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Ta có: \(\frac{-11}{3^7.7^3}=\frac{-11}{\frac{3^7.7^4.1}{7}}=-\frac{77}{3^7.7^4}=\frac{-78+1}{3^7.7^4}=-\frac{78}{3^7.7^4}+\frac{1}{3^7.7^4}\)
Do \(3^7.7^4>3^4.7^4\) => \(\frac{78}{3^7.7^4}< \frac{78}{3^4.7^4}\) => \(-\frac{78}{3^7.7^4}>-\frac{78}{3^4.7^4}\)=> \(-\frac{78}{3^7.7^4}+\frac{1}{3^7.7^4}>-\frac{78}{3^4.7^4}\)
=> \(-\frac{11}{3^7.7^4}>-\frac{78}{3^4.7^4}\)
Ta có: \(\frac{-1987}{-1986}=\frac{1986+1}{1986}=1+\frac{1}{1986}>1\)
\(\frac{-1984}{-1985}=\frac{1985-1}{1985}=1-\frac{1}{1985}< 1\)
=> \(\frac{-1987}{-1986}>\frac{-1984}{-1985}\)
Ta có: \(\frac{x}{5}< \frac{5}{4}< \frac{x+2}{5}\)
<=> \(x< \frac{25}{4}< x+2\)
Xét: \(x+2>\frac{25}{4}\) => \(x>\frac{17}{4}\)
=> \(\frac{17}{4}< x< \frac{25}{4}\)
Do x thuộc Z => x \(\in\){5; 6}
\(\left(\frac{1}{4}\right)^{44}.\left(\frac{1}{2}\right)^{12}=\left(\left(\frac{1}{2}\right)^2\right)^{44}.\left(\frac{1}{2}\right)^{12}=\left(\frac{1}{2}\right)^{88}.\left(\frac{1}{2}\right)^{12}=\left(\frac{1}{2}\right)^{100}\)
\(\frac{3^{17}.\left(3^4\right)^{11}}{\left(3^3\right)^{10}.\left(3^2\right)^{15}}=\frac{3^{17}.3^{44}}{3^{30}.3^{30}}=\frac{3^{61}}{3^{60}}=3\)
Bài 1: a) (2x+1)2 = 25
(2x+1)2 = 52
=> 2x + 1 = 5 hoặc 2x+1 = -5
=> x=2 hoặc x=-3
b) 2x+2 - 2x = 96
<=> 2x . 22 - 2x = 96
<=> 2x(4-1) =96
<=>2x = 96 :3 = 32 = 25
<=> x = 5
c) (x-1)3 = 125
<=> (x-1)3 = 53
<=> x-1=5
<=>x= 5 +1 = 6
Trả lời:
\(x=\frac{9^{11}+2}{9^{11}+3}=\frac{9^{11}+3-1}{9^{11}+3}=\frac{9^{11}+3}{9^{11}+3}-\frac{1}{9^{11}+3}=1-\frac{1}{9^{11}+3}\)
\(y=\frac{9^{12}+2}{9^{12}+3}=\frac{9^{12}+3-1}{9^{12}+3}=\frac{9^{12}+3}{9^{12}+3}-\frac{1}{9^{12}+3}=1-\frac{1}{9^{12}+3}\)
Ta có: \(9^{11}< 9^{12}\)
\(\Leftrightarrow9^{11}+3< 9^{12}+3\)
\(\Leftrightarrow\frac{1}{9^{11}+3}>\frac{1}{9^{12}+3}\)
\(\Leftrightarrow-\frac{1}{9^{11}+3}< -\frac{1}{9^{12}+3}\)
\(\Leftrightarrow1-\frac{1}{9^{11}+3}< 1-\frac{1}{9^{12}+3}\)
\(\Leftrightarrow x< y\)
Vậy x < y
2) a) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\)
\(\frac{21}{81^8}=\frac{21}{\left(3^4\right)^8}=\frac{21}{3^{32}}=\frac{21.3}{3^{33}}=\frac{63}{3^{33}}>\frac{1}{3^{33}}\)
=> \(\frac{21}{81^8}>\frac{1}{27^{11}}\)
b) Rõ ràng : 399 < 1121 => \(\frac{1}{399}>\frac{1}{11^{21}}\)
a) \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}\)=> \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=1-\sqrt[3]{\frac{4}{9}}\)
=> x = \(\frac{6}{5}-\frac{6}{5}.\sqrt[3]{\frac{4}{9}}\)
b) => \(\frac{1}{13}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\)
=> \(\left(\frac{1}{2}x-1\right)^4=\frac{13}{48}\)
=> \(\frac{1}{2}x-1=\sqrt[4]{\frac{13}{48}}\) hoặc \(\frac{1}{2}x-1=-\sqrt[4]{\frac{13}{48}}\)
=> \(x=2+2\sqrt[4]{\frac{13}{48}}\) hoặc \(x=2-2\sqrt[4]{\frac{13}{48}}\)
Đặt: \(\hept{\begin{cases}A=\frac{10^{10}+1}{10^{11}+1}\\B=\frac{10^{11}-1}{10^{12}-1}\end{cases}}\)
Ta có:
\(\hept{\begin{cases}10A=\frac{10^{11}+10}{10^{11}+1}=1+\frac{9}{10^{11}+1}\\10B=\frac{10^{12}-10}{10^{12}-1}=1-\frac{9}{10^{12}-1}\end{cases}}\)
\(\Rightarrow10A>10B\)
\(\Rightarrow A>B\)
a) 3^40= 3^4.10=(3^4)10=81^10
11^21> 11^20=11^2.10=(11^2)10=121^10
→ 3^40< 11^21
b) 2^195=2^15.13=(2^15)13=32768^13
3^130=3^10.13= (3^10)13=59049^13
→2^195<3^130
c) 2^90=2^5.18=(2^5)18= 32^18
5^36=5^2.18=(5^2)18=25^18
→2^90>5^36
a) \(3^{54}\)
\(2^{200}=4^{100}>3^{54}\)
\(\Rightarrow3^{54}< 2^{200}\)
b) \(15^{12}=3^{12}.5^{12}\)
\(1^3.125^3=\left(5^3\right)^3=5^9< 3^{12}.5^{12}\)
\(\Rightarrow15^{12}>1^3.125^3\)
c) \(78^{12}-78^{11}=78^{11}.\left(7-1\right)=78^{11}.6\)
\(78^{11}-78^{10}=78^{10}.\left(7-6\right)=78^{10}.6< 78^{11}.6\)
\(\Rightarrow78^{12}-78^{11}>78^{11}-78^{10}\)
d) \(72^{45}-72^{44}=72^{44}.\left(72-1\right)=72^{44}.72>27^{44}\)
\(\Rightarrow72^{45}-72^{44}>27^{44}\)
e) \(3^{39}=\left(3^3\right)^{13}=27^{13}>11^{11}\)
\(\Rightarrow3^{39}>11^{11}\)
Giải chi tiết dùm mik nhé.