\(\hept{\begin{cases}\frac{2}{x-y}+\sqrt{y+1}=4\\\frac{1}{x-y}-3\sqrt{y+1}=-5\end{cases}}\) 

ĐKXĐ : \(x\ne y,y\ge-1\)

Ta có : \(3.\left(\frac{2}{x-y}+\sqrt{y+1}\right)=12\)

Cộng với phương trình ( 2 ) ta có :

\(\frac{7}{x-y}=7\Rightarrow\frac{1}{x-y}=1\Rightarrow x=y=1\)

Thay vào hệ phương trình ta có :

\(\hept{\begin{cases}2+\sqrt{x}=4\\1-3.\sqrt{x}=-5\end{cases}}\)

Cộng pt 1 và pt 2 ta có : \(3-2.\sqrt{x}=-1\)\(\Rightarrow2.\sqrt{x}=4\)\(\Rightarrow\sqrt{x}=2\Rightarrow x=4\)

= > y = 3

-19684 nha!

x:3/7=13/15-2/5

x:3/7=7/15

      x=7/15x3/7

      x=1/5

chúc bạn học tốt

\(x\times\frac{4}{7}+\frac{1}{3}\times x=\frac{2}{3}\)

\(x\times\left(\frac{4}{7}+\frac{1}{3}\right)=\frac{2}{3}\)

\(x\times\frac{19}{21}=\frac{2}{3}\)

\(x=\frac{2}{3}:\frac{19}{21}\)

\(x=\frac{2}{3}\times\frac{21}{19}=\frac{14}{19}\)

cảm ơn bé sói nha

999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999999

câu hỏi đâu bạn

`Answer:`

Sửa đề phần c: Chứng minh KF//BC.

C H B A F K

a. Xét `\triangleAHB` và `\triangleAHC`

`AH` chung

`\hat{AHB}=\hat{AHC}=90^o`

`AB=AC`

`=>\triangleAHB=\triangleAHC(ch-cgv)`

b. Xét `\triangleFAH` và `\triangleKAH`

`AH` chung

`\hat{FAH}=\hat{KAH}`

`\hat{AFH}=\hat{AKH}=90^o`

`=>\triangleFAH=\triangleKAH(ch-gn)`

`=>HK=HF`

c. Theo phần b. `\triangleFAH=\triangleKAH`

`=>AF=AK`

`=>\triangleAFK` cân ở `A`

Ta có: `\triangleAFK` cân ở `A` và `\triangleABC` cân ở `A`

`=>\hat{AFK}=\hat{ABC}` mà hai góc này ở vị trí đồng vị \(\Rightarrow KF//BC\)

hình tự vẽ nhé.

xét: \(\Delta AHB\) VÀ   \(\Delta AHC\) CÓ:

\(\widehat{ABH}=\widehat{ACH}\)(DO TAM GIÁC ABC CÂN TẠI A)

\(AB=AC\)(DO TAM GIÁC ABC CÂN TẠI A)
\(\widehat{AHB}=\widehat{AHC}=90^0\)

\(\Rightarrow\Delta AHB=\Delta AHC\left(ch-gn\right)\left(1\right)\)

b) TỪ (1)\(\Rightarrow BH=CH\)(2 cạnh tương ứng)

XÉT: \(\Delta KBH\)VÀ    \(\Delta FCH\) CÓ:​​

\(BH=CH\left(cmt\right)\)

​​\(\widehat{BKH}=\widehat{CFH}=90^0\)

\(\widehat{KBH}=\widehat{FCH}\left(\widehat{B}=\widehat{C}\right)\)

​\(\Rightarrow\Delta KBH=\Delta FCH\left(ch-gn\right)\)

\(\Rightarrow HK=HF;BK=FC\)(2 cạnh tương ứng)(đpcm)

c) ta có:  \(AB=AC;;BK=FK\left(cmt\right)\)

\(\Rightarrow AB-BK=AC-FC\)

\(\Rightarrow AK=AF\Rightarrow\Delta AKF\) cân tại A

\(\Rightarrow\widehat{AKF}=\frac{180^0-\widehat{A}}{2}\left(2\right)\)

lại có \(\Delta ABC\)cân tại A\(\Rightarrow\widehat{ABC}=\frac{180^0-\widehat{A}}{2}\left(3\right)\)

TỪ (2)VÀ (3)\(\Rightarrow\widehat{AKF}=\widehat{ABC}\left(=\frac{180^0-\widehat{A}}{2}\right)\)

​mà 2 góc này ở vị trí đồng vị \(\Rightarrow KF\\ BC\left(đpcm\right)\)