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\(=5\left(\dfrac{5}{1\cdot6}+\dfrac{5}{6\cdot11}+...+\dfrac{5}{101\cdot106}\right)\\ =5\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+...+\dfrac{1}{101}-\dfrac{1}{106}\right)\\ =5\left(1-\dfrac{1}{106}\right)=5\cdot\dfrac{105}{106}=\dfrac{525}{106}\)
đơn giản câu trả lời là kêu mod Toán hoặc mạng và 1 cách chứng minh Th1 đề sai thì khỏi
Th2 đè đúng thì đề bảo cm thì chắc chắn nó đúng
nếu thấy dk
15/26 + x/16 = 46/52
x/16 = 46/52 - 15/26
x/16 = 1/4
x/16 = 4/16
Suy ra: x = 4
Vậy x = 4
\(\frac{15}{26}+\frac{x}{16}=\frac{46}{52}\)
\(\Rightarrow\frac{x}{16}=\frac{45}{52}-\frac{15}{26}\)
\(\Rightarrow\frac{x}{16}=\frac{4}{13}\Rightarrow\frac{64}{208}=\frac{x}{16}=\frac{64}{208}=\frac{x}{208}\Rightarrow x=64\)
\(B=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+\frac{1}{18\cdot19\cdot20}\)
\(B=\frac{1}{2}\left(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+\frac{2}{18\cdot19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+\frac{1}{3\cdot4}-\frac{1}{4\cdot5}+...+\frac{1}{18\cdot19}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\left(\frac{1}{1\cdot2}-\frac{1}{19\cdot20}\right)\)
\(B=\frac{1}{2}\cdot\frac{189}{380}=\frac{189}{760}\)
\(C=\frac{52}{1\cdot6}+\frac{52}{6\cdot11}+\frac{52}{11\cdot16}+...+\frac{52}{31\cdot36}\)
\(C=\frac{52}{5}\left(\frac{5}{1\cdot6}+\frac{5}{6\cdot11}+\frac{5}{11\cdot16}+...+\frac{6}{31\cdot36}\right)\)
\(C=\frac{52}{5}\left(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+\frac{1}{11}-\frac{1}{16}+...+\frac{1}{31}-\frac{1}{36}\right)\)
\(C=\frac{52}{5}\cdot\left(1-\frac{1}{36}\right)\)
\(C=\frac{91}{9}\)
Đặt \(A=1+5+5^2+5^3+...+5^{402}+5^{403}+5^{404}\)
\(\Rightarrow A=\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{399}+5^{400}+5^{401}\right)+\left(5^{402}+5^{403}+5^{404}\right)\)
\(\Rightarrow A=31.1+31.5^3+...+31.5^{402}\)
\(\Rightarrow A=31\left(1+5^3+5^6+...+5^{402}\right)\)
\(\Rightarrow A⋮31\left(đpcm\right)\)
\(\left(1+5+5^2\right)+\left(5^3+5^4+5^5\right)+...+\left(5^{402}+5^{403}+5^{404}\right)\\ =31+5^3.\left(1+5+5^2\right)+...+5^{402}.\left(1+5+5^2\right)\\ =31+5^3.31+...+5^{402}.31\\ =31.\left(1+5^3+...+5^{402}\right)⋮31\left(DPCM\right)\)
\(B=5+5^2+5^3+...+5^{88}+5^{89}+5^{90}\)
\(=\left(5+5^2+5^3\right)+\left(5^4+5^5+5^6\right)+...+\left(5^{88}+5^{89}+5^{90}\right)\)
\(=5\left(1+5+5^2\right)+5^4\left(1+5+5^2\right)+...+5^{88}\left(1+5+5^2\right)\)
\(=31\left(5+5^4+...+5^{88}\right)⋮31\)
=(2.26).(-31)+26.162
=26(-31.2)+26.162
=26.(-62)+26.162
=26(-62+162)
=26.100
=2600
( - 31 ) . 52 + ( - 26 ) . ( - 162 )
= ( - 31 ) . 52 + 26 . 162
= ( - 31 ) . 52 + 26 . 2 . 81
= ( - 31 ) . 52 + 52 . 81
= 52 . [ ( - 31 ) + 81 ]
= 52 . 50
= 2600
(-31) x 52 + (-26)x ( -162)
=(-31) x (-2) x (-26) + (- 26) x (-162)
= 62 x (-26) + ( - 26) x ( - 162)
=( -26)x [62 + (- 162)
=( - 26) x ( -100)
= 2600
A=52/1.6+52/6.11+....+52/26.31
A=5.(5/1.6+5/6.11+....+5/26.31)
A>5/1.6+5/6.11+...+5/26.31
A>1-1/6+1/6-1/11+...+1/26-1/31
a>1-1/31>1
VẬY A>1