a) Ta có: \(\dfrac{4}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5}{6}-\dfrac{y}{3}\)

\(\Rightarrow\dfrac{4}{x}=\dfrac{5-2y}{6}\)

\(\Rightarrow\left(5-2y\right)x=24\)

Vì \(x,y\in Z\Rightarrow\left[{}\begin{matrix}5-2y\in Z\\x\in Z\end{matrix}\right.\)

\(\Rightarrow5-2y\inƯ\left(24\right);x\inƯ\left(24\right)\)

Tự lập bảng xét các giá trị của \(x,y\) nhé.

b) Lại có: \(\dfrac{5}{x}-\dfrac{y}{3}=\dfrac{1}{6}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1}{6}+\dfrac{y}{3}\)

\(\Rightarrow\dfrac{5}{x}=\dfrac{1+2y}{6}\)

\(\Rightarrow\left(1+2y\right)x=30\)

Lí luận rồi lập bảng như câu \(a\)).

c) \(\dfrac{x}{6}-\dfrac{2}{y}=\dfrac{1}{30}\)

\(\Rightarrow\dfrac{2}{y}=\dfrac{x}{6}-\dfrac{1}{30}\)

\(\Rightarrow\dfrac{2}{y}=\dfrac{5x-1}{30}\)

\(\Rightarrow\left(5x-1\right)y=60\)

\(......Tương\) \(tự\) \(như\) \(câu\) \(a\))\(b\)).

a, \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}\)

ta có: \(\dfrac{3}{x}+\dfrac{y}{3}=\dfrac{5}{6}=>\dfrac{3}{x}=\dfrac{5}{6}-\dfrac{y}{3}=\dfrac{5-2y}{6}\)

=>\(\dfrac{3}{x}=\dfrac{5-2y}{6}=>x.\left(5-2y\right)=3.6=18\)

=> x và 5-2y thuộc Ư của 18={1,-1,2,-2,3,-3,6,-6}

vì 5-2y là số lẻ=> 5-2y= +-1 hoặc 5-2y=+-3

xét bảng

vậy giá trị x,y cần tìm là: {x=18.y=2}

{x=-18.y=3}

{x=6, y=1}Ư

{x=-6,y=4}

các ý a,b,c c chỉ cần nhan chéo cho nhau

a.

\(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow x\left(x+1\right).\left(x-1\right)\left(x+2\right)-24=0\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)-24=0\)

Đặt \(a=x^2+x-1\) , ta có pt:

\(\left(a+1\right)\left(a-1\right)-24=0\)

\(\Leftrightarrow a^2-1-24=0\)

\(\Leftrightarrow a^2-25=0\)

\(\Leftrightarrow\left(a-5\right)\left(a+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}a=5\\a=-5\end{matrix}\right.\)

*Với a = 5 ta được:

\(x^2+x-1=5\)

\(\Leftrightarrow x^2+x-6=0\)

\(\Leftrightarrow x^2+3x-2x-6=0\)

\(\Leftrightarrow\left(x^2+3x\right)-\left(2x+6\right)=0\)

\(\Leftrightarrow x\left(x+3\right)-2\left(x+3\right)=0\)

\(\Leftrightarrow\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\)

*Với a = -5 ta được:

\(x^2+x-1=-5\)

\(\Leftrightarrow x^2+x+4=0\)

\(\Leftrightarrow x^2+2.x.\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{15}{4}=0\)

\(\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{15}{4}=0\) ( loại)

Vậy pt có tập nghiệm là: \(s=\left\{-3;2\right\}\)

c)(ĐKXĐ: x khác 30;29)

\(\Leftrightarrow\dfrac{x-29}{30}-1+\dfrac{x-30}{29}-1=\dfrac{29}{x-30}-1+\dfrac{30}{x-29}-1\)

\(\Leftrightarrow\dfrac{x-59}{30}+\dfrac{x-59}{29}=\dfrac{x-59}{30-x}+\dfrac{x-59}{29-x}\)

\(\Leftrightarrow x=59\)(tm) or \(\dfrac{1}{30}+\dfrac{1}{29}-\dfrac{1}{30-x}-\dfrac{1}{29-x}=0\)

\(\Leftrightarrow\dfrac{-x}{30\left(30-x\right)}+\dfrac{-x}{29\left(29-x\right)}=0\)

\(\Leftrightarrow x=0\)(tm) or \(\dfrac{1}{30\left(30-x\right)}+\dfrac{1}{29\left(29-x\right)}=0\)

\(\Leftrightarrow1741-59x=0\)

\(\Leftrightarrow x=\dfrac{1741}{59}\left(tm\right)\)

Vậy S={0;\(\dfrac{1741}{59}\);59}

b)(ĐKXĐ:x khác 2;3;4;5;6)

\(\Leftrightarrow\dfrac{1}{\left(x-2\right)\left(x-3\right)}+\dfrac{1}{\left(x-3\right)\left(x-4\right)}+\dfrac{1}{\left(x-4\right)\left(x-5\right)}+\dfrac{1}{\left(x-5\right)\left(x-6\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-3}-\dfrac{1}{x-2}+\dfrac{1}{x-4}-\dfrac{1}{x-3}+\dfrac{1}{x-5}-\dfrac{1}{x-4}+\dfrac{1}{x-6}-\dfrac{1}{x-5}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{1}{x-6}-\dfrac{1}{x-2}=\dfrac{1}{8}\)

\(\Leftrightarrow\dfrac{4}{\left(x-6\right)\left(x-2\right)}=\dfrac{1}{8}\)

\(\Leftrightarrow x^2-8x+12=32\)

\(\Leftrightarrow x^2-8x-20=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-10\right)=0\)

\(\Leftrightarrow x=-2\) or x=10(đều thỏa)

Vậy ...

Bài 1 :

\(\dfrac{1}{2}x+\dfrac{5}{6}=\dfrac{-3}{4}\\ \Rightarrow\dfrac{1}{2}x=\dfrac{-19}{12}\\ \Rightarrow x=\dfrac{-19}{12}\cdot2=-\dfrac{19}{6}\)

Bài 2 :

\(a)\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{x+y}{2+3}=\dfrac{30}{5}=6\\ \Rightarrow\dfrac{x}{2}=6\Rightarrow x=12\\ \dfrac{y}{3}=6\Rightarrow y=18\\ b)\dfrac{x}{3}=\dfrac{y}{7}=\dfrac{x-y}{3-7}=\dfrac{15}{-4}\\ \Rightarrow\dfrac{x}{3}=\dfrac{-15}{4}\Rightarrow x=\dfrac{-45}{4}\\ \dfrac{y}{7}=\dfrac{-15}{4}\Rightarrow4y=-105\Rightarrow y=\dfrac{-105}{4}\)

Bài 1 :

a,\(\dfrac{1}{2}x\)+\(\dfrac{5}{6}\)=\(\dfrac{-3}{4}\)

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

\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-18}{24}\)-\(\dfrac{20}{24}\)

\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-38}{24}\)

\(\Rightarrow\)\(\dfrac{1}{2}x\)=\(\dfrac{-19}{12}\)

\(\Rightarrow\)x =\(\dfrac{-19}{12}\):\(\dfrac{1}{2}\)

\(\Rightarrow\)x =\(\dfrac{-19}{12}\).2

\(\Rightarrow\)x=\(\dfrac{-19}{6}\)

Vậy x=\(\dfrac{-19}{6}\)

Bài 2:

a,x+y=30 và \(\dfrac{x}{2}=\dfrac{y}{3}\)

Áp dụng tính chất của dãy tỉ số bằng nhau ta có :

\(\dfrac{x}{2}=\dfrac{y}{3}\)=\(\dfrac{x+y}{2+3}\)=\(\dfrac{30}{5}\)=6

\(\dfrac{x}{2}\)=6\(\Rightarrow\)x=2.6=12

\(\dfrac{y}{3}\)=6\(\Rightarrow\)y=3.6=18

Vậy x=12,y=18

b,x-y=15 và \(\dfrac{x}{3}=\dfrac{y}{7}\)

Đặt \(\dfrac{x}{3}\),\(\dfrac{y}{7}\)=k

\(\Rightarrow\)x=3k,y=7k

Thay x=3k,y=7k vào x-y=15 ta có :

3k-7k=15

\(\Rightarrow\)-4k=15

\(\Rightarrow\)k=\(\dfrac{-15}{4}\)

x=3k\(\Rightarrow\)x=3.\(\dfrac{-15}{4}\)=\(\dfrac{-45}{4}\)

y=7k\(\Rightarrow\)y=7.\(\dfrac{-15}{4}\)=\(\dfrac{-105}{4}\)

Vậy x=\(\dfrac{-45}{4}\),y=\(\dfrac{-105}{4}\)

Nếu đúng thì tick cho mk nha

Bài 1

a) {EQ /F(1,2)}

Ta có: \(\dfrac{x-1}{6}=\dfrac{-2y+3}{30}\)

\(\Leftrightarrow\dfrac{3x-3}{18}=\dfrac{-8y+12}{120}\)

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{3x-3}{18}=\dfrac{-8y+12}{120}=\dfrac{3x-3+8y-12}{18-120}=\dfrac{2-15}{-102}=\dfrac{13}{102}\)

Do đó: 

\(\left\{{}\begin{matrix}\dfrac{x-1}{6}=\dfrac{13}{102}\\\dfrac{3-2y}{30}=\dfrac{13}{102}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-1=\dfrac{13}{17}\\-2y+3=\dfrac{65}{17}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{17}\\-2y=\dfrac{14}{17}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{30}{17}\\y=\dfrac{-7}{17}\end{matrix}\right.\)

Ta có: 5x - 5 = 3 - 2y

=> 5x+2y = 8

=> 20x + 8y = 32

Mà 3x +8y = 2

=> 17x = 30

=> x = \(\dfrac{30}{7}\)

=> y = ... giải tiếp nha bạn.

Xin 1 like nha bạn. Thx bạn

\(a,=\dfrac{5x+30+x^2-30}{x\left(x+6\right)}=\dfrac{x\left(x+5\right)}{x\left(x+6\right)}=\dfrac{x+5}{x+6}\\ b,=\dfrac{3x^2+4x+1-x^2+2x-1-x^2-2x+3}{\left(x-1\right)^2\left(x+1\right)}\\ =\dfrac{x^2+4x+3}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{\left(x+1\right)\left(x+3\right)}{\left(x-1\right)^2\left(x+1\right)}=\dfrac{x+3}{\left(x-1\right)^2}\)

\(c,=\dfrac{3x^2+2x+1+x^2-2x+1-2x^2-2x-2}{\left(x-1\right)\left(x^2+x+1\right)}\\ =\dfrac{2x^2-2x}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{2x}{x^2+x+1}\)

\(\Rightarrow x+\dfrac{1}{1\times2}+\dfrac{1}{2\times3}+\dfrac{1}{3\times4}+\dfrac{1}{4\times5}+\dfrac{1}{5\times6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x+1-\dfrac{1}{6}=\dfrac{47}{42}\\ \Rightarrow x=\dfrac{47}{42}-\dfrac{5}{6}=\dfrac{2}{7}\)

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

2/7 nha

hok tốt

k mik đi