Bài 4:
a:
a>b>0
=>a-b>0
=>b-a<0
a>b>0
=>ab>0
\(\dfrac{1}{a}-\dfrac{1}{b}=\dfrac{b-a}{ab}< 0\)(Vì b-a<0; ab>0)
b: \(\dfrac{2022}{2023}=1-\dfrac{1}{2023};\dfrac{2023}{2024}=1-\dfrac{1}{2024}\)
Vì 2023<2024
nên \(\dfrac{1}{2023}>\dfrac{1}{2024}\)
=>\(-\dfrac{1}{2023}< -\dfrac{1}{2024}\)
=>\(-\dfrac{1}{2023}+1< -\dfrac{1}{2024}+1\)
=>\(\dfrac{2022}{2023}< \dfrac{2023}{2024}\)
Bài 5:
a: \(x^2+y^2-2xy=\left(x-y\right)^2>=0\forall x,y\)
=>\(x^2+y^2>=2xy\forall x,y\)
b: \(x^2+2x+5\)
\(=x^2+2x+1+4\)
\(=\left(x+1\right)^2+4>=4>0\forall x\)
c: \(-x^2+x-1\)
\(=-\left(x^2-x+1\right)\)
\(=-\left(x^2-x+\dfrac{1}{4}+\dfrac{3}{4}\right)\)
\(=-\left(x-\dfrac{1}{2}\right)^2-\dfrac{3}{4}< =-\dfrac{3}{4}< 0\forall x\)
d: \(x^2+xy+y^2\)
\(=x^2+2\cdot x\cdot\dfrac{1}{2}y+\dfrac{1}{4}y^2+\dfrac{3}{4}y^2\)
\(=\left(x+\dfrac{1}{2}y\right)^2+\dfrac{3}{4}y^2>=0\forall x,y\)
Bài 4
a)
Xét hiệu \(\dfrac{1}{a}-\dfrac{1}{b}\)
\(=\dfrac{b-a}{ab}\)
Do \(a>b>0\)
\(\Rightarrow\left\{{}\begin{matrix}b-a< 0\\ab>0\end{matrix}\right.\)
\(\Rightarrow\dfrac{b-a}{ab}< 0\)
\(\Rightarrow\dfrac{1}{a}< \dfrac{1}{b}\)
b)
Ta có:
\(\dfrac{2022}{2023}=1-\dfrac{1}{2023}\)
\(\dfrac{2023}{2024}=1-\dfrac{1}{2024}\)
Do \(2024>2023\)
\(\Rightarrow\dfrac{1}{2024}< \dfrac{1}{2023}\)
\(\Rightarrow1-\dfrac{1}{2024}>1-\dfrac{1}{2023}\)
Hay \(\dfrac{2022}{2023}< \dfrac{2023}{2024}\)
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giúp em với ạ. Em cảm ơn ạ
