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Bài 1:
a) \(\dfrac{2}{5}\cdot x-\dfrac{1}{4}=\dfrac{1}{10}\)
\(\dfrac{2}{5}\cdot x=\dfrac{1}{10}+\dfrac{1}{4}\)
\(\dfrac{2}{5}\cdot x=\dfrac{7}{20}\)
\(x=\dfrac{7}{20}:\dfrac{2}{5}\)
\(x=\dfrac{7}{8}\)
Vậy \(x=\dfrac{7}{8}\).
b) \(\dfrac{3}{5}=\dfrac{24}{x}\)
\(x=\dfrac{5\cdot24}{3}\)
\(x=40\)
Vậy \(x=40\).
c) \(\left(2x-3\right)^2=16\)
\(\left(2x-3\right)^2=4^2\)
\(\circledast\)TH1: \(2x-3=4\\ 2x=4+3\\ 2x=7\\ x=\dfrac{7}{2}\)
\(\circledast\)TH2: \(2x-3=-4\\ 2x=-4+3\\ 2x=-1\\ x=\dfrac{-1}{2}\)
Vậy \(x\in\left\{\dfrac{7}{2};\dfrac{-1}{2}\right\}\).
Bài 2:
a) \(25\%-4\dfrac{2}{5}+0.3:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}:\dfrac{6}{5}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{3}{10}\cdot\dfrac{5}{6}\)
\(=\dfrac{1}{4}-\dfrac{22}{5}+\dfrac{1}{4}\)
\(=\dfrac{5}{20}-\dfrac{88}{20}+\dfrac{5}{20}\)
\(=\dfrac{5-88+5}{20}\)
\(=\dfrac{78}{20}=\dfrac{39}{10}\)
b) \(\left(\dfrac{1}{6}-\dfrac{1}{5^2}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{25}\cdot5+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{1}{6}-\dfrac{1}{5}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5}{30}-\dfrac{6}{30}+\dfrac{1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=\left(\dfrac{5-6+1}{30}\right)\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\cdot\left(\dfrac{2011}{2010}+\dfrac{2010}{1009}+\dfrac{2009}{2008}\right)\)
\(=0\)
Bài 3:
a) \(\dfrac{4}{19}\cdot\dfrac{-3}{7}+\dfrac{-3}{7}\cdot\dfrac{15}{19}\)
\(=\dfrac{-3}{7}\left(\dfrac{4}{19}+\dfrac{15}{19}\right)\)
\(=\dfrac{-3}{7}\cdot1\)
\(=\dfrac{-3}{7}\)
b) \(7\dfrac{5}{9}-\left(2\dfrac{3}{4}+3\dfrac{5}{9}\right)\)
\(=\dfrac{68}{9}-\dfrac{11}{4}-\dfrac{32}{9}\)
\(=\dfrac{68}{9}-\dfrac{32}{9}-\dfrac{11}{4}\)
\(=4-\dfrac{11}{4}\)
\(=\dfrac{16}{4}-\dfrac{11}{4}\)
\(\dfrac{5}{4}\)
Bài 4:
\(\dfrac{4}{12\cdot14}+\dfrac{4}{14\cdot16}+\dfrac{4}{16\cdot18}+...+\dfrac{4}{58\cdot60}\)
\(=2\left(\dfrac{1}{12\cdot14}+\dfrac{1}{14\cdot16}+\dfrac{1}{16\cdot18}+...+\dfrac{1}{58\cdot60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{18}+...+\dfrac{1}{58}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{1}{12}-\dfrac{1}{60}\right)\)
\(=2\left(\dfrac{5}{60}-\dfrac{1}{60}\right)\)
\(=2\cdot\dfrac{1}{15}\)
\(=\dfrac{2}{15}\)
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
a) \(\frac{2^{12}.13+2^{12}.65}{2^{10}.104}+\frac{3^{10}.11+3^{10}.5}{3^9.2^4}\)
\(=\frac{2^{10}.\left(13.4+65.4\right)}{2^{10}.104}+\frac{3^9.\left(3.11+3.5\right)}{3^9.16}\)
\(=\frac{312}{104}+\frac{48}{16}\)
=3+3=6
b) \(\frac{1.5.6+2.10.12+4.20.24+9.45.54}{1.3.5+2.6.10+4.12.20+9.27.45}\)
\(=\frac{1.5.6\left(1+2.2.2+4.4.4+9.9.9\right)}{1.3.5\left(1+2.2.2+4.4.4+9.9.9\right)}\)
\(=\frac{1.5.6}{1.3.5}\)
\(=2\)
c) 1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013
Nhận xét:Giá trị tuyệt đối của hai số liền nhau hơn kém nhau 1 đơn vị
=> Tổng trên có 2013-1+1=2013(Số hạng)
Nhóm 4 số vào một nhóm, ta được 2013:4=503 nhóm (thừa 1 số)
=>1+2-3-4+5+6-7-8+...+2009+2010-2011-2012+2013
=1+(2-3-4+5)+(6-7-8+9)+...+(2010-2011-2012+2013)
=1+0+0+...+0 (có 503 số 0)
=1+0.503
=1+0
=1
1.
ta có: 2009A= (2009^2010+ 2009)/ (2009^2010+1)= (2009^10+1+2008)/(2009^2010+1)=1+ [2008/(2009^2010+1)]
làm tương tự như trên ta được :
2009B=1-[4016/(2009^2011-2)]
lại có:
2009A= .............(nt) > 1
2009B=...........<1
=>2009A>2009B
=>A>B
\(b)\) Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(\frac{2009^{2010}-2}{2009^{2011}-2}< \frac{2009^{2010}-2+2011}{2009^{2011}-2+2011}=\frac{2009^{2010}+2009}{2009^{2011}+2009}=\frac{2009\left(2009^{2009}+1\right)}{2009\left(2009^{2010}+1\right)}=\frac{2009^{2009}+1}{2009^{2010}+1}\)
Vậy \(\frac{2009^{2009}+1}{2009^{2010}+1}>\frac{2009^{1010}-2}{2009^{2011}-2}\)
Chúc bạn học tốt ~
Àk mình còn thiếu một điều kiện nữa xin lỗi nhé :
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\)\(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Bạn thêm vào nhé
2, ta thấy:
\(\dfrac{2008}{2009}< \dfrac{2008}{2009+2010}\left(1\right)\)
\(\dfrac{2009}{2010}< \dfrac{2009}{2009+20010}\left(2\right)\)
từ (1) và (2) cộng vế với vế ta đc :\(\dfrac{2008}{2009}+\dfrac{2009}{20010}< \dfrac{2008}{2009+2010}+\dfrac{2009}{2009+2010}=\dfrac{2008+2009}{2009+2010}\)
\(I=\left(1+5+...+2009\right)+\left(2+6+...+2010\right)-\left(3+7+...+2007\right)-\left(4+8+...+2008\right)\)
\(=\frac{\left(2009+1\right).\left[\left(2009-1\right):4+1\right]}{2}+\frac{\left(2010+2\right)\left[\left(2010-2\right):4+1\right]}{2}\)
\(-\frac{\left(2007+3\right).\left[\left(2007-3\right):4+1\right]}{2}+\frac{\left(2008+4\right)\left[\left(2008-4\right):4+1\right]}{2}\)
\(=\frac{2010.503}{2}+\frac{2012.503}{2}-\frac{2010.502}{2}-\frac{2012.502}{2}\)
\(=1005+1006=2011\)