Sai đề nhé.

Sửa: Cho \(x^2=yz\). CMR:\(\frac{x^2+y^2}{x^2+z^2}=\frac{y}{z}\left(x;y;z\ne0\right)\)

Ta có: \(x^2=yz\)

\(\Rightarrow\frac{yz+y^2}{yz+z^2}=\frac{y\left(y+z\right)}{z\left(y+z\right)}=\frac{y}{z}\)( vì \(y+z\ne0\))

                                          đpcm

Tham khảo nhé~

Tu B ve duong thang Bz song song voi Ax

nen \(\widehat{ABz}=\widehat{A}\)(so le trong; Bz//Ax)

ma \(\widehat{A}+\widehat{C}=\widehat{ABC}\)(gt)

nen \(\widehat{ABz}+\widehat{C}=\widehat{ABC}\)

mat khac \(\widehat{ABz}+\widehat{CBz}=\widehat{ABC}\)(BZ nam giua BA va BC)

nen \(\widehat{C}=\widehat{CBz}\)

ma 2 goc do o vi tri so le trong

nen Bz//Cy

ma Bz//Ax

vay Ax//Cy

\(a.A=2x-3-\left|3x-2\right|.\)

\(=2x-3-\left(3x-2\right)\)

\(=2x-3-3x+2\)

\(=-x-1\)

\(b.A=-x-1=5\)

\(-x=5+1=6\)

\(-x=6\)

\(\Rightarrow x=-6\)

Ta có :

\(\frac{x}{3}=\frac{y}{5}\)  (1)

\(\left(x+2\right)+\left(y+3\right)=21\)

\(\Leftrightarrow x+2+y+3=21\)

\(\Leftrightarrow x+y=16\)

\(\Leftrightarrow x=16-y\)

Thay x = 16 - y vào (1) ta được :

\(\frac{16-y}{3}=\frac{y}{5}\)

\(\Rightarrow5\left(16-y\right)=3y\)

\(\Leftrightarrow80-5y=3y\)

\(\Leftrightarrow80=8y\)

\(\Leftrightarrow y=10\)

\(\Rightarrow x=16-10=6\)

Vậy x = 6 và y = 10