ai chả lời mình tích cho.

\(\dfrac{4}{5};\dfrac{5}{6};\dfrac{7}{6};\dfrac{5}{4}\)

Answer:

ĐK: \(x\ne0;x\ne-1\)

\(\frac{1}{x^2}+\frac{3}{x+1}-\frac{2}{\left(x+1\right)^2}=2\)

\(\Leftrightarrow\frac{1}{x^2}-\frac{1}{\left(x+1\right)^2}+\frac{3}{x+1}-\frac{1}{\left(x+1\right)^2}-2=0\)

\(\Leftrightarrow\frac{\left(x+1\right)^2-x^2}{x^2.\left(x+1\right)^2}+\frac{3\left(x+1\right)-1-2\left(x+1\right)^2}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow\frac{2x+1}{x^2.\left(x+1\right)^2}-\frac{x\left(2x+1\right)}{\left(x+1\right)^2}=0\)

\(\Leftrightarrow\frac{2x+1}{x^2.\left(x+1\right)^2}-\frac{x^3.\left(2x+1\right)}{x^2.\left(x+1\right)^2}=0\)

\(\Leftrightarrow\frac{\left(2x+1\right)\left(1-x^3\right)}{x^2.\left(x+1\right)^2}=0\)

\(\Leftrightarrow\left(2x+1\right)\left(1-x^3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\1-x^3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}}\)

 y x 11,27= 11,25 x 8,4

 y x 11,27= 94,5

y             =94,5:11,27

y             = 8,39(xấp xỉ nhé)

HT

 y x 11,27= 11,25 x 8,4

 y x 11,27= 94,5

y             =94,5:11,27

y             = 8,39(làm tròn)

HT

a: Xét tứ giác AEHF có

\(\widehat{AEH}+\widehat{AFH}=180^0\)

Do đó: AEHF là tứ giác nội tiếp

b: Xét ΔADC vuông tại D và ΔAEH vuông tại E có 

\(\widehat{EAH}\) chung

DO đó: ΔADC\(\sim\)ΔAEH

Suy ra: AD/AE=AC/AH

hay \(AD\cdot AH=AE\cdot AC\)