K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
13 tháng 8 2015
a))\(\frac{2^3.3^3.3^3.3^4.5^4.5^4.3^7.2^7.2^7.2^7.2^6}{2^5.2^5.3^5.2^6.2^6.3^6.3^6.3^4.5^4}=\frac{5^4.2^7}{3^4}=\frac{625.128}{81}=\frac{80000}{81}\)
13 tháng 8 2015
\(\frac{18^3.75^4.24^7.2^6}{12^5.36^6.15^4}=\frac{2^3.3^6.3^4.5^8.2^{21}.3^7.2^6}{2^{10}.3^5.2^{12}.3^{12}.3^4.5^4}=\frac{2^{30}.3^{17}.5^8}{2^{22}.3^{21}.5^4}=\frac{2^8.5^4}{3^4}\)
mình làm đúng đó không tin thì bấm máy tính thử đi phạm thủy linh làm sai rồi
cho mình đúng hen
14 tháng 8 2018
a)
\(\left[\dfrac{3}{8}+\left(-\dfrac{3}{4}+\dfrac{7}{12}\right)\right].\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{5}{24}.\dfrac{5}{6}+\dfrac{1}{2}\)
\(=\dfrac{25}{144}+\dfrac{1}{2}\)
\(=\dfrac{97}{144}\)
b)
\(\dfrac{1}{2}+\dfrac{3}{4}-\left(\dfrac{3}{4}+\dfrac{1}{2}\right)\)
\(=\dfrac{1}{2}+\dfrac{3}{4}-\dfrac{3}{4}-\dfrac{1}{2}\)
\(=0\)
c)
\(\dfrac{6}{\dfrac{5}{12}}:\dfrac{2}{\dfrac{3}{4}}+\dfrac{11}{\dfrac{1}{4}}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{27}{5}+\dfrac{11}{\dfrac{1}{4}}.\dfrac{2}{15}\)
\(=\dfrac{27}{5}+\dfrac{88}{15}\)
\(=\dfrac{169}{15}\)
9 tháng 8 2020
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
Khử mẫu : \(12x-12+6x-6=4x+3x-7\)
\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)
KN
9 tháng 8 2020
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)
\(\Leftrightarrow18x-18=7x-7\)
\(\Leftrightarrow18x+7x=18+7\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
mk đg cần nhưng còn nhìu bài để làm cho nên giúp mk vs nha Thanks trước
1 tháng 5 2017
a)
\(\dfrac{2}{3}-\dfrac{5}{12}x=\dfrac{-8}{3}\)\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}-\left(-\dfrac{8}{3}\right)\)
\(\Rightarrow\dfrac{5}{12}x=\dfrac{2}{3}+\dfrac{8}{3}=\dfrac{10}{3}\)
\(\Rightarrow x=\dfrac{10}{3}:\dfrac{5}{12}=8\)
b) \(3x-2\left(2x-1\right)=1\dfrac{1}{3}\)\(\Rightarrow3x-4x+2=\dfrac{4}{3}\)
\(\Rightarrow3x-4x=\dfrac{4}{3}-2\)
\(\Rightarrow-x=-\dfrac{2}{3}\)\(\Rightarrow x=\dfrac{2}{3}\)
c) \(\dfrac{x+4}{20}=\dfrac{5}{x+4}\Rightarrow\left(x+4\right)\left(x+4\right)=20.5\)
\(\Rightarrow\left(x+4\right)^2=100\)
\(\Rightarrow\left(x+4\right)^2=10^2\) hoặc \(\left(x+4\right)^2=\left(-10\right)^2\)
=> x+4=10 => x+4=-10
=> x=6 => x=-14
14 tháng 3 2017
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)
\(\Rightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)
\(\Rightarrow2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)
\(\Rightarrow A=\left(2-1\right)+\dfrac{3}{2^2}+\left(\dfrac{4}{2^3}-\dfrac{3}{2^3}\right)+....\left(\dfrac{99}{2^{98}}-\dfrac{98}{2^{98}}\right)-\dfrac{100}{2^{100}}\)
\(\Rightarrow A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{100}{2^{100}}\)
\(\Rightarrow A=1+\dfrac{3}{2^2}+\left(\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}\right)-\dfrac{100}{2^{100}}\)
\(\Rightarrow A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 1 2019
\(VT=\left|2x+3\right|+\left|1-2x\right|\ge\left|2x+3+1-2x\right|=4\) \(\Rightarrow VT\ge4\) (1)
Lại có \(3\left(x+1\right)^2\ge0\Rightarrow3\left(x+1\right)^2+2\ge2\)
\(\Rightarrow\dfrac{8}{3\left(x+1\right)^2+2}\le\dfrac{8}{2}=4\) \(\Rightarrow VP\le4\) (2)
Từ (1), (2) \(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi và chỉ khi \(\left\{{}\begin{matrix}\left|2x+3\right|+\left|2x-1\right|=4\\\dfrac{8}{3\left(x+1\right)^2+2}=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(2x+3\right)\left(1-2x\right)\ge0\\3\left(x+1\right)^2=0\end{matrix}\right.\) \(\Rightarrow x=-1\)
Vậy pt có nghiệm duy nhất \(x=-1\)
HD
4
1 tháng 11 2018
\(\frac{1}{2}+\frac{2}{3}x=\frac{1}{4}\)
\(\frac{2}{3}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{2}{3}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}:\frac{2}{3}\)
\(x=-\frac{3}{8}\)
1 tháng 11 2018
\(\frac{2}{3}x\)\(=\)\(\frac{1}{4}\)\(-\)\(\frac{1}{2}\)
\(\frac{2}{3}x\)\(=\)\(\frac{-1}{4}\)
\(x\)\(=\)\(\frac{-1}{4}\)\(:\)\(\frac{2}{3}\)
\(x\)\(=\)\(\frac{-3}{8}\)
\(\dfrac{12^3\cdot28^2}{18^4\cdot24^3}=\dfrac{2^6\cdot3^3\cdot2^4\cdot7^2}{2^4\cdot3^8\cdot2^9\cdot3^3}=\dfrac{2^{10}\cdot7^2}{2^{13}\cdot3^8}=\dfrac{1}{8}\cdot\dfrac{49}{6561}=\dfrac{49}{52488}\)