1/2+ y/1=3/x

1/2+2y/2=3/x

1+2y/2=3/x

1+2y.x=2.3

vì 1+2y là lẻ suy ra thuộc ước lẻ của 6 thuộc -+1,-+3

nếu x=1, y= 0

nếu x=-1 thì y=-2

nếu x=3 thì y= 1

nếu x=-3 thì y= -2

mk ko chắc lắm, tóm lại cách làm là vậy đó

leeeeeeewjaénhgi

\(3\left(x-2\right)+4\left(x-5\right)=23\)

\(\Rightarrow3x-6+4x-20-23=0\)

\(\Rightarrow7x-49=0\)

\(\Rightarrow x=7\)

    3(x-2)+4(x-5)=23

<=>3x-6+4x-20=23

<=>7x-26=23

<=>7x=49

<=>x=7

Vậy x=7

3 ( x -2 ) + 4(x-5)= 23

=> 3x-6+4x-20   =23

=> 7x-26            =23

=>  7x                 =23+26=49

=>         x            =\(\dfrac{49}{7}\)=7

Vậy x=7.

\(\dfrac{3}{7}+\dfrac{a}{b}+\dfrac{2}{3}=\dfrac{1}{2}\)

\(\dfrac{3}{7}+\dfrac{a}{b}=\dfrac{1}{2}-\dfrac{2}{3}\)

\(\dfrac{3}{7}+\dfrac{a}{b}=-\dfrac{1}{6}\)

\(\dfrac{a}{b}=-\dfrac{1}{6}-\dfrac{3}{7}\)

\(\dfrac{a}{b}=-\dfrac{25}{42}\)

_____________

\(\dfrac{a}{b}-\dfrac{4}{9}+\dfrac{1}{10}=\dfrac{1}{7}\)

\(\dfrac{a}{b}-\dfrac{4}{9}=\dfrac{1}{7}-\dfrac{1}{10}\)

\(\dfrac{a}{b}-\dfrac{4}{9}=\dfrac{3}{70}\)

\(\dfrac{a}{b}=\dfrac{3}{70}+\dfrac{4}{9}\)

\(\dfrac{a}{b}=\dfrac{307}{630}\) 

\(\dfrac{x-2}{4}=\dfrac{9}{x-2}\)

=>(x-2)²=36

=>x-2=6 hoặc x-2=-6

=>x=8 hoặc x=-4

a)  ĐKXĐ:  \(x\ge0;x\ne9\)

\(P=\frac{\sqrt{x}}{\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}+\frac{5\sqrt{x}+3}{x-9}\)

  \(=\frac{\sqrt{x}.\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

  \(=\frac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\frac{2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\frac{5\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)

Bài 1: 

a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)

\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)

\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)

\(\Leftrightarrow-12x^2+14x+13=0\)

\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)

Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:

\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)

b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)

\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)

hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)

ai giúp mik vs

Bài 3: 

a) Ta có: \(2x-3=x+\dfrac{1}{2}\)

\(\Leftrightarrow2x-x=\dfrac{1}{2}+3\)

\(\Leftrightarrow x=\dfrac{7}{2}\)

b) Ta có: \(4x-\left(x+\dfrac{1}{2}\right)=2x-\left(\dfrac{1}{2}-5\right)\)

\(\Leftrightarrow3x-\dfrac{1}{2}-2x+\dfrac{1}{2}-5=0\)

\(\Leftrightarrow x=5\)

=\(\dfrac{n-6+6+9}{n-6}\)

=\(\dfrac{\left(n-6\right)+15}{n-6}\)

=1+\(\dfrac{15}{n-6}\)

⇒n-6ϵ Ư(15)=-15,-5,-3,-1,1,3,5,15

⇒n=-9,1,3,5,7,9,11,21