Ta có :

\(x\times\left(x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x+1=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=0\\x=-1\end{cases}}\)

Vậy \(x\in\left\{0;-1\right\}\)

=> x=0 hoặc x+1=0

     x=0 hoặc x=-1

chúc e học tốt

\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

ĐKXĐ : \(x\ne\pm2\)

\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\)

\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{x^2+3x+2}{\left(x-2\right)\left(x+2\right)}+\frac{x^2+5x-14}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow12-\left(x^2+3x+2\right)+x^2+5x-14=0\)

\(\Leftrightarrow12-x^2-3x-2+x^2+5x-14=0\)

\(\Leftrightarrow2x-4=0\)

\(\Leftrightarrow2x=4\)

\(\Leftrightarrow x=2\)( không tmđk )

=> Phương trình vô nghiệm

\(\frac{12}{x^2-4}-\frac{x+1}{x-2}+\frac{x+7}{x+2}=0\left(đk:x\ne2;-2\right)\)

\(\Leftrightarrow\frac{12}{\left(x-2\right)\left(x+2\right)}-\frac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{\left(x+7\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}=0\)

\(\Leftrightarrow12-\left(x^2+3x+3\right)+\left(x^2+5x-14\right)=0\)

\(\Leftrightarrow12-x^2+x^2-3x+5x-3-14=0\)

\(\Leftrightarrow2x-17+12=0\Leftrightarrow2x-5=0\Leftrightarrow x=\frac{5}{2}\left(tmđk\right)\)

giúp mik với các bn, mai mik phải nộp rồi ...huuuuhuhh

Mik hứa ai nhanh nhất mik luôn nha ok okokok ko

1) How deep is this well?

2) how much does the suitcase weigh?

lỡ mà sai bn đừng trách mình nha

105/150<105/145<105/135

105/150<105/140<105/135

tk cho mình nha

\(\frac{105}{150}\)< \(\frac{105}{140}\)< \(\frac{105}{135}\)

8*9 lớn hơn 9*8

tk mình nha

8.9=9.8

easy

a)\(3^x=9\Leftrightarrow x^x=3^2\Leftrightarrow x=2\)

b)\(6^{x-1}=36\Leftrightarrow6^{x-1}=6^2\Leftrightarrow x-1=2\Leftrightarrow x=3\)

c)\(5^x=125\Leftrightarrow5^x=5^3\Leftrightarrow x=3\)

d)\(3^{2x+1}=27\Leftrightarrow3^{2x+1}=3^3\Leftrightarrow2x+1=3\Leftrightarrow2x=2\Leftrightarrow x=1\)

e) \(3^{x+1}=9\Leftrightarrow3^{x+1}=3^2\Leftrightarrow x+1=2\Leftrightarrow x=1\)

f) \(x^{50}=x\Leftrightarrow\hept{\begin{cases}x=1\\x=0\end{cases}}\)

\(3^x=9\)

\(\Rightarrow3^x=3^2\)

\(\Rightarrow x=2\)

Nhanh mấy bạn ơi

\(\left(x-2020\right)\left(x-2021\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-2020=0\\x-2021=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0+2020=2020\\x=0+2021=2021\end{cases}}}\)

Vậy \(x=2020;2021\)

\(\left(x-2020\right)\left(x-2021\right)=0\)

\(\Leftrightarrow\hept{\begin{cases}x-2020=0\\x-2021=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2020\\x=2021\end{cases}}\)

vậy x=2020 hoặc x=2021

Tổng trên có số số hạng là:

     (50-20) : 1 + 1= 31 (số hạng)

Tổng trên là:

     (20+50) x 31 : 2= 1085

            Đáp số: 1085