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Áp dụng tính chất tỉ lệ thức và tính chất dãy tỉ số bằng nhau:
\(ay^2=bx^2\Leftrightarrow\dfrac{x^2}{a}=\dfrac{y^2}{b}=\dfrac{x^2+y^2}{a+b}=\dfrac{1}{a+b}\)
\(\Rightarrow\left(\dfrac{x^2}{a}\right)^{1000}=\left(\dfrac{y^2}{b}\right)^{1000}=\dfrac{1}{\left(a+b\right)^{1000}}\)
\(\Rightarrow\dfrac{x^{2000}}{a^{1000}}+\dfrac{y^{2000}}{b^{1000}}=\dfrac{2}{\left(a+b\right)^{1000}}\)
\(A=\left(-1\right)^{2n}.\left(-1\right)^n.\left(-1\right)^{n+1}\)
\(A=\left(-1\right)^{2n+n+n+1}\)
\(A=\left(-1\right)^{4n+1}\)
\(B=\left(10000-1^2\right).\left(10000-2^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-100^2\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...\left(10000-10000\right)...\left(10000-1000^2\right)\)
\(B=\left(10000-1^2\right)\left(10000-2^2\right)...0\left(10000-1000^2\right)\)
\(B=0\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{5^3}\right)...\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=\left(\dfrac{1}{125}-\dfrac{1}{1^3}\right)\left(\dfrac{1}{125}-\dfrac{1}{2^3}\right)...0....\left(\dfrac{1}{125}-\dfrac{1}{25^3}\right)\)
\(C=0\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-10^3\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...\left(1000-1000\right)}\)
\(D=1999^{\left(1000-1^3\right)\left(1000-2^3\right)...0}\)
\(D=1999^0\)
\(D=1\)
Ta có \(\left(\dfrac{a}{b}\right)^3=\dfrac{1}{1000}\)
\(\Rightarrow\left(\dfrac{a}{b}\right)^3=\left(\dfrac{1}{10}\right)^3\)
\(\Rightarrow\dfrac{a}{b}=\dfrac{1}{10}\)
\(\Rightarrow\dfrac{a}{1}=\dfrac{b}{10}\)
Áp dụng tính chất dãy tỉ số bằng nhau
\(\dfrac{a}{1}=\dfrac{b}{10}=\dfrac{b-a}{10-1}=\dfrac{36}{9}=4\)
\(\Rightarrow\dfrac{a}{1}=4\Rightarrow a=4.1=4\)
\(\dfrac{b}{10}=4\Rightarrow b=4.10=40\)
Vậy a=4, b=40
\(x^2+\left(y-\dfrac{1}{10}\right)^{2018}=0\\ \Leftrightarrow x^2+\left[\left(y-\dfrac{1}{10}\right)^{1009}\right]^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^{1009}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
\(A=\dfrac{1001}{1000^2+1}+\dfrac{1001}{1000^2+2}+....+\dfrac{1001}{1000^2+1000}\)
\(CMR:1< A^2< 4\)
Đặt 761=a; 139=b
\(M=\left(3+\dfrac{1}{3b}\right)\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\left(4+\dfrac{a}{a+1}\right)-\dfrac{4}{\left(a+1\right)\cdot3b}+\dfrac{5}{b}\)
\(=\dfrac{9b+1}{3b}\cdot\dfrac{1}{a+1}-\dfrac{1}{b}\cdot\dfrac{4a+4}{a+1}-\dfrac{4}{3b\left(a+1\right)}+\dfrac{5}{b}\)
\(=\dfrac{9b+1-3\left(4a+4\right)-4+15\left(a+1\right)}{3b\left(a+1\right)}\)
\(=\dfrac{9b+1-12a-12-4+15a+15}{3b\left(a+1\right)}\)
\(=\dfrac{3a+9b}{3b\left(a+1\right)}=\dfrac{3\left(a+3b\right)}{3b\left(a+1\right)}=\dfrac{a+3b}{b\left(a+1\right)}\)
\(=\dfrac{761+3\cdot139}{139\left(761+1\right)}=\dfrac{589}{52959}\)
a: \(\dfrac{877-160a}{1000-90a}=\dfrac{35.5}{100}\)
\(\Leftrightarrow87700-16000a=35500-3195a\)
\(\Leftrightarrow-12805a=-52200\)
hay a=10440/2561
b: \(\dfrac{642-120a}{1000-108a}=44.5\)
\(\Leftrightarrow642-120a=44500-4806a\)
=>4686a=43858
hay a=21929/2343