<=> 2(x^2-25) - 2x^2+3x-4x+6 + x^3-8x = x^3+1

=>2x^2-50 - 2x^2 -9x+6+x^3-x^3-1 = 0

<=>-9x - 45 =0

<=>-9x=45

<=>x=-5

Còn phần b và c bạn cứ khai triển ra,mình phải đi học nên không có thời gian giải cho bạn

`Answer:`

\(3\left(\frac{5}{3}x-7\right)-2\left(1.5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow3\left(\frac{5x}{3}-7\right)-2\left(5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow5x-21-10x-12-5x-20+4x=80\)

\(\Leftrightarrow-6x-53=80\)

\(\Leftrightarrow-6x=133\)

\(\Leftrightarrow x=-\frac{133}{6}\)

\(\frac{4}{5}x^2\left(\frac{x}{3}-\frac{1}{2}\right)-\left(\frac{1}{5}x-\frac{2}{3}\right)\left(4\frac{x^2}{3}+1\right)=\frac{22}{45}x^2\)

\(\Leftrightarrow36x^2\left(\frac{x}{3}-\frac{1}{2}\right)-45\left(\frac{x}{5}-\frac{2}{3}\right)\left(\frac{4x^2}{3}+1\right)=22x^2\)

\(\Leftrightarrow12x^3-18x^2-12x^3-9x+40x^2+30=22x^2\)

\(\Leftrightarrow22x^2-9x+30=22x^2\)

\(\Leftrightarrow-9x+30=0\)

\(\Leftrightarrow-9x=-30\)

\(\Leftrightarrow x=\frac{10}{3}\)

a: =>5x-21-3x-12+(x-5)(x+4)=80+x²

\(\Leftrightarrow x^2-x-20+2x-33=x^2+80\)

=>x-53=80

hay x=133

b: \(\Leftrightarrow\left(\dfrac{1}{5}x-\dfrac{2}{3}\right)\cdot\left(\dfrac{4}{3}x^2+1\right)\cdot\dfrac{1}{6}=\dfrac{22}{45}:\dfrac{4}{5}=\dfrac{11}{18}\)

\(\Leftrightarrow\left(\dfrac{1}{5}x-\dfrac{2}{3}\right)\left(\dfrac{4}{3}x^2+1\right)=\dfrac{11}{3}\)

\(\Leftrightarrow\dfrac{4}{15}x^3+\dfrac{1}{5}x-\dfrac{8}{9}x^2-\dfrac{2}{3}-\dfrac{11}{3}=0\)

\(\Leftrightarrow\dfrac{4}{15}x^3-\dfrac{8}{9}x^2+\dfrac{1}{5}x-\dfrac{13}{3}=0\)

\(\Leftrightarrow12x^3-40x^2+9x-195=0\)

hay \(x\in\left\{\dfrac{10+\sqrt{685}}{6};\dfrac{10-\sqrt{685}}{6}\right\}\)

a, \(3.\left(\dfrac{5}{3}x-7\right)-2\left(1,5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\)

\(\Rightarrow5x-21-3x-12-\left(5x+20-x^2-4x\right)-x^2=80\)

\(\Rightarrow5x-21-3x-12-5x-20+x^2+4x-x^2=80\)

\(\Rightarrow5x-3x-5x+4x+x^2-x^2=80+21+12+20\)

\(\Rightarrow x=133\)

Câu b tương tự! Cứ tách ra!

a) \(3\left(\dfrac{5}{3}x-7\right)-2\left(1,5x+6\right)-\left(5-x\right)\left(x+4\right)=80+x^2\) (1)

\(\Leftrightarrow\left(5x-21\right)-\left(3x+12\right)-\left(5x+20-x^2-4x\right)=80+x^2\)

\(\Leftrightarrow5x-21-3x-12-5x-20+x^2+4x=80+x^2\)

\(\Leftrightarrow x-53+x^2=80+x^2\)

\(\Leftrightarrow x+x^2-x^2=80+53\)

\(\Leftrightarrow x=133\)

Vậy tập nghiệm phương trình (1) là \(S=\left\{133\right\}\)

b) chưa rõ đề.

a, \(\left(x+2\right)^3-x\left(x^2+6x-3\right)=0\Leftrightarrow x^3+4x^2+4x+2x^2+8x+8-x^3-6x^2+3x=0\)

\(\Leftrightarrow15x+8=0\Leftrightarrow x=-\frac{8}{15}\)

b, \(\left(x+4\right)^3-x\left(x+6\right)^2=7\Leftrightarrow12x+64=0\Leftrightarrow x=-\frac{19}{4}\)làm tắt:P 

Tự làm nốt nhé 

Bài giải @ lớp 8 hiểu được thực sự bái phục.

\(\left(x+1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)=45x^2\)\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left(x-2\right)\left(x-3\right)=45x^2\)

\(\left(x^2+7x+6\right)\left(x^2-5x+6\right)=45x^2\)

đặt x^2+6=t

\(\left(t+7x\right)\left(t-5x\right)=45x^2\Leftrightarrow t^2-2tx-35x^2=45x^2\)

\(t^2-2tx+x^2=81x^2\Leftrightarrow\left(t-x\right)^2=\left(9x\right)^2\)

\(\orbr{\begin{cases}t-x=9x\\t-x=-9x\end{cases}\Leftrightarrow\orbr{\begin{cases}t=10x\\t=8x\end{cases}}}\)\(\Leftrightarrow\orbr{\begin{cases}x^2+6=10x\\x^2+6=-8x\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2-10x+25=25-6=19\\x^2+8x+16=16-6=10\end{cases}}}\)

\(\orbr{\begin{cases}\left(x-5\right)^2=19\left(1\right)\\\left(x+4\right)^2=10\left(2\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow\orbr{\begin{cases}x=5-\sqrt{19}\\x=5+\sqrt{19}\end{cases}}\) (2)\(\Leftrightarrow\orbr{\begin{cases}x=4-\sqrt{10}\\x=4+\sqrt{10}\end{cases}}\)

\(\left(x+1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)=45x^2\)

\(\Leftrightarrow\left(x^2-8x+6\right)\left(x^2+10x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-8x+6=0\\x^2+10x+6=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\Delta=\left(-8\right)^2-4\left(1\cdot6\right)=40\\\Delta=10^2-4\left(1\cdot6\right)=76\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x_{1,2}=\frac{8\pm\sqrt{40}}{2}\\x_{3,4}=\frac{-10\pm\sqrt{76}}{2}\end{cases}}\)

(2) \(\orbr{\begin{cases}x=-4-\sqrt{10}\\x=-4+\sqrt{10}\end{cases}}\)