\(\dfrac{2}{1^2}\cdot\dfrac{6}{2^2}\)\(\cdot\dfrac{12}{3^2}\cdot\dfrac{20}{4^2}\cdot\cdot\cdot\cdot\dfrac{110}{10^2}\)\(\cdot\left(x-2\right)=-20\left(x+1\right)+60\)
giải phương trình trên
Giải phương trình
\(1,\dfrac{x^2-2x-3}{x-1}+\dfrac{x^2-8x+20}{x-4}=\dfrac{x^2-4x+6}{x-2}+\dfrac{x^2-6x+12}{x-3}\)
\(2,\left(1+\dfrac{1}{1\cdot3}\right)\cdot\left(1+\dfrac{1}{2\cdot4}\right)\cdot\left(1+\dfrac{1}{3\cdot5}\right)\cdot...\cdot[1+\dfrac{1}{x\cdot\left(x+2\right)}]=\dfrac{31}{16}\left(x\in N\right)\)
1/S=\(\left(1+\dfrac{1}{2}\right)\cdot\left(1+\dfrac{1}{3}\right)\cdot\left(1+\dfrac{1}{4}\right)\cdot...\cdot\left(1+\dfrac{1}{100}\right)\)
2/B=\(\left(1-\dfrac{1}{2}\right)\cdot\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{4}\right)\cdot...\cdot\left(1-\dfrac{1}{2007}\right)\)
3/C=\(\dfrac{2^2}{1\cdot3}\cdot\dfrac{3^2}{2\cdot4}\cdot\dfrac{4^2}{3\cdot5}\cdot...\cdot\dfrac{100^2}{99\cdot101}\)
1: \(S=\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot\dfrac{5}{4}\cdot...\cdot\dfrac{101}{100}=\dfrac{101}{2}\)
2: \(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{2006}{2007}=\dfrac{1}{2007}\)
1: rút gọn rồi tính
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right)\) : \(\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
2: tìm x: \(3\cdot\left(4-x\right)+\left(x+2\right)\cdot\left(1+2x\right)=7\cdot\left(1+x\right)-2x\cdot\left(2-x\right)\)
3: tìm x: \(\dfrac{2\cdot\left(1+x\right)}{3}-\dfrac{5\cdot\left(2-x\right)}{6}=1\dfrac{1}{3}-\dfrac{3\cdot\left(2x+3\right)}{4}-1\dfrac{1}{2}\cdot\left(x+1\right)\)
4: cho a= \(3+3^{2^3}+3^3+3^4+...+3^{360}\)
Bài 1:
\(\left(-\dfrac{72}{40}-\dfrac{144}{60}-2\dfrac{1}{3}\right):\left(\dfrac{45}{100}-\dfrac{25}{60}+-\dfrac{75}{25}\right)\)
\(=\left(-\dfrac{9}{5}-\dfrac{12}{5}-\dfrac{7}{3}\right):\left(\dfrac{9}{20}-\dfrac{5}{12}+-3\right)\)
\(=\left(-\dfrac{27}{15}-\dfrac{36}{15}-\dfrac{21}{15}\right):\left(\dfrac{27}{60}-\dfrac{25}{60}+-3\right)\)
\(=\left(-\dfrac{28}{5}\right):\left(-\dfrac{89}{30}\right)\)
\(=\left(-\dfrac{28}{5}\right).\left(-\dfrac{30}{89}\right)\)
\(=\dfrac{168}{89}\)
1: \(\dfrac{2\cdot\left(x+2\right)}{3}-\dfrac{5\cdot\left(x-1\right)}{4}=\dfrac{3\cdot\left(5-x\right)}{2}-1\dfrac{1}{2}\cdot\left(2x+3\right)\)
\(\Leftrightarrow\dfrac{2}{3}x+\dfrac{4}{3}-\dfrac{5}{4}x+\dfrac{5}{4}=\dfrac{15}{2}-\dfrac{3}{2}x-\dfrac{3}{2}\left(2x+3\right)\)
\(\Leftrightarrow x\cdot\dfrac{-7}{12}+\dfrac{31}{12}=\dfrac{-15}{2}x+3\)
=>83/12x=5/12
hay x=5/83
Tính:
\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)\(N=\left(0,25\right)^{-1}\cdot\left(\dfrac{1}{4}\right)^{-2}\cdot\left(\dfrac{4}{3}\right)^{-2}\cdot\left(\dfrac{5}{4}\right)^{-1}\cdot\left(\dfrac{2}{3}\right)^{-3}\)
\(N=4\cdot16\cdot\dfrac{9}{16}\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}=4\cdot9\cdot\dfrac{4}{5}\cdot\dfrac{27}{8}\)
\(=\dfrac{16}{5}\cdot\dfrac{243}{8}=\dfrac{486}{5}\)
Tính giá trị các biểu thức sau theo cách hợp lí nhất.
a) $\mathrm{A}=\left(\dfrac{2}{7} \cdot \dfrac{1}{4}-\dfrac{1}{3} \cdot \dfrac{2}{7}\right):\left(\dfrac{2}{7} \cdot \dfrac{3}{9}-\dfrac{2}{7} \cdot \dfrac{2}{5}\right)$;
b) $\mathrm{B}=\dfrac{\left(\dfrac{1}{5}-\dfrac{2}{7}\right) \cdot \dfrac{3}{4}-\dfrac{3}{4} \cdot\left(\dfrac{1}{3}-\dfrac{2}{7}\right)}{\dfrac{1}{5} \cdot \dfrac{2}{7}-\dfrac{1}{3} \cdot\left(\dfrac{2}{7}+\dfrac{3}{9}\right)+\dfrac{3}{9} \cdot \dfrac{1}{5}} .$
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,Tìm x, biết:
a/\(^{\left(x-1\right)^{x+2}}=^{\left(x-1\right)^{x+4}}\)
b/\(\dfrac{1}{4}\cdot\dfrac{2}{6}\cdot\dfrac{3}{8}\cdot\dfrac{4}{10}\cdot...\cdot\dfrac{30}{62}\cdot\dfrac{31}{64}=2x\)
c/\(4^x+3^x=2^x+6^x\left(STN\right)\)
GIÚP MÌNH NHANH NHA MÌNH TICK CHO
a)x=1;2;-2(bạn nên tự giải)
b)=>\(\dfrac{1\cdot2\cdot3\cdot4\cdot...\cdot30\cdot31}{4\cdot6\cdot8\cdot10\cdot...\cdot62\cdot64}\)=2x
=>\(\dfrac{2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31}{60\left(2\cdot3\cdot4\cdot5\cdot...\cdot30\cdot31\right)\cdot64}=2x\)
=>\(\dfrac{1}{60\cdot64}=2x\)=> 1/3840 =2x
=>x = 1/7680
c)=>4x - 2x = 6x - 3x
=>2x (2x-1)= 3x(2x-1)
=> 2x = 3x
=>x = 0
\(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
Ta có: \(\left(\dfrac{4}{13}\cdot\dfrac{6}{5}+\dfrac{4}{13}\cdot\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\left(\dfrac{6}{5}+\dfrac{2}{5}\right)\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{4}{13}\cdot\dfrac{8}{5}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\dfrac{32}{65}\cdot\left(2x+1\right)^2=\dfrac{10}{13}\)
\(\Leftrightarrow\left(2x+1\right)^2=\dfrac{10}{13}:\dfrac{32}{65}=\dfrac{10}{13}\cdot\dfrac{65}{32}=\dfrac{25}{16}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=\dfrac{5}{4}\\2x+1=-\dfrac{5}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{5}{4}-1=\dfrac{1}{4}\\2x=-\dfrac{5}{4}-1=-\dfrac{9}{4}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{4}:2=\dfrac{1}{8}\\x=-\dfrac{9}{4}:2=-\dfrac{9}{8}\end{matrix}\right.\)
Vậy: \(x\in\left\{\dfrac{1}{8};-\dfrac{9}{8}\right\}\)