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

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

\(\Leftrightarrow x=5\)

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

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

\(\Rightarrow7x-49=0\)

\(\Rightarrow x=7\)

    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.

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

\(\left(5-x\right)^6\)=\(\left(x-2\right)^6\)

ko tồn tại x

\(\left(5-x\right)^6=\left(x-2\right)^6\)

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

\(\Leftrightarrow\orbr{\begin{cases}0=-3\left(L\right)\\2x=7\end{cases}}\Leftrightarrow x=\frac{7}{2}\)

\(\left(5-x\right)^6=\left(x-2\right)^6\)

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

\(\Rightarrow\orbr{\begin{cases}5+2=2x\\x\in\varnothing\end{cases}}\)

\(\Rightarrow x=\frac{7}{2}\)

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

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

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

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

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

_____________

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

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

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

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

\(\dfrac{a}{b}=\dfrac{307}{630}\) 

(x-6)³=(x-6)²

=> (x-6) \(\in\){0;1}

=> x-6=0              hoặc x-6=1

=> x=0+6             hoặc x=1+6

=> x=6                 hoặc x=7

=> x \(\in\){6; 7}

x thuộc(6;7;8;9;...)

vì x-6 phải là 1 số tn

\(\left|x-6\right|=x-6\)

\(\Leftrightarrow\orbr{\begin{cases}x-6=x-6\\x-6=6-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}0x=0\\2x=12\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x\in\forall Z\\x=6\end{cases}}\)