\(\frac{8}...">
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13 tháng 9 2019

\(A-B=\left(ax+by\right)^2-\left(a^2+b^2\right)\left(x^2+y^2\right)\)

\(=a^2x^2+2axby+b^2y^2-a^2x^2-a^2y^2-b^2x^2-b^2y^2\)

\(=-\left(a^2y^2-2axby+b^2x^2\right)\)

\(=-\left(ay-bx\right)^2\le0\)

\(\Rightarrow A\le B\) dấu "=" xảy ra \(\frac{a}{x}=\frac{b}{y}\)

Xét \(\frac{a}{x}=\frac{2}{\left(\frac{8}{11}\right)}=\frac{11}{4};\frac{b}{y}=\frac{\left(-1\right)}{\left(-\frac{5}{11}\right)}=\frac{11}{5}\Rightarrow\frac{a}{x}\ne\frac{b}{y}\)

Vậy \(A< B\)

\(A=\left(2\cdot\dfrac{8}{11}-1\cdot\dfrac{-5}{11}\right)^2=\left(\dfrac{16}{11}+\dfrac{5}{11}\right)^2=\left(\dfrac{21}{11}\right)^2=\dfrac{441}{121}\)

\(B=\left(4+1\right)\left(\dfrac{64}{121}+\dfrac{25}{121}\right)=5\cdot\dfrac{89}{121}\)

mà \(441< 5\cdot89\)

nên A<B

7 tháng 6 2020

Mình cần gấp

1 tháng 7 2021

\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B