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.

`Answer:`

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

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

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

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

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

\(\Leftrightarrow-6x=133\)

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

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

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

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

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

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

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

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

<=> 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

Noob ơi, bạn phải đưa vào máy tính ý solve cái là ra x luôn, chỉ tội là đợi hơi lâu

a, 4.(18 - 5x) - 12(3x - 7) = 15(2x - 16) - 6(x + 14) 

=> 72 - 20x - 36x + 84 = 30x - 240 - 6x - 84

=> (72 + 84) + (-20x - 36x) = (30x - 6x) + (-240 - 84) 

=> 156 -  56x = 24x - 324 

=>  24x + 56x = 324 + 156 

=> 80x = 480 

=> x = 480 : 80 =  6 

Vậy x = 6 

b, 5(3x + 5) - 4(2x - 3) = 5x + 3(2x + 12) + 1 

=> 15x + 25 - 8x + 12 = 5x + 6x + 36 + 1 

=> (15x - 8x) + (25 + 12) = 11x + 37 

=> 7x + 37 = 11x + 37 

=> 11x - 7x = 0 

=>  x = 0 

\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)

\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)

Bài 4:

a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)

\(\Leftrightarrow6x-9-2x+4=-3\)

\(\Leftrightarrow4x=2\)

hay \(x=\dfrac{1}{2}\)

b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

hay \(x=\dfrac{13}{3}\)

c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

hay x=1

a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)

\(\Leftrightarrow6x-9+4-2x=-3\)

\(\Leftrightarrow4x=2\)

\(\Leftrightarrow x=\dfrac{1}{2}\)

Vậy: \(x=\dfrac{1}{2}\)

===========

b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)

\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)

\(\Leftrightarrow3x=13\)

\(\Leftrightarrow x=\dfrac{13}{3}\)

Vậy: \(x=\dfrac{13}{3}\)

==========

c/  \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)

\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)

\(\Leftrightarrow-8x=-8\)

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)

\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)

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

\(\Leftrightarrow x=1\)

Vậy: \(x=1\)

==========

e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)

\(\Leftrightarrow10x-16-12x+15=12x-16+11\)

\(\Leftrightarrow-14x=-4\)

\(\Leftrightarrow x=\dfrac{2}{7}\)

Vậy: \(x=\dfrac{2}{7}\)

==========

f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)

\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)

\(\Leftrightarrow-x^3=8\)

\(\Leftrightarrow x=-2\)

Vậy: \(x=-2\)