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Vì 18/91 < 18/90 =1/5
23/114>23115=1/5
vậy 18/91<1/5<23/114
suy ra 18/91<23/114
vì 21/52=210/520
Mà 210/520=1-310/520
213/523=1-310/523
310/520>310/523
vậy 210/520<213/523
suy ra 21/52<213/523
BÀi 1
Để A \(\in\) Z
=>\(\left(n+2\right)⋮\left(n-5\right)\)
=>\([\left(n-5\right)+7]⋮\left(n-5\right)\)
=>\(7⋮\left(n-5\right)\)
=>\(n-5\in\left\{1;7;-1;-7\right\}\)
=>\(n\in\left\{6;13;4;-2\right\}\)
Vậy \(n\in\left\{6;13;4;-2\right\}\)
a, \(A-B=\frac{3}{8^3}+\frac{7}{8^4}-\frac{7}{8^3}-\frac{3}{8^4}==\left(\frac{7}{8^4}-\frac{3}{8^4}\right)-\left(\frac{7}{8^3}-\frac{3}{8^3}\right)=\frac{4}{8^4}-\frac{4}{8^3}< 0\)
Vậy A < B
b, \(A=\frac{10^7+5}{10^7-8}=\frac{10^7-8+13}{10^7-8}=1+\frac{13}{10^7-8}\)
\(B=\frac{10^8+6}{10^8-7}=\frac{10^8-7+13}{10^8-7}=1+\frac{13}{10^8-7}\)
Vì \(10^7-8< 10^8-7\Rightarrow\frac{1}{10^7-8}>\frac{1}{10^8-7}\Rightarrow\frac{13}{10^7-8}>\frac{13}{10^8-7}\Rightarrow A>B\)
c,Áp dụng nếu \(\frac{a}{b}>1\Rightarrow\frac{a}{b}>\frac{a+n}{a+n}\) có:
\(B=\frac{10^{1993}+1}{10^{1992}+1}>\frac{10^{1993}+1+9}{10^{1992}+1+9}=\frac{10^{1993}+10}{10^{1992}+10}=\frac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}=\frac{10^{1992}+1}{10^{1991}+1}=A\)
Vậy A < B
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
a) \(\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{6}\right)\left(1-\dfrac{1}{10}\right)...\left(1-\dfrac{1}{780}\right)\)
\(=\dfrac{2}{3}.\dfrac{5}{6}.\dfrac{9}{10}.....\dfrac{779}{780}\)\(=\)
a , \(\left(\dfrac{-2}{3}+1\dfrac{1}{4}-\dfrac{1}{6}\right):\dfrac{-24}{10}\)
=\(\left(\dfrac{-2}{3}+\dfrac{5}{4}-\dfrac{1}{6}\right):\dfrac{-12}{5}\)
=\(\left(\dfrac{-8}{12}+\dfrac{15}{12}-\dfrac{2}{12}\right)\cdot\dfrac{-5}{12}\)
=\(\dfrac{5}{12}\cdot\dfrac{-5}{12}=\dfrac{-25}{144}\)
b , \(\dfrac{13}{15}\cdot0,25\cdot3+\left(\dfrac{8}{15}-1\dfrac{19}{60}\right)1\dfrac{23}{24}\)
=\(\dfrac{13}{15}\cdot\dfrac{1}{4}\cdot3+\left(\dfrac{8}{15}-\dfrac{79}{60}\right)\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{47}{60}\cdot\dfrac{57}{24}\)
=\(\dfrac{13}{20}-\dfrac{893}{480}=\dfrac{312}{480}-\dfrac{893}{480}=\dfrac{-581}{480}\)
c , \(\left(\dfrac{12}{32}+\dfrac{5}{-20}-\dfrac{10}{24}\right):\dfrac{2}{3}\)
=\(\left(\dfrac{180}{480}-\dfrac{120}{480}-\dfrac{200}{480}\right)\cdot\dfrac{3}{2}\)
= \(\dfrac{-7}{24}\cdot\dfrac{3}{2}=\dfrac{-7}{16}\)
d , \(4\dfrac{1}{2}:\left(2,5-3\dfrac{3}{4}\right)+\left(-\dfrac{1}{2}\right)\)
=\(\dfrac{9}{2}:\left(\dfrac{5}{2}-\dfrac{15}{4}\right)-\dfrac{1}{2}\)
=\(\dfrac{9}{2}:\dfrac{-5}{4}-\dfrac{1}{2}=\dfrac{9}{2}\cdot\dfrac{-4}{5}-\dfrac{1}{2}=\dfrac{-18}{5}-\dfrac{1}{2}=\dfrac{-41}{10}\)
e , \(\dfrac{-5}{2}:\left(\dfrac{3}{4}-\dfrac{1}{2}\right)=\dfrac{-5}{2}\left(\dfrac{3}{4}-\dfrac{2}{4}\right)\)
=\(\dfrac{-5}{2}:\dfrac{1}{4}=\dfrac{-5}{2}\cdot4=-10\)
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
2) Tinh nhanh:
a) \(\dfrac{5}{23}\) . \(\dfrac{17}{26}\) + \(\dfrac{5}{23}\) . \(\dfrac{10}{26}\) - \(\dfrac{5}{23}\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{17}{26}+\dfrac{10}{26}-1\right)\)
= \(\dfrac{5}{23}\) . \(\left(\dfrac{27}{26}-1\right)\) = \(\dfrac{5}{23}\) . \(\dfrac{1}{26}\)
= \(\dfrac{5}{598}\)
b) \(\dfrac{1}{7}.\dfrac{5}{9}+\dfrac{5}{9}.\dfrac{2}{7}+\dfrac{5}{9}.\dfrac{1}{7}+\dfrac{5}{9}.\dfrac{3}{7}\)
= \(\dfrac{5}{9}.\left(\dfrac{1}{7}+\dfrac{2}{7}+\dfrac{1}{7}+\dfrac{3}{7}\right)\)
= \(\dfrac{5}{9}\) . 1= \(\dfrac{5}{9}\)
1.
\(\dfrac{19.20}{19+20}=\dfrac{380}{39}=9\dfrac{29}{39}\)
\(\dfrac{\overline{aaa}}{\overline{aa}}=\dfrac{111.a}{11.a}=\dfrac{111}{11}=10\dfrac{1}{11}\)
\(\dfrac{\overline{ababa}}{\overline{aba}}=\dfrac{100.\overline{aba}+\overline{ba}}{\overline{aba}}=\dfrac{100.\overline{aba}}{\overline{aba}}+\dfrac{\overline{ba}}{\overline{aba}}=100\dfrac{\overline{ba}}{\overline{aba}}\)
2.
\(6\dfrac{23}{41}=\dfrac{6.41+23}{41}=\dfrac{269}{41}\)
\(a\dfrac{a}{99}=\dfrac{a.99+a}{99}=\dfrac{100.a}{99}=\dfrac{\overline{a00}}{99}\)
\(1\dfrac{a-b}{a+b}=\dfrac{a+b+a-b}{a+b}=\dfrac{2.a}{a+b}\)
3.
\(\dfrac{69}{1000}=0,069\)
\(8\dfrac{77}{100}=8,77\)
\(\dfrac{34567}{10^4}=\dfrac{34567}{10000}=3,4567\)
\(\dfrac{\overline{abc}}{10^n}=\dfrac{\overline{abc}}{10...0}=\overline{0,0...0abc}\)
n số hạng 0 n - 3 số hạng 0 ở phần thập phân
\(n\left(n+3\right)=n^2+3n\)
\(\left(n+2\right)\left(n+1\right)=n^2+3n+2\)
Vì \(n^2+3n< n^2+3n+2\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\left(n\in N\right)\)
b) \(\dfrac{n}{2n+1}=\dfrac{3n}{6n+3}< \dfrac{3n+1}{6n+3}\)
c) \(\dfrac{10^8+2}{10^8-1}=1+\dfrac{1}{10^8-1}\)
\(\dfrac{10^8}{10^8-3}=\left(1+\dfrac{3}{10^8-3}\right)\)
Vì \(\dfrac{1}{10^8-1}>\dfrac{3}{10^8-3}\Rightarrow\dfrac{10^8+2}{10^8-1}< \dfrac{10^8}{10^8-3}\)
Làm dần dần và làm từ từ, suy ra được nhiều cách giải.
a) \(\dfrac{n}{n+1}\) và \(\dfrac{n+2}{n+3}\)
+ Cách 1:
\(\dfrac{n}{n+1}=\dfrac{n+1-1}{n+1}=1-\dfrac{1}{n+1}\)
\(\dfrac{n+2}{n+3}=\dfrac{n+3-1}{n+3}=1-\dfrac{1}{n+3}\)
Vì \(\dfrac{1}{n+1}>\dfrac{1}{n+3}\) nên \(1-\dfrac{n}{n+1}< 1-\dfrac{1}{n+3}\)
\(\Rightarrow\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\)
+ Cách 2:
Ta so sánh: \(n\left(n+3\right)\) và \(\left(n+1\right)\left(n+2\right)\)
\(n\left(n+3\right)=nn+3n=n^2+3n\)
\(\left(n+1\right)\left(n+2\right)=\left(n+1\right)n+\left(n+1\right).2=n^2+n+2n+2=n^2+3n+2\)
Vì \(n^2+3n< n^2+3n+2\) nên \(\dfrac{n}{n+1}< \dfrac{n+2}{n+3}\)
b) \(\dfrac{n}{2n+1}\) và \(\dfrac{3n+1}{6n+3}\)
Ta so sánh: \(n\left(6n+3\right)\) và \(\left(2n+1\right)\left(3n+1\right)\)
\(n\left(6n+3\right)=n.6n+3n=6n^2+3n\)
\(\left(2n+1\right)\left(3n+1\right)=\left(2n+1\right)3n+\left(2n+1\right)=6n^2+3n+2n+1=6n^2+5n+1\)
Vì \(6n^2+3n< 6n^2+5n+1\) nên \(\dfrac{n}{2n+1}< \dfrac{3n+1}{6n+3}\)
c) \(\dfrac{10^8+2}{10^8-1}\) và \(\dfrac{10^8}{10^8-3}\)
\(\dfrac{10^8+2}{10^8-1}=\dfrac{10^8-1+3}{10^8-1}=1+\dfrac{3}{10^8-1}\)
\(\dfrac{10^8}{10^8-3}=\dfrac{10^8-3+3}{10^8-3}=1+\dfrac{3}{10^8-3}\)
Vì \(\dfrac{3}{10^8-1}>\dfrac{3}{10^8-3}\) nên \(\dfrac{10^8+2}{10^8-1}>\dfrac{10^8}{10^8-3}\)
d) \(\dfrac{3^{17}+1}{3^{20}+1}\) và \(\dfrac{3^{20}+1}{3^{23}+1}\)
(đang tìm cách làm, và thêm vài cách khác)