\(\frac{7}{4}-\left(\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8}\right):x=0\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
\(\Rightarrow\)TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\) TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\) TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\)
\(\frac{1}{7}x=\frac{2}{7}\) \(-\frac{1}{5}x=\frac{3}{5}\) \(\frac{1}{3}x=\frac{4}{3}\)
\(x=\frac{2}{7}\cdot7\) \(x=\frac{3}{5}\cdot-5\) \(x=\frac{4}{3}\cdot3\)
\(x=2\) \(x=-3\) \(x=4\)
Vậy x = 2 hoặc x = -3 hoặc x = 4
b) \(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{5}x+1=0\)
\(x\cdot\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{5}\right)=1\)
\(x\cdot\frac{5+3-24}{30}=1\)
\(x\cdot\frac{-8}{15}=1\)
\(x=1\cdot\frac{-15}{8}=\frac{-15}{8}\)
Vậy x = \(\frac{-15}{8}\)
a)\(\frac{11^4.6-11^5}{11^4-11^5}:\frac{9^8.3-9^9}{9^8.5+9^8.7}\)
\(=1.6:\frac{9^8.3-9^8.9}{9^8.\left(5+7\right)}\)
\(=6:\frac{9^8.\left(3-9\right)}{9^8.12}\)
\(=6:\frac{9^8.\left(-6\right)}{9^8.12}\)
\(=6:\left(-\frac{6}{12}\right)\)
\(=6:\left(-\frac{1}{2}\right)\)
\(=-12\)
b) 3/5 : ( -1/5-1/6)+3/5:(-1/3-16/15) ( mình chuyển về ps luôn )
=3/5: (-11/30) + 3/5 : (-7/5)
=3/5:[-11/30+(-7/5)]
=3/5:53/30
=18/53
c) (1/2-13/14):5/7-(-2/21+1/7):5/7
= -3/7:5/7-1/21:5/7
=(-3/7-1/21):5/7
=-10/21:5/7
=-2/3
câu b vá c mình làm tắt nha. chúc bạn học tốt
Câu 1: \(\frac{2^{-1}+3^{-1}}{2^{-1}-3^{-1}}+\frac{2^{-1}.1}{2^3}=\frac{\frac{1}{2}+\frac{1}{3}}{\frac{1}{2}-\frac{1}{3}}+\frac{\frac{1}{2}}{8}=\frac{\frac{5}{6}}{\frac{1}{6}}+\frac{1}{6}=\frac{30}{6}+\frac{1}{6}=\frac{81}{16}\)
Câu 2:\(\frac{-1}{3}-1+\frac{\frac{1}{4}}{2}=\frac{-4}{3}+\frac{1}{8}=\frac{-29}{24}\)
Câu 3:\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{10}\left(2.3^2-3\right)}=\frac{2^{11}.3^{10}\left(2+2.5\right)}{2^{11}.3^{10}\left(2.3^2-3\right)}=\frac{4}{5}\)
Câu 4: \(\frac{1}{1-\frac{1}{1-2^{-1}}}+\frac{1}{1+\frac{1}{1+2^{-1}}}=\frac{1}{1-\frac{1}{1-\frac{1}{2}}}+\frac{1}{1+\frac{1}{1+\frac{1}{2}}}=\frac{1}{1-\frac{1}{\frac{1}{2}}}+\frac{1}{1+\frac{1}{\frac{3}{2}}}=-1+\frac{3}{5}=\frac{-2}{5}\)
#Đoàn Thị Huyền Đan ơi: Câu 1 với câu 4 thì đúng rồi còn câu 2 với 3 thì sai k/q rồi nhé!
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
\(a,\left(\frac{1}{7}x-\frac{2}{7}\right)\left(-\frac{1}{5}x+\frac{3}{5}\right)\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)
TH1 : \(\frac{1}{7}x-\frac{2}{7}=0\Rightarrow\frac{x-2}{7}=0\Rightarrow x-2=0\Leftrightarrow x=2\)
TH2 : \(-\frac{1}{5}x+\frac{3}{5}=0\Rightarrow\frac{-x+3}{5}=0\Rightarrow-x+3=0\Leftrightarrow x=3\)
TH3 : \(\frac{1}{3}x+\frac{4}{3}=0\Rightarrow\frac{x+4}{3}=0\Rightarrow x+4=0\Leftrightarrow x=-4\)
\(\Rightarrow x\in\left\{2;3;-4\right\}\)
\(b,\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x+1=0\)
\(\Rightarrow\frac{5}{30}x+\frac{3}{30}x-\frac{8}{30}x+1=0\)
\(\Rightarrow\frac{5x+3x-8x}{30}+1=0\)
\(\Rightarrow1=0\)( vô lý )\(\Rightarrow x\in\varnothing\)
a) \(\frac{\left(5.2\right)}{3.2}-\frac{1}{2}x+\frac{1}{3}+\frac{1}{5}=\frac{\left(3.2\right)}{5}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{1}{2}x+\frac{8}{15}=\frac{6}{5}\)
\(\Leftrightarrow\)\(\frac{1}{2}-\frac{2}{3}=\frac{1}{2}x\)
\(\Leftrightarrow\)\(-\frac{1}{6}=\frac{1}{2}x\)
\(\Leftrightarrow\)x=-1/3
b) VT= \(\frac{\left(3.5.4.2\right)}{5.2.3}=4\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right):6+4=4:\frac{2}{3}=6\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right):6=2\)
\(\Leftrightarrow x-\frac{1}{2}=12\)
=> x= 12,5
\(\frac{7}{4}-\left(\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8}\right)\div x=0\)
\((\frac{1}{2.2}+\frac{1}{4.3}+\frac{1}{6.4}+\frac{1}{8.5}+\frac{1}{10.6}+\frac{1}{12.7}+\frac{1}{14.8})\div x=\frac{7}{4}\)
\((\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112})\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\right]\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\right]\div x=\frac{7}{4}\)
\(\left[\frac{1}{2}\left(1-\frac{1}{8}\right)\right]\div x=\frac{7}{4}\)
\(\left(\frac{1}{2}.\frac{7}{8}\right)\div x=\frac{7}{4}\)
\(\frac{7}{16}\div x=\frac{7}{4}\)
\(x=\frac{7}{16}\div\frac{7}{4}\)
\(x=\frac{7}{16}\times\frac{4}{7}\)
\(x=\frac{1}{4}\)
\(\frac{7}{4}-\left(\frac{1}{2\cdot2}+\frac{1}{4\cdot3}+\frac{1}{6\cdot4}+\frac{1}{8\cdot5}+\frac{1}{10\cdot6}+\frac{1}{12\cdot7}+\frac{1}{14\cdot8}\right)\)
\(=\frac{7}{4}-\left(\frac{1}{4}+\frac{1}{12}+\frac{1}{24}+\frac{1}{40}+\frac{1}{60}+\frac{1}{84}+\frac{1}{112}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\left(1-\frac{1}{8}\right)\)
\(=\frac{7}{4}-\frac{1}{2}\cdot\frac{7}{8}\)
\(=\frac{7}{4}-\frac{7}{16}=\frac{28}{16}-\frac{7}{16}=\frac{21}{16}\)