Bài 2 : Tìm a , biết :
\(\left(1,256\cdot a+1,743\cdot a\right)-2014=2015\)
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b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
a)
( 4x - 9 ) ( 2,5 + (-7/3) . x ) = 0
\(\Rightarrow\orbr{\begin{cases}4x-9=0\\2,5+\frac{-7}{3}x=0\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{9}{4}\\x=\frac{15}{14}\end{cases}}\)
P/s: đợi xíu làm câu b
b) \(\frac{1}{x\left(x+1\right)}\cdot\frac{1}{\left(x+1\right)\left(x+2\right)}\cdot\frac{1}{\left(x+2\right)\left(x+3\right)}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}-\frac{1}{x}=\frac{1}{2015}\)
\(\frac{-1}{x+3}=\frac{1}{2015}\)
\(\Leftrightarrow x+3=-2015\)
\(\Leftrightarrow x=-2018\)
Vậy,.........
\(A=\left(\frac{1}{2^2}-1\right)\cdot\left(\frac{1}{3^2}-1\right)\cdot\left(\frac{1}{4^2}-1\right)\cdot...\cdot\left(\frac{1}{2014^2}-1\right)\)
\(A=\frac{-3}{2^2}\cdot\frac{-8}{3^2}\cdot\frac{-15}{4^2}\cdot...\cdot\frac{-2014^2+1}{2014^2}\)
\(A=\frac{1\cdot\left(-3\right)}{2^2}\cdot\frac{2\cdot\left(-4\right)}{3^2}\cdot\frac{3\cdot\left(-5\right)}{4^2}\cdot...\cdot\frac{2013\cdot\left(-2015\right)}{2014^2}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2013}{2\cdot3\cdot4\cdot...\cdot2014}\cdot\frac{\left(-3\right)\cdot\left(-4\right)\cdot\left(-5\right)\cdot...\cdot\left(-2015\right)}{2\cdot3\cdot4\cdot...\cdot2014}\)
\(A=\frac{1}{2014}\cdot\frac{-2015}{2}\)
\(A=\frac{-2015}{4028}\)
Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
\(\frac{242}{363}+\frac{1616}{2121}=\frac{2}{7}.\frac{2015}{A}\)
\(\frac{11.11.2}{11.11.3}+\frac{101.16}{101.21}=\frac{2}{7}.\frac{2015}{A}\)
\(\frac{2}{3}+\frac{16}{21}=\frac{2}{7}.\frac{2015}{A}\)
\(\Rightarrow\frac{2}{7}.\frac{2015}{A}=\frac{10}{7}\)
\(\frac{2015}{A}=\frac{10}{7}:\frac{2}{7}\)
\(\frac{2015}{A}=5\)
\(A=2015:5\)
\(A=403\)
\(\left(\frac{242}{363}+\frac{1616}{2121}\right)=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow\left(\frac{2}{3}+\frac{16}{21}\right)=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow\frac{10}{7}=\frac{2}{7}.\frac{2015}{A}\)
\(\Leftrightarrow5=\frac{2015}{A}\)
\(\Leftrightarrow A=403\)
\(\left(1,256.a+1,743.a\right)-2014=2015\)
\(\Leftrightarrow a\left(1,256+1,743\right)=2015+2014\)
\(\Leftrightarrow a.2,999=4029\)
\(\Leftrightarrow a=\frac{4029}{2,999}=1343,\left(44781593\right)\)
\(\left(1,256.a+1,743.a\right)-2014=2015\)
\(=\left(1,256.a+1,743.a\right)=2015+2014=4029\)
\(=\left(1,256.a+1,743.a\right)=4029\)
\(=\)gộp lại rùi tính