So sánh A và B
A=3^10+3^11+3^12+3^20+3^21+3^22
B=13x3^10x(3^10+1)
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A = \(\dfrac{10^{20}+3}{10^{21^{ }}+3}\)
B = \(\dfrac{10^{21}+4}{10^{22}+4}\) < 1
\(\Rightarrow\) B < \(\dfrac{10^{21}+4+6}{10^{22}+4+6}\)
\(\Rightarrow\) B < \(\dfrac{10^{21}+10}{10^{22}+10}\)
\(\Rightarrow\) B < \(\dfrac{10\left(10^{20}+1\right)}{10\left(10^{21}+1\right)}\)
\(\Rightarrow\) B < \(\dfrac{10^{20}+1}{10^{21}+1}\) < \(\dfrac{10^{21}+1+2}{10^{22}+1+2}\)
\(\Rightarrow\) B < \(\dfrac{10^{21}+3}{10^{22}+3}\)
\(\Rightarrow\) B < A
\(A=\left(\frac{20}{5}+\frac{27}{9}\right)\times\frac{21}{10}=\left(4+3\right)\times\frac{21}{10}=7\times\frac{21}{10}=\frac{147}{10}\)
\(B=\left(\frac{13}{6}-\frac{3}{8}\right)\times\frac{11}{22}\)
\(B=\left(\frac{52}{24}-\frac{9}{24}\right)\times\frac{11}{22}\)
\(B=\frac{43}{24}\times\frac{1}{2}=\frac{43}{48}\)
Dễ thấy \(A=\frac{147}{10}>1\)
Mà \(B=\frac{43}{48}< 1\)
=> tự so sánh
Bài 2 : So sánh:
107500 = 1012.625 = ( 1012)625
115000 = 118.625 = ( 118)625
Dễ thấy 1012 > 118 nên 107500 > 115000
Chúc bạn học tốt!!!
a: =(21/11+12/11)-(9/19+3/19+6/19)=3-18/19=39/19
b: =(2/3+4/3)+(3/4+7/4)+(4/5+16/5)
=2+4+10/4
=6+5/2
=17/2
a)
Có: \(2>1>0\)
\(\Rightarrow\sqrt{2}>1\Rightarrow1+\sqrt{2}>1+1\\ \Leftrightarrow1+\sqrt{2}>2\)
b) Có: \(0< \sqrt{3}< 3\)
\(\Rightarrow3+1>\sqrt{3}+1\\ \Rightarrow4>\sqrt{3}+1\)
c) Có: \(0< \sqrt{11}< \sqrt{25}\left(0< 11< 25\right)\)
\(\Rightarrow\sqrt{11}< 5\\ \Rightarrow-2\sqrt{11}>-2.5=-10\left(-2< 0\right)\)
d) Có: \(0< \sqrt{11}< \sqrt{16}=4\left(do.0< 11< 16\right)\)
\(\Rightarrow3\sqrt{11}< 3.4\\ \Leftrightarrow3\sqrt{11}< 12\)
a: 2=1+1<1+căn 2
b: 4=1+3>1+căn 3
c: -2căn 11=-căn 44
-10=-căn 100
mà 44<100
nên -2 căn 11>-10
d: 12=3*4=3*căn 16>3*căn 11
a) (x - 3)(y - 3) = 9 = 1.9 = 3.3
Lập bảng:
x - 3 | 1 | -1 | 3 | -3 | 9 | -9 |
y - 3 | 9 | -9 | 3 | -3 | 1 | -1 |
x | 4 | 2 | 6 | 0 | 12 | -3 |
y | 12 | -6 | 6 | 0 | 4 | 2 |
Vậy ...
b) A = \(\frac{10^{19}+1}{10^{20}+1}\) => 10A = \(\frac{10^{20}+10}{10^{20}+1}=1+\frac{9}{10^{20}+1}\)
B = \(\frac{10^{20}+1}{10^{21}+1}\) => 10B = \(\frac{10^{21}+10}{10^{21}+1}=1+\frac{9}{10^{21}+1}\)
Do \(10^{20}+1< 10^{21}+1\) => \(\frac{9}{10^{20}+1}>\frac{9}{10^{21}+1}\) => 10A > 10B => A > B