\(\dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{40}{24}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)

\(\dfrac{5}{3}+\dfrac{3}{x}=\dfrac{49}{24}\\ \dfrac{3}{x}=\dfrac{49}{24}-\dfrac{5}{3}\\ \dfrac{3}{x}=\dfrac{3}{8}\\ x=8\)

Lời giải:
$3xy-2x-2y=24$

$\Rightarrow (3xy-2x)-2y=24$

$\Rightarrow x(3y-2)-2y=24$

$\Rightarrow 3x(3y-2)-6y=72$

$\Rightarrow 3x(3y-2)-2(3y-2)=76$

$\Rightarrow (3x-2)(3y-2)=76$

Vì $x,y$ nguyên nên $3x-2, 3y-2$ cũng là số nguyên. Do đo $3x-2, 3y-2$ là ước của 76. 

Đến đây thì đơn giản rồi. Bạn chỉ cần xét các TH khác nhau của ước của 76.

\(\Leftrightarrow12x^3-12x^3-8x=24\)

hay x=-3

Tham khảo!

https://olm.vn/hoi-dap/detail/10185285054.html

a) \(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)-4=\left(x^2+6x+5\right)\left(x^2+6x+8\right)-4\)

Đặt \(t=x^2+6x+5\)

\(PT=t\left(t+3\right)-4=t^2+3t-4=\left(t-1\right)\left(t+4\right)\)

Thay t: \(PT=\left(x^2+6x+5-1\right)\left(x^2+6x+5+4\right)=\left(x^2+6x+4\right)\left(x^2+6x+9\right)=\left(x^2+6x+4\right)\left(x+3\right)^2\)

b)  Đặt \(t=\left(2x+1\right)^2\)

\(PT=t^2-3t+2=\left(t^2-3t+\dfrac{9}{4}\right)-\dfrac{1}{4}=\left(t+\dfrac{3}{2}\right)^2-\dfrac{1}{4}=\left(t+1\right)\left(t+2\right)\)

Thay t:

\(PT=\left[\left(2x+1\right)^2+1\right]\left[\left(2x+1\right)^2+2\right]=\left[4x^2+4x+2\right]\left[4x^2+4x+3\right]=2\left[2x^2+2x+1\right]\left[4x^2+4x+3\right]\)

a) \(x^2\left(x^2+4\right)-x^2-4=x^2\left(x^2+4\right)-\left(x^2+4\right)=\left(x^2+4\right)\left(x^2-1\right)=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)

b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)

a. \(x^2\left(x^2+4\right)-x^2-4\)

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

\(=\left(x^2-1\right)\left(x^2+4\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+4\right)\)

b. \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=x^4+2x^3+5x^2+4x-12\)

\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)

c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)

Đặt \(t=x^2+7x+10\), ta được

(*) \(=t\left(t+2\right)-24\)

\(=t^2+2t-24\)

\(=\left(t-4\right)\left(t+6\right)\)

hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)

a: Ta có: \(x^2\left(x^2+4\right)-x^2-4\)

\(=\left(x^2+4\right)\left(x^2-1\right)\)

\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)

b: Ta có: \(\left(x^2+x\right)^2+4x^2+4x-12\)

\(=\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)

\(=\left(x^2+x\right)^2+6\left(x^2+x\right)-2\left(x^2+x\right)-12\)

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

\(=\left(x^2+x+6\right)\left(x+2\right)\left(x-1\right)\)

c: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)

\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)

\(=\left(x^2+7x\right)^2+22\left(x^2+7x\right)+96\)

\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)

\(=\left(x^2+7x+16\right)\left(x+1\right)\left(x+6\right)\)

Sửa: \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\le0\)

Mà \(\left(\dfrac{1}{3}-2x\right)^{2020}+\left(3y-x\right)^{2022}\ge0\) với mọi x,y

Do đó \(\left\{{}\begin{matrix}\dfrac{1}{3}-2x=0\\3y-x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{6}\\y=\dfrac{1}{18}\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}=6+18=24\)

Bài làm

X x 2 + X x 3 + X x 4 + X = 2130

X x ( 2 + 3 + 4 + 1 )         = 2130

X x           10                    = 2130

X                                      = 2130 : 10

X                                      = 2130

Vậy X = 2130

# Học tốt #

\(x\times2+x\times3+x\times4+x=2130\)

\(\Rightarrow x\times2+x\times3+x\times4+x\times1=2130\)

\(\Rightarrow x\times\left(2+3+4+1\right)=2130\)

\(\Rightarrow10x=2130\)

\(\Rightarrow x=213\)

~Std well~

#Lâm

Xem chó https://olm.vn/thanhvien/kimmai123az, PTD/KM ? nè, thánh copy ,mời mn xem tcn của nó

`-3x=2y `

`=> x/2 = -y/3 `

AD t/c của dãy tỉ số bằng nhau ta có

`x/2 =-y/3 = (x-y)/(2+3) = 6/5`

`=>{(x=2*6/5 = 12/5),(y=-3*6/5 =-18/5):}`

a) `6/x =-3/2`

`=>x =6 :(-3/2) = 6*(-2/3)=-4`

`b)`\(-3x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)

Áp dụng t/c của DTSBN , ta đc :

\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{x-y}{2+3}=\dfrac{6}{5}\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{6}{5}\\\dfrac{y}{-3}=\dfrac{6}{5}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{12}{5}\\y=-\dfrac{18}{5}\end{matrix}\right. \)

`a)`

`6/x=-3/2`

`x=6:(-3/2)`

`x=6*(-2/3)`

`x=-4`

a, \(\dfrac{6}{x}=-\dfrac{3}{2}\)

\(\Rightarrow3x=6.\left(-2\right)\)

\(\Rightarrow3x=-12\)

\(\Rightarrow x=-\dfrac{12}{3}\)

=>x=-4

b, \(+\left(-3\right)x=2y\Rightarrow\dfrac{x}{2}=\dfrac{y}{-3}\)

+x-y=6

+Áp dụng tính chất dãy tỉ số bằng nhau có:

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

Suy ra \(\dfrac{x}{2}=\dfrac{6}{5}\Rightarrow x=\dfrac{6}{5}.2\Rightarrow x=\dfrac{12}{5}\)

           \(\dfrac{y}{-3}=\dfrac{6}{5}\Rightarrow y=\dfrac{6}{5}.\left(-3\right)=-\dfrac{18}{5}\)

Vậy \(x=\dfrac{12}{5};y=-\dfrac{18}{5}\)