5+2-2*2
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Những câu hỏi liên quan
1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2+1+2=
5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5+5=
...................=66
...................=110
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1+2+........+1+2 = 66
5+5+........+5+5 = 220
k mk nha
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C= 5+5³+5⁵+5⁷+...+5¹⁰¹
D=2¹⁰⁰-2⁹⁹+2⁹⁸-2⁹⁷+...+2²-2+1
\(C=5+5^3+5^5+...+5^{101}\)
\(5^2\cdot C=5^2\cdot\left(5+5^3+...+5^{101}\right)\)
\(25C=5^3+5^5+...+5^{103}\)
\(25C-C=\left(5^3+5^5+....+5^{103}\right)-\left(5+5^3+5^5+...+5^{101}\right)\)
\(24C=\left(5^3-5^3\right)+\left(5^5-5^5\right)+...+\left(5^{103}-5\right)\)
\(24C=5^{103}-5\)
\(C=\dfrac{5^{103}-5}{24}\)
_____________
\(D=2^{100}-2^{99}+2^{98}-...+2^2-2+1\)
\(2D=2\cdot\left(2^{100}-2^{99}+2^{98}-...-2+1\right)\)
\(2D=2^{101}-2^{100}+2^{99}-...-2^2+2\)
\(2D+D=2^{101}-2^{100}+...-2^2+2+2^{100}-2^{99}+...-2+1\)
\(D=2^{101}+1\)
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wow gioir thaatj
Cho A= 2× 2× 2× 2×...×2 × 5× 5× 5×...×5
2×2×2×2...×2 và 5×5×5×...×5 đều có thừa số là 2010
Hỏi số Anh gồm bao nhiêu chữ số
\(A=2^{2010}\cdot5^{2010}\)
\(A=10^{2010}\)
suy ra có 2011 chữ số
Nhớ bấm L I K E cho mk nhá :))))
đấy tóan lớp 5 á :)))
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Rút gọn : a . P = 3+2√3 / √3 + 2+√2 / √2+1 - ( √2 + √3 ) ; b. N = ( 1 - 5 + √5 / 1 + √5 ) ( 5 - √5 / 1- √5 - 1 ) ; c. Q = ( 5 - 2√5 / 2 - √5 - 2 ) ( 3+3 √5 / 3 + √5 - 2 ). Giúp mik vs ạ
a: \(P=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{1}-\sqrt{3}-\sqrt{2}\)
\(=2+\sqrt{3}+2-\sqrt{2}-\sqrt{3}-\sqrt{2}\)
\(=4-2\sqrt{2}\)
b: \(N=\left(1-\dfrac{\sqrt{5}\left(\sqrt{5}+1\right)}{\sqrt{5}+1}\right)\left(\dfrac{-\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}-1\right)\)
\(=\left(1-\sqrt{5}\right)\left(-\sqrt{5}-1\right)\)
\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)
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a) \(P=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}+1}-\left(\sqrt{2}+\sqrt{3}\right)\)
\(P=\dfrac{\sqrt{3}\cdot\left(\sqrt{3}+2\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\cdot\left(\sqrt{2}+1\right)}{\sqrt{2}+1}-\sqrt{2}-\sqrt{3}\)
\(P=\sqrt{3}+2+\sqrt{2}-\sqrt{2}-\sqrt{3}\)
\(P=2\)
b) \(N=\left(1-\dfrac{5+\sqrt{5}}{1+\sqrt{5}}\right)\left(\dfrac{5-\sqrt{5}}{1-\sqrt{5}}-1\right)\)
\(N=\left[1-\dfrac{\sqrt{5}\left(1+\sqrt{5}\right)}{1+\sqrt{5}}\right]\left[1+\dfrac{\sqrt{5}\left(1-\sqrt{5}\right)}{1-\sqrt{5}}\right]\)
\(N=\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)\)
\(N=1^2-\left(\sqrt{5}\right)^2\)
\(N=-4\)
c) \(Q=\left(\dfrac{5+2\sqrt{5}}{2-\sqrt{5}}-2\right)\left(\dfrac{5+3\sqrt{5}}{3+\sqrt{5}}-2\right)\)
\(Q=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-2\right)}{\sqrt{5}-2}+2\right]\left[\dfrac{\sqrt{5}\left(\sqrt{5}+3\right)}{\sqrt{5}+3}-2\right]\)
\(Q=\left(\sqrt{5}+2\right)\left(\sqrt{5}-2\right)\)
\(Q=\left(\sqrt{5}\right)^2-2^2\)
\(Q=1\)
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A= 2+2^2+2^3+...+2^19+2^20
b=2+2^3+2^5+...2^97+2^99
C=5+5^2+5^3+...+5^50
D=1+3+3^2+3^3+...+3^100
\(2A-A=\left(2^2+2^3+...+2^{21}\right)-\left(2+2^2+...+2^{20}\right)\)
\(A=2^{21}-2\)
B tương tự câu A
\(5C-C=\left(5^2+5^3+...+5^{51}\right)-\left(5+5^2+...+5^{50}\right)\)
\(C=\dfrac{5^{51}-5}{4}\)
\(3D-D=3+3^2+...+3^{101}-\left(1+3+...+3^{100}\right)\)
\(D=\dfrac{3^{101}-1}{2}\)
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\(A=2^1+2^2+2^3+...+2^{20}\)
\(2\cdot A=2^2+2^3+2^4+...+2^{21}\)
\(A=2^{21}-2\)
\(B=2^1+2^3+2^5+...+2^{99}\)
\(4\cdot B=2^3+2^5+2^7+...+2^{101}\)
\(B=\)\(\left(2^{101}-2\right):3\)
\(C=5^1+5^2+5^3+...+5^{50}\)
\(5\cdot C=5^2+5^3+5^4+...+5^{51}\)
\(C=(5^{51}-5):4\)
\(D=3^0+3^1+3^2+...+3^{100}\)
\(3\cdot D=3^1+3^2+3^3+...+3^{101}\)
\(D=(3^{101}-1):2\)
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2/5*2+4*2/5+2/5*5-2/5
2/5x2+4x2/5+2/5x5-2/5
=2/5x(2+4+5-1)
=2/5x10
=4
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\(\dfrac{2}{5}\times2+4\times\dfrac{2}{5}+\dfrac{2}{5}\times5-\dfrac{2}{5}\\ =\left(2+4+5-1\right)\times\dfrac{2}{5}\\ =10\times\dfrac{2}{5}\\ =4\)
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2 ^ 15 : 2 ^ 13 + 3 ^ 2. 2 ^ 3 - 5 ^ 2 - 2 ^ 0
81 . ( 27 + 9 ^ 15) : (3 ^ 5 + 3 ^ 32)
(4 ^ 42 .123 - 23 . 4 ^ 42) : (4 ^ 15 . 4 ^ 27)
(15 . 3 ^ 7 - 3 ^ 8) : (4 . 9 ^ 4)
(100 . 5 ^ 7 + 5 ^ 11 : 25 ) : ( 5 . 5 ^ 13 : 25 ^ 2)
( 2 ^ 5 . 3 ^ 6 . 5 ^ 7 + 2 ^ 6 . 3 ^ 7 . 5 ^ 8 ) : ( 2 ^ 4 . 3 ^ 5 . 5 ^ 6)
(4 ^ 5 . 10 . 5 ^ 6 + 2 ^ 8 . 25 ^ 5) : ( 2 ^ 8 . 5 ^ 4 + 5 ^ 4 + 5 ^ 7 . 2 ^ 5)
tặng 2 like cho người giải !!!
khó quá tui chưa học
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Tìm chữ số tận cùng của 2^2^2^2^5 ; 3^2^2^2^5 ; 2^4^2^5 ; 7^4^2^5.
2^4.x-3.5x=5^2-2
7^2x-14x=7^2.10-70
2^3x+5^2x=2(5^2+2^3)-33
320:(x-1)=(5^3-5^2):4+15
75:(x+2)=5^3-(3^2+4^2).2^2
(2^2+1)(x-12)+35=5^2+279:3^2
(2^2+1)(x+14)=5^2.4+(2^5+3^2+7^2):2
GIÚP MÌNH VS MỌI NGƯỜI MÌNH ĐANG CẦN GẤP
T.T
1. dfrac{-2}{sqrt{3}-1}2. dfrac{5}{1-sqrt{6}}3. dfrac{2+sqrt{5}}{2-sqrt{5}}4. dfrac{1}{5+2sqrt{6}}5. dfrac{sqrt{5}+2}{sqrt{5}-2}6. dfrac{5sqrt{2}-2sqrt{5}}{sqrt{2}-sqrt{5}}7. dfrac{sqrt{20}-3sqrt{10}}{3-sqrt{2}}8. dfrac{6-2sqrt{5}}{3+sqrt{5}}9. dfrac{9+4sqrt{5}}{sqrt{5}+2}
Đọc tiếp
1. \(\dfrac{-2}{\sqrt{3}-1}\)
2. \(\dfrac{5}{1-\sqrt{6}}\)
3. \(\dfrac{2+\sqrt{5}}{2-\sqrt{5}}\)
4. \(\dfrac{1}{5+2\sqrt{6}}\)
5. \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\)
6. \(\dfrac{5\sqrt{2}-2\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)
7. \(\dfrac{\sqrt{20}-3\sqrt{10}}{3-\sqrt{2}}\)
8. \(\dfrac{6-2\sqrt{5}}{3+\sqrt{5}}\)
9. \(\dfrac{9+4\sqrt{5}}{\sqrt{5}+2}\)
