1) 1515 và 813 . 1253
2) 7815 - 7812 và 7812 - 789
3) 2333 và 3 222
4) ( 985) và 9815
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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) \(3^{39}=\left(3^3\right)^{13}=27^{13}>11^{11}\)
\(\Rightarrow3^{39}>11^{11}\)
\(a,x\in\left\{15;20;...;70;75\right\}\\ b,x\in\left\{6;8;12\right\}\)
\(x\) \(⋮\) \(5\) và \(13 < x \) \(\le\) \(78\)
\(x = \) \(\left\{15;20;25;30;35;40;45;50;55;60;65;70;75\right\}\)
\(12\) \(⋮\) \(x\) và \(x > 4\)
\(x = \) \(\left\{6;12\right\}\)
Ta có: \(2^{333}=2^{3.111}=\left(2^3\right)^{111}=8^{111}\)
\(3^{222}=3^{2.111}=\left(3^2\right)^{111}=9^{111}\)
Vì \(8^{111}< 9^{111}\)
Nên \(2^{333}< 3^{222}\)
\(\frac{2224}{6290}=\frac{1112}{3145}\)
Ta có \(\frac{1112}{3145}< \frac{1112+618}{3145+618}=\frac{1730}{3763}< \frac{2223}{3763}\)
Vậy \(\frac{2223}{3763}>\frac{2224}{6290}\)
Ta có \(\frac{2224}{6290}=\frac{2223}{6290}+\frac{1}{6290}\)
Vì:\(\frac{2223}{3763}>\frac{2223}{6290}\)
Nên\(\frac{2223}{3763}>\frac{2224}{6290}\)
1) \(x:y:z=2:3:4\) ⇒ \(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\)
⇒ x=4;y=6;z=8
\(1,\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{x+y+z}{2+3+4}=\dfrac{18}{9}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot2=4\\y=2\cdot3=6\\z=2\cdot4=8\end{matrix}\right.\)
\(2,\) Áp dụng t/c dtsbn
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{3y}{-9}=\dfrac{2z}{8}=\dfrac{4x-3y-2z}{8-\left(-9\right)-8}=\dfrac{81}{9}=9\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot\left(-3\right)=-6\\z=2\cdot4=8\end{matrix}\right.\)
\(3,4y=3z\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{6}=\dfrac{z}{8};\dfrac{x}{3}=\dfrac{y}{2}\Rightarrow\dfrac{x}{9}=\dfrac{y}{6}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}\)
Áp dụng t/c dtsbn
\(\dfrac{x}{9}=\dfrac{y}{6}=\dfrac{z}{8}=\dfrac{x+y+z}{9+6+8}=\dfrac{46}{23}=2\\ \Rightarrow\left\{{}\begin{matrix}x=2\cdot9=18\\y=2\cdot6=12\\z=2\cdot8=16\end{matrix}\right.\)
\(4,5x=3y\Rightarrow\dfrac{x}{3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{9}=\dfrac{y}{15};\dfrac{y}{z}=\dfrac{3}{2}\Rightarrow\dfrac{y}{3}=\dfrac{z}{2}\Rightarrow\dfrac{y}{15}=\dfrac{z}{10}\\ \Rightarrow\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}\)
Áp dụng t/c dtsbn:
\(\dfrac{x}{9}=\dfrac{y}{15}=\dfrac{z}{10}=\dfrac{2x}{18}=\dfrac{3y}{45}=\dfrac{4z}{40}=\dfrac{2x+3y-4z}{18+45-40}=\dfrac{34}{23}\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{34}{23}\cdot9=\dfrac{306}{23}\\y=\dfrac{34}{23}\cdot15=\dfrac{510}{23}\\z=\dfrac{34}{23}\cdot10=\dfrac{340}{23}\end{matrix}\right.\)
Câu 4 là ( 985 ) 3 nha các bạn mình viết lộn đề.
Giải :
3) 2333 va 3222
2333 = ( 23 )111 = 8111
3222 = ( 32 )111 = 9111
Vì 8111 < 9111 => 2333 < 3222
Vậy 2333 < 3222
4) ( 985 )3 và 9815
( 985 )3 = 9815
Vì 9815 = 9815 => ( 985 )3 = 9815
Vậy ( 985 )3 = 9815