Question

[Phong Hà Nguyên]

Toán 8 mn giúp mk

Answer 1

[Nguyễn Việt Lâm]

a.

\(x^2+x+3=y^2\)

\(\Leftrightarrow4x^2+4x+12=4y^2\)

\(\Leftrightarrow\left(2x+1\right)^2+11=\left(2y\right)^2\)

\(\Leftrightarrow\left(2y\right)^2-\left(2x+1\right)^2=11\)

\(\Leftrightarrow\left(2y+2x+1\right)\left(2y-2x-1\right)=11\)

2y+2x+1	1	11	-1	-11
2y-2x-1	11	1	-11	-1
x	-3	2	2	-3
y	3	3	-3	-3

Vậy pt có các cặp nghiệm:

\(\left(x;y\right)=\left(-3;3\right);\left(2;3\right);\left(2;-3\right);\left(-3;-3\right)\)

Answer 2

[Nguyễn Việt Lâm]

b.

\(A=\dfrac{a+b}{2022a-b}+\dfrac{b+c}{2022c-b}=\dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{\dfrac{2022}{b}-\dfrac{1}{a}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{c}}{\dfrac{2022}{b}-\dfrac{1}{c}}\)

\(=\dfrac{\dfrac{1}{a}+\dfrac{1}{b}}{\dfrac{1}{c}}+\dfrac{\dfrac{1}{b}+\dfrac{1}{c}}{\dfrac{1}{a}}=\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c+a}{b}\)

Do \(\dfrac{1}{a}+\dfrac{1}{c}=\dfrac{2022}{b}\Rightarrow\dfrac{a+c}{2022ac}=\dfrac{1}{b}\Rightarrow\dfrac{c+a}{b}=\dfrac{\left(c+a\right)^2}{2022ac}\)

\(\Rightarrow A=\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{\left(c+a\right)^2}{2022ac}\ge2\sqrt{\dfrac{ca}{ac}}+\dfrac{4ac}{2022ac}=\dfrac{2024}{1011}\)

Dấu "=" xảy ra khi \(a=c=\dfrac{b}{1011}\)