2\(\dfrac{1}{3}\) - 125% . x = \(\dfrac{1}{4}\)
Tìm x, biết:
a) \(\dfrac{2}{3}\)x - \(\dfrac{1}{2}\)x = \(\left(-\dfrac{7}{12}\right)\) . \(1\dfrac{2}{5}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2\) = \(\dfrac{9}{4}\)
c) (1,25 - \(\dfrac{4}{5}\)x)3 = -125
a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)
\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)
\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)
\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)
\(\Rightarrow x=-\dfrac{49}{10}\)
b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)
\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)
+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)
\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)
\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)
\(\Rightarrow x=-\dfrac{13}{15}\)
+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)
\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)
\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)
\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)
\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)
\(\Rightarrow x=\dfrac{125}{16}\)
a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)
\(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)
\(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)
\(x\) = - \(\dfrac{49}{60}\).6
\(x\) = -\(\dfrac{49}{10}\)
1/ (\(\left(-\dfrac{2}{3}\right)\)\(^2\) x \(\dfrac{-9}{8}\) - 25% x \(\dfrac{-16}{5}\)
2/ -1\(\dfrac{2}{5}\) x 75% + \(\dfrac{-7}{5}\) x 25%
3/ -2\(\dfrac{3}{7}\) x (-125%) + \(\dfrac{-17}{7}\) x 25%
4/ (-2)\(^3\) x (\(\dfrac{3}{4}\) x 0.25) : (2\(\dfrac{1}{4}\) - 1\(\dfrac{1}{6}\))
1) Ta có: \(\left(-\dfrac{2}{3}\right)^2\cdot\dfrac{-9}{8}-25\%\cdot\dfrac{-16}{5}\)
\(=\dfrac{4}{9}\cdot\dfrac{-9}{8}-\dfrac{1}{4}\cdot\dfrac{-16}{5}\)
\(=\dfrac{-1}{2}+\dfrac{4}{5}\)
\(=\dfrac{-5}{10}+\dfrac{8}{10}=\dfrac{3}{10}\)
2) Ta có: \(-1\dfrac{2}{5}\cdot75\%+\dfrac{-7}{5}\cdot25\%\)
\(=\dfrac{-7}{5}\cdot\dfrac{3}{4}+\dfrac{-7}{5}\cdot\dfrac{1}{4}\)
\(=\dfrac{-7}{5}\left(\dfrac{3}{4}+\dfrac{1}{4}\right)=-\dfrac{7}{5}\)
3) Ta có: \(-2\dfrac{3}{7}\cdot\left(-125\%\right)+\dfrac{-17}{7}\cdot25\%\)
\(=\dfrac{-17}{7}\cdot\dfrac{-5}{4}+\dfrac{-17}{7}\cdot\dfrac{1}{4}\)
\(=\dfrac{-17}{7}\cdot\left(\dfrac{-5}{4}+\dfrac{1}{4}\right)\)
\(=\dfrac{17}{7}\)
4) Ta có: \(\left(-2\right)^3\cdot\left(\dfrac{3}{4}\cdot0.25\right):\left(2\dfrac{1}{4}-1\dfrac{1}{6}\right)\)
\(=\left(-8\right)\cdot\left(\dfrac{3}{4}\cdot\dfrac{1}{4}\right):\left(\dfrac{9}{4}-\dfrac{7}{6}\right)\)
\(=\left(-8\right)\cdot\dfrac{3}{16}:\dfrac{54-28}{24}\)
\(=\dfrac{-3}{2}\cdot\dfrac{24}{26}\)
\(=\dfrac{-72}{52}=\dfrac{-18}{13}\)
a.,\(\dfrac{4}{5}+5\dfrac{1}{2}\text{x }\left(4,5-2\right)=\dfrac{7}{10}\) b,125%x\(\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
c,\(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\) d, \(\dfrac{3}{17}+\dfrac{11}{4}+\dfrac{5}{8}+\dfrac{14}{17}+\dfrac{3}{8}\)
`a)4/5+5 1/2 xx (4,5-2)+7/10`
`=4/5+11/2*2,5+7/10`
`=0,8+2,2+0,7`
`=3+0,7=3,7`
`b)125%xx 17/4:(1 5/16-0,5)+2008`
`=1,25xx4,25:13/16+2008`
`=85/13+2008`
`=2014 7/13`
`c)5/11+(16/11+1)`
`=5/11+1+5/11+1`
`=2+10/11=32/11`
`d)3/17+11/4+5/8+14/17+3/8`
`=3/17+14/17+5/8+3/8+11/4`
`=1+1+11/4`
`=19/4`
a)
\(\dfrac{4}{5}+5\dfrac{1}{2}x\left(4,5-2\right)=\dfrac{7}{10}\)
<=> \(\dfrac{11}{2}x\times2,5=\dfrac{7}{10}-\dfrac{4}{5}=\dfrac{-1}{10}\)
<=> \(\dfrac{55}{4}x=\dfrac{-1}{10}< =>x=\dfrac{-2}{275}\)
b) \(125\%\times\dfrac{17}{4}:\left(1\dfrac{5}{16}-0,5\right)+2008\)
= \(\dfrac{85}{16}:\left(\dfrac{21}{16}-\dfrac{1}{2}\right)+2008=\dfrac{85}{16}:\dfrac{13}{16}+2008=\dfrac{26189}{13}\)
c) \(\dfrac{5}{11}+\left(\dfrac{16}{11}+1\right)\)
= \(\dfrac{21}{11}+1=\dfrac{32}{11}\)
d) \(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{5}{8}+\dfrac{3}{8}\right)+\dfrac{11}{4}\)
= 1 + 1 + \(\dfrac{11}{4}\) = \(\dfrac{19}{4}\)
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 nhanh A=\(\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{3^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right).....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)....0......\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(A=0\)
Tính :
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{4}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}=\dfrac{4}{4}=1\)
Thuc hien phep tinh;
a/ \(1\dfrac{4}{23}+\dfrac{5}{21}-\dfrac{4}{23}+0,5+\dfrac{16}{21}\)
b/ \(\left(\dfrac{1}{25}+\dfrac{1}{5}+1\right):\left(\dfrac{1}{25}-\dfrac{1}{5}-1\right)\)
c/ \(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}\)+ \(\dfrac{0,6-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-0,16-\dfrac{4}{125}-\dfrac{4}{625}}\)
d/ \(\dfrac{2^{12}.3^5-4^6.9^2}{\left(2^2.3\right)^6+8^4.3^5}-\dfrac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
a: \(=\left(1+\dfrac{4}{23}-\dfrac{4}{23}\right)+\left(\dfrac{5}{21}+\dfrac{16}{21}\right)+\dfrac{1}{2}\)
\(=1+1+\dfrac{1}{2}=2+\dfrac{1}{2}=\dfrac{5}{2}\)
b: \(=\left(\dfrac{1}{25}+\dfrac{5}{25}+\dfrac{25}{25}\right):\left(\dfrac{1}{25}-\dfrac{5}{25}-\dfrac{25}{25}\right)\)
\(=\dfrac{31}{25}:\dfrac{-29}{25}=\dfrac{-31}{29}\)
c: \(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{\dfrac{3}{5}-\dfrac{3}{25}-\dfrac{3}{125}-\dfrac{3}{625}}{\dfrac{4}{5}-\dfrac{4}{25}-\dfrac{4}{125}-\dfrac{4}{625}}\)
=1/4+3/4
=1
Câu 1. (2,0 điểm) Thực hiện phép tính:
a) $\sqrt{20}+2 \sqrt{45}-3 \sqrt{80}+\sqrt{125}$.
b) $\dfrac{2 \sqrt{3}+3 \sqrt{2}}{\sqrt{3}+\sqrt{2}}-4 \sqrt{\dfrac{3}{2}}-\dfrac{5}{1-\sqrt6}$.
c) $\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4(\sqrt{x}-1)}{x-2 \sqrt{x}}$ với $x \neq 4, x>0$.
a) \(\sqrt{ }\)20 + 2\(\sqrt{ }\)45 - 3\(\sqrt{ }\)80 + \(\sqrt{ }\)125
= \(\sqrt{ }\)4.5 +2\(\sqrt{ }\)9.5 - 3\(\sqrt{16.5}\)
= 2\(\sqrt{5}\) + 6\(\sqrt{5}\) - 12\(\sqrt{5}\)
= -4\(\sqrt{5}\)
b) \(\dfrac{2\sqrt{3}+3\sqrt{2}}{\sqrt{3}+\sqrt{2}}\) - \(4\sqrt{\dfrac{3}{2}}\)- \(\dfrac{5}{1-\sqrt{6}}\)
= \(\dfrac{2\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}\)- \(\sqrt{16.\dfrac{3}{2}}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{\left(1-\sqrt{6}\right)\left(1+\sqrt{6}\right)}\)
= 2 - \(\sqrt{24}\) - \(\dfrac{5\left(1+\sqrt{6}\right)}{1-6}\)
= 2 - \(\sqrt{4.6}\) + 1+\(\sqrt{ }\)6
= 2 - 2\(\sqrt{ }\)6 + 1+\(\sqrt{ }\)6
= 3 - \(\sqrt{ }\)6
c) (đề bài) với x khác 4...
= \(\dfrac{\sqrt{x}}{\sqrt{x}-2}\)- \(\dfrac{4\left(\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{x}{\sqrt{x}\left(\sqrt{x}-2\right)}\)- ....
= \(x-4\sqrt{x}+4\)/ \(\sqrt{x}\left(\sqrt{x}-2\right)\)
= (căn -2)2/ căn x(căn x -2)
= căn x-2/căn x
Bài 1:
a) \(\sqrt{20}+2\sqrt{45}-3\sqrt{80}+\sqrt{125}\)
= \(2\sqrt{5}+6\sqrt{5}-12\sqrt{5}+5\sqrt{5}\)
= \(\sqrt{5}\)
b) \(\dfrac{2\sqrt{3}+3\sqrt{2}}{\sqrt{3}+\sqrt{2}}-4\sqrt{\dfrac{3}{2}}-\dfrac{5}{1-\sqrt{6}}\)
= \(\left(2\sqrt{3}+3\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)-2\sqrt{6}+1+\sqrt{6}\)
= \(\sqrt{6}-\sqrt{6}+1=1\)
c) \(\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\left(\sqrt{x}-1\right)}{x-2\sqrt{x}}\) (ĐKXĐ: x > 0; x ≠ 4)
= \(\dfrac{x-4\sqrt{x}+4}{x-2\sqrt{x}}\)
= \(\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}\)
= \(\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
tinh
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3}{4}\)
\(=\dfrac{1}{4}+\dfrac{3}{4}\)
\(=1\)
\(\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{\dfrac{4}{9}-\dfrac{4}{7}-\dfrac{4}{11}}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}\)
\(=\dfrac{\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}}{4\left(\dfrac{1}{9}-\dfrac{1}{7}-\dfrac{1}{11}\right)}+\dfrac{3\left(\dfrac{1}{5}-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}{4\left(-\dfrac{1}{25}-\dfrac{1}{125}-\dfrac{1}{625}\right)}=\dfrac{1}{4}+\dfrac{3}{4}=1\)