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!

a: \(\Leftrightarrow\sqrt{6}\left(x+1\right)=5\sqrt{6}\)

=>x+1=5

=>x=4

b: =>x^2/10=1,1

=>x^2=11

=>x=căn 11 hoặc x=-căn 11

c: =>(4x+3)/(x+1)=9 và (4x+3)/(x+1)>=0

=>4x+3=9x+9

=>-5x=6

=>x=-6/5

d: =>(2x-3)/(x-1)=4 và x-1>0 và 2x-3>=0

=>2x-3=4x-4 và x>=3/2

=->-2x=-1 và x>=3/2

=>x=1/2 và x>=3/2

=>Ko có x thỏa mãn

e: Đặt căn x=a(a>=0)

PT sẽ là a^2-a-5=0

=>\(\left[{}\begin{matrix}a=\dfrac{1+\sqrt{21}}{2}\left(nhận\right)\\a=\dfrac{1-\sqrt{21}}{2}\left(loại\right)\end{matrix}\right.\)

=>x=(1+căn 21)^2/4=(11+căn 21)/2

bài toán lớp 8 hả anh

a) \(A=\frac{x+3}{x-2}=\frac{x-2+5}{x-2}=1+\frac{5}{x-2}\)

A nguyên => \(\frac{5}{x-2}\)nguyên => x-2 \(\in\)Ư (5)={-5;-1;1;5}
=> x={-3;1;3;7}

a: \(\Leftrightarrow4\left(2x+1\right)-3\left(6x-1\right)=2x+1\)

=>8x+4-18x+3=2x+1

=>-10x+7=2x+1

=>-12x=-6

hay x=1/2

b: \(\Leftrightarrow4x^2-12x+7x-21-x^2=3x^2+6x\)

=>5x-21=6x

=>-x=21

hay x=-21

\(\dfrac{x^3+8}{x^2+2x+1}.\dfrac{x^2+3x+2}{1-x^2}\left(x\ne\pm1\right)\\ =\dfrac{x^3+2^3}{\left(x+1\right)^2}.\dfrac{\left(x^2+x\right)+\left(2x+2\right)}{1^2-x^2}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{x\left(x+1\right)+2\left(x+1\right)}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{\left(x+2\right)\left(x^2-2x+4\right)}{\left(x+1\right)^2}.\dfrac{\left(x+2\right)\left(x+1\right)}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{\left(x+2\right)^2\left(x^2-2x+4\right)}{\left(1-x\right)\left(x+1\right)^2}\)

\(a,ĐK:...\\ PT\Leftrightarrow x^2-6x=x^2-7x+10\\ \Leftrightarrow x=10\left(tm\right)\\ b,ĐK:...\\ PT\Leftrightarrow2x\left(4-x\right)-\left(2-2x\right)\left(8-x\right)=\left(8-x\right)\left(4-x\right)\\ \Leftrightarrow8x-2x^2+16+18x-2x^2=32-12x+x^2\\ \Leftrightarrow3x^2-38x+16=0\left(casio\right)\\ c,ĐK:...\\ PT\Leftrightarrow2x\left(x-4\right)-4x=0\\ \Leftrightarrow2x^2-12x=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)