39/38 >1 mà 93/94 < 1

nên 39/38>93/94

\(\frac{39}{38}>\frac{93}{94}\)

\(\frac{43}{49}< \frac{93}{94}\)

a)Ta có : \(\frac{39}{38}>1>\frac{93}{94}\)

Vậy \(\frac{39}{38}>\frac{93}{94}\)

b) Ta có : MSC là 4606

      \(\frac{43}{49}=\frac{4042}{4606}va\frac{93}{94}=\frac{4557}{4606}\)

Mà \(\frac{4042}{4606}< \frac{4557}{4606}\)

Vậy \(\frac{43}{49}< \frac{93}{94}\)

Chúc bạn học giỏi

Đừng quên tích chi mình nha ! ^-^

A=[(38+17)+125]+29

A=175+29

A=204

B=312.3

B=936

C=1+1+1+1+1

C=5

D=(125:5).(36:9)

D=25.4

D=100

28,16+7,93+4,05 = 40,14

6,7+19,74+38=64,44

0,92+0,77+0,64=2,33

28 ,16 + 7 ,93 + 4 , 05=36,09+4,05=40,14

6 ,7 + 19 ,74 + 38=25,44+38=64,44

0 ,92 + 0 ,77 + 0 ,64 =1,69+0,64=2,33

Rút gọn.

\(B=\dfrac{x^{39}x^{36}x^{33}...x^31}{x^{40}x^{38}x^{36}...x^21}=\dfrac{x^{\left(39+36+33+...+3\right)}}{x^{\left(40+38+36+...+2\right)}}\)

ta có: \(39+36+33+...+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

=> \(B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

Tương tự như B => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Ta có:

\(B=\dfrac{x^{\left(39+36+33+....+3\right)}}{x^{\left(40+38+36+....+2\right)}}\)

\(39+36+33+....+3=\dfrac{\left(39+3\right)\left(\dfrac{39-3}{3}+1\right)}{2}=273\)

\(40+38+36+....+2=\dfrac{\left(40+2\right)\left(\dfrac{40-2}{2}+1\right)}{2}=420\)

\(\Rightarrow B=\dfrac{x^{273}}{x^{420}}=\dfrac{1}{x^{147}}\)

tương tự => \(A=\dfrac{x^{4560}}{x^{496}}=x^{4064}\)

Ta có : \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)

\(\Leftrightarrow\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+93}{7}-\dfrac{2x+94}{6}-\dfrac{2x+95}{5}=0\)

\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1-\dfrac{2x+93}{7}-1-\dfrac{2x+94}{6}-1-\dfrac{2x+95}{5}-1=0\)

\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}-\dfrac{2x+100}{7}-\dfrac{2x+100}{6}-\dfrac{2x+100}{5}=0\)

\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)

Thấy : \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)

\(\Rightarrow2x+100=0\)

\(\Leftrightarrow x=-50\)

Vậy ...

Ta có: \(\dfrac{2x+5}{95}+\dfrac{2x+6}{94}+\dfrac{2x+7}{93}=\dfrac{2x+93}{7}+\dfrac{2x+94}{6}+\dfrac{2x+95}{5}\)

\(\Leftrightarrow\dfrac{2x+5}{95}+1+\dfrac{2x+6}{94}+1+\dfrac{2x+7}{93}+1=\dfrac{2x+93}{7}+1+\dfrac{2x+94}{6}+1+\dfrac{2x+95}{5}+1\)

\(\Leftrightarrow\dfrac{2x+100}{95}+\dfrac{2x+100}{94}+\dfrac{2x+100}{93}=\dfrac{2x+100}{7}+\dfrac{2x+100}{6}+\dfrac{2x+100}{5}\)

\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}\right)=\left(2x+100\right)\left(\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\right)\)

\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}\right)-\left(2x+100\right)\left(\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{5}\right)=0\)

\(\Leftrightarrow\left(2x+100\right)\left(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\right)=0\)

mà \(\dfrac{1}{95}+\dfrac{1}{94}+\dfrac{1}{93}-\dfrac{1}{7}-\dfrac{1}{6}-\dfrac{1}{5}\ne0\)

nên 2x+100=0

\(\Leftrightarrow2x=-100\)

hay x=-50

Vậy: S={-50}

\(=\dfrac{9^3\cdot\left(2^3\cdot3+45\right)}{9^3}=8\cdot3+45=69\)

hợp lí nhất là tính trong ngoặc trước:)