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\(999993^{1999}-555557^{1997}=\left(999993^4\right)^{499}.999993^3-\left(555557^4\right)^{499}.555557\)
\(=\left(....1\right)^{499}.999993-\left(.....1\right)^{499}.555557=\left(....3\right)-\left(.....7\right)=\left(.....6\right)\)
\(\frac{1}{41}+\frac{1}{42}+....+\frac{1}{80}=\left(\frac{1}{41}+\frac{1}{42}+....+\frac{1}{60}\right)+\left(\frac{1}{61}+\frac{1}{62}+....+\frac{1}{80}\right)\)
\(< \left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\left(20\text{ số hạng}\right)\right)+\left(\frac{1}{60}+\frac{1}{60}+....+\frac{1}{60}\left(20\text{ số hạng}\right)\right)=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60 => (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60 và 1/61> 1/62> ... >1/79> 1/80 => (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy: 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12 => 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 7/12 => ĐPCM
Ta có : 1/41 + 1/42 + ... + 1/60 > 1/60 * 20 = 1/3 .
1/61 + 1/62 + ... + 1/80 > 1/80 * 20 = 1/4 .
⇒ 1/41 + 1/42 + ... + 1/80 > 1/3 + 1/4 = 4/12 + 3/12 .
= 7/12 .
Do đó : A > 7/12 .
Vậy bài toán được chứng minh .
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 \(x\in\left\{\dfrac{3}{2};1;2\right\}\)
đó giúp mk đi mà
à, mk quên chưa nói là ai giúp mk sẽ được luôn 2SP đó
giúp mk nha
cảm ơn nhiều!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Đặt \(A=\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+\dfrac{1}{44}+...+\dfrac{1}{80}\)
\(=\left(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}\right)+\) \(\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}\right)\)
Nhận xét:
\(\dfrac{1}{41}+\dfrac{1}{42}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\) \(=\dfrac{1}{3}\)
\(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{80}>\dfrac{1}{80}+\dfrac{1}{80}+...+\dfrac{1}{80}\) \(=\dfrac{1}{4}\)
\(\Rightarrow A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}>\dfrac{1}{12}\)
Vậy \(\dfrac{1}{41}+\dfrac{1}{42}+\dfrac{1}{43}+...+\dfrac{1}{80}>\dfrac{1}{12}\) (Đpcm)
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
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)
\(\Rightarrow B=1\left(2^1+2^2+...+2^6\right)+2^6\left(2^1+2^2+...+2^6\right)+.....+2^{24}\left(2^1+2^2+...+2^6\right)\)
\(\Rightarrow B=1.126+2^6.126+2^{12}.126+...+2^{24}.126\)
\(\Rightarrow B=126.\left(1+2^6+2^{12}+...+2^{24}\right)\)
\(\Rightarrow B=21.6.\left(1+2^6+2^{12}+...+2^{24}\right)⋮21\)
\(\Rightarrow B⋮21\)
Ta có:
A=9999931999−5555571997
A=9999931998.999993−5555571996.555557
A=(9999932)999.999993 − (5555572)998.555557
A=\(\overline{\left(....9\right)}^{999}\) . 999993 - \(\overline{\left(...1\right)}.\text{555557}\)
A=\(\overline{\left(...7\right)}-\overline{\left(...7\right)}\)
A= \(\overline{\left(...0\right)}\)
Vì A có tận cùng là 0 nên \(A⋮5\)