cho 2 số hữu tỉ a/b và c/d (b,d > 0). chứng minh rằng nếu a / b < c / d thì a/b < a+c / b+d < c / d
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MD
0
H9
HT.Phong (9A5)
CTVHS
25 tháng 7
\(9^x:3^x=3\\ =>\left(9:3\right)^x=3\\ =>3^x=3\\ =>3^x=3^1\\ =>x=1\)
Vậy: ...
DM
1
H9
HT.Phong (9A5)
CTVHS
25 tháng 7
\(2,8\cdot\dfrac{-6}{13}-7,2-2,8\cdot\dfrac{7}{13}\\ =\left(2,8\cdot\dfrac{-6}{13}-2,8\cdot\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\left(\dfrac{-6}{13}-\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\dfrac{-13}{13}-7,2\\=-2,8-7,2\\ =-10\)
DM
2
H9
HT.Phong (9A5)
CTVHS
25 tháng 7
\(2,8\cdot\dfrac{-6}{13}-7,2-2,8\cdot\dfrac{7}{13}\\ =\left(2,8\cdot\dfrac{-6}{13}-2,8\cdot\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\left(\dfrac{-6}{13}-\dfrac{7}{13}\right)-7,2\\ =2,8\cdot\dfrac{-13}{13}-7,2\\ =-2,8-7,2\\ =-10\)
H9
HT.Phong (9A5)
CTVHS
25 tháng 7
\(a.5\cdot3^x=5\cdot3^4\\ =>3^x=\dfrac{5\cdot3^4}{5}=3^4\\ =>x=4\\ b.7\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{7}=4^3\\ =>x=3\\ c.\dfrac{3}{5}\cdot4^x=7\cdot4^3\\ =>4^x=\dfrac{7\cdot4^3}{\dfrac{3}{5}}=\dfrac{35}{3}\cdot4^3\\ =>\dfrac{4^x}{4^3}=\dfrac{35}{3}\\ =>4^{x-3}=\dfrac{35}{3}\\ =>x-3=log_4\dfrac{35}{3}\\ =>x=log_4\dfrac{35}{3}+3\\ d.\dfrac{3}{2}\cdot5^x=\dfrac{3}{2}\cdot5^{12}\\ =>5^x=\dfrac{5^{12}\cdot\dfrac{3}{2}}{\dfrac{3}{2}}=5^{12}\\ =>x=12\)
25 tháng 7
e: \(9\cdot5^x=6\cdot5^6+3\cdot5^6\)
=>\(9\cdot5^x=9\cdot5^6\)
=>\(5^x=5^6\)
=>x=6
f: \(5\cdot3^x=7\cdot3^5-2\cdot3^5\)
=>\(5\cdot3^x=5\cdot3^5\)
=>\(3^x=3^5\)
=>x=5
g: \(5\cdot3^{x+6}=2\cdot3^5+3\cdot3^5\)
=>\(5\cdot3^{x+6}=5\cdot3^5\)
=>\(3^{x+6}=3^5\)
=>x+6=5
=>x=-1
H9
HT.Phong (9A5)
CTVHS
25 tháng 7
\(e.\left(\dfrac{-13}{3}-\dfrac{4}{9}\right)-\left(\dfrac{-10}{3}-\dfrac{4}{9}\right)\\ =\dfrac{-13}{3}-\dfrac{4}{9}+\dfrac{10}{3}+\dfrac{4}{9}\\ =\left(\dfrac{-13}{3}+\dfrac{10}{3}\right)+\left(\dfrac{4}{9}-\dfrac{4}{9}\right)\\ =-\dfrac{3}{3}=-1\\ d.\dfrac{-4}{12}-\left(-0,25-\dfrac{13}{39}\right)+0,75\\ =\dfrac{-1}{3}-\left(-\dfrac{1}{4}-\dfrac{1}{3}\right)+\dfrac{3}{4}\\ =-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{3}+\dfrac{3}{4}\\ =\left(-\dfrac{1}{3}+\dfrac{1}{3}\right)+\left(\dfrac{1}{4}+\dfrac{3}{4}\right)\\ =0+\dfrac{4}{4}\\ =1\)
HM
1
H9
HT.Phong (9A5)
CTVHS
25 tháng 7
\(a.\left(\dfrac{-1}{2}\right)^2\cdot\left(\dfrac{2}{5}\right)^2 \\ =\left(\dfrac{-1}{2}\cdot\dfrac{2}{5}\right)^2\\ =\left(\dfrac{-1}{5}\right)^2\\ =\dfrac{1}{25}\\ b.\left(\dfrac{1}{9}\right)^2:\left(\dfrac{1}{3}\right)^3\\ =\left[\left(\dfrac{1}{3}\right)^2\right]^2:\left(\dfrac{1}{3}\right)^3\\ =\left(\dfrac{1}{3}\right)^4:\left(\dfrac{1}{3}\right)^3\\ =\dfrac{1}{3}\\ c.\left(\dfrac{-1}{2}\right)^3\cdot\left(\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-1}{2}\cdot\dfrac{3}{2}\right)^3\\ =\left(\dfrac{-3}{4}\right)^3\\ =\dfrac{-27}{64}\)