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a) => \(\left(\frac{1}{3}-\frac{5}{6}x\right)^3=\frac{5}{6}-\frac{21}{54}=\frac{24}{54}=\frac{4}{9}\)
=> \(\frac{1}{3}-\frac{5}{6}x=\sqrt[3]{\frac{4}{9}}\) => \(\frac{5}{6}x=\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\) => \(x=\frac{6}{5}.\left(\frac{1}{3}-\sqrt[3]{\frac{4}{9}}\right)\)
b) \(\frac{1}{3}\left(\frac{1}{2}x-1\right)^4=\frac{1}{12}-\frac{1}{16}=\frac{1}{48}\) => \(\left(\frac{1}{2}x-1\right)^4=\frac{3}{48}=\frac{1}{16}\)
=> \(\frac{1}{2}x-1=\frac{1}{2}\) hoặc \(\frac{1}{2}x-1=-\frac{1}{2}\)
=> \(\frac{1}{2}x=\frac{3}{2}\) hoặc \(\frac{1}{2}x=\frac{1}{2}\) => x = 3 hoặc x = 1
c) \(\left(1+5\right).\left(\frac{3}{5}\right)^{x-1}=\frac{54}{25}\) => \(\left(\frac{3}{5}\right)^{x-1}=\frac{9}{25}=\left(\frac{3}{5}\right)^2\)
=> x - 1= 2 => x = 3
d) \(\left(1+\left(\frac{2}{3}\right)^2\right).\left(\frac{2}{3}\right)^x=\frac{101}{243}\) => \(\frac{13}{9}.\left(\frac{2}{3}\right)^x=\frac{101}{243}\)
=> \(\left(\frac{2}{3}\right)^x=\frac{101}{243}:\frac{13}{9}=\frac{101}{351}\) (có lẽ đề sai)
2) \(\frac{1}{27^{11}}=\frac{1}{\left(3^3\right)^{11}}=\frac{1}{3^{33}}\); \(\frac{1}{81^8}=\frac{1}{\left(3^4\right)^8}=\frac{1}{3^{32}}\)
Vì 333 > 332 => \(\frac{1}{3^{33}}\) < \(\frac{1}{3^{32}}\) => \(\frac{1}{27^{11}}\) < \(\frac{1}{81^8}\)
b) \(\frac{1}{3^{99}}=\frac{1}{\left(3^3\right)^{33}}=\frac{1}{27^{33}}
\(\dfrac{11}{8}:x-\dfrac{2}{5}+-\dfrac{1}{6}=-\dfrac{1}{5}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}-\left(-\dfrac{1}{6}\right)\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{5}+\dfrac{1}{6}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{6}{30}+\dfrac{5}{30}\\ =>\dfrac{11}{8}:x-\dfrac{2}{5}=-\dfrac{1}{30}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{2}{5}\\ =>\dfrac{11}{8}:x=-\dfrac{1}{30}+\dfrac{12}{30}\\ =>\dfrac{11}{8}:x=\dfrac{11}{30}\\ =>x=\dfrac{11}{8}:\dfrac{11}{30}\\ =>x=\dfrac{11}{8}.\dfrac{30}{11}\\ =>x=\dfrac{30}{8}\\ =>x=\dfrac{15}{4}\\ \dfrac{4}{7}x-\dfrac{1}{3}x+\left(-\dfrac{16}{21}\right)=-\dfrac{2}{3}\\ =>\left(\dfrac{4}{7}-\dfrac{1}{3}\right)x=-\dfrac{2}{3}-\left(-\dfrac{16}{21}\right)\\ =>\left(\dfrac{12}{21}-\dfrac{7}{21}\right)x=-\dfrac{2}{3}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=-\dfrac{14}{21}+\dfrac{16}{21}\\ =>\dfrac{5}{21}x=\dfrac{2}{21}\\ =>x=\dfrac{2}{21}:\dfrac{5}{21}\)
\(=>x=\dfrac{2}{21}.\dfrac{21}{5}\\ =>x=\dfrac{2}{5}\\ -\dfrac{11}{12}x+\dfrac{15}{2}\left(x+-\dfrac{1}{5}\right)=\dfrac{67}{8}\\ =>-\dfrac{11}{12}x+\dfrac{15}{2}.x-\dfrac{1}{5}=\dfrac{67}{8}\\ =>\left(-\dfrac{11}{12}+\dfrac{15}{2}\right)x=\dfrac{67}{8}+\dfrac{1}{5}\\ =>\left(-\dfrac{11}{12}+\dfrac{90}{12}\right)x=\dfrac{335}{40}+\dfrac{8}{40}\\ =>\dfrac{79}{12}x=\dfrac{343}{40}\\ =>x=\dfrac{343}{40}:\dfrac{79}{12}\\ =>x=\dfrac{343}{40}.\dfrac{12}{79}\\ =>x=\dfrac{343.12}{40.79}\\ =>x=\dfrac{343.3}{10.79}\\ =>x=\dfrac{1029}{790}\)
1.a,=(54+45+1).113
=100.113
=11300
b,=(3/7+8/14)+(4/9+10/18)
=1+1
=2
2.a,=13/10+1/3
=49/30
b,=12/9.(1/12+1/6)
=12/9.1/4
=1/3
c,=3/4.3/2
=9/8
d,=3/2-1/3
=7/6
1:tính bằng cách thuận tiện nhất:
a)54 x 113 + 45 x 113 + 113
= 54 x 113 + 45 x 113 + 113x1
=113 x(54+45+1)
= 113x100
=1300
b)3/7 + 4/9 + 8/14 + 10/18
=(3/7+8/14)+(4/9+10/18)
= 1 + 1
=2
1) \(\left|4-2x\right|.\dfrac{1}{3}=\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}:\dfrac{1}{3}\)
\(\left|4-2x\right|=\dfrac{1}{3}.3\)
\(\left|4-2x\right|=1\)
=>\(4-2x=\pm1\)
+)\(TH1:4-2x=1\) +)\(TH2:4-2x=-1\)
\(2x=4-1\) \(2x=4-\left(-1\right)\)
\(2x=3\) \(2x=4+1\)
\(x=3:2\) \(2x=5\)
\(x=1,5\) \(x=5:2\)
Vậy x=1,5 \(x=2,5\)
Vậy x=2,5
2) \(\left(-3\right)^2:\left|x+\left(-1\right)\right|=-3\)
\(9:\left|x+\left(-1\right)\right|=-3\)
\(\left|x+\left(-1\right)\right|=9:\left(-3\right)\)
\(\left|x+\left(-1\right)\right|=-3\)
=> \(x+\left(-1\right)\) sẽ không có giá trị nào ( Vì giá trị tuyệt đối luôn luôn lớn hơn hoặc bằng 0 )
Vậy x = \(\varnothing\)
a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
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 \(x=2-2\sqrt[4]{\frac{13}{48}}\)