A=1/2*1008+7.4*201.6+2016/100
Tính nhanh:1/2*1008+7.4*201.6+2016:100 Chú ý * là kí hiệu dấu nhân Đây là dạng toán lớp 5
cho bx^2=ay^2 va x^2+y^2=1 cm x^2016/a^1008+y^2016/b^1008=2/(a+b)^1008
\(Cho bx^2=ay^2\) và \(x^2+y^2=1.CMRa,\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{\left(a+b\right)^{1008}} b, bx^2=ay^2\)
Cho \(a,b,x,y\) là các số thực thỏa mãn: \(x^2+y^2=1\) và \(\dfrac{x^4}{a}+\dfrac{y^4}{b}=\dfrac{1}{a+b}\) Chứng minh rằng: \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{\left(a+b\right)^{1008}}\)
Cho \(bx^2=ay^2\) và \(x^2+y^2=1.CMR\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{\left(a+b\right)^{1008}}\)
\(bx^2=ay^{2^{ }}=\dfrac{x^2}{\dfrac{1}{b}}=\dfrac{y^2}{\dfrac{1}{a}}=\dfrac{x^2+y^2}{\dfrac{a+b}{ab}}=\dfrac{ab}{a+b}.\)
\(\Leftrightarrow\dfrac{x^2}{a}=\dfrac{1}{a+b}=\dfrac{y^2}{b}.\)
\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}=2.\left(\dfrac{1}{a+b}\right)^{1008}=\dfrac{2}{\left(a +b\right)^{1008}}\left(dpcm\right)\)
Theo bài ra ta có:
\(bx^2=ay^2\) \(\Rightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}\)
\(x^2+y^2=1\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{1}{a+b}\) \(\left(1\right)\)
Từ \(\left(1\right)\) suy ra :
\(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}\) \(=\dfrac{\left(x^2\right)^{1008}}{a^{1008}}+\dfrac{\left(y^2\right)^{1008}}{b^{1008}}\)
\(=\left(\dfrac{x^2}{a}\right)^{1008}+\left(\dfrac{y^2}{b}\right)^{1008}\)
\(=\left(\dfrac{1}{a+b}\right)^{1008}+\left(\dfrac{1}{a+b}\right)^{1008}\)
\(=2\cdot\left(\dfrac{1}{a+b}\right)^{1008}\)
\(=2\cdot\dfrac{1^{1008}}{\left(a+b\right)^{1008}}\)
\(=2\cdot\dfrac{1}{\left(a+b\right)^{1008}}\)
\(=\dfrac{2}{a+b}^{1008}\)
Vậy \(\dfrac{x^{2016}}{a^{1008}}+\dfrac{y^{2016}}{b^{1008}}=\dfrac{2}{a+b}^{1008}\)
a^2016+b^2016+c^2016=a^1008. b^1008+b^1008. c^1008+c^1008. a^1008.
Tính A=(a-b)^3+(b-c)^4+(c-a)^2015
cho 4 số a,b,c,d > o thỏa mãn a^4/b+c^4/d=1/b+d và a^2+c^2=1. chứng minh rằng a^2016/b^1006+c^2016/d^1008=2/(b+d)^1008
Cho a,b,c thỏa mãn:
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}a^{1008}\); a,b,c > 0
Tính biểu thức \(A=\left(a-b\right)^{15}+\left(b-c\right)^{2015}\left(a-c\right)^{2016}\)
\(a^{2016}+b^{2016}+c^{2016}=a^{1008}b^{1008}+b^{1008}c^{1008}+c^{1008}a^{1008}\)
\(\Rightarrow2a^{2016}+2b^{2016}+2c^{2016}=2a^{1008}b^{1008}+2b^{1008}c^{1008}+2c^{1008}a^{1008}\)
\(\Rightarrow\left(a^{2016}-2a^{1008}b^{1008}+b^{1008}\right)+\left(b^{2016}-2b^{1008}c^{1008}+c^{1008}\right)\)\(+\left(c^{2016}-2c^{1008}a^{1008}+a^{2016}\right)=0\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)=0\)
Vì \(\hept{\begin{cases}\left(a^{1008}-b^{1008}\right)^2\ge0\\\left(b^{1008}-c^{1008}\right)^2\ge0\\\left(c^{1008}-a^{1008}\right)^2\ge0\end{cases}}\)
\(\Rightarrow\left(a^{1008}-b^{1008}\right)^2+\left(b^{1008}-c^{1008}\right)^2+\left(c^{1008}-a^{1008}\right)^2\ge0\)
Dấu " = " xảy ra: \(\Leftrightarrow\hept{\begin{cases}a^{1008}-b^{1008}=0\\b^{1008}-c^{1008}=0\\c^{1008}-a^{1008}=0\end{cases}\Leftrightarrow}\hept{\begin{cases}a^{1008}=b^{1008}\\b^{1008}=c^{1008}\\c^{1008}=a^{1008}\end{cases}\Leftrightarrow}a=b=c\)
Thay a=b=c vào A ta có: \(A=\left(a-a\right)^{15}+\left(a-a\right)^{2015}+\left(a-a\right)^{2016}=0\)
Cho a,b,x,y là các số thực thỏa mãn: \(x^2+y^2=1\) và \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\). Chứng minh rằng:
\(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)
Ta có: \(\hept{\begin{cases}x^2+y^2=1\\\frac{x^4}{a}+\frac{y^4}{b}=\frac{\left(x^2+y^2\right)^2}{a+b}\end{cases}}\)
\(\Leftrightarrow b\left(a+b\right)x^4+a\left(a+b\right)y^4=ab\left(x^4+2x^2y^2+y^4\right)\)
\(\Leftrightarrow b^2x^4+a^2y^4-2abx^2y^2=0\)
\(\Leftrightarrow\left(bx^2-ay^2\right)^2=0\)
\(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)
\(\Rightarrow\frac{x^{2016}}{a^{1008}}=\frac{y^{2016}}{b^{1008}}=\frac{1}{\left(a+b\right)^{1008}}\)
\(\Rightarrow\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{21008}}=\frac{2}{\left(a+b\right)^{1008}}\)
Em vào câu hỏi tương tự tham khảo:
Ta có: \(x^2+y^2=1\Leftrightarrow x^4+2x^2y^2+y^4=1\)
Khi đó: \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{x^4+2x^2y^2+y^4}{a+b}\)
<=> \(\left(a+b\right)\left(\frac{x^4}{a}+\frac{y^4}{b}\right)=x^4+2x^2y^2+y^4\)
<=> \(\frac{b}{a}x^4+\frac{a}{b}y^4=2x^2y^2\)
<=> \(\frac{x^4}{a^2}+\frac{y^4}{b^2}-\frac{2x^2y^2}{ab}=0\)
<=> \(\left(\frac{x^2}{a}-\frac{y^2}{b}\right)^2=0\)
<=> \(\frac{x^2}{a}=\frac{y^2}{b}=\frac{x^2+y^2}{a+b}=\frac{1}{a+b}\)( dãy tỉ số bằng nhau)
Khi đó: \(\frac{x^{2016}}{a^{1008}}+\frac{y^{2016}}{b^{1008}}=2\frac{x^{2016}}{a^{1008}}=\frac{2}{\left(a+b\right)^{1008}}\)