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a))\(\frac{2^3.3^3.3^3.3^4.5^4.5^4.3^7.2^7.2^7.2^7.2^6}{2^5.2^5.3^5.2^6.2^6.3^6.3^6.3^4.5^4}=\frac{5^4.2^7}{3^4}=\frac{625.128}{81}=\frac{80000}{81}\)
\(\frac{18^3.75^4.24^7.2^6}{12^5.36^6.15^4}=\frac{2^3.3^6.3^4.5^8.2^{21}.3^7.2^6}{2^{10}.3^5.2^{12}.3^{12}.3^4.5^4}=\frac{2^{30}.3^{17}.5^8}{2^{22}.3^{21}.5^4}=\frac{2^8.5^4}{3^4}\)
mình làm đúng đó không tin thì bấm máy tính thử đi phạm thủy linh làm sai rồi
cho mình đúng hen
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)
\(\dfrac{6}{\dfrac{5}{12}}:\dfrac{2}{\dfrac{3}{4}}+\dfrac{11}{\dfrac{1}{4}}\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\)
\(=\dfrac{27}{5}+\dfrac{11}{\dfrac{1}{4}}.\dfrac{2}{15}\)
\(=\dfrac{27}{5}+\dfrac{88}{15}\)
\(=\dfrac{169}{15}\)
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
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
Khử mẫu : \(12x-12+6x-6=4x+3x-7\)
\(\Leftrightarrow18x-18=7x-7\Leftrightarrow11x=11\Leftrightarrow x=1\)
\(\frac{x-1}{1}+\frac{x-1}{2}=\frac{x}{3}+\frac{x}{4}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12}{12}+\frac{6x-6}{12}=\frac{4x}{12}+\frac{3x}{12}-\frac{7}{12}\)
\(\Leftrightarrow\frac{12x-12+6x-6}{12}=\frac{4x+3x-7}{12}\)
\(\Leftrightarrow18x-18=7x-7\)
\(\Leftrightarrow18x+7x=18+7\)
\(\Leftrightarrow25x=25\)
\(\Leftrightarrow x=1\)
\(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)
\(\Rightarrow2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\)
\(\Rightarrow2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)
\(\Rightarrow A=\left(2-1\right)+\dfrac{3}{2^2}+\left(\dfrac{4}{2^3}-\dfrac{3}{2^3}\right)+....\left(\dfrac{99}{2^{98}}-\dfrac{98}{2^{98}}\right)-\dfrac{100}{2^{100}}\)
\(\Rightarrow A=1+\dfrac{3}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}-\dfrac{100}{2^{100}}\)
\(\Rightarrow A=1+\dfrac{3}{2^2}+\left(\dfrac{1}{2^3}+...+\dfrac{1}{2^{98}}\right)-\dfrac{100}{2^{100}}\)
\(\Rightarrow A=1+\dfrac{3}{2^2}+\dfrac{1}{2^2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)
1) a.Từ\(\frac{x}{y}=\frac{11}{7}\Rightarrow\frac{x}{11}=\frac{y}{7}\)
Áp dụng t/c của dãy tỉ số bằng nhau, ta có:
\(\frac{x}{11}=\frac{y}{7}=\frac{x-y}{11-7}=\frac{12}{4}=3\)
\(\Rightarrow x=3.11=33;y=3.7=21\)
b) \(\sqrt{2x-3}=5\)
\(2x-3=25\)
\(2x=28\)
\(x=14\)
2) a) \(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2+\sqrt{4}=\frac{3}{2}-\frac{5}{6}:\frac{1}{4}+2\)
\(=\frac{3}{2}-\frac{10}{3}+2\)
\(=\frac{1}{6}\)
_Học tốt nha_
1. a, \(\frac{x}{y}=\frac{11}{7}\)và x-y=12
\(\Rightarrow\frac{x}{11}=\frac{y}{7}\)và x-y=12
Áp dung tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{11}=\frac{y}{7}=\frac{x-y}{11-7}=\frac{12}{4}=3\)
\(\Rightarrow\hept{\begin{cases}\frac{x}{11}=3\\\frac{y}{7}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}x=33\\y=21\end{cases}}\)
Vậy
b,\(\sqrt{2x-3}\)=5
\(\Rightarrow2x-3=25\)
\(\Rightarrow2x=28\)
\(\Rightarrow x=14\)
c,\(\frac{3}{2}-\frac{5}{6}:\left(\frac{1}{2}\right)^2+\sqrt{4}\)
\(=\frac{3}{2}-\frac{5}{6}:\frac{1}{4}+2\)
\(=\frac{3}{2}-\frac{10}{3}+2\)
\(=\frac{9}{6}-\frac{20}{6}+2\)
\(=\frac{-11}{6}+2\)
\(=\frac{1}{6}\)
2) -12:\(\left(-\dfrac{5}{6}\right)^2\)=\(-12:\dfrac{25}{36}=-12\cdot\dfrac{36}{25}=-\dfrac{432}{25}\)
s) \(-\dfrac{1}{12}-\left(2\dfrac{5}{8}-\dfrac{1}{3}\right)=-\dfrac{1}{12}-\left(\dfrac{21}{8}-\dfrac{1}{3}\right)\)
= \(-\dfrac{1}{12}-\dfrac{55}{24}=-\dfrac{2}{24}-\dfrac{55}{24}=-\dfrac{57}{24}=-\dfrac{19}{8}\)
t) \(-1,75-\left(-\dfrac{1}{9}-2\dfrac{1}{18}\right)=-1,75-\left(-\dfrac{2}{18}-\dfrac{37}{18}\right)\)
= -1,75-(\(-\dfrac{13}{6}\)) = \(-\dfrac{7}{4}+\dfrac{13}{6}=\dfrac{5}{12}\)
c) \(\left(\sqrt{\dfrac{1}{9}}-0,5\right)^3+\dfrac{-1}{3}=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)^3-\dfrac{1}{3}\)
= \(\left(-\dfrac{1}{6}\right)^3-\dfrac{1}{3}=\dfrac{-1}{216}-\dfrac{1}{3}=-\dfrac{73}{216}\)
d) \(\left(\dfrac{1}{2}-\sqrt{\dfrac{4}{25}}\right)^2-2\dfrac{1}{2}=\left(\dfrac{1}{2}-\dfrac{2}{5}\right)^2-\dfrac{5}{2}\)
= \(\left(\dfrac{1}{10}\right)^2-\dfrac{5}{2}=\dfrac{1}{100}-\dfrac{250}{100}=-\dfrac{249}{100}=-2,49\)
1. Tính:
a. \(\dfrac{\text{−1 }}{\text{4 }}+\dfrac{\text{5 }}{\text{6 }}=\dfrac{-3}{12}+\dfrac{10}{12}=\dfrac{7}{12}\)
b. \(\dfrac{\text{5 }}{\text{12 }}+\dfrac{\text{-7 }}{8}=\dfrac{10}{24}+\dfrac{-21}{24}=\dfrac{-11}{24}\)
c. \(\dfrac{-7}{6}+\dfrac{-3}{10}=\dfrac{-35}{30}+\dfrac{-9}{30}=\dfrac{-44}{30}=\dfrac{-22}{15}\)
d.\(\dfrac{-3}{7}+\dfrac{5}{6}=\dfrac{-18}{42}+\dfrac{35}{42}=\dfrac{17}{42}\)
2. Tính :
a. \(\dfrac{2}{14}-\dfrac{5}{2}=\dfrac{2}{14}-\dfrac{35}{14}=\dfrac{-33}{14}\)
b.\(\dfrac{-13}{12}-\dfrac{5}{18}=\dfrac{-39}{36}-\dfrac{10}{36}=\dfrac{49}{36}\)
c.\(\dfrac{-2}{5}-\dfrac{-3}{11}=\dfrac{-2}{5}+\dfrac{3}{11}=\dfrac{-22}{55}+\dfrac{15}{55}=\dfrac{-7}{55}\)
d. \(0,6--1\dfrac{2}{3}=\dfrac{6}{10}--\dfrac{5}{3}=\dfrac{3}{5}+\dfrac{5}{3}=\dfrac{9}{15}+\dfrac{25}{15}=\dfrac{34}{15}\)
3. Tính :
a.\(\dfrac{-1}{39}+\dfrac{-1}{52}=\dfrac{-4}{156}+\dfrac{-3}{156}=\dfrac{-7}{156}\)
b.\(\dfrac{-6}{9}-\dfrac{12}{16}=\dfrac{2}{3}-\dfrac{3}{4}=\dfrac{8}{12}-\dfrac{9}{12}=\dfrac{-17}{12}\)
c. \(\dfrac{-3}{7}-\dfrac{-2}{11}=\dfrac{-3}{7}+\dfrac{2}{11}=\dfrac{-33}{77}+\dfrac{14}{77}=\dfrac{-19}{77}\)
d.\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...\dfrac{1}{8.9}+\dfrac{1}{9.10}\)
\(=\dfrac{1}{1}+\dfrac{1}{10}\)
\(=\dfrac{10}{10}-\dfrac{1}{10}\)
= \(\dfrac{9}{10}\)
Chế Kazuto Kirikaya thử tham khảo thử đi !!!
Mấy câu trên kia dễ rồi mình chữa mình câu \(c\) bài \(3\) thôi nhé Kazuto Kirikaya
d) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
\(\dfrac{12^3\cdot28^2}{18^4\cdot24^3}=\dfrac{2^6\cdot3^3\cdot2^4\cdot7^2}{2^4\cdot3^8\cdot2^9\cdot3^3}=\dfrac{2^{10}\cdot7^2}{2^{13}\cdot3^8}=\dfrac{1}{8}\cdot\dfrac{49}{6561}=\dfrac{49}{52488}\)