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Ta có:
7/12 = 4/12 + 3/12 = 1/3 + 1/4 = 20/60 + 20/80
và 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 = (1/41 + 1/42 + 1/43 + ...+ 1/60) + (1/61 + 1/62 +...+ 1/79 + 1/80)
Do 1/41> 1/42 > 1/43 > ...>1/59 > 1/60
=> (1/41 + 1/42 + 1/43 + ...+ 1/60) > 1/60 + ...+ 1/60 = 20/60
và 1/61> 1/62> ... >1/79> 1/80
=> (1/61 + 1/62 +...+ 1/79 + 1/80) > 1/80 + ...+ 1/80 = 20/80
Vậy 1/41 + 1/42 + 1/43 +...+ 1/79 + 1/80 > 20/60 + 20/80 = 7/12
dễ thấy B=\(\frac{2015+2016}{2016+2017}\)<1
A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)=1-\(\frac{1}{2016}\)+1-\(\frac{1}{2017}\)=(1+1)-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))=2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))
vì (\(\frac{1}{2016}\)+\(\frac{1}{2017}\))<0,5+0,5=1 suy ra 2-(\(\frac{1}{2016}\)+\(\frac{1}{2017}\))>1 mà b<1suy ra A>B
Ta thấy: B=\(\frac{2015+2016}{2016+2017}\)=\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)
A=\(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)
Mà\(\frac{2015}{2016+2017}\)<\(\frac{2015}{2016}\); \(\frac{2016}{2016+2017}\)<\(\frac{2016}{2017}\)
Suy ra: \(\frac{2015}{2016}\)+\(\frac{2016}{2017}\)>\(\frac{2015}{2016+2017}\)+\(\frac{2016}{2016+2017}\)=\(\frac{2015+2016}{2016+2017}\)
Hay A>B
Từ dãy tỉ số bằng nhau đó, ta được:
\(\frac{2a+b+c+d}{a}-1=\frac{a+2b+c+d}{b}-1=\frac{a+b+2c+d}{c}-1=\frac{a+b+c+2d}{d}-1\)
hay \(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta được:
\(\frac{a+b+c+d}{a}=\frac{a+b+c+d}{b}=\frac{a+b+c+d}{c}=\frac{a+b+c+d}{d}=\frac{4\left(a+b+c+d\right)}{a+b+c+d}=4\)
Do đó, \(\frac{a+b+c+d}{a}=4\) => a=\(\frac{a+b+c+d}{4}\)
\(\frac{a+b+c+d}{b}=4\) =>b=\(\frac{a+b+c+d}{4}\)
\(\frac{a+b+c+d}{c}=4\) =>c=\(\frac{a+b+c+d}{4}\)
\(\frac{a+b+c+d}{d}=4\) => d=\(\frac{a+b+c+d}{4}\)
=>a=b=c=d
a+bc+d
Do đó, M=\(\frac{a+b}{c+d}+\frac{b+c}{c+d}+\frac{c+d}{a+b}+\frac{d+a}{b+c}=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}=1+1+1+1=4\)
Vậy M có giá trị là 4
cho 2014=2013+1 thay vào ta có:\(B=x^{2013}-\left(2013+1\right)x^{2012}+\left(2013+1\right)x^{2011}-...-\left(2013+1\right)x^2+\left(2013+1\right)x-1\)
\(=x^{2013}-\left(x+1\right)x^{2012}+\left(x+1\right)x^{2011}-...-\left(x+1\right)x^2+\left(x+1\right)x-1\)
\(=x^{2013}-x^{2013}-x^{2012}+x^{2012}+x^{2011}-...-x^3-x^2+x^2+x-1\)
\(=x-1=2013-1=2012\)
\(\frac{30}{43}\)=\(\frac{1}{\frac{43}{30}}\)= \(\frac{1}{1+\frac{13}{30}}\)=\(\frac{1}{1+\frac{1}{2+\frac{4}{13}}}\)=\(\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)
=> a=1,b=2,c=3,d=4.
Suy nghĩ đi, chỗ nào ko hiểu hỏi mình, lát mình quay lại giờ mình bận.
Bài 3:
\(\left(\dfrac{1}{32}\right)^7=\dfrac{1^7}{32^7}=\dfrac{1}{32^7}=\dfrac{1}{\left(2^5\right)^7}=\dfrac{1}{2^{35}}\\ \left(\dfrac{1}{16}\right)^9=\dfrac{1^9}{16^9}=\dfrac{1}{16^9}=\dfrac{1}{\left(2^4\right)^9}=\dfrac{1}{2^{36}}\)
Vì \(2^{35}< 2^{36}\) nên \(\dfrac{1}{2^{35}}>\dfrac{1}{2^{36}}\) hay \(\left(\dfrac{1}{32}\right)^7>\left(\dfrac{1}{16}\right)^9\)