Question

b1 cho \(a\ge1;b\ge1\)

cm \(a\sqrt{b-1}+b\sqrt{a-1}\le ab\)

b2 tìm các cặp số nguyên x thoả mãn \(y\left(x-1\right)=x^2+2\)

 

Answer 1

Áp dụng bđt AM - GM ta có :

\(\sqrt{b-1}\le\frac{b-1+1}{2}=\frac{b}{2}\Rightarrow a\sqrt{b-1}\le\frac{ab}{2}\)

\(\sqrt{a-1}\le\frac{a-1+1}{2}=\frac{a}{2}\Rightarrow b\sqrt{a-1}\le\frac{ba}{2}\)

\(\Rightarrow a\sqrt{b-1}+b\sqrt{a-1}\le\frac{ab}{2}+\frac{ab}{2}=ab\)(đpcm)

b2 dễ tự lm

Answer 2

b2 x2 là x mũ 2. y2 là y mũ 2 .

yx−y=x​2​​+2

yx−y−x​2​​−2=0

x=​−2​​−y+√​y​2​​−4y−8​​​​​,​−2​​−y−√​y​2​​−4y−8​​​​​

x=​−2​​−y+√​y​2​​−4y−8​​​​​,​−2​​−y−√​y​2​​−4y−8​​​​​

x=−​2​​−y+√​y​2​​−4y−8​​​​​,−​2​​−y−√​y​2​​−4y−8​​​​​

k sau giúp tiếp 

Answer 3

b1 ml làm đc òi nha

Answer 4

80m2 nha!