1. Với một khẩu phần ăn dầy đủ chất dinh dưỡng dược chứa là gì?

Khẩu phần ăn đầy đủ chất dinh dưỡng bao gồm: - Cacbohydrat 

                                                                       - protein 

                                                                       - lipit 

                                                                       - axit nucleic 

                                                                       - vitamin 

                                                                       - muối khoáng

                                                                       - chất xơ

2. Ở ruột non đã có những enzim gì để tiêu hóa những chất đó?

Enzyme ở ruột non tiêu hóa các chất:

- Amylase tuyến tụy: tiêu hóa các disaccarit

- Trypsin, chymotripsin, Carboxypeptitdase: tiêu hóa các chuỗi polipeptit

- Nuclease: Tiêu hóa axit nucleic

- Lipase: Tiêu hóa lipit

ahihi

a, \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}\right):\frac{2x}{5x-5}\)

\(=\left(\frac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x}{5x-5}\)

\(=\left(\frac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}\right):\frac{2x}{5\left(x-1\right)}\)

\(=\frac{4x}{\left(x-1\right)\left(x+1\right)}.\frac{5\left(x-1\right)}{2x}=\frac{10}{x+1}\)

b, Thay x = 3/4 ta cos : 

\(\frac{10}{\frac{3}{4}+1}=\frac{10}{\frac{7}{4}}=\frac{5}{14}\)

Căng =( mình làm mấy ý ez thôi nhá, còn mấy ý kia tương tự 

a, \(3x-2\left(5+2x\right)=45-2x\)

\(\Leftrightarrow3x-10-4x=45-2x\)

\(\Leftrightarrow-x-10=45-2x\Leftrightarrow x=55\)

d, \(\left(x-1\right)\left(5x+3\right)=\left(3x-8\right)\left(x-1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)-\left(3x-8\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+3-3x+8\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x+11\right)=0\Leftrightarrow x=1;-\frac{11}{2}\)

e, \(\left(x-1\right)\left(x^2+5x-2\right)-\left(x^3-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2\right)-\left(x-1\right)\left(x^2+x+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+5x-2-x^2-x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(4x-3\right)=0\Leftrightarrow x=1;\frac{3}{4}\)

a) 3x - 2(5 + 2x) =45 - 2x

=> 3x - 10 - 4x = 45 - 2x

=> 3x - 4x + 2x = 45 + 10

=> x = 55

b) \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

=> \(\frac{x-3}{5}=\frac{2x+17}{3}\)

=> 5(2x + 17) = 3(x - 3)

=> 10x + 85 = 3x - 9

=> 7x = -94

=> x = -94/7

c) \(\frac{5\left(x-1\right)+2}{6}-\frac{7x-1}{4}=\frac{2\left(2x+1\right)}{7}-5\)

=> \(\frac{5x-3}{6}-\frac{7x-1}{4}=\frac{4x-33}{7}\)

=> \(\frac{10x-6}{12}-\frac{21x-3}{12}=\frac{4x-33}{7}\)

=> \(\frac{-11x-3}{12}=\frac{4x-33}{7}\)

=> (-11x - 3).7 = (4x - 33).12

= -77x - 21 = 48x - 396

=> x = 3

d) (x - 1)(5x + 3) = (3x - 8)(x - 1)

=> (x - 1)(5x + 3) - (3x - 8)(x -1) = 0

=> (x - 1)(2x + 11) = 0

=> \(\orbr{\begin{cases}x-1=0\\2x+11=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=-5,5\end{cases}}\) 

e) (x - 1)(x² + 5x - 2) - (x³ - 1) = 0

=> (x - 1)(x² + 5x - 2) - (x - 1)(x² + x + 1) = 0

=> (x - 1)(4x - 3) = 0

=> \(\orbr{\begin{cases}x-1=0\\4x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=1\\x=0,75\end{cases}}\)

f) \(\frac{x-17}{33}+\frac{x-21}{29}+\frac{x}{25}=4\) 

=> \(\left(\frac{x-17}{33}-1\right)+\left(\frac{x-21}{29}-1\right)+\left(\frac{x}{25}-2\right)=0\)

=> \(\frac{x-50}{33}+\frac{x-50}{29}+\frac{x-50}{25}=0\)

=> \(\left(x-50\right)\left(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\right)=0\)

=> x - 50 = 0 (Vì \(\frac{1}{33}+\frac{1}{29}+\frac{1}{25}\ne0\))

=> x = 50

b, \(\frac{x-3}{5}=6-\frac{1-2x}{3}\)

\(\Leftrightarrow\frac{x-3}{5}=\frac{17+2x}{3}\Leftrightarrow3x-9=85+10x\)

\(\Leftrightarrow-7x=94\Leftrightarrow x=-\frac{94}{7}\)

f, sửa : \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)

\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)

\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)

\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\right)=0\)

\(\Leftrightarrow x=-66\)