Cho a+b+c=2106 và 1÷a +1÷b +1÷c=1÷2016 tính bt A=(a^2016-b^2016)(a^2016-b^2016)(b^2016-c^2016)
Những câu hỏi liên quan
Cho \(\frac{a^{2106}+b^{2016}}{c^{2016}+d^{2016}}\)= \(\frac{a^{2016}-b^{2016}}{c^{2016}-d^{2016}}\)Chứng minh rằng \(\frac{a}{b}\)= \(\pm\frac{c}{d}\)
\(\Leftrightarrow\left(a^{2016}+b^{2016}\right).\left(c^{2016}-d^{2016}\right)=\left(a^{2016}-b^{2016}\right).\left(c^{2016}+d^{2016}\right)\)
\(\Leftrightarrow ac^{2016}-ad^{2016}+bc^{2016}-bd^{2016}=ac^{2016}+ad^{2016}-bc^{2016}-bd^{2016}\)
\(\Leftrightarrow-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\)
nếu \(-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}=0\)
\(\Rightarrow ad^{2016}-bc^{2016}=0\Rightarrow ad=bc\Rightarrow\frac{a}{b}=\frac{c}{d}\left(1\right)\)
nếu \(\text{}-\left(ad^{2016}-bc^{2016}\right)=ad^{2016}-bc^{2016}\ne0\Rightarrow ad=-bc\Rightarrow\frac{a}{b}=-\frac{c}{d}\left(2\right)\)
từ (1) và (2) => đpcm
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Cho x+y+z=2016 và 1/x+1/y+1/z=1/2016. Tính giá trị biểu thức B=(x^2012+y^2012)(y^2014+z^2014)(z^2016+x^2106)
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\frac{1}{x+y+z}=0\)
\(\Leftrightarrow\frac{yz\left(x+y+z\right)+xz\left(x+y+z\right)+xy\left(x+y+z\right)-xyz}{xyz\left(x+y+z\right)}=0\)
\(\Leftrightarrow\)\(xyz+y^2z+yz^2+x^2z+xyz+xz^2+x^2y+xy^2+xyz-xyz=0\)
\(\Leftrightarrow\)\(\left(xyz+y^2z\right)+\left(xyz+x^2z\right)+\left(xz^2+yz^2\right)+\left(xy^2+x^2y\right)=0\)
\(\Leftrightarrow yz\left(x+y\right)+xz\left(x+y\right)+z^2\left(x+y\right)+xy\left(x+y\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(yz+xz+xy+z^2\right)=0\)
\(\Leftrightarrow\left(x+y\right)\left(x+z\right)\left(y+z\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+y\\x+z=0\end{cases}}=0\) hoặc y+z=0
Do đó ta có B=0
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Cho 3 số a,b,c thỏa mãn: a+b+c=2016 và \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{2016}\) Tính : A=(a2016-b2016)(b2016-c2016)(c2016-a2016)
cho a,b,c>0 thõa mãn abc=1. CM 1/(a^2016+b^2016+1)+1/(b^2016+c^2016+1)+1(c^2016+a^2016+1)≤1
cho a,b,c>0 thõa mãn abc=1. CM \(\frac{1}{a^{2016}+b^{2016}+1}+\frac{1}{b^{2016}+c^{2016}+1}+\frac{1}{c^{2016}+a^{2016}+1}\le1\)
e ơi e nên tải tài liệu của võ quốc bá cẩn đi
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cho a,b,c>0 thỏa mãn abc=1.chứng minh \(\frac{1}{a^{2016}+b^{2016}+1}+\frac{1}{b^{2016}+c^{2016}+1}+\frac{1}{c^{2016}+a^{2016}+1}\le1\)
So sánh A và B biết:
\(A=\frac{2016^{2106}+1}{2016^{2017}+1}\) và \(B=\frac{2016^{2015}+1}{2016^{2016}+1}\)
Giúp mình với!
Ta có
\(2016A=\frac{2016^{2017}+2016}{2016^{2017}+1}=\frac{2016^{2017}+1}{2016^{2017}+1}+\frac{2015}{2016^{2017}+1}=1+\frac{2015}{2016^{2017}+1}\)
\(2016B=\frac{2016^{2016}+2016}{2016^{2016}+1}=\frac{2016^{2016}+1}{2016^{2016}+1}+\frac{2015}{2016^{2016}+1}=1+\frac{2015}{2016^{2016}+1}\)
Do \(\frac{2015}{2016^{2017}+1}< \frac{2015}{2016^{2016}+1}\Rightarrow2016A< 2016B\Rightarrow A< B.\)
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B = \(\frac{2016^{2015}+1}{2016^{2016}+1}\)< A =\(\frac{2016^{2016}+1}{2016^{2017}+1}\)
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Cho \(\frac{a}{b}=\frac{c}{d}\) Chứng minh \(\frac{a^{2016}+b^{2016}}{a^{2016}-b^{2016}}.\frac{c^{2016}-d^{2016}}{c^{2016}+d^{2016}}=1\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{d}{b}=\frac{c}{a}\Leftrightarrow\frac{d^{2016}}{b^{2016}}=\frac{c^{2016}}{a^{2016}}=\frac{c^{2016}-d^{2016}}{a^{2016}-b^{2016}}=\frac{c^{2016}+d^{2016}}{a^{2016}+b^{2016}}\)
(áp dụng tính chất dãy tỉ số bằng nhau)
Suy ra \(\frac{a^{2016}+b^{2016}}{a^{2016}-b^{2016}}.\frac{c^{2016}-d^{2016}}{c^{2016}+d^{2016}}=\frac{a^{2016}+b^{2016}}{c^{2016}+d^{2016}}.\frac{c^{2016}-d^{2016}}{a^{2016}-b^{2016}}\)
\(=\frac{b^{2016}}{d^{2016}}.\frac{d^{2016}}{b^{2016}}=1\)
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cho a,b,c thoả mãn a^2016+b^2016+c^2016=a^2017+b^2017+c^2017=1. Tính B=a^2015+b^2016+c^2017