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Ta có :
\(D=\frac{5}{5.8.11}+\frac{5}{8.11.14}+...+\frac{5}{302.305.308}\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{6}{5.8.11}+\frac{6}{8.11.14}+...+\frac{6}{302.305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{8.11}+\frac{1}{8.11}\frac{1}{11.14}+...+\frac{1}{302.305}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\left(\frac{1}{5.8}-\frac{1}{305.308}\right)\)
\(\Rightarrow D=\frac{5}{6}.\frac{1}{40}-\frac{5}{6}.\frac{1}{305.308}\)
\(\Rightarrow D=\frac{1}{48}-\frac{5}{6.305.308}< \frac{1}{48}\) (đpcm )
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\Rightarrow x=-2004\)
\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}\)
\(\Rightarrow\dfrac{x+4}{2000}+\dfrac{x+3}{2001}-\dfrac{x+2}{2002}-\dfrac{x+1}{2003}=0\)
\(\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1-\dfrac{x+2}{2002}-1-\dfrac{x+1}{2003}-1=0\)
\(\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\)
\(\Rightarrow x+2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
Vậy \(x=-2004\)
Ta có:\(C=\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)
\(\Rightarrow C< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}\)
\(\Rightarrow C^2< \dfrac{2}{3}.\dfrac{4}{5}.....\dfrac{200}{201}.\dfrac{1}{2}.\dfrac{3}{4}.....\dfrac{199}{200}\)
\(\Rightarrow C^2< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.....\dfrac{199}{200}.\dfrac{200}{201}\)
\(\Rightarrow C^2< \dfrac{1}{201}\) (đpcm)
\(\dfrac{a+5}{a-5}=\dfrac{b+6}{b-6}\Leftrightarrow\left(a+5\right)\left(b-6\right)=\left(b+6\right)\left(a-5\right)\)
nhân ra ik ròi suy ra đpcm :D
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
\(A=\dfrac{1}{2}+\dfrac{1}{12}+\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\)
\(A=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{30}+...+\dfrac{1}{9506}+\dfrac{1}{9900}\right)\)
\(A>\dfrac{1}{2}+\dfrac{1}{12}\Rightarrow A>\dfrac{7}{12}\left(1\right)\)
\(A=\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{97.98}+\dfrac{1}{99.100}\)
\(A=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\left(1-\dfrac{1}{2}+\dfrac{1}{3}\right)-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A=\dfrac{5}{6}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{99}-\dfrac{1}{100}\)
\(A< \dfrac{5}{6}\left(2\right)\)
\(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
\(\rightarrowđpcm\)
Chúc bạn học tốt!
\(1,\\ a,3^{2^3}=3^8>3^6=\left(3^2\right)^3\\ b,\left(-8\right)^9=\left(-2\right)^{27}< \left(-2\right)^{25}=\left(-32\right)^5\\ c,2^{21}=8^7< 9^7=3^{14}\\ 2,\)
\(a,\) Áp dụng tcdtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)
\(b,\) Sửa: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow a=bk;c=dk\)
\(\Leftrightarrow\dfrac{ab}{cd}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2};\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\dfrac{b^2}{d^2}\\ \LeftrightarrowĐpcm\)
\(C=\dfrac{5}{5\cdot8\cdot11}+\dfrac{5}{8\cdot11\cdot14}+...+\dfrac{5}{302\cdot305\cdot308}\\ =\dfrac{5}{6}\cdot\left(\dfrac{6}{5\cdot8\cdot11}+\dfrac{6}{8\cdot11\cdot14}+...+\dfrac{6}{302\cdot305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{5\cdot8}-\dfrac{1}{8\cdot11}+\dfrac{1}{8\cdot11}-\dfrac{1}{11\cdot14}+...+\dfrac{1}{302\cdot305}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\left(\dfrac{1}{40}-\dfrac{1}{305\cdot308}\right)\\ =\dfrac{5}{6}\cdot\dfrac{1}{40}-\dfrac{5}{6}\cdot\dfrac{1}{305\cdot308}\\ =\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}\\ \dfrac{5}{6\cdot305\cdot308}>0\Rightarrow\dfrac{1}{48}-\dfrac{5}{6\cdot305\cdot308}< \dfrac{1}{48}\)