\(\dfrac{1}{2}-\dfrac{2}{3}+\dfrac{3}{4}\)
Tính:
a/\(\dfrac{1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}{1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}}:\dfrac{3+\dfrac{3}{2}+\dfrac{3}{3}+\dfrac{3}{4}}{2-\dfrac{2}{2}+\dfrac{2}{3}-\dfrac{2}{4}}\)
b/\(\dfrac{1+\dfrac{1}{4}+\dfrac{1}{1+\dfrac{1}{4}}}{1-\dfrac{1}{4}-\dfrac{1}{1-\dfrac{1}{4}}}\)
c/\(\dfrac{\dfrac{2}{5}-\dfrac{7}{5}}{\dfrac{2}{5}-\dfrac{\dfrac{3}{4}}{\dfrac{3}{4}.\dfrac{3}{7}-1}}-\dfrac{1}{\dfrac{3}{7}\left(\dfrac{3}{4}.\dfrac{3}{7}.\dfrac{2}{5}-\dfrac{2}{5}-\dfrac{3}{4}\right)}\)
d/\(\left(\dfrac{\dfrac{4}{3}}{2+\dfrac{4}{3}}+\dfrac{2-\dfrac{4}{3}}{\dfrac{4}{3}}\right).\left(\dfrac{\dfrac{2}{3}}{4+\dfrac{2}{3}}-\dfrac{4-\dfrac{2}{3}}{\dfrac{2}{3}}\right)\)
Giúp mik với các bạn ơi 1 bài thôi cug đc.
a
= { 1*( 1+1/2+1/3+1/4) } / { 1 * ( 1-1/2 +1/3-1/4)} : { 3*(1+1/2+1/3+1/4)} / { 2*( 1-1/2 +1/3-1/4)}
Sau đó bn tự tính ra nhé cứ tính nhu bình thường sẽ ra.
Mà mình thấy máy câu này yêu cầu tính chứ có bảo tính theo cách hợp lí đâu? Vì thế bn cứ lấy máy tính tính như bình thường là được .
\(3\dfrac{3}{3}.\dfrac{1}{3}-\dfrac{3}{4}.\dfrac{1}{3}\)
\(\left[\dfrac{11}{3}\right]-\left(\dfrac{-1}{2}\right)^2-4\dfrac{1}{2}\)
\(\left(\dfrac{3}{2}-\dfrac{5}{4}+\dfrac{1}{3}\right):\left(\dfrac{4}{3}+2\dfrac{3}{2}-\dfrac{3}{4}\right)\)
\(5\dfrac{5}{27}+\dfrac{7}{23}+0,5+\dfrac{-5}{27}+\dfrac{16}{23}\)
\(2\dfrac{5}{4}+\left(-2018\right)^0-\left[\dfrac{-1}{4}\right]\)
\(\dfrac{19}{11}.\dfrac{6}{5}+\dfrac{6^2}{11}.\dfrac{6}{5}-\left(\dfrac{1}{2}\right)^0\)
1, \(\dfrac{3}{4}\). ( \(\dfrac{2}{5}\) - \(\dfrac{1}{15}\)) +\(\dfrac{3}{4}\)
2, \(\dfrac{4}{9}\). (\(\dfrac{-13}{3}\)) + \(\dfrac{4}{3}\). \(\dfrac{40}{9}\)
3, \(\dfrac{4}{9}\) - \(\dfrac{2}{3}\). ( \(\dfrac{4}{5}\)+\(\dfrac{1}{2}\) )
giúp mình nha cảm ơn
1, \(\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\left(\dfrac{2}{5}-\dfrac{1}{15}+1\right)\)
\(=\dfrac{3}{4}.\dfrac{6-1+15}{15}=\dfrac{3}{4}.\dfrac{20}{15}=\dfrac{3}{4}.\dfrac{4}{3}=1\)
2, \(\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{3}.\dfrac{40}{9}=\dfrac{4}{9}.\left(-\dfrac{13}{3}\right)+\dfrac{4}{9}.\dfrac{40}{3}\)
\(=\dfrac{4}{9}.\left[\left(-\dfrac{13}{3}\right)+\dfrac{40}{3}\right]=\dfrac{4}{9}.9=4\)
3, \(\dfrac{4}{9}-\dfrac{2}{3}.\left(\dfrac{4}{5}+\dfrac{1}{2}\right)=\dfrac{2}{3}\left(\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{1}{2}\right)\)
\(=\dfrac{2}{3}.\dfrac{20-24-15}{30}=\dfrac{2}{3}.\left(-\dfrac{19}{30}\right)=-\dfrac{19}{45}\)
1. \(\dfrac{3}{4}.\left(\dfrac{6}{15}-\dfrac{1}{15}\right)+\dfrac{3}{4}=\dfrac{3}{4}.\dfrac{1}{3}+\dfrac{3}{4}=\dfrac{1}{4}+\dfrac{3}{4}=1\)
1/ \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
2/ \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
3/ \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
4/ \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
5/ \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
1: Ta có: \(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow2x-8+12x=4x-2\)
\(\Leftrightarrow10x=6\)
hay \(x=\dfrac{3}{5}\)
2: Ta có: \(\dfrac{5x-2}{5}-2=\dfrac{1-2x}{3}\)
\(\Leftrightarrow15x-6-30=10-20x\)
\(\Leftrightarrow35x=46\)
hay \(x=\dfrac{46}{35}\)
3: Ta có: \(\dfrac{x-2}{2}-\dfrac{2}{3}=x-1\)
\(\Leftrightarrow3x-6-4=6x-6\)
\(\Leftrightarrow-3x=4\)
hay \(x=-\dfrac{4}{3}\)
1)\(\dfrac{x-4}{3}+2x=\dfrac{4x-2}{6}\)
\(\Leftrightarrow\dfrac{\left(x-4\right).2}{3.2}+\dfrac{2x.6}{6}=\dfrac{4x-2}{6}\)
\(\Rightarrow2x-8+12x=4x-2\\ \Leftrightarrow10x=6\\ \Leftrightarrow x=\dfrac{3}{5}\)
4: Ta có: \(\dfrac{2x-1}{3}+\dfrac{3x-2}{4}=\dfrac{4x-3}{5}\)
\(\Leftrightarrow40x-20+45x-30=48x-36\)
\(\Leftrightarrow37x=14\)
hay \(x=\dfrac{14}{37}\)
5: Ta có: \(\dfrac{x-3}{9}-\dfrac{x+2}{6}=\dfrac{x+4}{18}-\dfrac{1}{2}\)
\(\Leftrightarrow2x-6-3x-6=x+4-9\)
\(\Leftrightarrow-x-x=-5-12=-17\)
hay \(x=\dfrac{17}{2}\)
Thu gọn:
A = \(\dfrac{2^4.3^3+2^3.3^4}{2^5.3^3-2^4.3^2}\) D = \(\dfrac{\left(\dfrac{1}{2}-\dfrac{1}{3}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}\right)}{\left(\dfrac{1}{2}+\dfrac{1}{3}\right).\left(\dfrac{2}{3}+\dfrac{3}{4}\right)}\)
B = \(\dfrac{2^3-3^4-2^4.3^3}{2^5.3^4-2^6.3^3}\) E = \(\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{3}{4}\right).\left(\dfrac{2}{3}-\dfrac{3}{4}+\dfrac{5}{6}\right)\)
C = \(\dfrac{\dfrac{1}{2}-\dfrac{1}{2}:\dfrac{3}{4}-\dfrac{3}{4}}{\dfrac{2}{3}-\dfrac{2}{3}:\dfrac{5}{6}-\dfrac{5}{6}}\)
Giúp mình với ! 10k nha
\(A=\dfrac{2^4.3^3+2^3.3^4}{2^5.3^4-2^6.3^3}=\dfrac{2^3.3^3.\left(2+3\right)}{2^5.3^3.\left(3-2\right)}=\dfrac{2^3.3^3.5}{2^5.3^3.1}\)
\(=\dfrac{5}{2^2}=\dfrac{5}{4}\)
a) 0,25-\(\dfrac{2}{3}\)+1\(\dfrac{1}{4}\)
b) \(\dfrac{3^2}{2}\):\(\dfrac{1}{4}\)+\(\dfrac{3}{4}\).2010
c) {[(\(\dfrac{1}{25}\)-0,6)2:\(\dfrac{49}{125}\)].\(\dfrac{5}{6}\)}-[(\(\dfrac{-1}{3}\))+\(\dfrac{1}{2}\)]
d) (-\(\dfrac{1}{2}\)-\(\dfrac{1^{ }}{3}\))2:[(\(\dfrac{-5}{36}\))-(\(\dfrac{-5}{36}\))0]
Mn giúp mk nhé mk gấp quá tí đi học ai làm được mk thả tim và like nhé
a) \(0,25-\dfrac{2}{3}+1\dfrac{1}{4}\)
\(=\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{5}{4}\)
\(=\dfrac{3}{12}-\dfrac{8}{12}+\dfrac{15}{12}\)
\(=\dfrac{10}{12}\)
\(=\dfrac{5}{6}\)
\(---\)
b) \(\dfrac{3^2}{2}:\dfrac{1}{4}+\dfrac{3}{4}\cdot2010\)
\(=\dfrac{9}{2}\cdot4+\dfrac{3015}{2}\)
\(=18+\dfrac{3015}{2}\)
\(=\dfrac{36}{2}+\dfrac{3015}{2}\)
\(=\dfrac{3051}{2}\)
\(---\)
c) \(\left\{\left[\left(\dfrac{1}{25}-0,6\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-1}{3}\right)+\dfrac{1}{2}\right]\)
\(=\left\{\left[\left(-\dfrac{14}{25}\right)^2:\dfrac{49}{125}\right]\cdot\dfrac{5}{6}\right\}-\left[\left(\dfrac{-2}{6}\right)+\dfrac{3}{6}\right]\)
\(=\left\{\left[\dfrac{196}{625}\cdot\dfrac{125}{49}\right]\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\left\{\dfrac{4}{5}\cdot\dfrac{5}{6}\right\}-\dfrac{1}{6}\)
\(=\dfrac{4}{6}-\dfrac{1}{6}\)
\(=\dfrac{3}{6}\)
\(=\dfrac{1}{2}\)
\(---\)
d) \(\left(-\dfrac{1}{2}-\dfrac{1}{3}\right)^2:\left[\left(\dfrac{-5}{36}\right)-\left(\dfrac{-5}{36}\right)^0\right]\)
\(=\left(-\dfrac{3}{6}-\dfrac{2}{6}\right)^2:\left[-\dfrac{5}{36}-1\right]\)
\(=\left(-\dfrac{5}{6}\right)^2:\left[-\dfrac{5}{36}-\dfrac{36}{36}\right]\)
\(=\dfrac{25}{36}:\left(\dfrac{-41}{36}\right)\)
\(=\dfrac{25}{36}\cdot\left(\dfrac{-36}{41}\right)\)
\(=-\dfrac{25}{41}\)
#\(Toru\)
Tính tổng đại số
\(A=\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}-\dfrac{1}{5}-\dfrac{2}{5}-\dfrac{3}{5}-\dfrac{4}{5}+...+\dfrac{1}{10}+\dfrac{2}{10}+...+\dfrac{9}{10}\)
\(B=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{3}+\dfrac{1}{4}+\dfrac{2}{4}+\dfrac{3}{4}+...+\dfrac{1}{n}+\dfrac{2}{n}+...+\dfrac{n-1}{n}\)\(\left(n\in Z,n\ge2\right)\)
Tìm số nguyên x, biết:
a) \(-4\dfrac{3}{5}\). \(2\dfrac{4}{3}\) < x < \(-2\dfrac{3}{5}\) : \(1\dfrac{6}{15}\)
b) \(-4\dfrac{1}{3}\).(\(\dfrac{1}{2}\)-\(\dfrac{1}{6}\)) < x < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))
a) Ta có \(-4\dfrac{3}{5}\cdot2\dfrac{4}{3}=-\dfrac{23}{5}\cdot\dfrac{10}{3}=-\dfrac{46}{3}\) và \(-2\dfrac{3}{5}\div1\dfrac{6}{15}=-\dfrac{13}{5}\div\dfrac{7}{5}=-\dfrac{13}{7}\)
Do đó \(-\dfrac{46}{3}< x< -\dfrac{13}{7}\)
Lại có \(-\dfrac{46}{3}\le-15\) và \(-\dfrac{13}{7}\ge-2\)
Suy ra \(-15\le x\le-2\), x ϵ Z
b) Ta có \(-4\dfrac{1}{3}\left(\dfrac{1}{2}-\dfrac{1}{6}\right)=-\dfrac{13}{3}\cdot\dfrac{1}{3}=-\dfrac{13}{9}\) và \(-\dfrac{2}{3}\left(\dfrac{1}{3}-\dfrac{1}{2}-\dfrac{3}{4}\right)=-\dfrac{2}{3}\cdot\dfrac{-11}{12}=\dfrac{11}{18}\)
Do đó \(-\dfrac{13}{9}< x< \dfrac{11}{18}\)
Lại có \(-\dfrac{13}{9}\le-1\) và \(\dfrac{11}{18}\ge0\)
Suy ra \(-1\le x\le0\), x ϵ Z
b, -4\(\dfrac{1}{3}\).(\(\dfrac{1}{2}\) - \(\dfrac{1}{6}\)) < \(x\) < - \(\dfrac{2}{3}\).(\(\dfrac{1}{3}\) - \(\dfrac{1}{2}\) - \(\dfrac{3}{4}\))
- \(\dfrac{13}{3}\).\(\dfrac{1}{3}\) < \(x\) < - \(\dfrac{2}{3}\).(-\(\dfrac{11}{12}\))
- \(\dfrac{13}{9}\) < \(x\) < \(\dfrac{11}{18}\)
\(x\) \(\in\) { -1; 0; 1}
a, -4\(\dfrac{3}{5}\).2\(\dfrac{4}{3}\) < \(x\) < -2\(\dfrac{3}{5}\): 1\(\dfrac{6}{15}\)
- \(\dfrac{23}{5}\).\(\dfrac{10}{3}\) < \(x\) < - \(\dfrac{13}{5}\): \(\dfrac{21}{15}\)
- \(\dfrac{46}{3}\) < \(x\) < - \(\dfrac{13}{7}\)
\(x\) \(\in\) {-15; -14;-13;..; -2}
1/ \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
2/ \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
3/ \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
4/ \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
5/ \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
1: Ta có: \(\dfrac{5x+1}{8}-\dfrac{x-2}{4}=\dfrac{1}{2}\)
\(\Leftrightarrow5x+1-2\left(x-2\right)=4\)
\(\Leftrightarrow5x+1-2x+4=4\)
\(\Leftrightarrow3x=-1\)
hay \(x=-\dfrac{1}{3}\)
2: Ta có: \(\dfrac{x+3}{4}+\dfrac{1-3x}{3}=\dfrac{-x+1}{18}\)
\(\Leftrightarrow9x+27+12-36x=-2x+2\)
\(\Leftrightarrow-27x+2x=2-39\)
hay \(x=\dfrac{37}{25}\)
3: Ta có: \(\dfrac{x+2}{4}-\dfrac{5x}{6}=\dfrac{1-x}{3}\)
\(\Leftrightarrow3x+6-10x=4-4x\)
\(\Leftrightarrow-7x+4x=4-6=-2\)
hay \(x=\dfrac{2}{3}\)
4: Ta có: \(\dfrac{x-3}{2}-\dfrac{x+1}{10}=\dfrac{x-2}{5}\)
\(\Leftrightarrow5x-15-x-1=2x-4\)
\(\Leftrightarrow4x-2x=-4+16=12\)
hay x=6
5: Ta có: \(\dfrac{4x+1}{4}-\dfrac{9x-5}{12}+\dfrac{x-2}{3}=0\)
\(\Leftrightarrow12x+3-9x+5+4x-8=0\)
\(\Leftrightarrow7x=0\)
hay x=0
1) \(23\dfrac{1}{4}.\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
2) \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(0,8-\dfrac{3}{4}\right)^2\)
3) \(16\dfrac{2}{7}:\left(\dfrac{-3}{5}\right)+28\dfrac{2}{7}:\dfrac{3}{5}\)
4) \(\left(2^2:\dfrac{4}{3}-\dfrac{1}{2}\right).\dfrac{6}{5}-17\)
5) \(\left(\dfrac{1}{3}\right)^{50}.\left(-9\right)^{25}-\dfrac{2}{3}:4\)
1: Ta có: \(23\dfrac{1}{4}\cdot\dfrac{7}{5}-13\dfrac{1}{4}:\dfrac{5}{7}\)
\(=\dfrac{93}{4}\cdot\dfrac{7}{5}-\dfrac{53}{4}\cdot\dfrac{7}{5}\)
\(=\dfrac{7}{5}\cdot10=14\)
2: Ta có: \(\left(1+\dfrac{2}{3}-\dfrac{1}{4}\right)\left(\dfrac{4}{5}-\dfrac{3}{4}\right)^2\)
\(=\dfrac{12+8-3}{12}\cdot\dfrac{1}{400}\)
\(=\dfrac{17}{12}\cdot\dfrac{1}{400}=\dfrac{17}{4800}\)