dễ mà tự làm đi

minh hoc lop 12

thì trả lời đi chị

4,(09)=405/99 bạn nhé?

[4/5] 5 x = [-4/5]^7

\(\left(\frac{4}{5}\right)^5.x=\left(\frac{-4}{5}\right)^7\)

\(\Rightarrow x=\left(\frac{-1.4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)

\(\Rightarrow x=\left(-1\right)^7.\left(\frac{4}{5}\right)^7:\left(\frac{4}{5}\right)^5\)

\(\Rightarrow x=\left(-1\right).\left(\frac{4}{5}\right)^2\)

\(\Rightarrow x=\left(-1\right).\frac{16}{25}\)

\(\Rightarrow x=\frac{-16}{25}\)

Giải:Đặt x2 =y5 =z7 =k

⇒x=2k,y=5k,z=7k

Ta có: A=x−y+zx+2y−z 

⇒A=2k−5k+7k2k+2(5k)−7k =k(2−5+7)2k+10k−7k =4k(2+10−7)k =45 

Vậy A=45 

\(\frac{4}{5}\) nhé máy mik bị lỗi

Ta có: \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}\)

Đặt  \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)

\(\Rightarrow x=2k;y=5k;z=7k\)

Theo đề ta có:

\(A=\frac{x-y+z}{x+2y-z}=\frac{2k-5k+7k}{2k+2\left(5k\right)-7k}\)

\(A=\frac{\left(2-5+7\right)k}{2k+10k-7k}=\frac{\left(2-5+7\right)k}{\left(2+10-7\right)k}\)

\(A=\frac{4k}{5k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)

Đặt : \(\frac{x}{2}=\frac{y}{5}=\frac{z}{7}=k\)

\(\Rightarrow\hept{\begin{cases}x=2k\\y=5k\\z=7k\end{cases}}\)

Thay vào \(\frac{x-y+z}{x+2y-z}\)ta có :

\(A=\frac{2k-5k+7k}{2k+2.5k-7k}=\frac{\left(2-5+7\right)k}{2k+10k-7k}=\frac{4k}{\left(2+10-7\right)k}=\frac{4k}{5k}=\frac{4}{5}\)

Vậy \(A=\frac{4}{5}\)

mik chưa thấy hình vẽ ah