76. When did you start your hobby?

77. We find making model very interesting, because we should be creative.

78.I think in the future, people will take up more outdoor activities.

79.My best friend doesn't like mountain climbing because he is afraid of heights.

80. My sister enjoys cooking and making new dishes on her own.

Chúc bạn học tốt!

tự kể màk

`a)M=(x+2)/(xsqrtx-1)+(sqrtx+1)/(x+sqrtx+1)-1/(sqrtx-1)(x>=0,x ne 1)`

`M=(x+2)/((sqrtx-1)(x+sqrtx+1))+((sqrtx+1)(sqrtx-1))/((sqrtx-1)(x+sqrtx+1))-(x+sqrtx+1)/((sqrtx-1)(x+sqrtx+1))`

`M=(x+2+x-1-x-sqrtx-1)/((sqrtx-1)(x+sqrtx+1))`

`M=(x-sqrtx)/((sqrtx-1)(x+sqrtx+1))`

`M=(sqrtx(sqrtx-1))/((sqrtx-1)(x+sqrtx+1))`

`M=sqrtx/(x+sqrtx+1)`

`b)x=25(tmđk)`

`=>sqrtx=5`

`=>M=5/(25+5+1)`

`=>M=5/31`

`c)M=sqrtx/(x+sqrtx+1)`

`x=0=>M=0<1/3`

`x>0=>M=1/(sqrtx+1+1/sqrtx)`

Áp dụng bđt cosi:

`sqrtx+1/sqrtx>=2`

`=>sqrtx+1+1/sqrtx>=3>0`

`=>M<=1/3`

Dấu "=" xảy ra khi `sqrtx=1/sqrtx<=>x=1`(KTMĐKXĐ)

`=>M<1/3`

Vậy `M<1/3`

`d)M=2/7`

`<=>sqrtx/(x+sqrtx+1)=2/7`

`<=>2x+2sqrtx+2=7`

`<=>2x-5sqrtx+2=0`

`<=>2x-4sqrtx-sqrtx+2=0`

`<=>(sqrtx-2)(2sqrtx-1)=0`

`<=>[(sqrtx=2),(2sqrtx=1):}`

`<=>[(x=4),(x=1/4):}(TMĐK)`

`e)` Vì `x>=0=>sqrtx>=0`

`=>x+sqrtx+1>=1>0`

`=>M>=0`

Mặt khác:`M<1/3`(câu b)

`=>M<1=>M-1<0`

`=>M(M-1)<=0`

`<=>M^2-M<=0`

`<=>M^2<=M`

a: Ta có: \(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\)

\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\)

b: Thay x=25 vào M, ta được:

\(M=\dfrac{5}{25+5+1}=\dfrac{5}{31}\)

c: Ta có: \(M-\dfrac{1}{3}\)

\(=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{3}\)

\(=\dfrac{3\sqrt{x}-x-\sqrt{x}-1}{3\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{-\left(x-2\sqrt{x}+1\right)}{3\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{-\left(\sqrt{x}-1\right)^2}{3\left(x+\sqrt{x}+1\right)}< 0\forall x\) thỏa mãn ĐKXĐ

hay \(M< \dfrac{1}{3}\)

d: Để \(M=\dfrac{2}{7}\) thì \(2x+2\sqrt{x}+2=7\sqrt{x}\)

\(\Leftrightarrow2x-5\sqrt{x}+2=0\)

\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(2\sqrt{x}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{1}{4}\left(nhận\right)\end{matrix}\right.\)

a: Xét ΔAHB vuông tại H và ΔCHA vuông tại H có

góc HAB=góc HCA

=>ΔAHB đồng dạng với ΔCHA

b,c: góc FAE+góc FHE=180 độ

=>FAEH nội tiếp

=>góc HFE=góc HAE=góc C

Xét ΔHFE vuông tại H và ΔHCA vuông tại H có

góc HFE=góc HCA

=>ΔHFE đồng dạng với ΔHCA

=>HF/HC=HE/HA

=>HF*HA=HC*HE