\(\frac{x}{z}=\frac{z}{y}\Rightarrow z^2=xy\)

\(\Rightarrow\frac{x^2+z^2}{y^2+z^2}=\frac{x^2+xy}{y^2+xy}=\frac{x\left(x+y\right)}{y\left(x+y\right)}=\frac{x}{y}\)

\(\frac{2}{1.5}+\frac{2}{5.9}+\frac{2}{9.13}+...+\frac{2}{95.99}\)

\(=\frac{1}{2}.\left(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{95.99}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{95}-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\left(1-\frac{1}{99}\right)\)

\(=\frac{1}{2}.\frac{98}{99}\)

\(=\frac{49}{99}\)

Chúc cậu học tốt !!!

Vì 5^x +11^y =26

mà 5^x> 0 

=> 11^y<26 => y<1 mà y là stn => y=0

=> 5^x+11^0 =26

=> 5^x=25 => x=2

 Ta có: 5^x + 11^y = 26

=> 11^y < 26 => 11^y \(\in\){1; 11} => y \(\in\){0; 1}

Với y = 0 => 5^x + 11⁰ = 26

=> 5^x + 1 = 26

=> 5^x = 26 - 1

=> 5^x = 25

=> 5^x = 5² => x = 2

Với y = 1 => 5^x  + 11¹ = 26

=> 5^x = 26 - 11

=> 5^x = 15 => ko có giá trị x t/m 

Vậy x = 2 và y = 0 tm

\(x.x+y.y+z.z=12\)

\(\Leftrightarrow\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}=\frac{12}{3}=4\)

\(\Rightarrow x^2=1.4=4\Leftrightarrow x=2\)

\(y^2=1.4=4\Leftrightarrow y=2\)

\(z^2=1.4=4\Leftrightarrow z=2\)

Áp dụng BĐT Cauchy - schwarz:

\(x^2+y^2+z^2=\frac{x^2}{1}+\frac{y^2}{1}+\frac{z^2}{1}\ge\frac{\left(x+y+z\right)^2}{1+1+1}=\frac{36}{3}=12\)

(Dấu "="\(\Leftrightarrow x=y=z\))

\(pt\Leftrightarrow3x^2=12\Leftrightarrow x^2=4\Leftrightarrow x=\pm2\)

\(\Rightarrow\orbr{\begin{cases}x=y=z=2\\x=y=z=-2\left(L\right)\end{cases}}\)(Vì x + y + z = 6)

Vậy x = y = z = 2

Đặt \(\frac{a}{c}=\frac{c}{b}=k\Rightarrow\hept{\begin{cases}a=ck\\c=bk\end{cases}}\)

\(\Rightarrow\frac{a^2+c^2}{b^2+c^2}=\frac{k^2\left(b^2+c^2\right)}{b^2+c^2}=k^2\)

và \(\frac{a}{b}=\frac{ck}{\frac{c}{k}}=k^2\)

Vậy \(\frac{a^2+c^2}{b^2+c^2}=\frac{a}{b}\)