\(1,\)

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

\(\Leftrightarrow2x\left(x-3\right)+\left(x-3\right)=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)

\(2,\)

\(3x\left(x+5\right)-6\left(x+5\right)=0\)

\(\Leftrightarrow\left(3x-6\right)\left(x+5\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x-6=0\\x+5=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-5\end{cases}}\)

\(3,\)

\(x^4-x^2=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}x^2=0\\x^2-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm1\end{cases}}\)

\(4,\)

\(x^2-2x=0\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

\(5,\)

\(x\left(x+6\right)-10\left(x-6\right)=0\)

\(\Leftrightarrow x^2+6x-10x+60=0\)

\(\Leftrightarrow x^2-4x+60=0\)

\(\Leftrightarrow x^2-4x+4+56=0\)

\(\Leftrightarrow\left(x-2\right)^2=-56\)(Vô lý)

=> Phương trình vô nghiệm

1/3x + 2 = 2x - 1/2

=>1/3x + 2x = 2 + ( - 1/2 ) 

7/3x = 3/2

x = 3/2 : 7/3

x = 9/14

a. 1/3x +2=2x-1/2

ta có: 2x-1/3x =2-1/2

           (2-1/3)x =3/2

            5/3x=3/2

            x=3/2:5/3

            x=9/10

mình nghĩ thế vì mình đang học lớp 5

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

\(\Rightarrow2+\frac{1}{2}=2x-\frac{1}{3}x\)

\(\Rightarrow\frac{5}{2}=\frac{5}{3}x\)

\(\Rightarrow\frac{5}{3}x=\frac{5}{2}\)

\(\Rightarrow x=\frac{5}{2}:\frac{5}{3}\)

\(\Rightarrow x=\frac{3}{2}.\)

Vậy \(x=\frac{3}{2}.\)

Chúc bạn học tốt!

\(\left|x-5\right|+\left|x-11\right|=3x\) ⁽¹⁾

+, \(x< 5\) thì \(\left(1\right)\) trở thành:

\(-\left(x-5\right)+\left[-\left(x-11\right)\right]=3x\)

\(\Rightarrow-2x+16=3x\)

\(\Rightarrow-5x=-16\Leftrightarrow x=\dfrac{16}{5}\left(tm\right)\)

+, \(5\le x< 11\) thì (1) trở thành:

\(x-5-\left(x-11\right)=3x\)

\(\Rightarrow6=3x\Leftrightarrow x=2\left(ktm\right)\)

+, \(x\ge11\) thì (1) trở thành:

\(x-5+x-11=3x\)

\(\Rightarrow2x-16=3x\)

\(\Rightarrow-x=16\Leftrightarrow x=-16\left(ktm\right)\)

Vậy \(x=\dfrac{16}{5}\)

\(\text{a, 3(x+1)+4x=10}\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow7x+3=10\)

\(\Rightarrow7x=10-3=7\)

\(\Rightarrow x=1\)

c, x+1/10+x+2/9=x+3/8+x+4/7

=> (x+1/10 +1) +(x+2/9 +1)= ( x+3/8 +1) +(x+4/7 +1)

=> x+11/10 + x+11/9 = x+11/8 + x+11/7

...............

a) \(3\left(x+1\right)+4x=10\)

\(\Rightarrow3x+3+4x=10\)

\(\Rightarrow3x+4x=10-3\)

\(\Rightarrow7x=7\)

\(\Rightarrow x=7\)

3x.(x-2)-x²+2x=0

⇔3x²-6x-x²+2x=0

⇔2x²-4x=0

⇔2x(x-2)=0

\(\Leftrightarrow\left[{}\begin{matrix}2x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)

vậy x=0 và x=2

3x(x-2)-x^2+2x=0

<=>3x(x-2)-x(x-2)=0

<=>(3x-x)(x-2)=0

<=>2x(x-2)=0

<=>2x=0 hoặc x-2=0

<=>x=0 hoặc x=2

\(3x\left(x-2\right)-x^2+2x=0\Rightarrow3x\left(x-2\right)-x\left(x-2\right)=0\Rightarrow\left(x-2\right)\left(3x-x\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)

Vì |2x-3| - |3x+2| = 0

Suy ra |2x-3|=|3x+2|

Ta có 2 trường hợp:

+)Trường hợp 1: Nếu 2x-3=3x+2

2x-3=3x+2

-3-2=3x-2x

-2=x

+)Trường hợp 2: Nếu 2x-3=-(3x+2)

2x-3=-(3x+2)

2x-3=-3x-2

2x+3x=3-2

5x=1

x=1/5

Vậy x thuộc {-1,1/5}

(2x - 3) - ( 3x + 2) = 0

tính trong ngoặc trước ngoài ngoặc sau

2x - 3 ko phải là 2 nhân âm 3.

2x = 2 nhân x

( 2x - 3) - ( 3x + 2) = 0 có nghĩa là 2x -3 = 3x + 2

còn đâu tự giải nhé

à nhâmf

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

\(\Rightarrow\left|3x-4\right|+\left|3-y\right|=0\)

\(\Rightarrow\hept{\begin{cases}3x-4=0\\3-y=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{4}{3}\\y=3\end{cases}}}\)

sai bét:

x=4/3;y=3

(vừa nhanh vừa gọn)

\(\Leftrightarrow\left|3x-4\right|=\left|y+3\right|\Leftrightarrow\orbr{\begin{cases}3x-4=y+3\\3x-4=-\left(y+3\right)\end{cases}}\Leftrightarrow\orbr{\begin{cases}y=3x-7\\y=-3x+1\end{cases}}\)

\(\left(x-2\right)\left(3x-1\right)=0\)

\(\Rightarrow x-2=0\) hoặc \(3x-1=0\)

+) \(x-2=0\Rightarrow x=2\)

+) \(3x-1=0\Rightarrow x=\frac{1}{3}\)

Vậy \(x\in\left\{2;\frac{1}{3}\right\}\)