2 1
_ - _ =
3 12
K
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LM
10 tháng 12 2017
\(x^3-x^2-x=\frac{1}{3}\)
\(\Leftrightarrow x^3=x^2+x+\frac{1}{3}\)
\(\Leftrightarrow3x^2=3\left(x^2+x+\frac{1}{3}\right)\)
\(\Leftrightarrow3x^2=3x^2+3x+1\)
\(\Leftrightarrow3x^3+x^3=x^3+3x^3+3x+1\)
\(\Leftrightarrow4x^3=\left(x+1\right)^3\)
\(\Leftrightarrow\sqrt[3]{\left(4x^3\right)}=\sqrt[3]{\left(x+1\right)^3}\)
\(\Leftrightarrow\sqrt[3]{4.x}=x+1\)
\(\Leftrightarrow\sqrt[3]{4.x}-x=1\)
\(\Leftrightarrow x\left(\sqrt[3]{4}-1\right)=1\)
\(\Leftrightarrow x=\frac{1}{\left(\sqrt[3]{4}-1\right)}\)
TK
0
22 tháng 2 2018
(3/2)x+1 = (3/2)x . (3/2)
(3/2)x . ( (3/2) - 1) = 27/16
Xem lại công thức lũy thừa là làm đk nhé
Kết quả x = 3
21 tháng 2 2018
\(\left(\frac{3}{2}\right)^{x+1}-\left(\frac{3}{2}\right)^x=\frac{27}{16}\)
\(\left(\frac{3}{2}\right)^{\left(x+1\right):x}=\frac{27}{16}\)
\(\left(\frac{3}{2}\right)^{\left(x+1\right):x}=\frac{3^3}{2^4}\)
\(\Rightarrow\frac{x+1}{x}=\frac{3}{4}\)
\(\Rightarrow3x=4x+4\)
\(\Rightarrow3x-4x=4\)
\(\Rightarrow-x=4\)
\(\Rightarrow x=-4\)
MV
9 tháng 8 2017
2.
\(A=\dfrac{36}{1\cdot3\cdot5}+\dfrac{36}{3\cdot5\cdot7}+...+\dfrac{36}{25\cdot27\cdot29}\\ =9\cdot\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\\ =9\cdot\dfrac{1}{3}-9\cdot\dfrac{1}{783}\\ =3-\dfrac{1}{87}< 3\)
Vậy \(A< 3\)
b,
\(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{50}\\ B< 2-\dfrac{1}{50}< 2\)
Vậy \(B< 2\)
MV
10 tháng 8 2017
\(P=\dfrac{2}{60\cdot63}+\dfrac{2}{63\cdot66}+...+\dfrac{2}{117\cdot120}+\dfrac{2}{2011}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)\)
\(Q=\dfrac{5}{40\cdot44}+\dfrac{5}{44\cdot48}+...+\dfrac{5}{76\cdot80}+\dfrac{5}{2011}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\)
\(\dfrac{3}{2011}< \dfrac{4}{2011}\Rightarrow\dfrac{1}{2}+\dfrac{3}{2011}< \dfrac{1}{2}+\dfrac{4}{2011}\left(1\right)\)
\(\dfrac{2}{3}< \dfrac{5}{4}\left(2\right)\)
Từ (1) và (2) ta có: \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)< \dfrac{5}{4}\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\Leftrightarrow P< Q\)
Vậy P < Q
TH
1
23 tháng 10 2023
\(\int\limits^1_{\sqrt{ }3}\)\(\sqrt{\left(1+x^2\right)}\)\(dx\)
7 tháng 1 2017
2+2+1+2-1+5-3+6=14 nhe ban
tk minh di
chuc cac ban nam moi an khang thinh vuong!
TL :
2/3 - 1/12 = 8/12 - 1/12
= 7/12
HT
2 / 3 - 1 / 12 = 8 / 12 - 1 / 12 = 7 / 12