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bn đổi ngược hai vế cho nhau là ra 1 bài toán bình thường thôi
Bài nỳ tuy rất dài nhưng cũng dễ
Chí cần cậu cuyển VT sang VP rồi tìm x bình thường
Chúc cậu học tốt
\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=-\frac{16}{10}:\frac{5}{3}\)
\(A=-\frac{16.3}{10.5}=-\frac{48}{50}=-\frac{24}{25}\)
\(B=1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(B=\frac{14}{10}\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)
\(B=\frac{2.7.3.5}{2.5.7.7}-\left(\frac{12+10}{15}\right):\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}\times\frac{5}{11}=\frac{3}{7}-\frac{2.11.5}{3.5.11}\)
\(B=\frac{3}{7}-\frac{2}{3}=\frac{9-14}{21}=-\frac{5}{21}\)
Ủng hộ mk nha !!! ^_^
\(\frac{1}{3}+\frac{1}{6}+...+\frac{1}{x\left(x+1\right):2}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2009}{2011}\)
\(\Leftrightarrow2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2011}:2\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2009}{4022}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2011}\)
\(\Leftrightarrow x+1=2011\)
\(\Leftrightarrow x=2010\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.......+\frac{1}{x\times\left(x+1\right)\div2}=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2\times3}+\frac{1}{3\times4}+\frac{1}{4\times5}+.......+\frac{1}{x\times\left(x+1\right)}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{x}+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(2\times\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2009}{2011}\)
\(1-\frac{2}{x+1}=\frac{2009}{2011}\)
\(\frac{2}{x+1}=1-\frac{2009}{2011}\)
\(\frac{2}{x+1}=\frac{2}{2011}\)
\(x+1=2011\)
\(x=2011-1\)
\(\Rightarrow x=2010\)
Bài 1:
a: \(A=\dfrac{\left(85+\dfrac{7}{30}-83-\dfrac{5}{18}\right):\dfrac{8}{3}}{\dfrac{1}{25}}\)
\(=\left(2+\dfrac{7}{30}-\dfrac{5}{18}\right)\cdot\dfrac{3}{8}\cdot25\)
\(=\dfrac{180+21-25}{90}\cdot\dfrac{75}{8}\)
\(=\dfrac{176}{90}\cdot\dfrac{75}{8}=\dfrac{55}{3}\)
=>12,5% của A là 55/8x1/8=55/64
b: \(B=\dfrac{\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{36}{5}}{19.75:2.5}\)
\(=\dfrac{\left(3+\dfrac{27}{70}\right)\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{\dfrac{210+27}{70}\cdot\dfrac{36}{5}}{\dfrac{79}{10}}\)
\(=\dfrac{4266}{175}\cdot\dfrac{10}{79}=\dfrac{108}{35}\)
=>5% là 108/35x1/20=27/175
\((11:21)2×(32010−3):3+3=35n+5⇒32010−3+3=35n+5\)
\(⇒32010=35n+5⇒5n+5=2010⇒5n=2005⇒n=401\)
\(2\times\left(3^{2010}-3\right):3+3=3^{5n+5}\)
\(\Rightarrow3^{2010}-3+3=3^{5n+5}\)
\(\Rightarrow3^{2010}=3^{5n+5}\)
\(\Rightarrow5n+5=2010\)
\(\Rightarrow5n=2005\)
\(\Rightarrow n=401\)
ta có : \(\dfrac{3}{2}\)A= \(\dfrac{3}{4}+\)\(\left(\dfrac{3}{2}\right)^2+\left(\dfrac{3}{2}\right)^3+\)\(...+\left(\dfrac{3}{2}\right)^{2013}\) (1)
A= \(\dfrac{1}{2}+\dfrac{3}{2}\)\(+\left(\dfrac{3}{2}\right)^2+...+\)\(\left(\dfrac{3}{2}\right)^{2012}\) (2)
Lấy (1) trừ đi (2) vế theo vế:
\(\dfrac{3}{2}A-A=\dfrac{3}{4}-\dfrac{1}{2}-\dfrac{3}{2}+\left(\dfrac{3}{2}\right)^{2013}\)
\(\dfrac{1}{2}A=\left(\dfrac{3}{2}\right)^{2013}-\dfrac{5}{4}\Rightarrow A=\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\)
ta có : \(B=\left(\dfrac{3}{2}\right)^{2013}:2=\dfrac{3^{2013}}{2^{2013}}.\dfrac{1}{2}=\dfrac{3^{2013}}{2^{2014}}\)
Vậy \(A-B=\dfrac{3^{2013}}{2^{2014}}-\left(\dfrac{3^{2013}}{2^{2012}}-\dfrac{5}{2}\right)\)
1 3/7-4/5=10/7-4/5=50/35-28/35=22/35
6/13x-3/10+2/5x4/13
=-9/65+8/65
=-1/65
câu đầu mik tính ra số to mà cx ko chắc là đúng nên mik ko viết
*chúc bn học tốt đạt nhiều điểm cao*
\(6\div2\left(1+2\right)\)
\(=6\div2.3\)
\(=3.3\)
\(=9\)
\(6\div2\left(1+2\right)=6\div2.3\\ =6\div6\\ =1 \)
\(6\div2\times\left(1+2\right)=3\times\left(1+2\right)\\ =3\times3\\ =9\)