tính nhanh
a) \(498^2+996\cdot502+502^2\)b) \(126^2-52\cdot126+26^2\)
c)\(1995^2-1994\cdot1996\)d)\(2005^2-2004\cdot2006\)
e) \(2005^4-2004\cdot2006\cdot\left(2005^2+1\right)\)g) \(1999\cdot\left(2000^2+2001\right)-2001\left(2000^2-1999\right)\)
Tính giá trị của các biểu thức sau 1) \(A=1+2+2^2+...+2^{2015}\) 2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\) 3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\) 4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\) 5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\) 6) Cho 13+23+...+103=3025 Tính S= 23+43+63+...+203
a\(\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)
b, \(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
ý thứ 2 giả cả 2 cách hộ mình nhé
\(a,\left(\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right):\frac{1}{4}:\frac{1}{6}\)
\(=\left(\frac{1}{6}+\frac{1}{4}-\frac{1}{5}\right)\cdot\frac{1}{4}\cdot\frac{1}{6}\)
\(=\left(\frac{10}{60}+\frac{15}{60}-\frac{12}{60}\right)\cdot\frac{1}{24}\)
\(=\frac{13}{60}\cdot\frac{1}{24}\)
\(=\frac{13}{1440}\)
\(b,\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(\frac{2006\cdot2005-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot\left(2004+1\right)-1}{2004 \cdot2006+2005}\)
\(=\frac{2006\cdot2004+2006\cdot1-1}{2004\cdot2006+2005}\)
\(=\frac{2006\cdot2004+2005}{2004\cdot2006 +2005}=1\)
Mình nghĩ phần b, ko có cách 2 đâu bạn .
\(\left(1+\frac{1}{1\cdot3}\right)\cdot\left(1+\frac{1}{2\cdot4}\right)\cdot\cdot\cdot\cdot\left(1+\frac{1}{2004\cdot2006}\right)\)
\(\left(1+\frac{1}{1\cdot3}\right)\left(1+\frac{1}{2\cdot4}\right)\left(1+\frac{1}{3\cdot5}\right)...\left(1+\frac{1}{2004\cdot2006}\right)\)
\(=\frac{4}{1\cdot3}+\frac{9}{2\cdot4}+\frac{16}{3\cdot5}+...+\frac{420025}{2004\cdot2006}\)
\(=\frac{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2005\cdot2005\right)}{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2004\cdot2006\right)}\)
\(=\frac{\left(2\cdot3\cdot4\cdot...\cdot2005\right)\left(2\cdot3\cdot4\cdot...\cdot2005\right)}{\left(1\cdot2\cdot3\cdot...\cdot2004\right)\left(3\cdot4\cdot5\cdot...\cdot2006\right)}\)
\(=\frac{2005\cdot2}{1\cdot2006}\)
\(=\frac{4010}{2006}\)
\(\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{2004.2006}\right)\)
\(=\frac{1.3+1}{1.3}.\frac{2.4+1}{2.4}....\frac{2004.2006+1}{2004.2006}\)
\(=\frac{2^2}{1.3}.\frac{3^2}{2.4}....\frac{2005^2}{2004.2006}\)
\(=\frac{2.3....2005}{1.2....2004}.\frac{2.3...2005}{3.4....2006}\)
\(=2005.\frac{2}{2006}=\frac{2005}{1003}\)
\(=\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right)...\left(1+\frac{1}{2004.2006}\right)\)
\(=\left(\frac{2.2}{1.3}\right).\left(\frac{3.3}{2.4}\right)....\left(\frac{2005.2005}{2004.2006}\right)\)
\(=\frac{\left(2.3.4....2005\right).\left(2.3.4....2005\right)}{\left(1.2.3...2004\right).\left(3.4.5....2006\right)}\)
\(=\frac{2005}{2006}\)
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
Tính giá trị của các biểu thức sau
1) \(A=1+2+2^2+...+2^{2015}\)
2) \(B=\left(\dfrac{1}{4}-1\right)\cdot\left(\dfrac{1}{9}-1\right)\cdot\left(\dfrac{1}{16}-1\right)\cdot\cdot\cdot\cdot\cdot\left(\dfrac{1}{400}-1\right)\)
3) \(C=\left(\dfrac{1}{4\cdot9}+\dfrac{1}{9\cdot14}+\dfrac{1}{14\cdot19}+...+\dfrac{1}{44\cdot49}\right)\cdot\dfrac{1-3-5-7-...-49}{89}\)
4) \(D=\dfrac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^2\cdot3\right)^6+8^4\cdot3^5}-\dfrac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)
5) \(E=\dfrac{\dfrac{1}{2003}+\dfrac{1}{2004}-\dfrac{1}{2005}}{\dfrac{5}{2003}+\dfrac{5}{2004}-\dfrac{5}{2005}}-\dfrac{\dfrac{2}{2002}+\dfrac{2}{2003}-\dfrac{2}{2004}}{\dfrac{3}{2002}+\dfrac{3}{2003}-\dfrac{3}{2004}}\)
6) Cho 13+23+...+103=3025
Tính S= 23+43+63+...+203
1) \(A=1+2+2^2+2^3+......+2^{2015}\)
\(\Leftrightarrow2A=2+2^2+2^3+......+2^{2016}\)
\(\Leftrightarrow2A-A=\left(2+2^2+2^3+......+2^{2016}\right)-\left(1+2+2^2+2^3+......+2^{2015}\right)\)
\(\Leftrightarrow A=2^{2016}-1\)
Vậy \(A=2^{2016}-1\)
6)Ta có: \(13+23+33+43+.......+103=3025\)
\(\Leftrightarrow2.13+2.23+2.33+2.43+.......+2.103=2.3025\)
\(\Leftrightarrow26+46+66+86+.......+206=6050\)
\(\Leftrightarrow\left(23+3\right)+\left(43+3\right)+\left(63+3\right)+\left(83+3\right)+.......+\left(203+3\right)=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+3.10=6050\)
\(\Leftrightarrow23+43+63+83+.......+203+=6050-30\)
\(\Leftrightarrow23+43+63+83+.......+203+=6020\)
Vậy S=6020
b, B có 19 thừa số
=> \(-B=(1-\frac{1}{4})(1-\frac{1}{9})(1-\frac{1}{16})...(1-\frac{1}{400}) \)
<=>\(-B=\frac{(2-1)(2+1)(3-1)(3+1)(4-1)(4+1)...(20-1)(20+1)}{4.9.16...400} \)
<=>\(-B=\frac{(1.2.3.4...19)(3.4.5...21)}{(2.3.4.5.6...20)(2.3.4.5...20)} \)
<=>\(-B=\frac{21}{20.2} =\frac{21}{40} \)
<=>\(B=\frac{-21}{40} \)
Bài 1 tim GTLN
\(B=x^2-2xy+4y^2-2x-10y+2018\)
\(C=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
Bài 2 so sánh\(2004\cdot2006\left(2005^2+1\right)vs2005^4\)
giúp mk vs
\(B=\left(x-y-1\right)^2+3\left(y-2\right)^2+2005\text{ }\ge2005\)
\(C=\left(x^2+4x\right)^2-25\ge-25\)
\(2004.2006.\left(2005^2+1\right)=\left(2005-1\right)\left(2005+1\right)\left(2005^2+1\right)\)
\(=\left(2005^2-1\right)\left(2005^2+1\right)=2005^4-1< 2005^4\)
bạn có thể giải chi tiết ra hộ mk ko
So sánh:
\(\dfrac{2004\cdot2005-1}{2004\cdot2005}\) và \(\dfrac{2005\cdot2006-1}{2005\cdot2006}\)
Các bạn giúp mình với mình cảm ơn nhiều ạ!!!!
\(\dfrac{2004.2005-1}{2004.2005}=1-\dfrac{1}{2004.2005}\)
\(\dfrac{2005.2006-1}{2004.2006}=1-\dfrac{1}{2005.2006}\)
\(Vì\dfrac{1}{2004.2005}>\dfrac{1}{2005.2006}\Rightarrow1-\dfrac{1}{2004.2005}< 1-\dfrac{1}{2005.2006}\Rightarrow\dfrac{2004.2005-1}{2004.2005}< \dfrac{2005.2006-1}{2004.2006}\)
Tính Q=\(\frac{\left(2003^2\cdot2013+31\cdot2004-1\right)\cdot\left(2003\cdot2008+4\right)}{2004\cdot2005\cdot2006\cdot2007\cdot2008}\)
ae giải nhanh lên mình đag cần gấp mình ticks cho
tìm x,y biết
a,\(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
b,\(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
c,\(2024^{\left|x-1\right|=y^2-1}\cdot3^{2024}=9^{1012}\)
a: \(\left(2^3\right)^{1^{2005}}\cdot x+2005^0\cdot x=9915:3+1^{2025}\)
=>\(8\cdot x+1\cdot x=3305+1\)
=>\(9x=3306\)
=>\(x=\dfrac{3306}{9}=\dfrac{1102}{3}\)
b: \(2^x+2^{x+1}+2^{x+2}+2^{x+3}=480\)
=>\(2^x+2^x\cdot2+2^x\cdot4+2^x\cdot8=480\)
=>\(2^x\left(1+2+4+8\right)=480\)
=>\(2^x\cdot15=480\)
=>\(2^x=32\)
=>\(2^x=2^5\)
=>x+5