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LQ
1
AH
Akai Haruma
Giáo viên
4 tháng 9 2021
Lời giải:
Gọi biểu thức là $A$
\(A=\frac{2^{10}.3^8+5.(2^2)^5.3^8}{2^{10}.(3^3)^3-2^{10}.(3^2)^4}=\frac{2^{10}.3^8+5.2^{10}.3^8}{2^{10}.3^9-2^{10}.3^{8}}\)
\(=\frac{2^{10}.3^8(1+5)}{2^{10}.3^8(3-1)}=\frac{6}{2}=3\)
NN
1
30 tháng 4 2021
-1 5/27-(3x-7/9)3=-24/27
(3x-7/9)3 =-32/27-(-24/27)
(3x-7/9)3=-8/27
(3x-7/9)3=(-2/3)3
⇒3x-7/9=-2/3
3x =-2/3+7/9
3x =1/9
x =1/9:3
x =1/27
Chúc bạn học tốt!
20 tháng 3 2017
d, Vì B=10^1993+1/10^1992+1 > 1 =>10^1993+1/10^1992+1>10^1993+1+9/10^1992+1+9 = 10^1993+10/10^1992+10= 10. (10^1992+1)/10. (10^1991+1) = 10^1992+1/10^1991+1=A Vậy A=B
26 tháng 3 2017
cau d B>1 ta co tinh chat (\(\dfrac{a}{b}>\dfrac{a+m}{b+m}\) ) B> \(\dfrac{10^{1993}+1+9}{10^{1992}+1+9}\)\(=\dfrac{10^{1993}+10}{10^{1992}+10}\)=\(\dfrac{10\left(10^{1992}+1\right)}{10\left(10^{1991}+1\right)}\)=\(\dfrac{10^{1992}+1}{10^{1991}+1}\)=A
Suy ra B>A(chuc ban hoc goi nhe)
22 tháng 7 2021
a) Ta có: \(-3\dfrac{1}{4}\cdot x-75\%+\dfrac{3x}{2}=-1.2:\dfrac{-9}{10}-1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{-13x}{4}-\dfrac{3}{4}+\dfrac{3x}{2}=\dfrac{-6}{5}\cdot\dfrac{10}{-9}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-13x-3+6x}{4}=\dfrac{4}{3}-\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{-7x-3}{4}=\dfrac{1}{12}\)
\(\Leftrightarrow-7x-3=\dfrac{1}{3}\)
\(\Leftrightarrow-7x=\dfrac{10}{3}\)
hay \(x=-\dfrac{10}{21}\)
22 tháng 7 2021
b) Ta có: \(\dfrac{5}{3}+\dfrac{5}{15}+\dfrac{5}{35}+...+\dfrac{5}{x\left(x+2\right)}=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(\dfrac{2}{3}+\dfrac{2}{15}+\dfrac{2}{35}+...+\dfrac{2}{x\left(x+2\right)}\right)=2\dfrac{8}{17}\)
\(\Leftrightarrow\dfrac{5}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{x}-\dfrac{1}{x+2}\right)=2+\dfrac{8}{17}\)
\(\Leftrightarrow\left(1-\dfrac{1}{x+2}\right)=\dfrac{42}{17}:\dfrac{5}{2}\)
\(\Leftrightarrow\dfrac{x+1}{x+2}=\dfrac{42}{17}\cdot\dfrac{2}{5}=\dfrac{84}{85}\)
\(\Leftrightarrow85x+85=84x+168\)
\(\Leftrightarrow x=83\)
S
1 tháng 4 2021
\(\dfrac{2}{3}+\dfrac{1}{5}.\dfrac{10}{7}=\dfrac{2}{3}+\dfrac{10}{35}=\dfrac{70}{105}+\dfrac{30}{105}=\dfrac{100}{105}=\dfrac{50}{21}\)
1 tháng 4 2021
a) Ta có: \(\dfrac{2}{3}+\dfrac{1}{5}\cdot\dfrac{10}{7}\)
\(=\dfrac{2}{3}+\dfrac{2}{7}\)
\(=\dfrac{14}{21}+\dfrac{6}{21}\)
\(=\dfrac{20}{21}\)
NN
9 tháng 6 2017
\(a,\left(\dfrac{7}{20}+\dfrac{11}{15}-\dfrac{15}{12}\right):\left(\dfrac{11}{20}-\dfrac{26}{45}\right).\)
\(=\left(\dfrac{21}{60}+\dfrac{44}{60}-\dfrac{75}{60}\right):\left(\dfrac{99}{180}-\dfrac{104}{180}\right).\)
\(=\left(\dfrac{65}{60}-\dfrac{75}{60}\right):\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{10}{60}:\left(-\dfrac{5}{180}\right).\)
\(=-\dfrac{1}{6}:\left(-\dfrac{1}{36}\right).\)
\(=-\dfrac{1}{6}.\left(-36\right).\)
\(=\dfrac{-1.\left(-36\right)}{6}=\dfrac{36}{6}=6.\)
Vậy......
\(b,\dfrac{5-\dfrac{5}{3}+\dfrac{5}{9}-\dfrac{5}{27}}{8-\dfrac{8}{3}+\dfrac{8}{9}-\dfrac{8}{27}}:\dfrac{15-\dfrac{15}{11}+\dfrac{15}{121}}{16-\dfrac{16}{11}+\dfrac{16}{121}}.\)
\(=\dfrac{5\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}{8\left(1-\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{1}{27}\right)}:\dfrac{15\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}{16\left(1-\dfrac{1}{11}+\dfrac{1}{121}\right)}.\)
\(=\dfrac{5}{8}:\dfrac{15}{16}.\)
\(=\dfrac{5}{8}.\dfrac{16}{15}=\dfrac{5.16}{8.15}=\dfrac{1.2}{1.3}=\dfrac{2}{3}.\)
Vậy......
c, (làm tương tự câu b).
~ Học tốt!!! ~
21 tháng 9 2018
a) (3x-1)3 = 27 = 33
=> 3x - 1 = 3
3x = 4
x = 4/3
b) 32x-1 = 27 = 33
=> 2x - 1 = 3
2x = 4
x = 2
c) 68-2x = 36 = 62
=> 8 -2x = 2
=> 2x = 6
x = 3
d) (3x-2)10 = (3x-2)7
=> (3x-2)10 - (3x-2)7 = 0
(3x-2)7. [(3x-2)3 - 1] = 0
=> (3x-2)7 = 0 => 3x-2 = 0 => 3x = 2 => x = 2/3
(3x-2)3 - 1 = 0 => (3x-2)3 = 1 => 3x - 2 = 1 => 3x = 3 => x = 1
KL:...
K
1
PT
15 tháng 4 2017
\(-1\dfrac{5}{27}-\left(3x-\dfrac{7}{9}\right)^3=-\dfrac{24}{27}\)
\(\Rightarrow\left(3x-\dfrac{7}{9}\right)^3=\dfrac{-8}{27}\)
\(\Rightarrow3x-\dfrac{7}{9}=\dfrac{-2}{3}\)
\(\Rightarrow3x=\dfrac{1}{9}\Rightarrow x=\dfrac{1}{27}\)
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