bạn cần nói rõ đề hơn nhé

3 dấu giá trị tuyệt đối là sao

còn=)) kb đi:)

Đặt \(\frac{x}{z}=\frac{z}{y}=k\)

\(\Rightarrow\hept{\begin{cases}x=zk\\z=yk\end{cases}}\)

Khi đó : \(\frac{x^2+z^2}{y^2+z^2}=\frac{\left(zk\right)^2+z^2}{y^2+\left(yk\right)^2}=\frac{z^2\left(k^2+1\right)}{y^2\left(k^2+1\right)}=\frac{z^2}{y^2}=\frac{\left(y.k\right)^2}{y^2}=k^2\)

\(\frac{x}{y}=\frac{y.k^2}{y}=k^2\)

=> \(\frac{x^2+z^2}{y^2+z^2}=\frac{x}{y}\left(\text{đpcm}\right)\)

\(\frac{x}{z}=\frac{z}{y}\)

cmr: \(\left(\frac{x^2+z^2}{y^2+z^2}\right)=\frac{x}{y}\)

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

áp dụng t/c dãy tỉ số = nhau

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

vì \(\left(2\right)\frac{x}{z}=\frac{z}{y}\Rightarrow\frac{x}{y}=\frac{z}{z}\)

từ (1) và (2) =>\(\left(\frac{x^2+z^2}{y^2+z^2}\right)=\frac{x}{y}\)

\(26\frac{1}{5}-44\frac{1}{5}=26+\frac{1}{5}-\left(44+\frac{1}{5}\right)\)

\(=26+\frac{1}{5}-44-\frac{1}{5}=26-44=-18\)

\(=\frac{131}{5}-\frac{221}{5}\)

\(=-\frac{90}{5}\)

\(=-18\)

\(a^2=bc\)

\(\Leftrightarrow2a^2=2bc\)

\(\Leftrightarrow ac-ab+2a^2=2bc+ac-ab\)

\(\Leftrightarrow ac+bc-a^2-ab=a^2-ab+ac-bc\)

\(\Leftrightarrow c\left(a+b\right)-a\left(a+b\right)=a\left(a-b\right)+c\left(a-b\right)\)

\(\Leftrightarrow\left(c-a\right)\left(a+b\right)=\left(a+c\right)\left(a-b\right)\)

\(\Leftrightarrow\frac{a+b}{a-b}=\frac{a+c}{c-a}\)

Ta có: a²=b.c

->  a/c=b/a = a+b/c+a = a-b/c-a

-> a+b/a-b= c+a/c-a

\(\Rightarrow x.8=-2.5\)

\(\Rightarrow x.8=-10\)

\(\Rightarrow x=-\frac{5}{4}\)

\(\frac{x}{-2}=\frac{5}{8}\)

\(\Leftrightarrow8x=-10\)

\(\Leftrightarrow x=\frac{-5}{4}\)