\(\left(3x-1\right)^6=\left(3x-1\right)^4\\ =>\left(3x-1\right)^6-\left(3x-1\right)^4=0\\ =>\left(3x-1\right)^4\left[\left(3x-1\right)^2-1\right]=0\\ =>\left[{}\begin{matrix}\left(3x-1\right)^4=0\\\left(3x-1\right)^2-1=0\end{matrix}\right.=>\left[{}\begin{matrix}\left(3x-1\right)=0\\3x-1=1\\3x-1=-1\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=\dfrac{2}{3}\\x=0\end{matrix}\right.\)

Lời giải:
Gọi tổng trên là $A$
\(A=\frac{1}{\frac{1\times 2}{2}}+\frac{1}{\frac{2\times 3}{2}}+\frac{1}{\frac{3\times 4}{2}}+\frac{1}{\frac{3\times 4}{2}}+...+\frac{1}{\frac{9\times 10}{2}}\)

\(=\frac{2}{1\times 2}+\frac{2}{2\times 3}+\frac{2}{3\times 4}+....+\frac{2}{9\times 10}\)

\(=2\times (\frac{2-1}{1\times 2}+\frac{3-2}{2\times 3}+...+\frac{10-9}{9\times 10})\)

\(=2\times (1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{9}-\frac{1}{10})\)

\(=2\times (1-\frac{1}{10})=2\times \frac{9}{10}=\frac{9}{5}\)

5555

DĐeo

|7x-5y|, |2z-3x| ≥ 0

Để |7x-5y| + |2z-3x| = 0

=> 7x - 5y = 2z - 3x = 0

=> 7x = 5y ; 2z = 3x

=> \(\dfrac{x}{5}\)=\(\dfrac{y}{7}\)

      \(\dfrac{z}{3}\)=\(\dfrac{x}{2}\)

=> \(\dfrac{x}{10}\)=\(\dfrac{y}{14}\)=\(\dfrac{z}{15}\)

=> (\(\dfrac{x}{10}\))² = \(\dfrac{xy}{140}\)= \(\dfrac{yz}{210}\)=\(\dfrac{xz}{150}\)=\(\dfrac{xy+yz+xz}{140+150+210}\)= 4

=> x = 20, y = 28, z = 30

4x = 7,5 + 2/3 

4x = 15/2 + 2/3

4x = 49/6

x = 49/6 : 4

x = 49/24

cảm ơn bạn  nhé

đổi 250dm = 25m

đổi : 468750l = 468,75m³

thể tích của bể nước là: 468,75 : 75 x 100 = 625 (m³)

chiều cao của bể nước là : 625 : 25 : 12,5 = 2 (m)

kết luận bể sâu 2m

Ta có:

\(|x+4|=\left[{}\begin{matrix}x+4\Leftrightarrow x+4\ge0\Rightarrow x\ge-4\\-\left(x+4\right)\Leftrightarrow x+4\le0\Rightarrow x\le-4\end{matrix}\right.\)

\(|x-8|=\left[{}\begin{matrix}x-8\Leftrightarrow x-8\ge0\Rightarrow x\ge8\\-\left(x-8\right)\Leftrightarrow x-8\le0\Rightarrow x\le8\end{matrix}\right.\)

Xét \(x\le-4\) ta có:

\(|x+4|-|x-8|=6\\ \Leftrightarrow-\left(x+4\right)-\left[-\left(x-8\right)\right]=6\\ \Leftrightarrow-x-4+x-8=6\\ \Leftrightarrow-12=6\left(vô.nghiệm\right)\)

Xét \(-4\le x\le8\) ta có:

\(|x+4|-|x-8|=6\\ \Leftrightarrow x+4-\left[-\left(x-8\right)\right]=8\\ \Leftrightarrow x+4+x-8=6\\ \Leftrightarrow2x-4=6\\ \Leftrightarrow2x=10\\ \Rightarrow x=10:2=5\left(thõa.mãn\right)\)

Xét \(x\ge8\) ta có:

\(|x+4|-|x-8|=6\\ \Leftrightarrow x+4-\left(x-8\right)=6\\ \Leftrightarrow x+4-x+12=6\\ \Leftrightarrow16=6\left(vô.nghiệm\right)\)

Đs....