b) \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)

<=> \(\frac{13\left(x+1\right)}{12}-\frac{5x+3}{6}=\frac{x+7}{12}\)

<=> 13(x + 1) - 2(5x + 3) = x + 7

<=> 13x + 13 - 10x - 6 = x + 7

<=> 3x + 7 = x + 7

<=> 3x + 7 - x = 7

<=> 2x + 7 = 7

<=> 2x = 7 - 7

<=> 2x = 0

<=> x = 0

c) 2x + 4(x - 2) = 5

<=> 2x + 4x - 8 = 5

<=> 6x - 8 = 5

<=> 6x = 5 + 8

<=> 6x = 13

<=> x = 13/6

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

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

\(\Leftrightarrow x=5\)

\(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)

\(\Rightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)

\(\Rightarrow\left[x^2+4x-5\right]\left[x^2+4x-5-16\right]=297\)

Đặt \(x^2+4x-5=t\)

\(\Rightarrow t\left(t-16\right)=297\)

\(\Rightarrow t^2-16t+64=297+64\)

\(\Rightarrow\left(t+8\right)^2=361\)

\(\Rightarrow\left[{}\begin{matrix}t+8=19\\t+8=-19\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}t=11\\t=-27\end{matrix}\right.\)

Ta có : \(\left(x-1\right)\left(x-3\right)\left(x+5\right)\left(x+7\right)=297\)

\(\Leftrightarrow\left[\left(x-1\right)\left(x+5\right)\right]\left[\left(x-3\right)\left(x+7\right)\right]=297\)

\(\Leftrightarrow\left(x^2-x+5x-5\right)\left(x^2-3x+7x-21\right)=297\)

\(\Leftrightarrow\left(x^2+4x-5\right)\left(x^2+4x-21\right)=297\)

\(\Leftrightarrow\left(x^2+4x-13+8\right)\left(x^2+4x-13-8\right)=297\)

\(\Leftrightarrow\left(x^2+4x-13\right)^2-64=297\)

\(\Leftrightarrow\left(x^2+4x-13\right)^2=361\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x-13=19\\x^2+4x-13=-19\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+4x+4-17=19\\x^2+4x+4-17=-19\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2-17=19\\\left(x+2\right)^2-17=-19\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+2\right)^2=36\\\left(x+2\right)^2=-2\left(VL\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=4\\x=-8\end{matrix}\right.\)

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

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

\(\Rightarrow7x-49=0\)

\(\Rightarrow x=7\)

    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.

\(\frac{3}{7}:x+\frac{1}{4}=\frac{5}{9}\)

\(\frac{3}{7}:x=\frac{5}{9}-\frac{1}{4}=\frac{20-9}{36}=\frac{11}{36}\)

\(x=\frac{3}{7}:\frac{11}{36}=\frac{3}{7}\times\frac{36}{11}=\frac{3\times36}{7\times11}=\frac{108}{77}\)

Vậy \(x=\frac{108}{77}\)

\(\frac{3}{7}:x+\frac{1}{4}=\frac{5}{9}\)

\(\frac{3}{7}:x=\frac{5}{9}-\frac{1}{4}\)

\(\frac{3}{7}:x=\frac{11}{36}\)

\(x=\frac{3}{7}:\frac{11}{36}\)

\(x=\frac{108}{77}\)

3/7:x+1=5/9

3/7:x=5/9-1

3/7:x=-4/9

x=3/7:-4/9

x=âm27phần28

Nhớ k cho mình nha

a) \(\frac{35}{14}=\frac{5\times7}{2\times7}=\frac{5}{2}\)

\(\frac{125}{50}=\frac{5\times5\times5}{2\times5\times5}=\frac{5}{2}\)

b)\(\frac{4\times5+4\times11}{8\times7+4\times3}=\frac{4\times\left(5+11\right)}{4\times\left(2\times7+3\right)}=\frac{16}{17}\)

c) \(\frac{3\times11+7\times11}{22\times2+11\times6}=\frac{11\times\left(3+7\right)}{22\times\left(2+3\right)}=\frac{11\times10}{22\times5}=\frac{11\times2\times5}{11\times2\times5}=1\)

x + 2y = 0 

=> x = 0 - 2y

=> x = -2y

=> x = -2y

=> y = -2 : x

(x-1/2)=19/4:5/3

(x-1/2)=19/4x3/5

x-1/2=57/20

x=57/20+1/2

x=67/20

tk cho mk nha

\(\frac{19}{4}:\left(x-\frac{1}{2}\right)=\frac{5}{3}\)

\(x-\frac{1}{2}=\frac{19}{4}:\frac{5}{3}=\frac{57}{20}\)

\(x=\frac{57}{20}+\frac{1}{2}\)

\(x=\frac{67}{20}\)

thank kiu các bn nhìu nha