\(\frac{5}{17}.\frac{16}{33}+\frac{5}{33}.\frac{20}{17}-\frac{2}{17}.\frac{5}{33}\)

\(=\frac{5}{17}.\frac{16}{33}+\frac{5}{17}.\frac{20}{33}-\frac{5}{17}.\frac{2}{33}\)

\(=\frac{5}{17}.\left(\frac{16}{33}+\frac{20}{33}-\frac{2}{33}\right)\)

\(=\frac{5}{17}.\frac{34}{33}\)

\(=\frac{10}{33}\)

Chúc bạn học tốt !!! 

\(\left(1\frac{1}{4}-\frac{3}{5}\right):\frac{17}{20}< \frac{x}{17}< \left(5\frac{1}{3}-3\frac{1}{2}\right).\frac{12}{17}\)

= \(\left(\frac{5-3}{4}\right):\frac{17}{20}< \frac{x}{17}< \left(\frac{16}{3}-\frac{7}{2}\right).\frac{12}{17}\)

= \(\frac{1}{2}:\frac{17}{20}< \frac{x}{17}< \left(\frac{32-21}{6}\right).\frac{12}{17}\)

= \(\frac{10}{17}< \frac{x}{17}< \frac{3}{2}.\frac{12}{17}\)

= \(\frac{10}{17}< \frac{x}{17}< \frac{18}{17}\)

( Mik thấy mẫu giống nhau mik sẽ bỏ mẫu đi mik sẽ tìm tử )

=> 10 < 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 < 18

=> x = { 11 ; 12 ; 13 ; 14 ; 15 ; 16 ; 17 }

k mik nha làm ơn đó

áp dụng tính chất \(\frac{a}{b}< 1\Rightarrow\frac{a+m}{b+m}< 1\left(m\in N\right)\)

Ta có: \(A=\frac{17^{18}-1}{17^{20}-1}< \frac{17^{18}-1-16}{17^{20}-1-16}\)\(=\frac{17^{18}-17}{17^{20}-17}=\frac{17.\left(17^{17}-1\right)}{17.\left(17^{19}-1\right)}\)\(=\frac{17^{17}-1}{17^{19}-1}\)

\(\Rightarrow A< B\)

\(A=\frac{17^{18}-1}{17^{20}-1}\Rightarrow17^2A=\frac{17^{18}-1}{17^{18}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}\left(1\right)\)

\(B=\frac{17^{17}-1}{17^{19}-1}\Rightarrow17^2B=\frac{17^{17}-1}{17^{17}-\frac{1}{17^2}}=1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(2\right)\)

\(\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}< \frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\Rightarrow1-\frac{1-\frac{1}{17^2}}{17^{18}-\frac{1}{17^2}}>1-\frac{1-\frac{1}{17^2}}{17^{17}-\frac{1}{17^2}}\left(3\right)\)

Từ \(\left(1\right);\left(2\right)\&\left(3\right)\Rightarrow17^2A>17^2B\Leftrightarrow A>B.\)

\(A=\frac{17^{18}-1}{17^{20}-1}\)

\(17^2A=\frac{17^2\left(17^{18}-1\right)}{17^{20}-1}=\frac{17^{20}-17^2}{17^{20}-1}=\frac{17^{20}-1-288}{17^{20}-1}=1-\frac{288}{17^{20}-1}\)

\(B=\frac{17^{17}-1}{17^{19}-1}\)

\(17^2B=\frac{17^2\left(17^{17}-1\right)}{17^{19}-1}=\frac{17^{19}-17^2}{17^{19}-1}=\frac{17^{19}-1-288}{17^{19}-1}=1-\frac{288}{17^{19}-1}\)

Ta có : \(\frac{288}{17^{20}-1}< \frac{288}{17^{19}-1}\)nên \(-\frac{288}{17^{20}-1}>-\frac{288}{17^{19}-1}\)

\(\Rightarrow A>B\)

ta có A=\(\frac{17^{18}+1}{17^{19}+1}\)<\(\frac{17^{18}+1+16}{17^{19}+1+16}\) (nếu a/b<1 thì a+c/b+c>a/b)

A<\(\frac{17\left(17^{17}+1\right)}{17\left(17^{18}+1\right)}\)

A,<\(\frac{17^{17}+1}{17^{18}+1}\)=B

hay A<B

\(A=\frac{17^{18}+1}{17^{19}+1}\) với \(B=\frac{17^{17}+1}{17^{18}+1}\)

Ta có :B=\(\frac{17^{17}+1}{17^{18}+1}=\frac{17^{18}+17}{17^{19}+17}\)

Ta có:1-B=\(1-\frac{17^{18}+17}{17^{19}+17}=\frac{17^{19}+17-17^{18}-17}{17^{19}+17}=\frac{17^{19}-17^{18}}{17^{19}+17}\)

         1-A=1-\(\frac{17^{18}+1}{17^{19}+1}=\frac{17^{19}+1-17^{18}-1}{17^{19}+1}=\frac{17^{19}-17^{18}}{17^{19}+1}\)

Do \(17^{19}+1< 17^{19}+10\Rightarrow1-A>1-B\)

                  \(\Rightarrow A< B\)

Ta có:\(\frac{17}{21}+\frac{17}{20}+\frac{17}{19}>\frac{17}{21}+\frac{17}{21}+\frac{17}{21}\)

Mà :\(\frac{17}{21}+\frac{17}{21}+\frac{17}{21}=\frac{51}{21}>\frac{42}{21}=2\)

\(\Rightarrow\frac{17}{21}+\frac{17}{20}+\frac{17}{19}>2\left(đpcm\right)\)

Chúc Bạn Học Tốt (Tks PP)

Giải:

a) A=17¹⁸+1/17¹⁹+1

17A=17¹⁹+17/17¹⁹+1

17A=17¹⁹+1+16/17¹⁹+1

17A=1+16/17¹⁹+1

Tương tự:

B=17¹⁷+1/17¹⁸+1

17B=17¹⁸+17/17¹⁸+1

17B=17¹⁸+1+16/17¹⁸+1

17B=1+16/17¹⁸+1

Vì 16/17¹⁹+1<16/17¹⁸+1 nên 17A<17B

⇒A<B

b) A=10⁸-2/10⁸+2

    A=10⁸+2-4/10⁸+2

    A=1+-4/10⁸+2

Tương tự:

B=10⁸/10⁸+4

B=10⁸+4-4/10⁸+1

B=1+-4/10⁸+1

Vì -4/10⁸+2>-4/10⁸+1 nên A>B

c)A=20¹⁰+1/20¹⁰-1

   A=20¹⁰-1+2/20¹⁰-1

   A=1+2/20¹⁰-1

Tương tự:

B=20¹⁰-1/20¹⁰-3

B=20¹⁰-3+2/20¹⁰-3

B=1+2/20¹⁰-3

Vì 2/20¹⁰-3>2/20¹⁰-1 nên B>A

⇒A<B

Chúc bạn học tốt!

\(=\dfrac{57}{20}-\dfrac{26}{15}+6.45:3\)

\(=\dfrac{58\cdot3-26\cdot2}{60}+\dfrac{43}{20}\)

\(=\dfrac{122}{60}+\dfrac{129}{60}=\dfrac{251}{60}\)