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8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
a: \(=\left(\dfrac{5}{15}-\dfrac{12}{9}\right)+\left(\dfrac{14}{15}+\dfrac{11}{25}\right)+\dfrac{2}{7}\)
\(=\left(\dfrac{1}{3}-\dfrac{4}{3}\right)+\dfrac{70+33}{75}+\dfrac{2}{7}\)
\(=-1+\dfrac{2}{7}+\dfrac{103}{75}=\dfrac{-5}{7}+\dfrac{103}{75}=\dfrac{346}{525}\)
b: \(4\cdot\left(-\dfrac{1}{2}\right)^3+\dfrac{1}{2}\)
\(=4\cdot\dfrac{-1}{8}+\dfrac{1}{2}=\dfrac{-1}{2}+\dfrac{1}{2}=0\)
c: \(\dfrac{10^3+5\cdot10^2+5^3}{6^3+3\cdot6^2+3^3}=\dfrac{5^3\cdot8+5\cdot5^2\cdot2^2+5^3}{3^3\cdot2^3+3\cdot2^2\cdot3^2+3^3}\)
\(=\dfrac{5^3\left(8+4+1\right)}{3^3\left(8+4+1\right)}=\dfrac{125}{27}\)
e: \(\dfrac{2^8\cdot9^2}{6^4\cdot8^2}=\dfrac{2^8\cdot3^4}{3^4\cdot2^4\cdot2^6}=\dfrac{1}{4}\)
a, \(A=\dfrac{10^{15}+1}{10^6+1}>1\);\(B=\dfrac{10^6+1}{10^{17}+1}< 1\)
⇒\(A>B\)
b, \(D=\dfrac{2^{2007}+3}{2^{2006}-1}=\dfrac{2^{2008}+6}{2^{2007}-2}\)
Ta có : \(\dfrac{2^{2008}-3}{2^{2007}-1}< \dfrac{2^{2008}-3}{2^{2007}-2}< \dfrac{2^{2008}+6}{2^{2007}-2}\)
⇒ \(C< D\)
c, \(M=\dfrac{3}{8^3}+\dfrac{7}{8^4}=\dfrac{3}{8^3}+\dfrac{3}{8^4}+\dfrac{4}{8^4}\)
\(N=\dfrac{7}{8^3}+\dfrac{3}{8^4}=\dfrac{3}{8^3}+\dfrac{4}{8^3}+\dfrac{3}{8^4}\)
Vì \(\dfrac{4}{8^4}< \dfrac{4}{8^3}\)
⇒ \(M< N\)
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) \(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(=1-\dfrac{1}{10}\)
\(=\dfrac{9}{10}\)
a. \(x:3\dfrac{1}{15}=1\dfrac{1}{2}\)
\(x:\dfrac{46}{15}=\dfrac{3}{2}\)
\(x=\dfrac{3}{2}.\dfrac{46}{15}=\dfrac{23}{5}\)
b. \(x.\dfrac{3}{2}=-\dfrac{7}{6}\)
\(x=-\dfrac{7}{6}:\dfrac{3}{2}=-\dfrac{7}{9}\)
c. \(\dfrac{5}{6}+\dfrac{1}{4}:x=-\dfrac{2}{3}\)
\(\dfrac{13}{12}:x=-\dfrac{2}{3}\)
\(x=\dfrac{13}{12}:\left(-\dfrac{2}{3}\right)=-\dfrac{13}{8}\)
Còn lại tương tự thôi
\(\)
1: \(\Leftrightarrow3x+4=2\)
=>3x=-2
=>x=-2/3
2: \(\Leftrightarrow7x-7=6x-30\)
=>x=-23
3: =>\(5x-5=3x+9\)
=>2x=14
=>x=7
4: =>9x+15=14x+7
=>-5x=-8
=>x=8/5
a: \(A=\dfrac{2^{12}\cdot3^{10}+2^3\cdot2^9\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot7}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2}{3}\cdot\dfrac{6}{7}=\dfrac{12}{21}=\dfrac{4}{7}\)
b: \(B=\left(\dfrac{12}{105}+\dfrac{9^{15}}{3}\right)\cdot\dfrac{1}{3}\cdot\dfrac{6^8}{6^4\cdot2^4}\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot\dfrac{1}{3}\cdot3^4\)
\(=\dfrac{12+35\cdot9^{15}}{105}\cdot3^3=\dfrac{9\left(12+35\cdot9^{15}\right)}{35}\)
\(a)\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}\)
\(=\dfrac{18}{24}+\dfrac{12}{24}+\left(-\dfrac{5}{24}\right)\)
\(=\dfrac{18+12+\left(-5\right)}{24}\)
\(=\dfrac{25}{24}\)
\(b)\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}.\dfrac{-2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}\)
\(=\dfrac{5}{7}\left(\dfrac{-2}{13}+\dfrac{-11}{13}+\dfrac{13}{13}\right)\)
\(=\dfrac{5}{7}.0=0\)
\(c)\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}\)
\(=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}\)
\(=2\dfrac{1}{2}\)
\(d)\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{38}{51}+\dfrac{306}{714}\)
\(=\dfrac{315}{714}+\dfrac{238}{714}+\dfrac{532}{714}+\dfrac{306}{714}\)
\(=\dfrac{1391}{714}\)
a)\(\dfrac{3}{4}+\dfrac{6}{12}-\dfrac{5}{24}=\dfrac{18}{24}+\dfrac{12}{24}-\dfrac{5}{24}=\dfrac{25}{24}\)
b)\(\dfrac{-5}{7}.\dfrac{2}{13}-\dfrac{5}{7}.\dfrac{11}{13}+\dfrac{5}{7}=\dfrac{5}{7}\left(\dfrac{-2}{13}-\dfrac{11}{13}+1\right)=\dfrac{5}{7}.0=0\)
c)\(\dfrac{27}{23}+\dfrac{5}{21}+\dfrac{1}{2}-\dfrac{4}{23}+\dfrac{16}{21}=\left(\dfrac{27}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}=1+1+\dfrac{1}{2}=2,5\)
d)\(\dfrac{15}{34}+\dfrac{7}{21}+\dfrac{19}{34}.\dfrac{20}{15}+\dfrac{3}{7}=\dfrac{15}{34}+\left(\dfrac{1}{3}+\dfrac{38}{51}+\dfrac{3}{7}\right)=\dfrac{15}{34}+\dfrac{538}{357}=\dfrac{1391}{714}\)
xin lỗi mk sai
sửa lại:
\(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-6}{10}=\dfrac{20}{30}+\dfrac{-6}{30}=\dfrac{14}{30}=\dfrac{7}{15}\)
\(\Rightarrow\)đáp án đúng là B
ps:xin lỗi các bạn mk nhầm
ta có:\(\dfrac{2}{3}+\dfrac{1}{3}.\dfrac{-6}{10}=1.\dfrac{-6}{10}=\dfrac{-6}{10}\)
\(\Rightarrow\)đáp án đúng là A