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Sửa đề: 3^5+3^5+3^5; 2^x
=>\(2^x=\dfrac{4^5\cdot4}{3^5\cdot3}\cdot\dfrac{6^5\cdot6}{2^5\cdot2}\)
=>\(2^x=\left(\dfrac{4}{3}\right)^6\cdot\left(\dfrac{6}{2}\right)^6=4^6=2^{12}\)
=>x=12
5: \(=3-\dfrac{1}{4}+\dfrac{2}{3}-5+\dfrac{1}{3}+\dfrac{6}{5}-6+\dfrac{7}{4}-\dfrac{3}{2}\)
\(=3-5-6+\dfrac{-1}{4}+\dfrac{7}{4}+\dfrac{2}{3}+\dfrac{1}{3}+\dfrac{6}{5}-\dfrac{3}{2}\)
\(=-8+\dfrac{3}{2}+1+\dfrac{-3}{10}\)
\(=-7+\dfrac{15-3}{10}=-7+\dfrac{6}{5}=-\dfrac{29}{5}\)
6: \(=6-\dfrac{2}{3}+\dfrac{1}{2}-5-\dfrac{5}{3}+\dfrac{3}{2}-3+\dfrac{7}{3}-\dfrac{5}{2}\)
\(=6-5-3-\dfrac{2}{3}-\dfrac{5}{3}+\dfrac{7}{3}+\dfrac{1}{2}+\dfrac{3}{2}-\dfrac{5}{2}\)
\(=-2-\dfrac{1}{2}=-\dfrac{5}{2}\)
7: \(=\dfrac{5}{3}-\dfrac{3}{7}+9-2-\dfrac{5}{7}+\dfrac{2}{3}+\dfrac{8}{7}-\dfrac{4}{3}-10\)
\(=9-2-10+\dfrac{5}{3}+\dfrac{2}{3}-\dfrac{4}{3}+\dfrac{-3}{7}-\dfrac{5}{7}+\dfrac{8}{7}\)
=-3+1
=-2
8: \(=8-\dfrac{9}{4}+\dfrac{2}{7}+6+\dfrac{3}{7}-\dfrac{5}{4}-3-\dfrac{2}{4}+\dfrac{9}{7}\)
\(=8+6-3+\dfrac{2}{7}+\dfrac{3}{7}+\dfrac{9}{7}-1-\dfrac{2}{4}\)
\(=11+2-1-\dfrac{1}{2}\)
=11+1/2
=11,5
\(\frac{2^8.5^4+2^7.5^3}{2^7.5^5-2^8.5^4}=\frac{2^7.5^3\left(2.5+1\right)}{2^7.5^4\left(5-2\right)}=\frac{2^7.5^3.11}{2^7.5^4.3}=\frac{5^3.11}{5^3.5.3}=\frac{11}{5.3}=\frac{11}{15}\)
a,
\(5^{x+4}-3.5^{x+3}=2.5^{11}\)
\(\Rightarrow5^{x+3}\left(5-3\right)=2.5^{11}\)
\(\Rightarrow5^{x+3}2=2.5^{11}\)
\(\Rightarrow5^{x+3}=5^{11}\)
\(\Rightarrow x+3=11\)
\(\Rightarrow x=8\)
b, (Check lai xem de sai o dau khong nhe)
\(3.5^{x+2}+4.5^{x+3}=19.5^{10}\)
Dat 5x ra ben ngoai
\(\Rightarrow5^x.5^23+5^x:5^{-3}.4\)
\(\Rightarrow5^x\left(5^2.3+5^{-3}.4\right)\)
\(\Rightarrow5^x\left(5^{-3}.5^5.3+5^{-3}.4\right)\)
\(\Rightarrow5^x[5^{-3}\left(5^53+4\right)\)
\(\Rightarrow5^x[5^{-3}\left(3125.3+4\right)\)
\(\Rightarrow5^x\left(5^{-3}\right).9379\)
=> Khong tim duoc gia tri cua x \(\Rightarrow x\in\varnothing\)
1: =1/8*9/4=9/32
2: =8/27*243/32=9/4
3: =(5/4*4/5)^5*(4/5)^2=16/25
4: \(=\left(-\dfrac{5}{6}\cdot\dfrac{6}{5}\right)^2\cdot\left(\dfrac{6}{5}\right)^2=\dfrac{36}{25}\)
5: \(=\left(-\dfrac{4}{3}\right)^3\cdot\left(\dfrac{3}{4}\right)^{10}=\left(-1\right)\left(\dfrac{3}{4}\right)^7=-\left(\dfrac{3}{4}\right)^7\)
6: \(=\left(\dfrac{1}{3}\cdot\dfrac{-9}{2}\right)^4\left(-\dfrac{9}{2}\right)^2=\left(-\dfrac{3}{2}\right)^4\cdot\dfrac{81}{4}=\dfrac{9}{4}\cdot\dfrac{81}{4}=\dfrac{729}{16}\)
8: =(0,2*5)^4*5^2=25
10: =-0,5^5*2^10
=-0,5^5*2^5*2^5
=-32
13: =(0,5*2)^2*2^2=4
A=5+53+55+...+52015
52A=53+55+...+52017
25A-A=(53+55+...+52017)-(5+53+55+..+52015)
24A=52017-5
A=(52017-5)/24
B=12+22+32+..+20162
B=1/6.2016(2016+1).2....(2016+1)
B=
a) Ta có: \(\dfrac{1}{2022}-\dfrac{5}{2\cdot4}-\dfrac{5}{4\cdot6}-\dfrac{5}{6\cdot8}-...-\dfrac{5}{2020\cdot2022}\)
\(=\dfrac{1}{2022}-5\left(\dfrac{1}{2\cdot4}+\dfrac{1}{4\cdot6}+\dfrac{1}{6\cdot8}+...+\dfrac{1}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{2}{2\cdot4}+\dfrac{2}{4\cdot6}+\dfrac{2}{6\cdot8}+...+\dfrac{2}{2020\cdot2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{6}+...+\dfrac{1}{2020}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\left(\dfrac{1}{2}-\dfrac{1}{2022}\right)\)
\(=\dfrac{1}{2022}-\dfrac{5}{2}\cdot\dfrac{1010}{2022}\)
\(=\dfrac{1}{2022}-\dfrac{2025}{2022}=\dfrac{-1262}{1011}\)
b) Ta có: \(\dfrac{2^2}{1\cdot3}+\dfrac{2^2}{3\cdot5}+...+\dfrac{2^2}{197\cdot199}\)
\(=2\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{197\cdot199}\right)\)
\(=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{197}-\dfrac{1}{199}\right)\)
\(=2\left(1-\dfrac{1}{199}\right)\)
\(=2\cdot\dfrac{198}{199}=\dfrac{396}{199}\)
(-5) + 2 = - ( 5 - 2 ) = -3
\(-5+2=\left(-3\right)\)