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a: \(=5-2\cdot\dfrac{1}{4}=5-\dfrac{1}{2}=\dfrac{9}{2}\)
b: \(=\left(\dfrac{7}{2}\right)^3+\dfrac{1}{2}=\dfrac{343}{8}+\dfrac{1}{2}=\dfrac{347}{8}\)
c: \(=\left(5+\dfrac{5}{27}-\dfrac{5}{27}\right)+\left(\dfrac{7}{23}+\dfrac{16}{23}\right)-\dfrac{1}{2}=5+1-\dfrac{1}{2}=5+\dfrac{1}{2}=5.5\)
e: \(=\dfrac{-5}{4}\left(35+\dfrac{1}{6}-45-\dfrac{1}{6}\right)=\dfrac{-5}{4}\cdot\left(-10\right)=\dfrac{50}{4}=\dfrac{25}{2}\)
\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=\dfrac{140}{27}+\dfrac{7}{23}+\dfrac{1}{2}-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=\dfrac{140}{27}-\dfrac{5}{27}+\dfrac{7}{23}+\dfrac{16}{23}+\dfrac{1}{2}\)
\(=5+1+\dfrac{1}{2}=6+\dfrac{1}{2}=\dfrac{13}{2}\)
Giải:
\(5\dfrac{27}{5}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=5+\dfrac{27}{5}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=5+\dfrac{27}{5}+\dfrac{43}{23}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{52}{5}+\dfrac{43}{23}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{1411}{115}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{2937}{230}-\dfrac{5}{27}\)
\(\approx12,58\)
Vậy ...
\(5\frac{27}{5}+\frac{27}{23}+0,5-\frac{5}{27}+\frac{16}{23}\)
=\(\frac{52}{5}+\frac{27}{23}+\frac{1}{2}-\frac{5}{27}+\frac{16}{23}\)
=(\(\frac{52}{5}+\frac{1}{2}\)) + (\(\frac{27}{23}+\frac{16}{23}\)) - \(\frac{5}{27}\)
=\(\frac{109}{10}+\frac{43}{23}-\frac{5}{27}\)
=12,58438003
hay \(\approx12,58\)
a: \(=\dfrac{-3}{4}\left(31+\dfrac{11}{23}+8+\dfrac{12}{23}\right)=\dfrac{-3}{4}\cdot40=-30\)
b: \(=\left(\dfrac{7}{3}+\dfrac{7}{2}\right):\left(-\dfrac{25}{6}+\dfrac{22}{7}\right)+\dfrac{15}{2}\)
\(=\dfrac{35}{6}:\dfrac{-175+132}{42}+\dfrac{15}{2}\)
\(=\dfrac{35}{6}\cdot\dfrac{42}{-43}+\dfrac{15}{2}\)
\(=\dfrac{35\cdot7}{-43}+\dfrac{15}{2}\)
\(=\dfrac{-70\cdot7+15\cdot43}{86}=\dfrac{155}{86}\)
c: \(=\dfrac{-7}{5}\left(4+\dfrac{5}{9}+5+\dfrac{4}{9}\right)=\dfrac{-7}{5}\cdot10=-14\)
d: \(=4+\dfrac{25}{16}+25\cdot\left(\dfrac{9}{16}\cdot\dfrac{64}{125}\cdot\dfrac{-8}{27}\right)\)
\(=\dfrac{89}{16}+25\cdot\dfrac{-32}{375}\)
\(=\dfrac{89}{16}-\dfrac{32}{15}=\dfrac{823}{240}\)
e: \(=\dfrac{2}{3}-4\cdot\left(\dfrac{2}{4}+\dfrac{3}{4}\right)=\dfrac{2}{3}-5=-\dfrac{13}{3}\)
Mình chỉ giải câu a thôi,mấy câu còn lại dễ.
a)Ta có:\(\dfrac{x}{27}=\dfrac{-3}{x}\)
=>\(x^2=-3\cdot27=-81\)(Nhân chéo)
Mà x2>0 với mọi x nên :
Không có giá trị nào thỏa mãn điều kiện của x
Tìm x biết :
a) \(\dfrac{x}{27}=-\dfrac{3}{x}\) \(\Rightarrow2x=-3.27\Rightarrow2x=-81\Rightarrow x=-40,5\)
b) \(-\dfrac{9}{x}=-\dfrac{x}{\dfrac{4}{49}}\Rightarrow2x=-9.\left(-\dfrac{4}{9}\right)\Rightarrow2x=4\Rightarrow x=2\)
c) \(\left|7x-\dfrac{5}{3}\right|+\dfrac{7}{19}=-\dfrac{8}{15}\) ( mk nghĩ bn chép sai đề bài câu này )
\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{8}{15}-\dfrac{7}{19}\)
\(\Rightarrow\left|7x-\dfrac{5}{3}\right|=-\dfrac{257}{285}\)
\(\Rightarrow\left[{}\begin{matrix}7x-\dfrac{5}{3}=-\dfrac{257}{285}\\7x-\dfrac{5}{3}=\dfrac{257}{285}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{218}{1995}\\x=\dfrac{244.}{665}\end{matrix}\right.\)
d) \(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=\dfrac{18}{19}-1\dfrac{2}{5}\)
\(\left|\dfrac{1}{23}x\right|+\dfrac{18}{90}=-\dfrac{43}{95}\)
\(\left|\dfrac{1}{23}x\right|=-\dfrac{43}{95}-\dfrac{18}{90}\)
\(\left|\dfrac{1}{23}x\right|=-\dfrac{62}{95}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{23}x=\dfrac{62}{95}\\\dfrac{1}{23}x=-\dfrac{62}{95}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\dfrac{1}{95}\\x=-15\dfrac{1}{95}\end{matrix}\right.\)
\(1:5\dfrac{25}{7}+\dfrac{27}{23}+0,5-\dfrac{5}{27}+\dfrac{16}{23}\)
\(=\dfrac{60}{7}+\left(\dfrac{27}{23}+\dfrac{16}{23}\right)+0,5-\dfrac{5}{27}\)
\(=\dfrac{60}{7}+\dfrac{43}{23}+0,5-\dfrac{5}{27}\)
\(=\dfrac{1681}{161}+\dfrac{1}{2}-\dfrac{5}{27}\)
\(=\dfrac{3523}{322}-\dfrac{5}{27}\)
\(=10,7558086\)
\(2:\left(-0,75-\dfrac{1}{4}\right):\left(-5\right)+\dfrac{1}{15}-\left(-\dfrac{1}{5}\right):\left(-3\right)\)
\(=-1:\left(-0,5\right)+\dfrac{1}{15}-\dfrac{1}{15}\)
\(=2+0=2\)
a. \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1+0,5\)
\(=2,5\)
b. \(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}.\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)\)
\(=\dfrac{3}{7}.\left(-14\right)=-6\)
c. \(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(\dfrac{-5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)\)
\(=-10:\left(-\dfrac{5}{7}\right)\)
\(=14\)
d. \(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\dfrac{-5}{21}:\dfrac{4}{5}+\dfrac{5}{21}:\dfrac{4}{5}\)
\(=\left(\dfrac{-5}{7}+\dfrac{5}{7}\right):\dfrac{4}{5}\)
\(=0:\dfrac{4}{5}\)
\(=0\)
a,
\(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
\(=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5\)
\(=1+1-0,5=1,5\)
b,
\(\dfrac{3}{7}\cdot19\dfrac{1}{3}-\dfrac{3}{7}.33\dfrac{1}{3}\)
\(=\dfrac{3}{7}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{3}{7}.\left(-14\right)=-6\)
c,
\(15\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)-25\dfrac{1}{4}:\left(-\dfrac{5}{7}\right)\)
\(=\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right):\left(-\dfrac{5}{7}\right)=-10:\left(-\dfrac{5}{7}\right)=14\)
d,
\(\left(-\dfrac{2}{3}+\dfrac{3}{7}\right):\dfrac{4}{5}+\left(-\dfrac{1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left(-\dfrac{2}{3}+\dfrac{3}{7}+\dfrac{-1}{3}+\dfrac{4}{7}\right):\dfrac{4}{5}\)
\(=\left[\left(-\dfrac{2}{3}+\dfrac{-1}{3}\right)+\left(\dfrac{3}{7}+\dfrac{4}{7}\right)\right]:\dfrac{4}{5}\)
\(=\left(-1+1\right):\dfrac{4}{5}=0:\dfrac{4}{5}=0\)
a) \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}=\left(1\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+0,5=1+1+0,5=2,5\)b)
\(\dfrac{3}{7}.19\dfrac{1}{3}-\dfrac{7}{7}.33\dfrac{1}{3}=\dfrac{7}{3}\left(19\dfrac{1}{3}-33\dfrac{1}{3}\right)=\dfrac{7}{3}.\left(-14\right)=-\dfrac{1}{6}\)
c,
\(\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{5}{7}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-5}{7}\right)=\left(15\dfrac{1}{4}+2010\right):\left(-\dfrac{7}{5}\right)-\left(25\dfrac{1}{4}+2016\right):\left(\dfrac{-7}{5}\right)\)
\(\left(-\dfrac{7}{5}\right)\left(15\dfrac{1}{4}+2010-25\dfrac{1}{4}-2016\right)=\left(-\dfrac{7}{5}\right)\left(-10-6\right)=22,4\)
d,
\(\left(2017-\dfrac{3}{7}+\dfrac{9}{11}\right)-\left(2016-\dfrac{3}{7}+\dfrac{8}{17}\right)-\left(2015+\dfrac{9}{11}-\dfrac{8}{17}\right)=2017-\dfrac{3}{7}+\dfrac{9}{11}-2016+\dfrac{3}{7}-\dfrac{8}{17}-2015-\dfrac{9}{11}+\dfrac{8}{17}\)\(\left(2017-2016-2015\right)+\left(-\dfrac{3}{7}+\dfrac{3}{7}\right)+\left(\dfrac{9}{11}-\dfrac{9}{11}\right)+\left(-\dfrac{8}{17}+\dfrac{8}{17}\right)=-2014\)
Bạn ơi cho mình hỏi tại sao đề bài câu c là -5/7 mà bn lm -7/5
Dấu " / " là phân số nhé :
a) 5 5/27 + 7/23 + 0,5 - 5/27 + 16/23
= 140/27 + 7/23 + 1/2 - 5/27 + 16/23
= ( 140/27 - 5/27 ) + ( 7/23 + 16/23 ) + 1/2
= 5 + 1 + 1/2
= 6 + 1/2
= 13/2
b) 3/8 . 27 1/5 - 51 .1/5 . 3/8 + 19
= 3/8 . 136/5 - 51 . 1/5 . 3/8 + 19
= 3/8 . ( 136/5 - 1/5 ) - 51 + 19
= 3/8 . 27 - 51 + 19
= 81/8 - 52 + 19
= -183/8
Sorry nha mik sửa lại con B
b) 3/8 . 27 1/5 - 51 . 1/5 . 3/8 + 19
= 3/8 . 136/5 - 51/5 . 3/8 + 19
= 3/8 . ( 136/5 - 51/5 ) + 19
= 3/8 . 17 + 19
= 51/8 + 19
= 203/8