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Ta có :
\(A=\dfrac{34}{7.13}+\dfrac{51}{13.22}+\dfrac{85}{22.37}+\dfrac{68}{37.49}\)
\(\dfrac{A}{17}=\dfrac{2}{7.13}+\dfrac{3}{13.22}+\dfrac{5}{22.37}+\dfrac{4}{37.49}\)
\(A.\dfrac{3}{17}=\dfrac{6}{7.13}+\dfrac{9}{13.22}+\dfrac{15}{22.37}+\dfrac{12}{37.49}\)
\(A.\dfrac{3}{17}=\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\)
\(A.\dfrac{3}{17}=\dfrac{1}{7}-\dfrac{1}{49}\)
\(A.\dfrac{3}{17}=\dfrac{6}{49}\)
\(\Rightarrow A=\dfrac{6}{49}:\dfrac{3}{17}=\dfrac{34}{49}\)
\(B=\dfrac{39}{7.16}+\dfrac{65}{16.31}+\dfrac{52}{31.43}+\dfrac{26}{37.49}\)
\(\dfrac{B}{13}=\dfrac{3}{7.16}+\dfrac{5}{16.31}+\dfrac{4}{31.43}+\dfrac{2}{37.49}\)
\(B.\dfrac{3}{13}=\dfrac{9}{7.16}+\dfrac{15}{16.31}+\dfrac{12}{31.43}+\dfrac{6}{43.49}\)
\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{43}+\dfrac{1}{43}-\dfrac{1}{49}\)
\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{49}\)
\(B.\dfrac{3}{13}=\dfrac{1}{7}-\dfrac{1}{49}=\dfrac{6}{49}\)
\(\Rightarrow B=\dfrac{6}{49}:\dfrac{3}{13}=\dfrac{26}{49}\)
\(\Rightarrow\dfrac{A}{B}=\dfrac{34}{49}:\dfrac{26}{49}=\dfrac{17}{13}\)
Chúc bn học tốt!!!!!!!!!
Link này bạn: Câu hỏi của Quỳnh Anh Shuy - Toán lớp 7 | Học trực tuyến
a) \(2\dfrac{3}{4}.\left(-0,4\right)-1\dfrac{3}{5}.2,75+\left(-1,2\right):\dfrac{4}{11}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\dfrac{11}{4}\)
= \(2,75.\left(-0,4\right)-\left(1,6\right).\left(2,75\right)+\left(-1,2\right).\left(2,75\right)\)
= \(2,75.\left\{\left(-0,4\right)-\left(1,6\right)+\left(-1,2\right)\right\}\)
= \(2,75.\left(-3,2\right)\)
= \(-8,8\)
b) \(1,4.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):2\dfrac{1}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\left(\dfrac{4}{5}+\dfrac{2}{3}\right):\dfrac{11}{5}\)
= \(\dfrac{7}{5}.\dfrac{15}{49}-\dfrac{22}{15}.\dfrac{5}{11}\)
= \(\dfrac{3}{7}-\dfrac{2}{3}\)
= \(-\dfrac{5}{21}\)
c) \(\left(-3,2\right).\dfrac{15}{64}+\left(0,8-2\dfrac{4}{15}\right):3\dfrac{2}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(\dfrac{4}{5}-2\dfrac{4}{15}\right):\dfrac{11}{3}\)
= \(-\dfrac{16}{5}.\dfrac{15}{64}+\left(-\dfrac{22}{15}\right).\dfrac{3}{11}\)
= \(\left(-\dfrac{3}{4}\right)+\left(-\dfrac{2}{5}\right)\)
= \(-\dfrac{23}{20}\)
d) \(0,02.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-2\dfrac{9}{20}\right).\dfrac{2}{7}\)
= \(\dfrac{1}{50}.\dfrac{-25}{2}+\dfrac{3}{8}+\left(-\dfrac{49}{20}\right).\dfrac{2}{7}\)
=\(\left(-\dfrac{1}{4}\right)+\dfrac{3}{8}+\left(-\dfrac{7}{10}\right)\)
= \(\dfrac{1}{8}+\left(-\dfrac{7}{10}=\right)\)
= \(-\dfrac{23}{40}\)
e) \(34\%:\dfrac{51}{16}-3\dfrac{7}{9}.6,5-\left(0,4\right)^2\)
= \(\dfrac{17}{50}.\dfrac{16}{51}-\dfrac{34}{9}.\dfrac{13}{2}-\dfrac{4}{25}\)
= \(\dfrac{8}{75}-\dfrac{221}{9}-\dfrac{4}{15}\)
= \(-\dfrac{5501}{225}\)
\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)
=>A>0(1)
\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)
=>B<0(2)
C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)
=>C=0(3)
Từ 1;2;3 =>A>C>B
\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)
\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)
\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
2155-(174+2155)+(-68+174)=2155-174-2155-68+174
= -68
( 1 - \(\dfrac{1}{2}\) ) ( 1- \(\dfrac{1}{3}\)) ( 1 - \(\dfrac{1}{4}\)) ( 1 - \(\dfrac{1}{5}\)) = \(\dfrac{1}{2}.\dfrac{1}{3}.\dfrac{1}{4}.\dfrac{1}{5}\)
= \(\dfrac{1}{120}\)
Mình ps có 2 câu à ^.^!
a) \(\dfrac{-3}{7}+\dfrac{15}{26}-\left(\dfrac{2}{13}-\dfrac{3}{7}\right)\\ =\dfrac{-3}{7}+\dfrac{15}{26}-\dfrac{2}{13}+\dfrac{3}{7}\\ =\left(\dfrac{-3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{15}{26}-\dfrac{2}{13}\right)\\ =0+\left(\dfrac{15}{26}-\dfrac{4}{26}\right)\\ =0+\dfrac{11}{26}\\ =\dfrac{11}{26}\)
\(c)\dfrac{-11}{23}.\dfrac{6}{7}+\dfrac{8}{7}.\dfrac{-11}{23}-\dfrac{1}{23}\\=\dfrac{-1}{23}\left ( \dfrac{66}{7}+\dfrac{88}{7}+1 \right )\\ =\dfrac{-1}{23}.23=-1\)
Bài 1:
a, \(\left(x-2\right)^2=9\)
\(\Rightarrow x-2\in\left\{-3;3\right\}\Rightarrow x\in\left\{-1;5\right\}\)
b, \(\left(3x-1\right)^3=-8\)
\(\Rightarrow3x-1=-2\Rightarrow3x=-1\)
\(\Rightarrow x=-\dfrac{1}{3}\)
c, \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{1}{16}\)
\(\Rightarrow x+\dfrac{1}{2}\in\left\{-\dfrac{1}{4};\dfrac{1}{4}\right\}\)
\(\Rightarrow x\in\left\{-\dfrac{3}{4};-\dfrac{1}{4}\right\}\)
d, \(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
\(\Rightarrow\left(\dfrac{2}{3}\right)^x=\left(\dfrac{2}{3}\right)^2\)
Vì \(\dfrac{2}{3}\ne\pm1;\dfrac{2}{3}\ne0\) nên \(x=2\)
e, \(\left(\dfrac{1}{2}\right)^{x-1}=\dfrac{1}{16}\)
\(\Rightarrow\left(\dfrac{1}{2}\right)^{x-1}=\left(\dfrac{1}{2}\right)^4\)
Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(x-1=4\Rightarrow x=5\)
f, \(\left(\dfrac{1}{2}\right)^{2x-1}=8\) \(\Rightarrow\left(\dfrac{1}{2}\right)^{2x-1}=\left(\dfrac{1}{2}\right)^{-3}\) Vì \(\dfrac{1}{2}\ne\pm1;\dfrac{1}{2}\ne0\) nên \(2x-1=-3\) \(\Rightarrow2x=-2\Rightarrow x=-1\) Chúc bạn học tốt!!!
\(\left(\frac{3}{8}\right)^5=\frac{243}{8^5}>\frac{125}{8^5}>\frac{125}{9^5}=\frac{125}{\left(3^2\right)^5}=\frac{125}{3^{10}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\left(\frac{5}{243}\right)^3\)
\(=>\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
1. Ta có : (\(\dfrac{-3}{8}\))3 < 0
(\(\dfrac{8}{243}\))3 > 0
=> (\(\dfrac{-3}{8}\))3 < (\(\dfrac{8}{243}\))3
@Cuber Việt
\(\left(\dfrac{-3}{8}\right)^3< 0< \left(\dfrac{8}{243}\right)^3\)
Vậy \(\left(\dfrac{-3}{8}\right)^3< \left(\dfrac{8}{243}\right)^3\)
\(A=\dfrac{34}{7\cdot13}+\dfrac{51}{13\cdot22}+\dfrac{85}{22\cdot37}+\dfrac{68}{37\cdot49}\\ =\dfrac{17}{3}\cdot\dfrac{6}{7\cdot13}+\dfrac{17}{3}\cdot\dfrac{9}{13\cdot22}+\dfrac{17}{3}\cdot\dfrac{15}{22\cdot37}+\dfrac{17}{3}\cdot\dfrac{12}{37\cdot49}\\ =\dfrac{17}{3}\cdot\left(\dfrac{6}{7\cdot13}+\dfrac{9}{13\cdot22}+\dfrac{15}{22\cdot37}+\dfrac{12}{37\cdot49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{22}+\dfrac{1}{22}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\left(\dfrac{1}{7}-\dfrac{1}{49}\right)\\ =\dfrac{17}{3}\cdot\dfrac{6}{49}\\ =\dfrac{34}{49}\)