\(1,\dfrac{3x-1}{4}+\dfrac{6x-2}{8}=\dfrac{1-3x}{6}\\ \Leftrightarrow1,\dfrac{3x-1}{4}+\dfrac{3x-1}{4}=\dfrac{1-3x}{6}\\ \Leftrightarrow2.\dfrac{3x-1}{4}=\dfrac{1-3x}{6}\\ \Leftrightarrow\dfrac{3x-1}{4}=\dfrac{1-3x}{12}\\ \Leftrightarrow12\left(3x-1\right)=4\left(1-3x\right)\\ \Leftrightarrow3\left(3x-1\right)=1-3x\\ \Leftrightarrow9x-3-1+3x=0\\ \Leftrightarrow12x-4=0\\ \Leftrightarrow x=\dfrac{1}{3}\)

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

1,3x - (1-x)=10,5

<=>1,3x-1+x=10,5

<=>2,3x-1=10,5

<=>2,3x=10,5+1=11,5

<=>x=11,5:2,3

=>x=5

1 3x - { 1 -x ) = 10 ,5 

=> 13x - 1 +x = 10,5

=> 14x            = 11,5

=> x                 = 11,5/14 

\(3x\left(x-2020\right)-x+2020=0\)

\(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\left(3x-1\right)\left(x-2020\right)=0\)

\(\orbr{\begin{cases}x=\frac{1}{3}\left(TM\right)\\x=2020\left(TM\right)\end{cases}}\)

\(b,4-9x^2=0\)

\(2^2-\left(3x\right)^2=0\)

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

\(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}\orbr{\begin{cases}x=\frac{2}{3}\left(TM\right)\\x=-\frac{2}{3}\left(TM\right)\end{cases}}}\)

\(c,x^2-x+\frac{1}{4}=0\)

\(x^2-x+\left(\frac{1}{2}\right)^2=0\)

\(\left(x-\frac{1}{2}\right)^2=0\)

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

\(x=\frac{1}{2}\)

\(d,x\left(x-3\right)+\left(x-3\right)=0\)

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

\(\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}\orbr{\begin{cases}x=3\left(TM\right)\\x=-1\left(TM\right)\end{cases}}}\)

\(e,9x\left(x-7\right)-x+7=0\)

\(9x\left(x-7\right)-\left(x-7\right)=0\)

\(\left(9x-1\right)\left(x-7\right)=0\)

\(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}\orbr{\begin{cases}x=\frac{1}{9}\left(TM\right)\\x=7\left(TM\right)\end{cases}}}\)

a) 3x(x - 2020) - x + 2020 = 0 

<=> 3x(x - 2020) - (x - 2020) = 0

<=> (3x - 1)(x - 2020) = 0

<=> \(\orbr{\begin{cases}3x-1=0\\x-2020=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=2020\end{cases}}\)

Vậy tập nghiệm phương trình là \(S=\left\{\frac{1}{3};2020\right\}\)

b) \(4-9x^2=0\)

<=> \(\left(2-3x\right)\left(2+3x\right)=0\)

<=> \(\orbr{\begin{cases}2-3x=0\\2+3x=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-\frac{2}{3}\end{cases}}\)

Vậy \(x\in\left\{\frac{2}{3};-\frac{2}{3}\right\}\)là nghiệm phương trình 

c) \(x^2-x+\frac{1}{4}=0\)

<=> \(\left(x-\frac{1}{2}\right)^2=0\)

<=> \(x-\frac{1}{2}=0\)

<=> \(x=\frac{1}{2}\)

d) x(x - 3) + (x - 3) = 0

<=> (x + 1)(x - 3) = 0

<=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)

Vậy \(x\in\left\{-1;3\right\}\)là nghiệm phương trình

e) 9x(x - 7) - x + 7 = 0

<=> (9x - 1)(x - 7) = 0

<=> \(\orbr{\begin{cases}9x-1=0\\x-7=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{9}\\x=7\end{cases}}\)

Vậy \(x\in\left\{\frac{1}{9};7\right\}\)là nghiệm phương trình

1, \(3x\left(x-2020\right)-x+2020=0\)

\(\Leftrightarrow\)  \(3x\left(x-2020\right)-\left(x-2020\right)=0\)

\(\Leftrightarrow\left(x-2020\right)\left(3x-1\right)=0\)  

  \(\Leftrightarrow\)\(\orbr{\begin{cases}x-2020=0\\3x-1=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\3x=1\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=2020\\x=\frac{1}{3}\end{cases}}\)

Vậy phương trình có nghiệm x=2020 hoặc x=\(\frac{1}{3}\)

2, \(4-9x^2=0\)

\(\Leftrightarrow4=9x^2\)

\(\Leftrightarrow\frac{4}{9}=x^2\)

\(\Leftrightarrow x=\pm\frac{2}{3}\)

Vậy phương trình có nghiệm x=\(\pm\frac{2}{3}\)

3, \(x^2-x+\frac{1}{4}=0\)

\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)

\(\Leftrightarrow x-\frac{1}{2}=0\)

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

Vậy phương trình có nghiệm x=\(\frac{1}{2}\)

4, \(x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x+1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)

Vậy phương trình có nghiệm x=3 hoặc x= -1

5, \(9x\left(x-7\right)-x+7=0\)

\(\Leftrightarrow9x\left(x-7\right)-\left(x-7\right)=0\)

\(\Leftrightarrow\left(x-7\right)\left(9x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\9x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\9x=1\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=\frac{1}{9}\end{cases}}\)

Vậy phương trình có nghiệm x=7 hoặc x=\(\frac{1}{9}\)

1)\(3x=4y\Rightarrow\frac{x}{4}=\frac{y}{3}=\frac{2x+3y}{8+9}=\frac{34}{17}=2\)

\(\Rightarrow x=4.2=8;y=2.3=6\)

2)\(\frac{x}{2}=\frac{y}{-5}=\frac{x-y}{2+5}=\frac{7}{7}=1\)

\(\Rightarrow x=2;y=-5\)

a) Ta có :

\(\frac{x}{4}=\frac{y}{3}\)

\(=\frac{2x}{8}=\frac{3y}{9}\)

Theo tính chất của dãy tỉ số bằng nhau,ta có :

\(\frac{2x}{8}=\frac{3y}{9}=\frac{2x+3y}{8+9}=\frac{34}{17}=2\)

\(\Rightarrow\hept{\begin{cases}\frac{2x}{8}=2\\\frac{3x}{9}=2\end{cases}\Rightarrow}\hept{\begin{cases}x=2.8\div2\\y=2.9\div3\end{cases}\Rightarrow\hept{\begin{cases}x=8\\y=6\end{cases}}}\)

Vậy ....
b) Theo tính chất của dãy tỉ số bằng nhau,ta có :

\(\frac{x}{2}=\frac{y}{-5}=\frac{x-7}{2-\left(-5\right)}=\frac{7}{7}=1\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{2}=1\\\frac{y}{-5}=1\end{cases}\Rightarrow\hept{\begin{cases}x=1.2\\y=1.\left(-5\right)\Rightarrow\end{cases}}\hept{\begin{cases}x=2\\y=-5\end{cases}}}\)

Vậy ... 

\(a,x=8;y=6\)

\(b,x=2;y=-5\)

học tốt

\(1,3x+3\frac{1}{2}=x\)

\(1,3x+\frac{7}{2}=x\)

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

\(x.\left(1,3-1\right)=-\frac{7}{2}\)

\(x\cdot0,3=-\frac{7}{2}\)

\(x\cdot\frac{3}{10}=-\frac{7}{2}\)

\(x=-\frac{7}{2}:\frac{3}{10}=-\frac{7}{2}\cdot\frac{10}{3}=-\frac{35}{3}\)

|-1,3.x+1/2| = |-5/6+0,2.x|

TH1: -1,3.x+1/2 = -5/6+0,2.x

=> -1,3.x - 0,2.x = -5/6 - 1/2 

-1,5.x = -4/3

x = -8/9

TH2: -1,3.x+1/2 = 5/6-0,2.x

=> -1,3.x + 0,2.x = 5/6-1/2

-1,1.x = 1/3

x =  -10/33

KL:...

b => ,\(\left(\frac{2}{5}-x\right)^2=1\Rightarrow TH1:\frac{2}{5}-x=1\Rightarrow x=\frac{-3}{5}\)

                                         \(TH2:\frac{2}{5}-x=-1\Rightarrow x=\frac{7}{5}\)

c, = > \(\frac{x}{3}-\frac{1}{2}=1\Rightarrow\frac{x}{3}=\frac{3}{2}\Rightarrow x=\frac{9}{2}\)

e,  => \(\frac{1}{3}-\frac{1}{2}+\frac{3}{2}x=1\Rightarrow\frac{-1}{6}+\frac{3}{2}x=1\Rightarrow\frac{3}{2}x=\frac{7}{6}\Rightarrow x=\frac{7}{9}\)

còn câu c bạn tự làm nhé

​nhân dùng cho mk nha!!!!!!!!!!!!!!!!!!!!!!

mik ko bít

I don't now

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