A = 1/22 +1/32+1/42+.....+1/20192 . So Sánh A với 3/4
Giúp vs mn mai mik đi hc òi
A=1/22 + 1/32 + 1/42 + ...+ 1/302 .
giúp mik vs nhé . Thanks :)))
Tính nhanh
1, 1532+94.135+472
2,1,24-2,48.0,24=0,242
3, 2055-955
4,38.58-(154-1)(154+1)
5,12-22+32-42+...-20192+20202
6,(2+1)(22+1)(24+1)...(22020+1)+1
1.
$=153^2+2.47.153+47^2=(153+47)^2=200^2=40000$
2.
$=1,24^2-2.1,24.0,24+0,24^2=(1,24-0,24)^2=1^2=1$
3. Không phù hợp để tính nhanh
4.
$=15^8-(15^8-1)=1$
5.
$=(1^2-2^2)+(3^2-4^2)+(5^2-6^2)+...+(2019^2-2020^2)$
$=(1-2)(1+2)+(3-4)(3+4)+(5-6)(5+6)+...+(2019-2020)(2019+2020)$
$=(-1)(1+2)+(-1)(3+4)+(-1)(5+6)+....+(-1)(2019+2020)$
$=(-1)(1+2+3+4+....+2019+2020)=(-1).2020(2020+1):2=-2041210$
6:
\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^4-1\right)\left(2^4+1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^8-1\right)....\left(2^{2020}+1\right)+1\\ =\left(2^{2020}-1\right)\left(2^{2020}+1\right)+1\\ =2^{4040}-1+1=2^{4040}\)
Cho A=3/22+8/32+15/42+....+20232-1/20232 Không là số tự nhiên (lm nhanh giúp mik vs ạ )
A=(1/22 - 1)*(1/32 - 1)*(1/42 - 1)(1/52 - 1)*...*(1/1002 - 1)
So sánh với -1/2
nani "Doge"
a)22 (6x-32)-3=33
b)4(x+2)=3(x+1)+17
giúp mik vs ạ
thanks mn nhìu
a) 2²(6x - 3²) - 3 = 33
4(6x - 9) = 33 + 3
4(6x - 9) = 36
6x - 9 = 36 : 4
6x - 9 = 9
6x = 9 + 9
6x = 18
x = 18 : 6
x = 3
b) 4(x + 2) = 3(x + 1) + 17
4x + 8 = 3x + 3 + 17
4x - 3x = 3 + 17 - 8
x = 12
Chứng minh:
A=1/22+1/32+1/42+...+1/102<1
Giảii giúp mik vs ạ mik đag cần gấp
A=1/2^2+1/3^2+...+1/10^2
=>A<1-1/2+1/2-1/3+...+1/9-1/10=1-1/10<1
So sánh: A = \(\frac{2005^{2005}+1}{2005^{2006}+1}\) và B = \(\frac{2005^{2004}+1}{2005^{2005}+1}\)
GIÚP MIK NHA MAI THI ÒI
\(10A=\frac{2005^{2006}+10}{2005^{2006}+1}\)
\(10B=\frac{2005^{2005}+10}{2005^{2005}+1}\)
Rồi bạn so sánh 10A và 10B là ra.
Ai thấy đúng thì ủng hộ nha !!!, sai thì góp ý cho mink nha
Ta có
A <\(\frac{2005^{2005}+2005}{2005^{2006}+2005}=\frac{2005\left(2005^{2004}+1\right)}{2005\left(2005^{2005}+1\right)}\)=\(\frac{2005^{2004}+1}{2005^{2005}+1}\)
\(\RightarrowĐPCM\)
\(A=\frac{2005.\left(2005^{2005}+1\right)}{2005^{2006}+1}=\frac{2005^{2006}+2005}{2005^{2006}+1}=\frac{2005^{2006}+1+2004}{2005^{2006}+1}=1+\frac{2004}{2005^{2006}+1}\)
\(B=\frac{2005.\left(2005^{2004}+1\right)}{2005^{2005}+1}=\frac{2005^{2005}+2005}{2005^{2005}+1}=\frac{2005^{2005}+1+2004}{2005^{2005}+1}=1+\frac{2004}{2005^{2005}+1}\)
Vì \(2005^{2005}+1< 2005^{2006}+1\)
\(\Rightarrow\frac{2004}{2005^{2005}+1}>\frac{2004}{2005^{2006}+1}\Rightarrow2005B>2005A\Rightarrow B>A\)
So sánh A = 2019.2021.a và B = ( 2019 2 + 2 . 2019 + 1 ) a (với a > 0)
A. A= B
B. A ≥ B
C. A > B
D.A < B
Ta có A = 2019.2021.a = (2020 – 1)(2020 + 1)a = ( 2020 2 – 1)a
Và B = ( 2019 2 + 2 . 2019 + 1 ) a = ( 2019 + 1 ) 2 a = 2020 2 a
Vì 2020 2 – 1 < 2020 2 và a > 0 nên ( 2020 2 – 1 ) a < 2020 2 a hay A < B
Đáp án cần chọn là: D
giúp mk vs
so sánh:1/3+1/4+...+1/32 với 2