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
Tìm số tự nhiên n, biết rằng:
\(\dfrac {4^{5} + {4^{5}} +{4^{5}} + {4^{5}}}{{3^{5}} + {3^{5}} + {3^{5}}}\) . \(\dfrac{6^{5} + {6^{5}} + {6^{5}} + {6^{5}} + {6^{5}} + {6^{5}} }{2^{5} + 2^{5}} = 2^{n}\)
Tìm x, biết
a)\(\dfrac{1}{2}\)x\(x\)-\(\dfrac{7}{3}\)=\(\dfrac{-5}{6}\)+\(\dfrac{3}{4}\)x\(x\)
b)\(\dfrac{4}{5}\)x\(x\)-\(\dfrac{6}{5}\)=\(\dfrac{1}{2}\)+\(\dfrac{3}{2}\)x\(x\)
c)\(\dfrac{2}{5}\)x(3x\(x\)+\(\dfrac{3}{4}\))=\(1\dfrac{1}{5}\)-\(\dfrac{1}{3}\)x\(x\)
d)2x(3x\(x \)+\(\dfrac{3}{4}\))+\(\dfrac{4}{5}\)=\(\dfrac{1}{2}\)-2x\(x\)
Số tự nhiên n thỏa mãn:
\(\dfrac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\dfrac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^n\)
Tìm số nguyên x, biết:
a) \(-4\dfrac{3}{5}\). \(2\dfrac{4}{3}\) < x < \(-2\dfrac{3}{5}\) : \(1\dfrac{6}{15}\)
b) \(-4\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{6}\)) < x < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))
a)tìm x,y biết:x-1/y+2 và x+y=23
b)tìm x biết: 4^5+4^5+4^5+4^5/3^5+3^5+3^5.6^5+6^5+6^5+6^5+6^5+6^5/2^5+2^5=8^2x-6
1/Thực hiện phép tính:
𝒂) ( \(\dfrac{7}{5}\) − \(\dfrac{8}{7}\) ) + \(3\dfrac{2}{5}\) ∶ 𝟎,𝟔
b) \(\dfrac{11}{6}\) + \(\dfrac{7}{6}\) ∶ \(\dfrac{-5}{4}\)
c) \(\dfrac{7}{4}\) − \(\dfrac{5}{4}\) . \(\dfrac{8}{15}\)
2/ Tìm x biết
b) 𝒙 + 𝟐,𝟓 = \(\dfrac{-8}{3}\)
b) 𝟐𝒙 − 𝟐𝟎% = \(\dfrac{9}{5}\)
𝒄) ( \(\dfrac{5}{4}\) )2 + 𝒙 = \(2\dfrac{3}{4}\)
Tìm x biết:
a.
\(\frac{1}{4}.\frac{2}{6}.\frac{3}{8}.\frac{4}{10}.\frac{5}{12}...\frac{30}{62}.\frac{31}{64}=2^x\)
b.
\(\frac{4^5+4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=2^x\)
Tìm x biết: \(\frac{4^5+4^5+4^5}{3^5+3^5+3^5}.\frac{6^5+6^5+6^5+6^5+6^5+6^5}{2^5+2^5}=8^x\)
Bài 1: Tìm x; y ϵ \(ℤ\)
a) 2x - y\(\sqrt{6}\) = 5 + (x + 1)\(\sqrt{6}\)
b) 5x + y - (2x -1)\(\sqrt{7}\) = y\(\sqrt{7}\) + 2
Bài 2: So sánh M và N
M = \(\dfrac{\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{3}{7}-\dfrac{3}{11}}{\dfrac{6}{4}+\dfrac{6}{5}+\dfrac{6}{7}-\dfrac{6}{11}}\)
N = \(\dfrac{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}{\dfrac{6}{2}+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}\)
Bài 3: Chứng minh:
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{2023!}< 1\)
