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làm được bài nào hay bài ấy!
\(A=-\frac{9}{10^{2010}}-\frac{19}{10^{2011}}=-\frac{9}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}=-\frac{9}{10^{2010}}-\frac{9}{10^{2011}}-\frac{1}{10^{2010}}=\frac{8}{10^{2010}}-\frac{9}{10^{2011}}\)\(>B=-\frac{19}{10^{2010}}-\frac{9}{10^{2011}}\)
câu 1 x=0 y =6 hoặc x=4 y=2 or x=9 y=6
câu 2 cho A > B
\(\Leftrightarrow\dfrac{-9}{10^{2010}}-\dfrac{19}{10^{2011}}>\dfrac{-9}{10^{2011}}-\dfrac{19}{10^{2010}}\)
\(\Leftrightarrow\dfrac{-9}{10^{2010}}+\dfrac{19}{10^{2010}}+\dfrac{9}{10^{2011}}-\dfrac{19}{10^{2011}}>0\)
\(\Leftrightarrow\dfrac{10}{10^{2010}}-\dfrac{10}{10^{2011}}>0\)
\(\Leftrightarrow\dfrac{10}{10^{2010}}-\dfrac{1}{10^{2010}}>0\Leftrightarrow\dfrac{9}{10^{2010}}>0\) ( luôn đúng)
vậy A>B
\(\left(2^{19}.27^3+15.4^9.9^4\right):\left(6^9.2^{10}+12^{10}\right)\)
\(=\left[2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4\right]:\left[2^9.3^9.2^{10}+2^{10}.6^{10}\right]\)
\(=\left(2^{19}.3^9+3.5.2^{18}.3^8\right):\left(2^{19}.3^9+2^{10}.2^{10}.3^{10}\right)\)
\(=\left(2^{19}.3^9+5.3^9.2^{18}\right):\left(2^{19}.3^9+2^{20}.3^{10}\right)\)
\(=2^{18}.3^9.\left(1.2+5\right):2^{19}.3^9.\left(1+2.3\right)\)
\(=\left(2^{18}.3^9.7\right):\left(2^{18}.2.3^9.7\right)\)
\(=1:2\)
\(=0.5\)
Ta có:A-1=\(\dfrac{10^8+2}{10^8-1}-1=\dfrac{10^8+2-10^8+1}{10^8-1}=\dfrac{3}{10^8-1}\)
B-1=\(\dfrac{10^8}{10^8-3}-1=\dfrac{10^8-10^8+3}{10^8-3}=\dfrac{3}{10^8-3}\)
Do \(\dfrac{3}{10^8-1}>\dfrac{3}{10^8-3}\)
=>A-1>B-1
<=>A>B
Vậy...
Mình làm gọn nhé ,mình không có thời gian nhiều
\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.2^4.5^3.14}=\frac{-1.3^2.7.4}{7.2}=-18\)
câu kia đề bị sai rồi ,tính không ra
k câu đó mk ghi k sai đâu
hôm nay thầy giải cho mk oy
nhưng mà dù gì thì cx cảm ơn bn nhé!
Từ đề bài, ta có:
\(10^2A=10^2+10^4+10^6+...+10^{2018}\)
\(\Rightarrow100A-A=99A=10^{2018}-1\)
\(\Rightarrow A=\dfrac{10^{2018}-1}{99}\)
A = 1 + 102 + 104 + 106 + ... + 102016
\(\Rightarrow\)10A = 10 + 103 + 105 + ... + 102017
\(\Rightarrow\) 10A - A = 102017 - 1
\(\Rightarrow\) 9A = 102017 - 1
\(\Rightarrow\) A = \(\dfrac{10^{2017}-1}{9}\)
a, Ta có: \(\dfrac{32}{37}>\dfrac{32}{54}>\dfrac{19}{54}\Rightarrow\dfrac{32}{37}>\dfrac{19}{54}\)
b, Ta có: \(\dfrac{18}{53}>\dfrac{18}{54}=\dfrac{1}{3}\Rightarrow\dfrac{18}{53}>\dfrac{1}{3}\left(1\right)\)
\(\dfrac{26}{78}=\dfrac{1}{3}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{18}{53}>\dfrac{26}{78}\)
c, Ta thấy: \(\dfrac{25}{103}< \dfrac{25}{100}=\dfrac{1}{4}\left(1\right)\)
\(\dfrac{74}{295}>\dfrac{74}{296}=\dfrac{1}{4}\left(2\right)\)
Từ (1) và (2) ta suy ra \(\dfrac{25}{103}< \dfrac{74}{295}\)
để n thuộc z thì => 4 ⋮ 2n
=> 2n thuộc Ư(4) = {1;-1;2;-2;4;-4}
ta có bảng
| 2n | 1 | -1 | 2 | -2 | 4 | -4 |
| n | 1/2(loại) | -1/2(loại) | 1 | -1 | 2 | -2 |
vậy n= 1; -1 ;2 ;-2
+) \(\frac{3.4+3.7}{6.5+9}\)=\(\frac{3.4+3.7}{2.3.5+3.3}\)=\(\frac{3.\left(4+7\right)}{3.\left(2.5+3\right)}\)=\(\frac{3.11}{3.13}\)=\(\frac{11}{13}\)
+) \(\frac{6.9-2.17}{63.3-119}\)=\(\frac{2.3.3.3-2.17}{3.3.7-7.17}\)=\(\frac{2.\left(27-17\right)}{7.\left(9-17\right)}\)=\(\frac{2.10}{7.\left(-8\right)}\)=\(\frac{20}{-56}\)=\(\frac{5}{-14}\)=\(\frac{-5}{14}\)
Ta có 13=13; 14= 2.7
MC= BCNN (13;14) =2.7.13=182
\(\frac{11}{13}\)=\(\frac{11.14}{13.14}\)=\(\frac{154}{182}\)
\(\frac{-5}{14}=\frac{-5.13}{14.13}=\frac{-65}{182}\)
Bài 2:
Ta có: \(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-90}{10^{2011}}+\frac{-19}{10^{2011}}=\frac{-109}{10^{2011}}\)
\(B=\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-190}{10^{2011}}=\frac{-199}{10^{2011}}\)
Vì \(\frac{-109}{10^{2011}}>\frac{-199}{10^{2011}}\) nên A > B
Vậy A > B
Cám ơn bạn nhiều !