so sánh
1+[15 phần (2021^10)-8] và 1+[5 phần (2021^11)-5]
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1) `-3\sqrt13=-3\sqrt13`
`-9=-3\sqrt9`
`\sqrt13>\sqrt9`
`=> -3\sqrt13 < -3\sqrt9`
`=> -3\sqrt13 < 9`.
2) `\sqrt15 < \sqrt16`
`<=> \sqrt15-1 < \sqrt16-1`
`<=> \sqrt15-1 < 3 < \sqrt10`
`=> \sqrt15-1 <\sqrt10`
3) `5=4+1=\sqrt16+1`
`\sqrt8+1=\sqrt8+1`
`=> 5>\sqrt8+1`
1) \(-3\sqrt{13}=-\sqrt{117}< -\sqrt{81}=-9\)
3) Ta có: \(5^2=25=9+16\)
\(\left(2\sqrt{2}+1\right)^2=9+4\sqrt{2}\)
mà \(16>4\sqrt{2}\)
nên \(5>2\sqrt{2}+1\)
18:
a: \(S=3\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+...+\dfrac{2}{98\cdot100}\right)\)
=3*(1/2-1/4+1/4-1/6+...+1/98-1/100)
=3*49/100=147/100
b: Để A là số nguyên thì n-1 thuộc Ư(2)
=>n-1 thuộc {1;-1;2;-2}
=>n thuộc {2;0;3;-1}
\(A=\dfrac{2021^{10}-2021+2020}{2021^9-1}\\ =\dfrac{2021\left(2021^9-1\right)+2020}{2021^9-1}\\ =2021+\dfrac{2020}{2021^9-1}\\ B=\dfrac{2021^{11}-1}{2021^{10}-1}=2021+\dfrac{2020}{2021^{10}-1}\)
Ta có:
\(2021^9-1< 2021^{10}-1\\ \Rightarrow\dfrac{2020}{2021^9-1}>\dfrac{2020}{2021^{10}-1}\)
Do đó A > B.
Ta có \(b-a=9.10^{2019}-\dfrac{9}{10^{2021}}>0\Rightarrow b>a\).
A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) ; B = \(\dfrac{2020+2021}{2021+2022}\)
B = \(\dfrac{2020+2021}{2021+2022}\) = \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\)
\(\dfrac{2020}{2021}\) > \(\dfrac{2020}{2021+2022}\)
\(\dfrac{2021}{2022}\) > \(\dfrac{2021}{2021+2022}\)
Cộng vế với vế ta có:
A = \(\dfrac{2020}{2021}\) + \(\dfrac{2021}{2022}\) > \(\dfrac{2020}{2021+2022}\) + \(\dfrac{2021}{2021+2022}\) = B
Vậy A > B
A = \(\dfrac{10^{10}-1}{10^{11}-1}\)
A \(\times\) 10 = \(\dfrac{(10^{10}-1)\times10}{10^{11}-1}\) = \(\dfrac{10^{11}-10}{10^{11}-1}\) = 1 - \(\dfrac{9}{10^{11}-1}\) < 1
B = \(\dfrac{10^{10}+1}{10^{11}+1}\)
B \(\times\) 10 = \(\dfrac{(10^{10}+1)\times10}{10^{11}+1}\) = \(\dfrac{10^{11}+10}{10^{11}+1}\) = 1 + \(\dfrac{9}{10^{11}+1}\) > 1
Vì 10 A< 1< 10B
Vậy A < B
S=1+(2-3)+(-4+5)+(6-7)+(-8+9)+...+(-2020+2021)
S=1-1+1-1+1+...+1
S=1+0+0+...+0
S=1
a) Ta có: \(\left(-68+42\right)-\left(5042-6068\right)-\left(-2\right)^0\)
\(=-68+42-5042+6068-1\)
\(=6000-5000-1\)
\(=999\)
b) Ta có: \(29\cdot\left(19-37\right)-19\cdot\left(29-37\right)\)
\(=29\cdot19-29\cdot37-29\cdot19+19\cdot37\)
\(=-29\cdot37+19\cdot37\)
\(=37\cdot\left(-29+19\right)\)
\(=37\cdot\left(-10\right)=-370\)
c) Ta có: \(\left(-15\right)\cdot24+15\cdot\left(-75\right)-15\)
\(=\left(-15\right)\cdot24+\left(-15\right)\cdot75+\left(-15\right)\cdot1\)
\(=\left(-15\right)\cdot\left(24+75+1\right)\)
\(=-15\cdot100=-1500\)
d) Ta có: \(\frac{1}{5}+\frac{4}{7}-\frac{11}{35}\)
\(=\frac{5}{35}+\frac{20}{35}-\frac{11}{35}\)
\(=\frac{14}{35}=\frac{2}{5}\)
e) Ta có: \(\left(13\cdot95-73\right)-\left(13\cdot45+27\right)-\left(-1\right)^{2021}\)
\(=13\cdot95-73-13\cdot45-27-\left(-1\right)\)
\(=13\left(95-45\right)-\left(73+27\right)+1\)
\(=13\cdot50-100+1\)
\(=551\)
Bài 1: Tính hợp lý
a) ( - 68 + 42 ) - ( 5042 - 6068 ) - (-2)\(^0\)
= -26 - (-1026) - 1
= - 26 + 1026 - 1
= 1000 - 1
= 999
Mấy câu còn lại c hơi làm biếng, sr :<<