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đó giúp mk đi mà
à, mk quên chưa nói là ai giúp mk sẽ được luôn 2SP đó
giúp mk nha
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Bài 1:
a) Ta có:
\(3^{200}=\left(3^2\right)^{100}=9^{100}\)
\(2^{300}=\left(2^3\right)^{100}=8^{100}\)
Vì \(9^{100}>8^{100}\Rightarrow3^{200}>2^{300}\)
b) Ta có:
\(71^{50}=\left(71^2\right)^{25}=5041^{25}\)
\(37^{75}=\left(37^3\right)^{25}=50653^{25}\)
Vì \(5041^{25}< 50653^{25}\Rightarrow71^{50}< 37^{75}\)
c) Ta có:
\(\frac{201201}{202202}=\frac{201.1001}{202.1001}=\frac{201}{202}\)
\(\frac{201201201}{202202202}=\frac{201.1001001}{202.1001001}=\frac{201}{202}\)
\(\Rightarrow\frac{201201}{202202}=\frac{201201201}{202202202}\)
Bài 2:
a) \(A=\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+....+\frac{1}{50^2}\)
Ta có: \(\frac{1}{1^2}=1;\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\Rightarrow\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}< 1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
\(\Rightarrow A< 1+1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}\)
\(\Rightarrow A< 1+1-\frac{1}{50}\)
\(\Rightarrow A< 2-\frac{1}{50}< 2\)
b) \(B=2^1+2^2+2^3+...+2^{30}\) (Có 30 số hạng)
\(\Rightarrow B=\left(2^1+2^2+...+2^5+2^6\right)+\left(2^7+2^8+2^9+...+2^{12}\right)+...+\left(2^{25}+2^{26}+...+2^{29}+2^{30}\right)\)
(có \(30:6=5\) nhóm)
\(\Rightarrow B=1\left(2^1+2^2+...+2^6\right)+2^6\left(2^1+2^2+...+2^6\right)+.....+2^{24}\left(2^1+2^2+...+2^6\right)\)
\(\Rightarrow B=1.126+2^6.126+2^{12}.126+...+2^{24}.126\)
\(\Rightarrow B=126.\left(1+2^6+2^{12}+...+2^{24}\right)\)
\(\Rightarrow B=21.6.\left(1+2^6+2^{12}+...+2^{24}\right)⋮21\)
\(\Rightarrow B⋮21\)
a) M =1+3+32+33+......+3118+3119
M = ( 1+3+32 ) +...+ ( 3117 + 3118+3119 )
M = 1. ( 1+3+32 ) + ... + 3117 . ( 3117 + 3118+3119 )
M = ( 1+3+32 ) .( 1 + ... + 3117 )
M = 13 . ( 1 + ... + 3117 ) \(⋮\) 13 (đpcm )
b) Ta có:
\(\dfrac{1}{2^2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}< \dfrac{1}{2.3}\)
\(\dfrac{1}{4^2}< \dfrac{1}{3.4}\)
...
\(\dfrac{1}{2009^2}< \dfrac{1}{2008.2009}\)
\(\dfrac{1}{2010^2}< \dfrac{1}{2009.2010}\)
=> \(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\) (1)
Biến đổi vế trái:
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2008.2009}+\dfrac{1}{2009.2010}\)
= \(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2008}-\dfrac{1}{2009}+\dfrac{1}{2009}-\dfrac{1}{2010}\)
= \(1-\dfrac{1}{2010}\)
= \(\dfrac{2009}{2010}< 1\) (2)
Từ (1) và (2), suy ra :
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2009^2}+\dfrac{1}{2010^2}\) < 1 hay:
N < 1
2. Chứng tỏ:\(\dfrac{2}{5}< A< \dfrac{8}{9}.\)
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)
Giải:
Ta có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}.\)
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{8.9}.\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{8}-\dfrac{1}{9}.\)
\(A< 1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{8}-\dfrac{1}{8}\right)-\dfrac{1}{9}.\)
\(A< 1+0+0+0+...+0-\dfrac{1}{9}.\)
\(A< 1-\dfrac{1}{9}.\)
\(A< \dfrac{8}{9}_{\left(1\right)}.\)
Ta lại có:
\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{9^2}.\)
\(A=\dfrac{1}{2.2}+\dfrac{1}{3.3}+\dfrac{1}{4.4}+...+\dfrac{1}{9.9}.\)
\(A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}.\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}.\)
\(A>\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+\left(\dfrac{1}{5}-\dfrac{1}{5}\right)+...+\left(\dfrac{1}{9}-\dfrac{1}{9}\right)-\dfrac{1}{10}.\)
\(A>\dfrac{1}{2}+0+0+0+...+\dfrac{1}{10}.\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}.\)
\(A>\dfrac{4}{10}.\)
\(\Rightarrow A>\dfrac{2}{5}_{\left(2\right)}.\) (vì \(\dfrac{4}{10}=\dfrac{2}{5}.\))
Từ \(_{\left(1\right)}\) và \(_{\left(2\right)}\).
\(\Rightarrow A< \dfrac{8}{9}\) và \(A>\dfrac{2}{5}.\)
\(\Rightarrow\) \(\dfrac{8}{9}>A>\dfrac{2}{5}\) hay \(\dfrac{2}{5}< A< \dfrac{8}{9}.\)
Vậy ta thu được \(đpcm.\)
~ Học tốt!!!... ~ ^ _ ^
Câu 2 : Câu hỏi của Nguyễn Thu Hà - Toán lớp 6 | Học trực tuyến
Câu 2:
a: \(\Leftrightarrow12x-60=7x-5\)
=>5x=55
=>x=11
b: \(\Leftrightarrow\left(2x-3\right)^{2010}\left[\left(2x-3\right)^2-1\right]=0\)
=>(2x-3)(2x-2)(2x-4)=0
hay \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
a, \(\dfrac{x-2}{5}=\dfrac{x}{3}\)
\(\Leftrightarrow3\left(x-2\right)=5x\)
\(\Leftrightarrow3x-6=5x\)
\(\Leftrightarrow5x-3x=6\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=3\)
b, \(\dfrac{x+23}{x+40}=\dfrac{3}{4}\)
\(\Leftrightarrow4\left(x+23\right)=3\left(x+40\right)\)
\(\Leftrightarrow4x+92=2x+80\)
\(\Leftrightarrow4x-2x=80-92\)
\(\Leftrightarrow2x=-12\)
\(\Leftrightarrow x=-6\)
c, \(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...........+\dfrac{1}{2^{2017}}\)
\(\Leftrightarrow2A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...........+\dfrac{1}{2^{2016}}\)
\(\Leftrightarrow2A-A=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2016}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+........+\dfrac{1}{2^{2017}}\right)\)
\(\Leftrightarrow A=1-\dfrac{1}{2^{2017}}\)
d, \(B=1+2+2^2+........+2^{2017}\)
\(\Leftrightarrow2B=2+2^2+2^3+......+2^{2018}\)
\(\Leftrightarrow2B-B=\left(2+2^2+.....+2^{2018}\right)-\left(1+2+....+2^{2017}\right)\)
\(\Leftrightarrow B=2^{2018}-1\)
\(\dfrac{x-2}{5}=\dfrac{x}{3}=>3\left(x-2\right)=5x\)
\(< =>3x-6=5x=>x=-3\)
\(\dfrac{x+23}{x+40}=\dfrac{3}{4}=>4\left(x+23\right)=3\left(x+40\right)\)
\(4x+92=3x+120=>x=28\)
1. Tính tổng:
B = 2 - 4 - 6 + 8 + 10 - 12 - 14 + 16 + ... + 2002 - 2004 - 2006 + 2008
=> ( 2 - 4 - 6 + 8 )+ (10 - 12 - 14 + 16) + ... + (2002 - 2004 - 2006 + 2008)
=> (-8+ 8) +(-16+ 16) +.........+ ( -2008+ 2008)(1)
=> 0+0+...........+0
=> 0
Ta thấy (1) đều là những số đối nên kết quả đường nhiên bằng 0
\(A=1+4+4^2+4^3+...+4^{99}\\ \Rightarrow4A=4+4^2+4^3+...+4^{100}\\ \Rightarrow3.A=4^{100}-1\\ \Rightarrow A=\dfrac{4^{100}-1}{3}< \dfrac{4^{100}}{3}=\dfrac{B}{3}\\ \Rightarrow A< \dfrac{B}{3}\)
1.Tính hợp lý:
a. 1152 - (374 + 1152) + (374 - 65) = 1152 - 374 - 1152 + 374 - 65 = ( 1152 - 1152 ) + ( -65) + ( 374 - 374 ) = 0 + ( - 65) + 0 = -65
Bài 1 : Tính hợp lý : c. \(\dfrac{11.3^{22}.3^7-9^{15}}{\left(2.3^{14}\right)^2}\) = \(\dfrac{11.3^{29}-3^{30}}{2^2.3^{28}}\) = \(\dfrac{3^{29}.\left(11-3\right)}{2^2.3^{28}}\) = \(\dfrac{3^{29}.2^3}{2^2.3^{28}}\) = 6
b.ta chia B thành 10 nhóm mỗi nhóm có 6 hạng tử \(B=\left(2+2^2+2^3+2^4+2^5+2^6\right)+....+\left(2^{55}+2^{56}+2^{57}+2^{58}+2^{59}+2^{60}\right)\)
\(B\text{=}2\left(1+2+2^2+2^3+2^4+2^5\right)+...+2^{55}\left(1+2+2^2+2^3+2^4+2^5\right)\)
\(B\text{=}2.63+...+2^{56}.63\)
\(\Rightarrow B⋮63\)
\(\Rightarrow B⋮21\)