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a) \(\dfrac{5}{6}:x=30:3\)
\(\Leftrightarrow\dfrac{5}{6}:x=10\)
\(\Leftrightarrow x=\dfrac{5}{6}:10\)
\(\Leftrightarrow x=\dfrac{1}{12}\)
Vậy .......
b) \(x:2,5=0,003:0,75\)
\(\Leftrightarrow x:2,5=0,004\)
\(\Leftrightarrow x=0,004.2,5\)
\(\Leftrightarrow x=0,01\)
Vậy .......
c) \(3,8:\left(2x\right)=\dfrac{1}{4}:2\dfrac{2}{3}\)
\(\Leftrightarrow3,8:\left(2x\right)=\dfrac{1}{4}:\dfrac{8}{3}=\dfrac{3}{32}\)
\(\Leftrightarrow2x=3,8:\dfrac{3}{32}\)
\(\Leftrightarrow2x=\dfrac{698}{25}\)
\(\Leftrightarrow x=\dfrac{304}{15}\)
Vậy ...
d) \(\dfrac{2}{3}:0,4=x:\dfrac{4}{5}\)
\(\Leftrightarrow x:\dfrac{4}{5}=\dfrac{2}{3}\)
\(\Leftrightarrow x=\dfrac{8}{15}\)
Vậy ....
e) \(3\dfrac{4}{5}:40\dfrac{8}{15}=0,25:x\)
\(\Leftrightarrow0,25:x=\dfrac{19}{5}:\dfrac{608}{15}\)
\(\Leftrightarrow0,25x=\dfrac{57}{608}\)
\(\Leftrightarrow x=\dfrac{228}{608}\)
Vậy ...
e) \(\dfrac{x}{-15}=\dfrac{-60}{x}\)
\(\Leftrightarrow xx=\left(-60\right)\left(-15\right)\)
\(\Leftrightarrow x^2=900\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=30^2\\x^2=\left(-30\right)^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=30\\x=-30\end{matrix}\right.\)
Vậy ...
A= \(\dfrac{1}{3}-\dfrac{3}{5}+\dfrac{5}{7}-\dfrac{7}{9}+\dfrac{9}{11}-\dfrac{5}{7}+\dfrac{3}{5}-\dfrac{9}{11}=\dfrac{1}{3}-\dfrac{7}{9}=\dfrac{3}{9}-\dfrac{7}{9}=-\dfrac{4}{9}\)
\(B=\left(\dfrac{1}{5}+\dfrac{2}{15}+\dfrac{2}{3}\right)+\left(-\dfrac{2}{7}+\dfrac{1}{42}-\dfrac{13}{28}-\dfrac{1}{4}\right)\)
\(=\dfrac{3+2+10}{15}+\dfrac{-2\cdot12+2-13\cdot3-21}{84}\)
=1-82/84
=2/84=1/42
\(C=\dfrac{1}{50}-\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\right)\)
\(=\dfrac{1}{50}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\right)\)
\(=\dfrac{1}{50}-1+\dfrac{1}{50}=\dfrac{1}{25}-1=-\dfrac{24}{25}\)
\(D=\dfrac{3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
a. \(A=\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,75-2,2+\dfrac{11}{7}+\dfrac{11}{13}}=\dfrac{3\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11\left(0,25-0,2+\dfrac{1}{7}+\dfrac{1}{13}\right)}=\dfrac{3}{11}\)
Vậy \(A=\dfrac{3}{11}\)
b. \(B=\dfrac{2^{12}\cdot13+2^{12}\cdot65}{2^{10}\cdot104}+\dfrac{3^{10}\cdot11+3^{10}\cdot5}{3^9\cdot2^4}=\dfrac{2^{12}\left(13+65\right)}{2^{10}\cdot104}+\dfrac{3^{10}\left(11+5\right)}{3^9\cdot2^4}=\dfrac{2^{12}\cdot78}{2^{10}\cdot104}+\dfrac{3^{10}\cdot16}{3^9\cdot16}=\dfrac{2^2\cdot3}{1\cdot4}+3=\dfrac{12}{4}+3=3+3=6\)
Vậy \(B=6\)
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
a, \(\dfrac{0,75-0,6+\dfrac{3}{7}+\dfrac{3}{13}}{2,72-2,2+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}}{\dfrac{11}{4}-\dfrac{11}{5}+\dfrac{11}{7}+\dfrac{11}{13}}\)
= \(\dfrac{3.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}{11.\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)}\)
= \(\dfrac{3}{11}\)
b. \(\dfrac{0,357-0,3+\dfrac{3}{11}+\dfrac{3}{12}}{0,625-0,5+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{\dfrac{3}{8}-\dfrac{3}{10}+\dfrac{3}{11}+\dfrac{3}{12}}{\dfrac{5}{8}-\dfrac{5}{10}+\dfrac{5}{11}+\dfrac{5}{12}}\)
= \(\dfrac{3.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}{5.\left(\dfrac{1}{8}-\dfrac{1}{10}+\dfrac{1}{11}+\dfrac{1}{12}\right)}\)
= \(\dfrac{3}{5}\)
c, \(-\left|-1,5\right|.\left(1\dfrac{1}{3}-2\right)-\left|-\dfrac{2}{3}\right|\)
= \(-1,5.\left(\dfrac{4}{3}-2\right)-\dfrac{2}{3}\)
= \(-1,5.\left(\dfrac{-2}{3}\right)-\dfrac{2}{3}\)
= \(1-\dfrac{2}{3}=\dfrac{1}{3}\)
a) \(\dfrac{1,2}{x+3}=\dfrac{5}{4}\)
\(\Rightarrow\left(x+3\right).5=1,2.4\)
\(\Rightarrow\left(x+3\right).5=4,8\)
\(\Rightarrow x+3=4,8:5\)
\(\Rightarrow x+3=0,96\)
\(\Rightarrow x=-2,04\)
vậy \(x=-2,04\)
b)\(\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{1}{2}:\dfrac{4}{5}\)
\(\Rightarrow\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{5}{8}\)
\(\Rightarrow\dfrac{2x}{15}=\dfrac{3}{5}:\dfrac{5}{8}\)
\(\Rightarrow\dfrac{2x}{15}=\dfrac{24}{25}\)
\(\Rightarrow15.24=\left(2x\right).25\)
\(\Rightarrow360=\left(2x\right).25\)
\(\Rightarrow360:25=2x\)
\(\Rightarrow14,4=2x\)
\(\Rightarrow x=7,2\)
vậy \(x=7,2\)
\(a,\dfrac{1,2}{x+3}=\dfrac{5}{4}\\ \left(x+3\right).5=1,2.4\\ 5x+8=4,8\\ 5x=4,8-8\\ 5x=-3,2\\ x=-3,2:5=-0,64\)
\(b,\dfrac{3}{5}:\dfrac{2x}{15}=\dfrac{1}{2}:\dfrac{4}{5}\\ \dfrac{2x}{15}=\dfrac{3}{5}\cdot\dfrac{4}{5}:\dfrac{1}{2}\\ \dfrac{2x}{15}=\dfrac{12}{25}.2\\ \dfrac{2x}{25}=\dfrac{24}{25}\\ 2x=\dfrac{24}{25}.5\\ 2x=\dfrac{24}{5}\\ x=\dfrac{24}{5}\cdot\dfrac{1}{2}=\dfrac{12}{5}\)
\(c,-\dfrac{4}{2,5}:3,5=1,5:x\\ x=3,5.1,5:\left(-\dfrac{4}{25}\right)\\ x=\dfrac{21}{4}\cdot\left(-\dfrac{25}{4}\right)=-\dfrac{525}{16}\)
\(d,0,12:3=2x:\dfrac{3}{5}\\ 2x=0,12\cdot\dfrac{3}{5}:3\\ 2x=\dfrac{9}{125}\cdot\dfrac{1}{3}\\ 2x=\dfrac{3}{125}\\ x=\dfrac{3}{125}\cdot\dfrac{1}{2}=\dfrac{3}{250}\)
a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)
=0.75+0.25-2.5
=1-2.5=-1.5
b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)
=3.(-1.4)+3.08
=-4.2+3.08=-1.12
c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{11}{153}+\dfrac{12}{17}\)
=\(\dfrac{7}{9}\)
d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
=-0.44+\(\dfrac{127}{175}\)
=\(\dfrac{2}{7}\)
a) \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
\(\Leftrightarrow3-2x=-1\)
\(\Leftrightarrow-2x=-1-3\)
\(\Leftrightarrow-2x=-4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
\(\Leftrightarrow\dfrac{5}{3}x+\dfrac{1}{5}=x-\dfrac{7}{8}\)
\(\Leftrightarrow200x+24=120x-105\)
\(\Leftrightarrow80x=-129\)
\(\Leftrightarrow x=-\dfrac{129}{80}\)
Vậy \(x=-\dfrac{129}{80}\)
c) \(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
\(\Leftrightarrow-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
\(\Leftrightarrow-\left|x+0,5\right|=-\dfrac{11}{20}\)
\(\Leftrightarrow\left|x+0,5\right|=\dfrac{11}{20}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+0,5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{21}{20};x_2=\dfrac{1}{20}\)
d) \(\left(x+0,2\right)^2+0,75=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2+\dfrac{3}{4}=1\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=1-\dfrac{3}{4}\)
\(\Leftrightarrow\left(x+\dfrac{1}{5}\right)^2=\dfrac{1}{4}\)
\(\Leftrightarrow x+\dfrac{1}{5}=\pm\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=\dfrac{1}{2}\\x+\dfrac{1}{5}=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{10}\\x=-\dfrac{7}{10}\end{matrix}\right.\)
Vậy \(x_1=-\dfrac{7}{10};x_2=\dfrac{3}{10}\)
a, \(\dfrac{3}{4}-\dfrac{1}{2}x=-\dfrac{1}{4}\)
=>\(-\dfrac{1}{2}x=-\dfrac{1}{4}-\dfrac{3}{4}\)
=>\(-\dfrac{1}{2}x=-1\)
=>\(x=-1:(-\dfrac{1}{2})\)
=>\(x=2\)
vậy ...........
b,\(1\dfrac{2}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x+0,2=x-\dfrac{7}{8}\)
=>\(\dfrac{5}{3}x-x=-\dfrac{7}{8}-0,2\)
=>\(\dfrac{2}{3}x=-\dfrac{43}{40}\)
=>\(x=-\dfrac{43}{40}:\dfrac{2}{3}\)
=>\(x=-\dfrac{192}{80}\)
vậy...................
c,\(\dfrac{3}{4}-\left|x+0,5\right|=\dfrac{1}{5}\)
=>\(-\left|x+0,5\right|=\dfrac{1}{5}-\dfrac{3}{4}\)
=>\(-\left|x+0,5\right|=-\dfrac{11}{20}\)
=>\(\left|x+0,5\right|=\dfrac{11}{20}\)
=>\(\left[{}\begin{matrix}x+0.5=\dfrac{11}{20}\\x+0,5=-\dfrac{11}{20}\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=\dfrac{1}{20}\\x=-\dfrac{21}{20}\end{matrix}\right.\)
vậy ....... hoặc.....
d,\((x+0,2)^2+0,75=1\)
=>\(\left(x+0,2\right)^2=1-0,75\)
=>\(\left(x+0,2\right)^2=0,25\)
=>\(\left[{}\begin{matrix}x+0,2=0,5\\x+0,2=-0,5\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0,3\\x=-0,7\end{matrix}\right.\)
vậy..........................
HỌC TỐT NHA !!!!!
B = .................
Xét thừa số 63.1,2 - 21.3,6 = 0 nên B = 0
\(C=\left|\dfrac{4}{9}-\left(\dfrac{\sqrt{2}}{2}\right)^2\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{\dfrac{2}{3}-\dfrac{4}{5}-\dfrac{6}{7}}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}}{2\left(\dfrac{1}{3}-\dfrac{2}{5}-\dfrac{3}{7}\right)}\right|\)
\(C=\left|\dfrac{4}{9}-\dfrac{1}{2}\right|+\left|0,4+\dfrac{1}{2}\right|=\dfrac{1}{18}+\dfrac{9}{10}=\dfrac{43}{45}\)
Mình làm câu 1,2 trước, câu 3 sau
Câu 1:
\(\sqrt{x^2}=0\)
=> \(\left(\sqrt{x^2}\right)^2=0^2\)
\(\Leftrightarrow x^2=0\Leftrightarrow x=0\)
Câu 2:
\(A=\left(0,75-0,6+\dfrac{3}{7}+\dfrac{3}{12}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+2,75-2,2\right)\)
\(A=\left(\dfrac{3}{4}-\dfrac{3}{5}+\dfrac{3}{7}+\dfrac{3}{13}\right)\left(\dfrac{11}{7}+\dfrac{11}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=3\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{13}\right)\cdot11\left(\dfrac{1}{7}+\dfrac{1}{3}+\dfrac{11}{4}-\dfrac{11}{5}\right)\)
\(A=33\cdot\dfrac{491}{1820}\cdot\dfrac{221}{420}=\dfrac{3580863}{764400}\)