3x\right)\left(1-3x\right)}\)

\(\Leftrightarrow12\left(1+3x\right)\left(1-3x\right)=\left(1-3x\right)\left(1+3x\right)\left(1-3x-1-3x\right)\left(1-3x+1+3x\right)\)

\(\Leftrightarrow12=\left(-6x\right).2\Leftrightarrow6=-6x\)

\(\Leftrightarrow x=-1\left(TM\right)\)

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

ĐKXĐ: \(x\ge1\).

Phương trình đã cho tương đương:

\(\sqrt{x+3}+\sqrt{x-1}=\dfrac{8}{\sqrt{4x^4-12x^3+9x^2+16}-\left(2x^2-3x\right)}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x-1}=\dfrac{\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)}{2}\)

\(\Leftrightarrow\sqrt{4x^4-12x^3+9x^2+16}+\left(2x^2-3x\right)-2\sqrt{x+3}-2\sqrt{x-1}=0\)

\(\Leftrightarrow\left(\sqrt{4x^4-12x^3+9x^2+16}-2\sqrt{x+3}\right)+\left(2x^2-3x-2\sqrt{x-1}\right)=0\)

\(\Leftrightarrow\dfrac{4x^4-12x^3+9x^2-4x+4}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{4x^4-12x^3+9x^2-4x+4}{2x^2-3x+2\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x^3-4x^2+x-2\right)\left(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}\right)=0\).

Do \(x\ge1\) nên ta có \(\dfrac{1}{\sqrt{4x^4-12x^3+9x^2+16}+2\sqrt{x+3}}+\dfrac{1}{2x^2-3x+2\sqrt{x-1}}>0\).

Do đó \(\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\left(TMĐK\right)\\4x^3-4x^2+x-2=0\left(1\right)\end{matrix}\right.\).

Giải phương trình bậc 3 ở (1) ta được \(x=\dfrac{\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}{\sqrt[6]{279936}}+\dfrac{1}{\sqrt[6]{7776}\sqrt[3]{36\sqrt{13}+53\sqrt{6}}}+\dfrac{1}{3}\approx1,157298106\left(TMĐK\right)\).

Vậy...

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