\(\dfrac{4}{3.5}\) + \(\dfrac{8}{5.9}\) + \(\dfrac{12}{9.15}\)+.......+
\(\dfrac{32}{x(x+16)}\) = \(\dfrac{16}{15}\)
\(\dfrac{32}{x(x+16)}\) = \(\dfrac{16}{15}\)
\(\dfrac{HeLp}{me}\) \(\dfrac{nguyễn}{đức}{Trí}\)
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a) \(\left\{{}\begin{matrix}39=3.13⋮13\\143=11.13⋮13\\221=17.13⋮13\end{matrix}\right.\) \(\Rightarrow\left(39+143+221\right)⋮13\)
b) \(\left\{{}\begin{matrix}208=16.13⋮13\\169=13.13⋮13\end{matrix}\right.\) \(\Rightarrow\left(208-169\right)⋮13\)
c) \(37⋮̸13\) \(\Rightarrow\left(65+37+195\right)⋮̸13\)
c) \(37\) không chia hết 13 và 65;195 chia hết cho 13 nên tổng 65+37+195 không chia hết cho13
3) \(...\Rightarrow2^x\left(2^3+1\right)=36\)
\(\Rightarrow2^x.9=36\)
\(\Rightarrow2^x=4\)
\(\Rightarrow2^x=2^2\Rightarrow x=2\)
4) \(...\Rightarrow4^{x+1}-4^x=12\)
\(\Rightarrow4^x\left(4-1\right)=12\)
\(\Rightarrow4^x.3=12\)
\(\Rightarrow4^x=4=4^1\Rightarrow x=1\)
5) \(...\Rightarrow5^{x+1}\left(5^2-1\right)=3000\)
\(\Rightarrow5^{x+1}.24=3000\)
\(\Rightarrow5^{x+1}=125\)
\(\Rightarrow5^{x+1}=5^3\)
\(\Rightarrow x+1=3\)
\(\Rightarrow x=2\)
6) Bạn xem lại đề
a. \(2^x.2^3+2^x=36\)
\(2^x\left(2^3+1\right)=36\)
\(2^x.9=36\)
\(2^x=4\Rightarrow x=2\)
b. \(4^x.4^1-\left(2^2\right)^x=12\)
\(4^x.4-4^x=12\)
\(4^x\left(4-1\right)=12\)
\(4^x.3=12\)
\(4^x=4\)
x = 1
c. \(5^x.5^3-5^x.5^1=3000\)
\(5^x\left(5^3-5^1\right)=3000\)
\(5^x.120=3000\)
\(5^x=25\)
x = 2
d. \(4^{x+1}=2^{2x}\)
\(4^x.4=\left(2^2\right)^x\)
\(4^x.4=4^x\)
Có vẻ như câu 4 này để bài thiếu
\(1+\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{10}+...+\dfrac{2}{x\left(x+1\right)}=1\dfrac{1989}{1991}\)
\(\Rightarrow2\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{x\left(x+1\right)}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow2\left(1-\dfrac{1}{x+1}\right)=\dfrac{3980}{1991}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{3980}{1991}.\dfrac{1}{2}\)
\(\Rightarrow1-\dfrac{1}{x+1}=\dfrac{1990}{1991}\)
\(\Rightarrow\dfrac{1}{x+1}=1-\dfrac{1990}{1991}\)
\(\Rightarrow\dfrac{1}{x+1}=\dfrac{1}{1991}\)
\(\Rightarrow x+1=1991\)
\(\Rightarrow x=1990\)
\(\left(x-4\right)^9=49.\left(x-4\right)^7\\ =>\left(x-4\right)^9:\left(x-4\right)^7=49\\ =>\left(x-4\right)^2=49\\ =>\left[{}\begin{matrix}x-4=7\\x-4=-7\end{matrix}\right.\\ =>\left[{}\begin{matrix}x=11\\x=-3\end{matrix}\right.\)
(x - 4)9 = 49 . (x - 4)7
(x - 4)9 : (x - 4)7 = 49
(x - 4)2 = 72 = (-7)2
TH1 : TH2 :
(x - 4)2 = 72 (x - 4)2 = (-7)2
x - 4 = 7 x - 4 = -7
x = 7 + 4 x = -7 + 4
x = 11 x = -3
Vậy x = 11 Vậy x = -3
Vì M nằm giữa và OM = 1/2 ON nên M là trung điểm của ON.
\(\Rightarrow\) Độ dài MN là 3 cm
mà 3 cm = 3cm nên MN = OM
Vậy MN = OM.
\(\dfrac{4}{3.5}+\dfrac{8}{5.9}+\dfrac{12}{9.15}+...+\dfrac{32}{x\left(x+16\right)}=\dfrac{16}{15}\)
\(2.\left(\dfrac{2}{3.5}+\dfrac{4}{5.9}+\dfrac{6}{9.15}+..+\dfrac{16}{X.\left(X+16\right)}\right)=\dfrac{16}{15}\)
\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{15}+...+\dfrac{1}{X}-\dfrac{1}{X+16}=\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{1}{3}-\dfrac{8}{15}\)
\(\dfrac{1}{X+16}=\dfrac{-1}{5}\)
\(X+16=-5\)
\(X=-21\)