anh em copy link này lên youtube xem rồi đăng kí nhe cảm ơn

https://www.youtube.com/shorts/hhpTDItpePY

cảm ơn rất nhiều luôn

Ta thấy :   \(x^2+1\ge1\)  nên để   \(\left(3x-1\right)\left(x^2+1\right)< 0\)\(thì\) \(3x-1< 0\)\(hay\)  \(x< \frac{1}{3}\)

\(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-15-1\right)\left(2x-15+1\right)=0\)

\(\Rightarrow\left(2x-15\right)^3\left(2x-16\right)\left(2x-14\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\2x-16=0\\2x-14=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=15\\2x=16\\2x=14\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

a) => (2x - 15)⁵ - (2x - 15)³ = 0

(2x-15)³ [(2x-15)²-1]=0

\(\Rightarrow\left[{}\begin{matrix}2x-15=0\\\left(2x-15\right)^2=1\end{matrix}\right.\)  \(\Rightarrow\left[{}\begin{matrix}2x=15\\2x-15=1\\2x-15=-1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

b) x + 2x + 3x + ... + 2022x = 2022.2023

x(1 + 2 + 3 +...+ 2022)= 2022.2023 (1)

Đặt A = 1+2+3+...+2022

Số số hạng trong A là: (2022 - 1): 1 + 1 = 2022 (số)

Tổng A bằng: \(\dfrac{\left(2022+1\right).2022}{2}=\dfrac{2022.2023}{2}\)

Thay A vào (1) , ta  được:

x.\(\dfrac{2022.2023}{2}=2022.2023\)

=> x = 2

\(x+2x+3x+...+2022x=2022\cdot2023\)

\(\Rightarrow x\cdot\left(1+2+3+...+2022\right)=2022\cdot2023\)  

\(\Rightarrow x\cdot\left(2022+1\right)\cdot\left[\left(2022-1\right):1+1\right]:2=2022\cdot2023\)

\(\Rightarrow x\cdot2023\cdot\left(2022-1+1\right):2=2022\cdot2023\)

\(\Rightarrow x\cdot2023\cdot2022:2=2022\cdot2023\)

\(\Rightarrow x:2=2022\cdot2023:2022\cdot2023\)

\(\Rightarrow x:2=1\)

\(\Rightarrow x=1\cdot2\)

\(\Rightarrow x=2\)

Độ dài quãng đường là: 3,5*36=126km

Thời gian đi là 126/45=2,8h

về đến nơi lúc:

10h30+2h48=13h18

a: Xét ΔABC có

AM là đường trung tuyến

G là trọng tâm

Do đó: \(\dfrac{AG}{AM}=\dfrac{2}{3}\)

Xét ΔABM có DG//BM

nên \(\dfrac{AD}{AB}=\dfrac{AG}{AM}\)

=>\(\dfrac{AD}{AB}=\dfrac{2}{3}\)

b: Xét ΔAMC có GE//MC

nên \(\dfrac{AE}{AC}=\dfrac{AG}{AM}\)

=>\(\dfrac{AE}{AC}=\dfrac{2}{3}\)

=>\(AE=\dfrac{2}{3}AC\)

AE+EC=AC

=>\(EC+\dfrac{2}{3}AC=AC\)

=>\(EC=\dfrac{1}{3}AC\)

\(AE=\dfrac{2}{3}AC=2\cdot\dfrac{1}{3}\cdot AC=2\cdot EC\)