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1.
Có : 5^299 < 5^300 = (5^2)^150 = 25^150
3^501 > 3^450 = (3^3)^150 = 27^150
Mà 25^150 < 27^150 => 5^299 < 3^501
Tk mk nha
\(A=\frac{1+5+5^2+...+5^9}{1+5+5^2+...+5^8}=\frac{1+5\left(1 +5+5^2+...+5^8\right)}{1+5+5^2+...+5^8}=5+\frac{1}{1+5+5^2+...+5^8} \)
\(B=\frac{1+3+3^2+....+3^9}{1+3+3^2+....+3^8}=\frac{1+3\left(1+3+3^2+....+3^8\right)}{1+3+3^2+....+3^8}=3+\frac{1}{1+3+3^2+....+3^8}\)
\(=5+\frac{1}{1+3+3^2+....+3^8}-2\)
Có: \(\frac{1}{1+5+5^2+...+5^8}>0\) và \(\frac{1}{1+3+3^2+....+3^8}-2< 0\)
\(\Rightarrow A>B\)
\(5,\left(x\cdot0,5-\frac{3}{7}\right):\frac{1}{2}=1\frac{1}{7}\)
\(\Leftrightarrow x\cdot0,5:\frac{1}{2}-\frac{3}{7}:\frac{1}{2}=1\frac{1}{7}\)
\(\Leftrightarrow x-\frac{6}{7}=\frac{8}{7}\)
\(\Leftrightarrow x=2\)
\(6,x\cdot1,75=1\frac{3}{10}+45\%\)
\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{13}{10}+\frac{9}{20}\)
\(\Leftrightarrow x\cdot\frac{7}{4}=\frac{7}{4}\)
\(\Leftrightarrow x=1\)
\(7,\frac{5-x}{15}+\frac{5}{12}-\frac{1}{8}=\frac{3}{8}\)
\(\Leftrightarrow\frac{5-x}{15}=\frac{3}{8}-\frac{5}{12}+\frac{1}{8}\)
\(\Leftrightarrow\frac{5-x}{15}=\frac{1}{12}\)
\(\Leftrightarrow60-12x=15\)
\(\Leftrightarrow12x=45\)
\(\Leftrightarrow x=\frac{15}{4}\)
\(8,\left|x-\frac{25}{33}\right|-\frac{3}{11}=\frac{2}{3}\)
\(\Leftrightarrow\left|\frac{x-25}{33}\right|=\frac{31}{33}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{25}{33}=\frac{31}{33}\\x-\frac{25}{33}=-\frac{31}{33}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{56}{33}\\x=-\frac{2}{11}\end{cases}}\)
\(9,-\frac{9}{8}+\frac{-3}{8}\cdot x=-\frac{1}{8}\)
\(\Leftrightarrow\frac{-9}{8}+\frac{-3}{8}\cdot x+\frac{1}{8}=0\)
\(\Leftrightarrow-1-\frac{3}{8}x=0\)
\(\Leftrightarrow\frac{3}{8}x=-1\)
\(\Rightarrow x=-\frac{8}{3}\)
Câu 1:
A = -3/12 + 5/7 - (-1)/42
A = -21/84 + 60/84 + 2/84
A = 39/84 + 2/84
A = 41/84
Câu a:
33 ⋮ (x+ 1)
(x+ 1) ∈ Ư(33) = {-33; -11; -3; -1; 1; 3; 11}
x ∈ {-34; -12; -4; -2; 0; 2; 10}
Vậy: {-34; -12; -4; -2; 0; 2; 10}
Câu b:
x ∈ ƯC(250; 48)
250 = 2.5^3; 48 = 2^4.3
ƯCLN(250; 48) = 2
x ∈ ƯC(2) = {-2; -1}
Vậy x ∈ {-2; -1}
A = 1/3^2 + 1/4^2 + 1/5^2 + ... + 1/50^2
1/3^2 = 1/9
1/4^2 < 1/3.4 = 1/3 - 1/4
1/5^2 < 1/4.5 = 1/4 - 1/5
.............................................
1/50^2 < 1/49.50 = 1/49 - 1/50
Cộng vế với vế ta có:
A = 1/3^2+1/4^2+..+1/50^2 = 1/9 + 1/3 - 1/50
A = 4/9 - 1/50 < 4/9
1/3^2 = 1/9
1/4^2 > 1/4.5 = 1/4 - 1/5
1/5^2 > 1/5.6 = 1/5 - 1/6
............................................
1/50^2 > 1/49.50 = 1/49 - 1/50
Cộng vế với vế ta có:
A = 1/3^2+1/4^2+ ...+ 1/50^2 > 1/9+1/4-1/50
A > 1/4 + (1/9 - 1/50)
1/9 > 1/50
1/9 - 1/50 > 0
A > 1/4 + 1/9 - 1/50 > 1/4
Vậy 1/4 < A < 4/9 (đpcm)
Đầu tiên tìm số bị trừ:
\(D=\frac{2\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{11}\right)}{7\left(\frac{1}{-5}-\frac{1}{9}+\frac{1}{11}\right)}\)=\(-\frac{178}{763}\)
Xong tìm số trừ:
\(\frac{1\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)\(\frac{2}{7}\)
Công thức SBT-ST=H
mk nghĩ là A>B