ta có toz đối đỉnh t''oz''

=>toz=t''oz''=50

+ tOz = t'Oz' (vì là hai góc đối đỉnh)

Mà tOz = 50^o

=> t'Oz' = 50^o (t/c)

Ta có \(99^{20}=(99^2)^{10}=9801^{10}\)

Mà \(9801^{10}< 9999^{10}\)

\(\Rightarrow99^{20}< 9999^{10}\)

Từ đẳng thức : \(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

\(\Rightarrow1:\frac{40}{x-30}=1:\frac{20}{y-15}=1:\frac{28}{z-21}\)

\(\Rightarrow\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)

\(\Rightarrow\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)

\(\Rightarrow\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

Đặt \(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=k\Rightarrow\hept{\begin{cases}x=40k\\y=20k\\z=28k\end{cases}}\)

Khi đó : xyz = 22400

<=> 40k.20k.28k = 22400

=> 22400.k³ = 22400

=> k³ = 1

=> k³ = 1³

=> k = 1

Khi đó : x = 40.1 = 40 ; 

              y = 20.1 = 20;

              z = 28.1 = 28 

Vậy x = 40 ; y = 20 ; z = 28

Ta có:\(\frac{40}{x-30}=\frac{20}{y-15}=\frac{28}{z-21}\)

hay\(\frac{x-30}{40}=\frac{y-15}{20}=\frac{z-21}{28}\)

\(=\frac{x}{40}-\frac{3}{4}=\frac{y}{20}-\frac{3}{4}=\frac{z}{28}-\frac{3}{4}\)

\(\Rightarrow\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}\)

\(\Rightarrow\)\(\frac{x}{40}=\frac{y}{20}=\frac{z}{28}=\frac{x.y.z}{40.20.28}=\frac{22400}{22400}=1\)

\(\Rightarrow\)\(\hept{\begin{cases}\frac{x}{40}=1\\\frac{y}{20}=1\\\frac{z}{28}=1\end{cases}}\)\(\Rightarrow\)\(\hept{\begin{cases}x=40\\y=20\\z=28\end{cases}}\)

Vậy x=40; y=20; z=28

toán lớp mấy zậy

\(\left(-\frac{3}{7}+\frac{5}{11}\right):\left(-\frac{3}{5}\right)+\left(\frac{-4}{7}+\frac{6}{11}\right):\frac{3}{5}\)

\(=\left(-\frac{3}{7}+\frac{5}{11}\right).\left(-\frac{5}{3}\right)+\left(-\frac{4}{7}+\frac{6}{11}\right).\left(-\frac{5}{3}\right)\)

\(=\left(-\frac{3}{7}+\frac{5}{11}-\frac{4}{7}+\frac{6}{11}\right).\left(-\frac{5}{3}\right)\)

\(=\left(-1+1\right).\left(-\frac{5}{3}\right)\)

\(=0.\left(-\frac{5}{3}\right)=0\)