a. \(A=0.1+0.2+.......+0.19\)
b. \(B=\left(2017\cdot2016+2014\cdot2015\right)\cdot\left(1+\frac{1}{2}:1\frac{1}{2}+1\frac{1}{3}\right)\)
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\(\left(2013.2014+2014.2015+2015.2016\right)\left(1+\frac{1}{3}-1\frac{1}{3}\right)\)
\(=\left(2013.2014+2014.2015+2015.2016\right)\left(\frac{4}{3}-\frac{4}{3}\right)\)
\(=\left(2013.2014+2014.2015+2015.2016\right).0\)
\(=0\)
Ta có :
\(\left(1+\frac{1}{1.3}\right).\left(1+\frac{1}{2.4}\right).\left(1+\frac{1}{3.5}\right)....\left(1+\frac{1}{2014.2016}\right)\)
\(=\frac{4}{1.3}.\frac{9}{2.4}.\frac{16}{3.5}.....\frac{4060225}{2014.2016}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{2015.2015}{2014.2016}\)
\(=\frac{2.3.4....2015}{1.2.3....2014}.\frac{2.3.4....2015}{3.4.5....2016}\)
\(=\frac{2015}{1}.\frac{2}{2016}\)
\(=2015.\frac{1}{1008}=\frac{2015}{1008}\)
\(\Rightarrow\frac{2015}{1008}=\frac{x}{1008}\Rightarrow x=2015\)
Vậy \(x=2015\)
Ủng hộ mk nha !!! ^_^
\(a\left(b^2+c^2\right)+b\left(c^2+a^2\right)+c\left(a^2+b^2\right)+2abc=0\)
\(\Rightarrow ab^2+ac^2+bc^2+ba^2+c\left(a+b\right)^2=0\)
\(\Rightarrow ab\left(a+b\right)+c^2\left(a+b\right)+c\left(a+b\right)^2=0\)
\(\Rightarrow\left(a+b\right)\left(ab+c^2+ca+cb\right)=0\)
\(\Rightarrow\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]=0\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(a+c\right)=0\)
Từ đó a = -b hoặc b = -c hoặc c = -a
Nếu a = -b mà \(a^3+b^3+c^3=1\Rightarrow\left(-b\right)^3+b^3+c^3=1\Rightarrow c^3=1\Rightarrow c=1\)
Khi đó: \(A=\frac{1}{\left(-b\right)^{2017}}+\frac{1}{b^{2017}}+\frac{1}{1^{2017}}=0+1=1\)
Tương tự với các trường hợp b = -c và a = -c, ta tính được A = 1
b)
\(x-2.\left(\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\right)=\frac{16}{9}\)
\(x-2\cdot\left(\frac{1}{3}-\frac{1}{9}\right)=\frac{16}{9}\)
\(x-2=\frac{16}{9}:\left(\frac{1}{3}-\frac{1}{9}\right)\)
\(x-2=8\)
=> x = 10
a)
\(A=\frac{1}{2}.\frac{2}{3}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{2013}{2014}\cdot\frac{2014}{2015}\cdot\frac{2015}{2016}\)
\(A=\frac{1}{2016}\)
a. ta có (0.1+0.19)+(0.2+0.18)......+0.10
A=0.20+0.20++0.20+0.20+0.20+0.20+0.20+0.20+0.20+0.10
A=1.90
câu b mình pó tay
a ) \(A=0,1+0,2+...+0,19\)
\(A=\left(0,1+0,2+...+0,9\right)+\left(0,10+0,11+...+0,19\right)\)
\(A=0,1\times\left(1+2+...+9\right)+0,1\times\left(1+1,1+...+1,9\right)\)
\(A=0,1\times45+0,1\times14,5\)
\(A=0,1\times\left(45+14,5\right)\)
\(A=0,1\times59,5\)
\(A=5,95\)
b ) \(B=\left(2017\times2016+2014\times2015\right)\times\left(1+\frac{1}{2}\div1\frac{1}{2}+1\frac{1}{3}\right)\)
\(B=\left(2017\times2016+2014\times2015\right)\times\left(1+\frac{1}{2}\div\frac{3}{2}+\frac{4}{3}\right)\)
\(B=\left(2017\times2016+2014\times2015\right)\times\left(1+\frac{2}{6}+\frac{4}{3}\right)\)
\(B=\left(2017\times2016+2014\times2015\right)\times\frac{8}{3}\)