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

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

\(\Leftrightarrow x=5\)

a \(\dfrac{2}{3}x+\dfrac{1}{3}=\dfrac{1}{5}\\ \dfrac{2}{3}x=\dfrac{1}{5}-\dfrac{1}{3}\\ \dfrac{2}{3}x=\dfrac{-2}{15}\\ x=-\dfrac{2}{15}:\dfrac{2}{3}\\ x=-\dfrac{1}{5}\)   b) \(\dfrac{4}{5}-\dfrac{5}{3}x=-2\\ \dfrac{5}{3}x=\dfrac{4}{5}+2\\ \dfrac{5}{3}x=\dfrac{14}{5}\\ x=\dfrac{14}{5}:\dfrac{5}{3}\\ x=\dfrac{42}{25}\)c) \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\\ \dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}\\ \dfrac{5}{3}:x=\dfrac{3}{10}\\ x=\dfrac{5}{3}:\dfrac{3}{10}\\ x=\dfrac{50}{9}\)d) \(\dfrac{5}{7}:x-3=-\dfrac{2}{7}\\ \dfrac{5}{7}:x=3-\dfrac{2}{7}\\ \dfrac{5}{7}:x=\dfrac{19}{7}\\ x=\dfrac{5}{7}:\dfrac{19}{7}\\ x=\dfrac{5}{19}\)

Lời giải:
a.

$0< x< \frac{1}{4}+\frac{4}{5}$

$\Rightarrow 0< x< \frac{21}{20}$ hay $0< x< 1,05$

$\Rightarrow x=1$

b.

$\frac{4}{7}+\frac{3}{7}< x< \frac{5}{3}+\frac{2}{3}$
$\Rightarrow 1< x< \frac{7}{3}$
$\Rightarrow x=2$

cha loi di

\(a,\dfrac{3}{7}-x=\dfrac{1}{2}x-3\)

\(\Rightarrow-x-\dfrac{1}{2}x=-3-\dfrac{3}{7}\)

\(\Rightarrow-\dfrac{3}{2}x=-\dfrac{24}{7}\)

\(\Rightarrow x=-\dfrac{24}{7}:\left(-\dfrac{3}{2}\right)\)

\(\Rightarrow x=\dfrac{16}{7}\)

\(b,5x-\dfrac{2}{3}=\dfrac{5}{3}-2x\)

\(\Rightarrow5x+2x=\dfrac{5}{3}+\dfrac{2}{3}\)

\(\Rightarrow7x=\dfrac{7}{3}\)

\(\Rightarrow x=\dfrac{7}{3}:7\)

\(\Rightarrow x=\dfrac{1}{3}\)

#Toru

a: 3/7-x=1/2x-3

=>-3/2x=-3+3/7

=>-1/2x=-1+1/7=-6/7

=>1/2x=6/7

=>x=6/7*2=12/7

b: =>5x+2x=5/3+2/3

=>7x=7/3

=>x=1/3

a)4/3:x=-4/7                         b)5/3.x=7/12

⇒x=-16/21                             ⇒x=7/20

a. \(\dfrac{5}{7}+\dfrac{4}{3}:x=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3}.\dfrac{1}{x}=\dfrac{1}{7}\)

<=> \(\dfrac{5}{7}+\dfrac{4}{3x}=\dfrac{1}{7}\)         ĐKXĐ: x \(\ne\) 0

<=> \(\dfrac{15x}{21x}+\dfrac{28}{21x}=\dfrac{3x}{21x}\)

<=> 15x + 28 = 3x

<=> 15x - 3x = -28

<=> 12x = -28

<=> x = \(\dfrac{-28}{12}=-\dfrac{7}{3}\)

b. \(\dfrac{5}{3}x.\dfrac{-1}{4}=\dfrac{2}{6}\)

<=> \(\dfrac{-5x}{12}=\dfrac{2}{6}\)

<=> -5x . 6 = 12 . 2

<=> -30x = 24

<=> x = \(-\dfrac{4}{5}\)

\(a,\Leftrightarrow25x^2-70x+49-25x^2=32\\ \Leftrightarrow-70x=-17\Leftrightarrow x=\dfrac{17}{70}\\ b,\Leftrightarrow x^2-6x+9+x^2+2x+1-5=0\\ \Leftrightarrow2x^2-4x+5=0\\ \Leftrightarrow2\left(x^2-2x+1\right)+3=0\\ \Leftrightarrow2\left(x-1\right)^2=-3\Leftrightarrow\left(x-1\right)^2=-\dfrac{3}{2}\left(\text{vô lí}\right)\\ \Leftrightarrow x\in\varnothing\)

b: =>x(8-7)=-33

=>x=-33

c: =>-12x+60+21-7x=5

=>-19x=-76

hay x=4

d: =>-2x-2-x+5+2x=0

=>3-x=0

hay x=3

a, \(x=-\dfrac{5}{6}-\dfrac{7}{12}=\dfrac{-10-7}{12}=-\dfrac{17}{12}\)

b, \(\dfrac{2}{9}-x=-\dfrac{4}{3}.\dfrac{5}{6}=-\dfrac{24}{18}=-\dfrac{4}{3}\Leftrightarrow x=\dfrac{2}{9}+\dfrac{4}{3}=\dfrac{14}{9}\)

c, \(-3=x-1\Leftrightarrow x=-2\)

d, \(\dfrac{3}{5}x-\dfrac{2}{3}=\dfrac{4}{5}:\dfrac{1}{5}=4\Leftrightarrow\dfrac{3}{5}x=4+\dfrac{2}{3}=\dfrac{14}{3}\Leftrightarrow x=\dfrac{14}{3}:\dfrac{3}{5}=\dfrac{70}{9}\)

a, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{x+y}{3+5}=\dfrac{16}{8}=2\\ \Rightarrow\left\{{}\begin{matrix}x=6\\y=10\end{matrix}\right.\\ b,x:2=y:\left(-5\right)\Rightarrow\dfrac{x}{2}=\dfrac{y}{-5}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{-5}=\dfrac{x-y}{2-\left(-5\right)}=\dfrac{-7}{7}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-2\\y=5\end{matrix}\right.\)