1 + 2+ 3 +...+ 29 =
Tuyển Cộng tác viên Hoc24 nhiệm kì 28 tại đây: https://forms.gle/GrfwFgzveoKLVv3p6
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17 tháng 7 2021 lúc 20:46
(29^1) . (29^2) . (29^3) . (29^4)
\(29^1\cdot29^2\cdot29^3\cdot29^4=29^{10}\)
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291 . 292 . 293 . 294 = 291+2+3+4=2910
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TÍNH S :1*2+2*3+3*4+...+28*29+29*30
S = 1 * 2 + 2 * 3 + 3 * 4 + ...... + 28 * 29 + 29 * 30
3S = 1 x 2 x 3 - 1 x 2 x 3 + 2 x 3 x 4 - 2 x 3 x 4 + .............. + 29 x 30 x 31
3S = 29 x 30 x 31
S = 29 x 30 x 31 : 3 = 8990
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Ta có : S = 1.2 + 2.3 + 3.4 + ..... + 28.29 + 29.30
<=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ...... + 29.30.31
<=> 3S = 29.30.31
<=> S = 29.30.31 / 3 = 8990
Chúc học tốt nhé !
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s = 1.2 + 2.3 + 3.4 +...+ 29.29 + 29.30
=> 3S = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 +...+ 29.30.31
3S = 29.30.31
S = 29.30 . 31/3 = 8990
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7-(2x-1/3)^2=3
(2x+1/3)^2-3/8=1/8
12:[29-(x-2/3)^2]=3
(3x-1/2)^3+8/3=29/9-14/27
2(2x-1/3)^2+4/3=5/6+13/18
Tìm X
a: ta có: \(7-\left(2x-\frac13\right)^2=3\)
=>\(\left(2x-\frac13\right)^2=7-3=4\)
=>\(\left[\begin{array}{l}2x-\frac13=2\\ 2x-\frac13=-2\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=2+\frac13=\frac73\\ 2x=-2+\frac13=-\frac53\end{array}\right.\)
=>\(\left[\begin{array}{l}x=\frac73:2=\frac76\\ x=-\frac53:2=-\frac56\end{array}\right.\)
b: \(\left(2x+\frac13\right)^2-\frac38=\frac18\)
=>\(\left(2x+\frac13\right)^2=\frac38+\frac18=\frac48=\frac12\)
=>\(2x+\frac13=\sqrt{\frac12}=\frac{\sqrt2}{2}\)
=>\(2x=\frac{\sqrt2}{2}-\frac13=\frac{3\sqrt2-2}{6}\)
=>\(x=\frac{3\sqrt2-2}{12}\)
c: \(12:\left\lbrack29-\left(x-\frac23\right)^2\right\rbrack=3\)
=>\(29-\left(x-\frac23\right)^2=12:3=4\)
=>\(\left(x-\frac23\right)^2=29-4=25\)
=>\(\left[\begin{array}{l}x-\frac23=5\\ x-\frac23=-5\end{array}\right.\Rightarrow\left[\begin{array}{l}x=5+\frac23=\frac{17}{3}\\ x=-5+\frac23=-\frac{13}{3}\end{array}\right.\)
d: \(\left(3x-\frac12\right)^3+\frac83=\frac{29}{9}-\frac{14}{27}\)
=>\(\left(3x-\frac12\right)^3+\frac83=\frac{87}{27}-\frac{14}{27}=\frac{73}{27}\)
=>\(\left(3x-\frac12\right)^3=\frac{73}{27}-\frac83=\frac{73}{27}-\frac{72}{27}=\frac{1}{27}=\left(\frac13\right)^3\)
=>\(3x-\frac12=\frac13\)
=>\(3x=\frac12+\frac13=\frac56\)
=>\(x=\frac{5}{18}\)
e: \(2\left(2x-\frac13\right)^2+\frac43=\frac56+\frac{13}{18}\)
=>\(2\left(2x-\frac13\right)^2=\frac{15}{18}+\frac{13}{18}-\frac43=\frac{28}{18}-\frac{24}{18}=\frac{4}{18}=\frac29\)
=>\(\left(2x-\frac13\right)^2=\frac19\)
=>\(\left[\begin{array}{l}2x-\frac13=\frac13\\ 2x-\frac13=-\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac23\\ 2x=0\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac13\\ x=0\end{array}\right.\)
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Tìm x
\(\frac{-3}{29}+\frac{-7}{29}\le\frac{x}{29}\le\frac{-3}{29}-\frac{5}{29}\)
Tìm \(n\in Z\)để \(\frac{n^2+n-5}{n+2}\)là số nguyên
Tính \(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+......+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(1,-\frac{3}{29}+\frac{-7}{29}\le\frac{x}{29}\le-\frac{3}{29}-\frac{5}{29}\)
\(\Rightarrow-\frac{10}{29}\le\frac{x}{29}\le-\frac{8}{29}\Rightarrow-10\le x\le-8\)
\(\Rightarrow x=\left\{-8;-9;-10\right\}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)
\(S=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}+\frac{1}{2^{100}}\)
\(\Rightarrow2S-S=S=1-\frac{1}{2^{100}}\)
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cmr: (29^n + 1) . (29^n + 2) . (29^n + 3). (29^n +4) chia hết cho 5 với mọi n
A=((2-2/19+2/23-1/1010)/(3-3/19+3/23-3/2020))x((4-4/29+4/41-1/505)/(5-5/29+5/41-1/404))
giúp mình với
(4 5/58 - 3 1/2 + 8 15/29)-(3 5/58 - 7 14/29)
\(=4+\dfrac{5}{58}-3-\dfrac{1}{2}+8+\dfrac{15}{29}-3-\dfrac{5}{58}+7+\dfrac{14}{29}=13+\dfrac{1}{2}=\dfrac{27}{2}\)
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\(S=\dfrac{\sqrt{2}-\sqrt{1}}{1+2}+\dfrac{\sqrt{3}-\sqrt{2}}{2+3}+\dfrac{\sqrt{4}-\sqrt{3}}{3+4}+...+\dfrac{\sqrt{25}-\sqrt{29}}{29+25}< \dfrac{2}{5}\)
Xét :\(\dfrac{\sqrt{n+1}-\sqrt{n}}{n+\left(n+1\right)}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}=\dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}< \dfrac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}=\dfrac{1}{2}\left(\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)\)
Do đó :
\(S< \dfrac{1}{2}\left(\dfrac{1}{\sqrt{1}}-\dfrac{1}{\sqrt{2}}+\dfrac{1}{\sqrt{2}}-\dfrac{1}{\sqrt{3}}+...+\dfrac{1}{\sqrt{24}}-\dfrac{1}{\sqrt{25}}\right)=\dfrac{1}{2}\left(1-\dfrac{1}{5}\right)=\dfrac{2}{5}\)
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Cho \(A=3+3^2+3^3+...+3^{29}\) và \(B=\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{29}}\). Tính \(\dfrac{A}{B}\)
Ta có :
\(A=3+3^2+3^3+........+3^{29}\)
\(\Rightarrow3A=3^2+3^3+...............+3^{29}+3^{30}\)
\(\Rightarrow3A-A=\left(3^2+3^3+........+3^{30}\right)-\left(3+3^3+................+3^{29}\right)\)
\(\Rightarrow2A=3^{30}-3\)
\(\Rightarrow A=\dfrac{3^{30}-3}{2}\)
Lại có :
\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+................+\dfrac{1}{3^{29}}\)
\(\Rightarrow3B=1+\dfrac{1}{3}+\dfrac{1}{3^2}+.............+\dfrac{1}{3^{28}}\)
\(\Rightarrow3B-B=\left(1+\dfrac{1}{3}+\dfrac{1}{3^2}+.......+\dfrac{1}{3^{28}}\right)-\left(\dfrac{1}{3}+\dfrac{1}{3^2}+..........+\dfrac{1}{3^{29}}\right)\)
\(\Rightarrow2B=1-\dfrac{1}{3^{29}}\)
\(\Rightarrow B=\dfrac{1-\dfrac{1}{3^{29}}}{2}\)
\(\dfrac{\Rightarrow A}{B}=\dfrac{\dfrac{3^{30}-3}{2}}{\dfrac{1-\dfrac{1}{3^{29}}}{2}}\)
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\(B=\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{29}}\)
\(3^{30}.B=3^{29}+3^{28}+...+3=A\)
\(\dfrac{A}{B}=\dfrac{3^{30}.B}{B}=3^{30}\)
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Ai giải giúp mik bài này với:
Bài 1: Tính giá trị biểu thức
A=29 1/2 x 2/3 + 39 1/3 x 3/4 + 5/6
( số 29 1/2 ý mik nói là 29 và 1 phần 2 nhé số 39 1/3 cũng thế nhé)
Bài 2;
Tìm số tự nhiên x biết 100,1 < 6x + x6 < 111,1
A=29 1/2 * 2/3 +39 1/3 * 3/4 + 5/6
A=29 1/2 * 39 1/3 * (2/3 + 3/4 + 5/6)
A=29 1/2 * 39 1/3 * (1/2 + 5/6)
A=29 1/2 * 39 1/3 * 4/3
A=29 1/2 * 52
A=1534
Dấu * là dấu nhân nha !!!
Còn bài 2 mình ko biết