Sửa đề : \(\dfrac{2x}{3y}=\dfrac{-1}{3}\) và 2x+3y=7

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

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

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

\(\Rightarrow\left\{{}\begin{matrix}2x=-1.\dfrac{7}{2}\\3y=3.\dfrac{7}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}2x=\dfrac{-7}{2}\\y=\dfrac{7}{2}\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{-7}{4}\\y=\dfrac{7}{2}\end{matrix}\right.\)

Nhầm :

Dấu "=" xảy ra \(\left\{{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

(x+1)²+(y-2)¹⁰=0

Vì \(\left\{{}\begin{matrix}\left(x+1\right)^2≥0∀x\\\left(y-2\right)^{10}≥0∀y\end{matrix}\right.\)\(\Rightarrow\left(x+1\right)^2+\ge\left(y-2\right)^{10}0∀x,y\)

Dáu "=" xảy ra <=> \(\left[{}\begin{matrix}x+1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\y=2\end{matrix}\right.\)

`(x+7).(x+15)=0`

`=>x+7=0`

`x=0-7`

`x=-7`

`=>x+15=0`

`x=0-15`

`x=-15`

(x+7).(x+15)=0

 \(\Rightarrow\left[{}\begin{matrix}x+7=0\\x+15=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=-15\end{matrix}\right.\)

Ta có :

(x-20).(x-8).(x-2022) = 0

\(\Rightarrow\left[{}\begin{matrix}x-20=0\\x-8=0\\x-2022=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=20\\x=8\\x=2022\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=20\\x=8\\x=2022\end{matrix}\right.\)

`(x-20).(x-8).(x-2022)=0`

`=>` \(\left[{}\begin{matrix}x-20=0\\x-8=0\\x-2022=0\end{matrix}\right.\)

`=>`\(\left[{}\begin{matrix}x=0+20=20\\x=0+8=8\\x=0+2022=2022\end{matrix}\right.\)

Vậy `x = {20;8;2022}`

(x-20).(x-8).(x-2022)=0           

x =   

(a): `3x+1` \(⋮\) `(x-2)`

`=>3(x-2)+7` \(⋮\) `(x-2)`

`=>7` \(⋮\) `(x-2)`

`=>x-2\in {1;7;-1;-7}`

`=>x\in {3;9;1;-5}`

(b): `2x+xy+y=4`

`=>x(2+y)+(2+y)=4+2`

`=>(y+2)(x+1)=6=1.6=2.3=(-1).(-6)=(-2).(-3)`

Lập bảng ....

câu b mik thấy nó đã ra KQ đâu

(d) : \(16< 2^n:2< 64\\ =>2^4< 2^{n-1}< 2^6\\ =>n-1=5\\ =>n=6\)

(e) : \(3^{3n-3}=9^n.3^3\\ =>3^{3n-3}:3^3=\left(3^2\right)^n\\ =>3^{3n-6}=3^{2n}\\ =>3n-6=2n\\ =>n=6\)

d, n = 6

e, n = 6

(5x + 3)²⁸ = 16⁷

(5x+3)²⁸ = (2⁴)⁷ 

(5x+3)²⁸ = 2²⁸

5x + 3 = 2 

5x  = 2 - 3

5x = -1

x = -1/5

(5x+3)²⁸ = (2⁴)⁷ = -(2⁴)⁷

\(\Rightarrow\) 5x+3 = 2 = -(2)

\(\left[{}\begin{matrix}5x+3=2\\5x+3=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}5x=-1\\5x=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-1}{5}\\x=-1\end{matrix}\right.\)

Vậy x = \(\dfrac{-1}{5}\) hoặc x = -5