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19 tháng 7 2019

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

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

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

\(\Leftrightarrow x=-32\)

7 tháng 7 2017

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!

7 tháng 7 2017

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

8 tháng 3 2020

1)2x-3=11-5x

2x+5x=11+3

7x =14

x=2.

a) Ta có: \(\frac{7}{8}x-5\left(x-9\right)=\frac{20x+1,5}{6}\)

\(\Leftrightarrow\frac{7x}{8}-5x+45-\frac{20x+1,5}{6}=0\)

\(\Leftrightarrow\frac{21x}{24}-\frac{120x}{24}+\frac{1080}{24}-\frac{4\left(20x+1,5\right)}{24}=0\)

\(\Leftrightarrow-99x+1080-4\left(20x+1,5\right)=0\)

\(\Leftrightarrow-99x+1080-80x-6=0\)

\(\Leftrightarrow1074-179x=0\)

\(\Leftrightarrow179x=1074\)

hay x=6

Vậy: x=6

b) Ta có: \(4\left(0,5-1,5x\right)=-\frac{5x-6}{3}\)

\(\Leftrightarrow2-6x=\frac{6-5x}{3}\)

\(\Leftrightarrow\frac{3\left(2-6x\right)}{3}-\frac{6-5x}{3}=0\)

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

\(\Leftrightarrow-13x=0\)

mà -13≠0

nên x=0

Vậy: x=0

c) Ta có: \(\frac{x+4}{5}-x+4=\frac{x}{3}-\frac{x-2}{2}\)

\(\Leftrightarrow\frac{6\left(x+4\right)}{30}+\frac{30\left(-x+4\right)}{30}-\frac{10x}{30}+\frac{15\left(x-2\right)}{30}=0\)

\(\Leftrightarrow6\left(x+4\right)+30\left(4-x\right)-10x+15\left(x-2\right)=0\)

\(\Leftrightarrow6x+24+120-30x-10x+15x-30=0\)

\(\Leftrightarrow-19x+114=0\)

\(\Leftrightarrow-19x=-114\)

hay x=6

Vậy: x=6

d) Ta có: \(\frac{4x+3}{5}-\frac{6x-2}{7}=\frac{5x+4}{3}+3\)

\(\Leftrightarrow\frac{21\left(4x+3\right)}{105}-\frac{15\left(6x-2\right)}{105}-\frac{35\left(5x+4\right)}{105}-\frac{315}{105}=0\)

\(\Leftrightarrow84x+63-90x+30-175x-140-315=0\)

\(\Leftrightarrow-181x-362=0\)

\(\Leftrightarrow-181x=362\)

hay x=-2

Vậy: x=-2

e) Ta có: \(\frac{1}{4}\left(x+3\right)=3-\frac{1}{2}\left(x+1\right)-\frac{1}{3}\left(x+2\right)\)

\(\Leftrightarrow\frac{x+3}{4}=3-\frac{x+1}{2}-\frac{x+2}{3}\)

\(\Leftrightarrow\frac{3\left(x+3\right)}{12}-\frac{36}{12}+\frac{6\left(x+1\right)}{12}+\frac{4\left(x+2\right)}{12}=0\)

\(\Leftrightarrow3x+9-36+6x+6+4x+8=0\)

\(\Leftrightarrow13x-13=0\)

\(\Leftrightarrow13x=13\)

hay x=1

Vậy: x=1

10 tháng 3 2022

`Answer:`

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

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

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

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

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

\(\Leftrightarrow-6x=133\)

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

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

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

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

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

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

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

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