17/20= 1/...+ 1/...+1/...
So sánh:
a/ \(A=\dfrac{17^{18}+1}{17^{19}+1};B=\dfrac{17^{17}+1}{17^{18}+1}\)
b/ \(A=\dfrac{10^8-2}{10^8+2};B=\dfrac{10^8}{10^8+4}\)
c/ \(A=\dfrac{20^{10}+1}{20^{10}-1};B=\dfrac{20^{10}-1}{20^{10}-3}\)
GIÚP MÌNH VỚI
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!
So sánh
M=\(\dfrac{17^{20}+1}{17^{19}+1}\) và N=\(\dfrac{17^{17}+1}{17^{16}+1}\)
so sánh A=: \(\frac{17^{18}-1}{17^{20}-1}\)Và B= \(\frac{17^{17}-1}{17^{19}-1}\)
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 \(-\frac{288}{17^{20}-1}>-\frac{288}{17^{19}-1}\)
\(\Rightarrow A>B\)
\(A=\frac{17^{20}+2}{17^{20}-1}v\text{à }B=\frac{17^{20}-2}{17^{20}-5}\)
\(A=\frac{17^{20}+2}{17^{20}-1}=\frac{17^{20}-1+3}{17^{20}-1}=1+\frac{3}{17^{20}-1}\)
\(B=\frac{17^{20}-2}{17^{20}-5}=\frac{17^{20}-5+3}{17^{20}-5}=1+\frac{3}{17^{20}-5}\)
Vì \(17^{20}-1>17^{20}-5\)
\(=>\frac{3}{17^{20}-1}1+\frac{3}{17^{20}-1}
Tìm x biết
d) 32%-0,25:x=-17/5
e)(x+1/5)^2+17/25=26/25
f)-32/27-(3x-7/9)^3=-24/27
g)60%x+0,4x+x:3=2
h)|20/9-x|=1/12+1/20+1/30+1/42+1/56+1/72
i)8/5+(2/7+2/17+2/37/5/7+5/17+5/37).x=16/5
Lưu ý: câu i 2/7+2/17+2/37 phần(vế trên) 5/7+5/17+5/37(vế dưới)
d) Ta có: \(32\%-0.25:x=-\dfrac{17}{5}\)
\(\Leftrightarrow0.25:x=\dfrac{8}{25}+\dfrac{17}{5}=\dfrac{93}{25}\)
hay \(x=\dfrac{25}{372}\)
Vậy: \(x=\dfrac{25}{372}\)
e) Ta có: \(\left(x+\dfrac{1}{5}\right)^2+\dfrac{17}{25}=\dfrac{26}{25}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{9}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{3}{5}\\x+\dfrac{1}{5}=-\dfrac{3}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{5}\\x=-\dfrac{4}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{2}{5};-\dfrac{4}{5}\right\}\)
f) Ta có: \(-\dfrac{32}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Leftrightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Leftrightarrow3x-\dfrac{7}{9}=-\dfrac{2}{3}\)
\(\Leftrightarrow3x=\dfrac{1}{9}\)
hay \(x=\dfrac{1}{27}\)
g) Ta có: \(60\%\cdot x+0.4x+x:3=2\)
\(\Leftrightarrow\dfrac{4}{3}x=2\)
hay \(x=\dfrac{3}{2}\)
Vậy: \(x=\dfrac{3}{2}\)
h) PT \(\Leftrightarrow\left|\dfrac{20}{9}-x\right|=\dfrac{2}{9}\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{20}{9}-x=\dfrac{2}{9}\\x-\dfrac{20}{9}=\dfrac{2}{9}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{22}{9}\end{matrix}\right.\)
Vậy ...
i) PT \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}x=\dfrac{16}{5}\) \(\Leftrightarrow\dfrac{2}{5}x=\dfrac{8}{5}\) \(\Leftrightarrow x=4\)
Vậy ...
A=1/6×13+1/13×20+1/20×27+1/17×34
chứng minh rằng S=1/5+1/13+1/25+....+1/19^20^2 nhỏ hơn 17/20
\(\dfrac{1}{10}+\dfrac{17}{20}+\dfrac{3}{10}+\dfrac{2}{20}+\dfrac{6}{10}+\dfrac{1}{20}\)
\(\dfrac{1}{10}+\dfrac{17}{20}+\dfrac{3}{10}+\dfrac{2}{20}+\dfrac{6}{10}+\dfrac{1}{20}\)
= \(\left(\dfrac{1}{10}+\dfrac{3}{10}+\dfrac{6}{10}\right)+\left(\dfrac{17}{20}+\dfrac{2}{20}+\dfrac{1}{20}\right)\)
= 1 + 1 = 2
= ( 1/10 + 3/10 + 6/10 ) + ( 17/20 + 2/20 + 1/20 ) = 1+1 = 2
8. Tính bằng cách thuận tiện nhất
b) 13/17 : 8/3 - 5/17 : 8/3
c) 1/6 + 1/12 + 1/20 + 1/20
viết với tính rõ ra cho mình nhé cảm ơn
b) 13/17 : 8/3 - 5/17 : 8/3
= ( 13/17 - 5/17 ) : 8/3
= 7/17 * 3/8
=21/136
c) 1/6 + 1/12 + 1/20 + 1/20
= (1/6 + 1/12) + (1/20 + 1/20)
= (2/12 + 1/12) + (1/20 + 1/20)
= 3/12 + 2/20
= 1/4 +2/20
= 5/20 + 2/20
= 7/20
\(b)\frac{13}{17}\div\frac{8}{3}-\frac{5}{17}\div\frac{8}{3}\)
\(=\left(\frac{13}{17}-\frac{5}{17}\right)\div\frac{8}{3}\)
\(=\frac{8}{17}\div\frac{8}{3}\)
\(=\frac{3}{17}\)
\(c)\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{20}\)
\(=\left(\frac{2}{12}+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}\right)\)
\(=\frac{1}{4}+\frac{2}{20}\)
\(=\frac{5}{20}+\frac{2}{20}=\frac{7}{20}\)
(13/17 - 5/17) : 8/3 = 8/17 : 8/3 = 3/17
1/6+1/12+1/20 nhân 1=(1/6+1/12)+1/20=1/4 +1/20=6/20=3/10
đây nha em.
Chứng tỏ S=1/16+1/17+1/18+1/29+1/20<1/3
\(S< \dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}+\dfrac{1}{15}=\dfrac{5}{15}=\dfrac{1}{3}\)