Do \(\left(2y+1\right)^2\) luôn lẻ

\(\Rightarrow x^2+117\) lẻ

\(\Rightarrow x^2\) chẵn

\(\Rightarrow x\) chẵn

Mà x là số nguyên tố \(\Rightarrow x=2\)

Thế vào pt ban đầu:

\(\left(2y+1\right)^2=117+2^2=121\)

\(\Rightarrow2y+1=11\)

\(\Rightarrow y=5\) là SNT (tm)

Vậy \(\left(x;y\right)=\left(2;5\right)\)

câu trả lời chính là D

D

vô QANDA CHẢ LỜI CHO

1234.2341

\(x+y+z=xyz\Rightarrow\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)

Đặt \(\left(\dfrac{1}{x};\dfrac{1}{y};\dfrac{1}{z}\right)=\left(a;b;c\right)\Rightarrow ab+bc+ca=1\)

\(P=\dfrac{2a}{\sqrt{1+a^2}}+\dfrac{b}{\sqrt{1+b^2}}+\dfrac{c}{\sqrt{1+c^2}}=\dfrac{2a}{\sqrt{ab+bc+ca+a^2}}+\dfrac{b}{\sqrt{ab+bc+ca+b^2}}+\dfrac{c}{\sqrt{ab+bc+ca+c^2}}\)

\(P=\dfrac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}+\dfrac{b}{\sqrt{\left(a+b\right)\left(b+c\right)}}+\dfrac{c}{\sqrt{\left(a+c\right)\left(b+c\right)}}\)

\(P=\sqrt{\dfrac{2a}{a+b}.\dfrac{2a}{a+c}}+\sqrt{\dfrac{2b}{a+b}.\dfrac{b}{2\left(b+c\right)}}+\sqrt{\dfrac{2c}{c+a}.\dfrac{c}{2\left(c+b\right)}}\)

\(P\le\dfrac{1}{2}\left(\dfrac{2a}{a+b}+\dfrac{2a}{a+c}+\dfrac{2b}{a+b}+\dfrac{b}{2\left(b+c\right)}+\dfrac{2c}{c+a}+\dfrac{c}{2\left(c+b\right)}\right)=\dfrac{9}{4}\)

\(P_{max}=\dfrac{9}{4}\) khi \(\left(a;b;c\right)=\left(\dfrac{7}{\sqrt{15}};\dfrac{1}{\sqrt{15}};\dfrac{1}{\sqrt{15}}\right)\) hay \(\left(x;y;z\right)=\left(\dfrac{\sqrt{15}}{7};\sqrt{15};\sqrt{15}\right)\)

Vì \(158< 160< 161,9< 162,1< 162,5\)

Vậy các số thập phân theo thứ tự tăng dần là :

\(158;160;161,9;162,1;162,5\)

_HT_

158;  160;  161,9;  162,1;  162,5.

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