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Bài 1: a) Ta có : \(\dfrac{-3}{x}=\dfrac{x}{-27}\Leftrightarrow\left(-3\right).\left(-27\right)=x.x\Leftrightarrow81=x^2\Leftrightarrow9^2=x^2\Leftrightarrow x=9\)
b) Do \(\dfrac{2}{3}\) của x là -150 nên x là: (-150) : \(\dfrac{2}{3}\) = -225
c) \(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{2}-\dfrac{4}{9}\)
\(\Leftrightarrow\dfrac{1}{x+2}=\dfrac{1}{18}\)
\(\Leftrightarrow x+2=18\)
\(\Leftrightarrow x=16\)
Bài 2:
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right)\left(\dfrac{1}{6}-\dfrac{1}{6}\right)\)
\(A=\left(\dfrac{1}{99}+\dfrac{12}{999}+\dfrac{123}{999}\right).0\)
\(A=0\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)
\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Rightarrow x=12\) (nh)
\(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+...+\dfrac{1}{\left(2x-2\right)2x}=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{\left(2x-2\right)2x}\right)=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{11}{48}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{11}{24}\)\(\Leftrightarrow\dfrac{1}{2x}=\dfrac{1}{24}\)
\(\Leftrightarrow2x=24\Leftrightarrow x=12\) (thỏa mãn)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2x-\left(2x-2\right)}{\left(2x-2\right)2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+..+\dfrac{1}{2x-2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2x}=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-1}{2x}=\dfrac{-1}{4}\)
\(\Rightarrow x=2\)
Ta có: \(\dfrac{1}{2.4}+\dfrac{1}{4.6}+...+\dfrac{1}{\left(2x-2\right).2x}=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2x-2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\)
\(\Rightarrow\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2x}\right)=\dfrac{1}{8}\Rightarrow\dfrac{1}{2}.\dfrac{x-1}{2x}=\dfrac{1}{8}\Rightarrow\dfrac{x-1}{4x}=\dfrac{1}{8}\)
\(\Rightarrow8\left(x-1\right)=4x\Rightarrow8x-8=4x\Rightarrow4x=8\Rightarrow x=2\)
\(A=\dfrac{2^2}{1.3}+\dfrac{3^2}{2.4}+\dfrac{4^2}{3.5}+\dfrac{5^2}{4.6}+\dfrac{6^2}{5.7}\)
\(A=\dfrac{2.2.3.3.4.4.5.5.6.6}{1.3.2.4.3.5.4.6.5.7}\)
\(A=\dfrac{2.3.4.5.6}{1.2.3.4.5}.\dfrac{2.3.4.5.6}{3.4.5.6.7}\)
\(A=\dfrac{6}{1}.\dfrac{2}{7}=\dfrac{12}{7}\)
\(B=\left(1+\dfrac{1}{1.3}\right)\left(1+\dfrac{1}{2.4}\right)\left(1+\dfrac{1}{3.5}\right)\left(1+\dfrac{1}{9.11}\right)\)
\(B=\dfrac{4}{3}.\dfrac{9}{8}.\dfrac{16}{15}.\dfrac{100}{99}\)
\(B=\dfrac{4.9.16.100}{3.8.15.99}\)
\(B=\dfrac{2.2.3.3.4.4.10.10}{1.3.2.4.3.5.9.11}\)
\(B=\dfrac{2.3.4.10}{1.2.3.9}.\dfrac{2.3.4.10}{3.4.5.11}\)
\(B=10.\dfrac{2}{11}=\dfrac{20}{11}\)
mình ko biết mình làm đúng hay sai bạn nhé, mong mọi người góp ý
= 1/2.( 1/2.4+1/4.6+....+1/(2x-2)2x)=1/8
= 1/2.(1/2-1/4+1/4-1/6+....+1/(2x-2)-1/2x)=1/8
= 1/2.( 1/2-1/2x)=1/8
( 1/2-1/2x)=1/8:1/2
1/2-1/2x=1/4
1/2x =1/2-1/4
1/2x =1/4
2x = 4
x =4:2
x =2
B = \(\dfrac{12}{\left(2.4\right)^2}+\dfrac{20}{\left(4.6\right)^2}+...+\dfrac{388}{\left(96.98\right)^2}+\dfrac{396}{\left(98.100\right)^2}\)
= \(\dfrac{4^2-2^2}{2^{2^{ }}.4^{2^{ }}}+\dfrac{6^{2^{ }}-4^2}{4^2.6^2}+...+\dfrac{98^2-96^2}{96^2.98^2}+\dfrac{100^2-98^2}{98^2.100^2}\)
=\(\dfrac{1}{2^{2^{ }}}-\dfrac{1}{4^{2^{ }}}+\dfrac{1}{4^2}-\dfrac{1}{6^2}+\dfrac{1}{6^2}+....-\dfrac{1}{98^2}+\dfrac{1}{98^2}-\dfrac{1}{100^2}\)
= \(\dfrac{1}{2^2}-\dfrac{1}{100^2}=\dfrac{1}{4}-\dfrac{1}{100^2}< \dfrac{1}{4}\)
Vậy B < \(\dfrac{1}{4}\)
B = 12(2.4)2+20(4.6)2+...+388(96.98)2+396(98.100)212(2.4)2+20(4.6)2+...+388(96.98)2+396(98.100)2
= 42−2222.42+62−4242.62+...+982−962962.982+1002−982982.100242−2222.42+62−4242.62+...+982−962962.982+1002−982982.1002
=122−142+142−162+162+....−1982+1982−11002122−142+142−162+162+....−1982+1982−11002
= 122−11002=14−11002<14122−11002=14−11002<14
Vậy B < 14
ta có
\(2.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(5.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}+\dfrac{1}{11}\right)\)
_______________________ X ________________________
\(4.\left(\dfrac{1}{3}+\dfrac{1}{13}+\dfrac{1}{11}+\dfrac{1}{6}\right)\) \(9.\left(\dfrac{1}{4}+\dfrac{1}{7}+\dfrac{1}{6}\dfrac{1}{11}\right)\)
= \(\dfrac{2}{4}X\dfrac{5}{9}\)= \(\dfrac{10}{36}\)= \(\dfrac{5}{18}\)
\(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{1}{2}-\dfrac{1}{x+2}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x+2-2}{2\left(x+2\right)}=\dfrac{4}{9}\)
\(\Rightarrow\dfrac{x}{2x+4}=\dfrac{4}{9}\)
\(\Rightarrow9x=8x+16\)
\(\Rightarrow x=16\)
Vậy x = 16
2/2.4+2/4.6+......+2/x.(x+2)=4/9
1/2-1/4+1/4-1/6+....+1/x-1/x+2=4/9
1/2-1/x+2=4/9
1/x+2=1/2-4/9
1/x+2=1/18
=>x+2=18
x=18-2=16