a=1/1*21+1/2+22+.....+1/80*100
b=1/1*81+1/2+82+.....+1/20*100
tính tỉ số a/b
cần gấp
Tìm x:4x(1/1*81+1/2*82+1/3*83+...+1/20*100)=1/1*21+1/2*22+1/3*23+...+1/80*100
So sánh A và B biết :
a) A = \(\frac{2^{40}+1}{2^{41}+1}\) và B = \(\frac{2^{41}+1}{2^{42}+1}\)
b) A = \(\frac{3^{60}+2}{3^{61}+2}\) và B = \(\frac{3^{80}+2}{3^{81}+2}\)
c) A = \(\frac{5^{20}+1}{5^{22}+1}\) và B = \(\frac{5^{22}+1}{5^{24}+1}\)
d) A = \(\frac{4^{80}-3}{4^{81}-3}\) và B = \(\frac{4^{81}-3}{4^{82}-3}\)
e) A = \(\frac{6^{10}-1}{6^9-1}\) và B = \(\frac{6^9-1}{6^8-1}\)
Các bạn giúp mình với, mình đang cần gấp. Các bạn giải chi tiết giùm mình rồi mình tick cho.
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Cho A=\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{80},CMR:1< A< 2\)
Ta có
\(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{60}+...+\frac{1}{80}\right)\) \(A>\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)+\left(\frac{1}{80}+\frac{1}{80}+...+\frac{1}{80}\right)\)
\(A>\frac{20}{40}+\frac{20}{60}+\frac{20}{80}\Rightarrow A>\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\Rightarrow A>\frac{13}{12}\Rightarrow A>1\) (1)
LẠi có \(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{60}+...+\frac{1}{80}\right)\)
\(A< \left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)+\left(\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\right)\)
\(A< \frac{20}{20}+\frac{20}{40}+\frac{20}{60}\Rightarrow A< 1+\frac{1}{2}+\frac{1}{3}\Rightarrow A< \frac{11}{6}< \frac{12}{6}\Rightarrow A< 2\) (2)
Từ (1) và (2) suy ra điều phải CM
So sánh : A=1/20^2+1/21^2+1/22^2+...+1/30^2 và B=1/19
Ta có:
\(\dfrac{1}{20^2}< \dfrac{1}{20\cdot19}=\dfrac{1}{19}-\dfrac{1}{20}\)
\(\dfrac{1}{21^2}< \dfrac{1}{20\cdot21}=\dfrac{1}{20}-\dfrac{1}{21}\)
\(...\)
\(\dfrac{1}{30^2}< \dfrac{1}{29\cdot30}=\dfrac{1}{29}-\dfrac{1}{30}\)
\(\Rightarrow A< \dfrac{1}{19}-\dfrac{1}{30}< \dfrac{1}{19}\)
A=(1 / 21)+(1 / 2 x 22)+. . . .+{1 / n x (n+20)}+. . .+(1 / 1993 x 2013)
dấu "/" là dấu phân số
1/22 + 1/42 + 1/62 + 1/82 + ... + 1/20222 < 1/2 (gần gấp)
Lời giải:
Gọi vế trái là $A$
$2A=\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+...+\frac{2}{2022^2}$
Xét số hạng tổng quát:
$\frac{2}{n^2}$. Ta sẽ cm $\frac{2}{n^2}< \frac{1}{(n-1)n}+\frac{1}{n(n+1)}(*)$
$\Leftrightarrow \frac{2}{n^2}< \frac{n+1+n-1}{n(n-1)(n+1)}$
$\Leftrightarrow \frac{2}{n^2}< \frac{2}{(n-1)(n+1)}$
$\Leftrightarrow \frac{2}{n^2}< \frac{2}{n^2-1}$ (luôn đúng)
Thay $n=2,4,...., 2022$ vào $(*)$ ta có:
$\frac{2}{2^2}< \frac{1}{1.2}+\frac{1}{2.3}$
$\frac{2}{4^2}< \frac{1}{3.4}+\frac{1}{4.5}$
.......
Suy ra: $2A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2022.2023}$
$2A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{2022}-\frac{1}{2023}$
$2A< 1-\frac{1}{2023}< 1$
$\Rightarrow A< \frac{1}{2}$
Các bn ghi rõ cách giải nha nhanh mk tích cho mk đang cần gấp lắm cảm ơn
1: Tính nhanh
A= 47*36+64*47+15
B= 27+35+65+73+75
C= 37+37*15+84*37
D= 1/20*21+1/21*22+1/22*23+1/23*24
E= 1/1*2+1/2*3+1/3*4+...+1/49.50
\(A=47.36+64.47+15\)
\(A=47.\left(36+64\right)+15\)
\(A=47.100+15\)
\(A=4700+15\)
\(A=4715\)
\(B=27+35+65+73+75\)
\(B=\left(27+73\right)+\left(35+65\right)+75\)
\(B=100+100+75\)
\(B=275\)
\(C=37+37.15+84.37\)
\(C=37.\left(1+15+84\right)\)
\(C=37.100\)
\(C=3700\)
\(D=\frac{1}{20.21}+\frac{1}{21.22}+\frac{1}{22.23}+\frac{1}{23.24}\)
\(D=\frac{1}{20}-\frac{1}{21}+\frac{1}{21}-\frac{1}{22}+\frac{1}{22}-\frac{1}{23}+\frac{1}{23}-\frac{1}{24}\)
\(D=\frac{1}{20}-\frac{1}{24}\)
\(D=\frac{24}{480}-\frac{20}{480}\)
\(D=\frac{4}{480}=\frac{1}{120}\)
\(E=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(E=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(E=1-\frac{1}{50}\)
\(E=\frac{49}{50}\)
Rút gọn giá trị biểu thức: a) 3^8×7^8-20×22×(21^2+1)×(21^4+1) b) (x^2+3x+1)^2+(3x-1)^2-2(x^2+3x+1)×(3x-1)
b: \(=\left(x^2+3x+1-3x+1\right)^2=\left(x^2+2\right)^2\)