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

\(\Leftrightarrow-16-3x-12-20+x^2-x=80+x^2\)

\(\Leftrightarrow-4x-128=0\)

\(\Leftrightarrow x=-32\)

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

1)2x-3=11-5x

2x+5x=11+3

7x =14

x=2.

a: =>10x-14=15-9x

=>19x=29

hay x=29/19

b: \(\Leftrightarrow3\left(10x+3\right)=36+4\left(8x+6\right)\)

=>30x+9=36+32x+24

=>30x+9=32x+60

=>-2x=51

hay x=-51/2

c: \(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)

=>35x-5+60x=96-6x

=>101x=101

hay x=1

d: \(\Leftrightarrow12\left(\dfrac{1}{2}-\dfrac{3}{2}x\right)=-5x+6\)

\(\Leftrightarrow6-18x+5x-6=0\)

=>-13x=0

hay x=0

\(a,\dfrac{5x-7}{3}=\dfrac{5-3x}{2}\\ \Leftrightarrow2\left(5x-7\right)=3\left(5-3x\right)\\ \Leftrightarrow10x-14=15-9x\\ \Leftrightarrow10x-14-15+9x=0\\ \Leftrightarrow19x-19=0\\ \Leftrightarrow x=1\)

\(b,\dfrac{10x+3}{12}=1+\dfrac{6+8x}{9}\\ \Leftrightarrow\dfrac{3\left(10x+3\right)}{36}=\dfrac{36}{36}+\dfrac{4\left(6+8x\right)}{36}\\ \Leftrightarrow30x+9=36+24+32x\\ \Leftrightarrow36+24+32x-30x-9=0\\ \Leftrightarrow2x+51=0\\ \Leftrightarrow x=-\dfrac{51}{2}\)

\(c,\dfrac{7x-1}{6}+2x=\dfrac{16-x}{5}\\ \Leftrightarrow\dfrac{7x-1+12x}{6}=\dfrac{16-x}{5}\\ \Leftrightarrow5\left(19x-1\right)=6\left(16-x\right)\\ \Leftrightarrow95x-5=96-6x\\ \Leftrightarrow95x-5-96+6x=0\\ \Leftrightarrow101x-101=0\\ \Leftrightarrow x=1\)

\(d,4\left(0,5-1,5x\right)=-\dfrac{5x-6}{3}\\ \Leftrightarrow12\left(0,5-1,5x\right)=6-5x\\ \Leftrightarrow6-18x=6-5x\\ \Leftrightarrow6-5x-6+18x=0\\ \Leftrightarrow13x=0\\ \Leftrightarrow x=0\)

Bài 2: 

a: (x+3)/5=5/7

=>x+3=25/7

hay x=4/7

b: ||x-5|-4|=5

=>|x-5|-4=5 hoặc |x-5|-4=-5

=>|x-5|=9

=>x-5=9 hoặc x-5=-9

=>x=14 hoặc x=-4

c: \(\left(-\dfrac{4}{3}\right)^{3x+1}=\dfrac{256}{81}\)

nên 3x+1=4

=>3x=3

hay x=1