tính nhanh
4 phần 5 + 19 phần 18 - 1 phần 2 + 1 phần 5 - 10 phần 8
ai bt lm thì lm đầy đủ giúp mik nhanh ạ
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3 x \(\frac{4}{11}\) = \(\frac{3}{1}\) x \(\frac{4}{11}\) = \(\frac{12}{11}\)
1 : \(\frac{5}{4}\) = \(\frac{1\times4}{5}\) = \(\frac{4}{5}\)
\(\frac{4}{5}\) x \(\frac{6}{7}\) + \(\frac{2}{3}\) = \(\frac{24}{35}\) + \(\frac{2}{3}\) = \(\frac{142}{105}\)
\(\frac{3}{4}\) + \(\frac{1}{2}\) : \(\frac{5}{6}\) = \(\frac{3}{4}\) + \(\frac{1}{2}\) x \(\frac{6}{5}\) = \(\frac{3}{4}\)+ \(\frac{6}{10}\) \(\frac{27}{20}\)
\(\frac{3}{4}\)+ \(\frac{1}{2}\) = \(\frac{3}{4}\) + \(\frac{2}{4}\)= \(\frac{5}{4}\)
1) \(\left(2x+3\right)^2=4x^2+12x+9\)
\(\left(3x+2\right)^2=9x^2+12x+4\)
\(\left(2x+5\right)^2=4x^2+20x+25\)
\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)
2) \(\left(2x-3\right)^2=4x^2-12x+9\)
\(\left(3x-2\right)^2=9x^2-12x+4\)
\(\left(2x-5\right)^2=4x^2-20x+25\)
\(\left(2x-\dfrac{1}{3}\right)^2=4x^2-\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x-\dfrac{1}{3}\right)^2=9x^2-2x+\dfrac{1}{9}\)
3) \(\left(2x-3\right)\left(2x+3\right)=4x^2-9\)
\(\left(3x-4\right)\left(3x+4\right)=9x^2-16\)
\(\left(2x-5\right)\left(2x+5\right)=4x^2-25\)
\(\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=x^2-\dfrac{1}{4}\)
\(\left(2x-\dfrac{1}{3}\right)\left(2x+\dfrac{1}{3}\right)=4x^2-\dfrac{1}{9}\)
1: \(\left(2x+3\right)^2=4x^2+12x+9\)
\(\left(3x+2\right)^2=9x^2+12x+4\)
\(\left(2x+5\right)^2=4x^2+20x+25\)
\(\left(2x+\dfrac{1}{3}\right)^2=4x^2+\dfrac{4}{3}x+\dfrac{1}{9}\)
\(\left(3x+\dfrac{1}{3}\right)^2=9x^2+2x+\dfrac{1}{9}\)
\(a.\frac{2}{3}+\frac{1}{5}\cdot\frac{10}{7}\)
\(=\frac{2}{3}+\frac{1\cdot2}{1\cdot7}\)
\(=\frac{2}{3}+\frac{2}{7}=\frac{2\cdot7}{21}+\frac{2\cdot3}{21}=\frac{14}{21}+\frac{6}{21}=\frac{20}{21}\)
\(b.\frac{2}{7}\cdot\frac{4}{7}+\frac{2}{7}\cdot\frac{3}{7}\)
\(=\frac{2}{7}\cdot\left(\frac{4}{7}+\frac{3}{7}\right)\)
\(=\frac{2}{7}\cdot\frac{7}{7}=\frac{2}{7}\cdot1=\frac{2}{7}\)
\(c.\left[-\frac{1}{4}+\frac{3}{10}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\left[-\frac{5}{20}+\frac{6}{20}\right]:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\frac{1}{20}:\left(-\frac{3}{5}\right)-\frac{7}{6}\)
\(=\frac{1}{20}\cdot\left(-\frac{5}{3}\right)-\frac{7}{6}\)
\(=\frac{1\cdot\left(-1\right)}{4\cdot3}-\frac{7}{6}\)
\(=\left(-\frac{1}{12}\right)-\frac{7}{6}=\left(-\frac{1}{12}\right)-\frac{14}{12}=-\frac{15}{12}\)
bài 2 :a) \(2x-2\frac{2}{7}=2\frac{5}{7}\)
\(2x=2\frac{5}{7}-2\frac{2}{7}\)
\(2x=\left(2-2\right)+\left(\frac{5}{7}-\frac{2}{7}\right)\)
\(2x=0+\frac{3}{7}\)
\(2x=\frac{3}{7}\)
\(x=\frac{3}{7}:2=\frac{3}{7}\cdot\frac{1}{2}=\frac{3}{14}\)
\(b.\frac{17}{13x}=\frac{4}{39}\)
\(\Rightarrow13x\cdot4=17\cdot39\)
\(\Rightarrow13x\cdot4=663\)
\(\Rightarrow13x=663:4\)
\(\Rightarrow13x=165,75\)
\(\Rightarrow x=165,75:13\)
\(\Rightarrow x=12,75\)
\(****nha!!!!!!!!!!!!\)
1,
\(\frac{25}{12}+\left(\frac{-4}{12}\right)=\frac{7}{4}\)
\(\frac{-10}{8}+\frac{15}{4}=\frac{5}{2}\)
\(\frac{3}{8}+\frac{-14}{6}=\frac{-47}{24}\)
\(\frac{350}{150}+\left(\frac{-200}{360}\right)=\frac{16}{9}\)
\([\frac{5}{8}+\left(\frac{-3}{4}\right)]+\frac{15}{6}=\frac{-1}{8}+\frac{15}{6}=\frac{19}{8}\)
\(\frac{7}{3}+[\left(\frac{-5}{6}\right)+\left(\frac{-2}{3}\right)]=\frac{7}{3}+\left(\frac{-3}{2}\right)=\frac{5}{6}\)
Ta có 19/18=1+1/18
2005/2004=1+1/2004
Vì 1/18>1/2004 nên 1+1/18 >1/2004 +1 hay 19/18>2005/2004
1. Ta có: \(\frac{19}{18}>\frac{18}{18}=1=\frac{2005}{2005}>\frac{2004}{2005}\)
hay: 19/18 > 1 > 2004/2005
Vậy 19/18 > 2004/2005
2. \(\frac{3}{6}=\frac{1}{2}=\frac{7}{14};\frac{2}{7}=\frac{4}{14};\frac{4}{7}=\frac{8}{14}\)
Vì 4/14 < 7/14 < 8/14
nên: 2/7 < 3/6 < 4/7.
3. \(\frac{12}{25};\frac{9}{5}=\frac{45}{25};\frac{3}{5}=\frac{15}{25};\frac{1}{25}\)
Vì 45/25 > 15/25 > 12/25 > 1/25
nên: 9/5 > 3/5 > 12/25 > 1/25
4. \(\frac{35}{42}+\frac{14}{42}=\frac{35+14}{42}=\frac{49}{42}\)
5. \(\frac{6}{16}+\frac{32}{16}=\frac{6+32}{16}=\frac{38}{16}=\frac{19}{8}\).
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}\)
\(=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)-\left(\dfrac{4}{8}+\dfrac{10}{8}\right)+\dfrac{19}{18}\)
\(=\dfrac{5}{5}-\dfrac{14}{8}+\dfrac{19}{18}\)
\(=1-\dfrac{7}{4}+\dfrac{19}{18}\)
\(=-\dfrac{3}{4}+\dfrac{19}{18}=\dfrac{11}{36}\)
\(\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{10}{8}=\dfrac{4}{5}+\dfrac{19}{18}-\dfrac{1}{2}+\dfrac{1}{5}-\dfrac{5}{4}=\left(\dfrac{4}{5}+\dfrac{1}{5}\right)+\left(\dfrac{19}{18}-\dfrac{1}{2}\right)-\dfrac{5}{4}=1+\dfrac{5}{9}-\dfrac{5}{4}=\dfrac{36}{36}+\dfrac{20}{36}-\dfrac{45}{36}=\dfrac{11}{36}\)