46093721

46093721

Áp dụng định lý Bơ-du ta có:

f(x) ⋮ (2x+5) ⇔ f(\(\dfrac{-5}{2}\))=0 (Kiến thức nâng cao lớp 8)

                    ⇔ (\(\dfrac{-5}{2}\))⁴+(\(\dfrac{-5}{2}\))³+12.(\(\dfrac{-5}{2}\))²-m=0

                    ⇔ \(\dfrac{1575}{16}\)-m=0

                    ⇔ m= \(\dfrac{1575}{16}\)

Vậy m=\(\dfrac{1575}{16}\) thì x⁴+x³+12x²-m chia hết cho 2x+5

\(B=\dfrac{x^2-x+1+4}{x^2-x+1}=1+\dfrac{4}{x^2-x+1}=1+\dfrac{4}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)

Do \(\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4};\forall x\)

\(\Rightarrow B\le1+\dfrac{4}{\dfrac{3}{4}}=\dfrac{19}{3}\)

Vậy \(B_{max}=\dfrac{19}{3}\) khi \(x-\dfrac{1}{2}=0\Rightarrow x=\dfrac{1}{2}\)

\(\left(1+x^2\right)\left(1+y^2\right)+4xy+2\left(x+y\right)\left(1+xy\right)\)

\(=1+x^2+y^2+x^2y^2+4xy+2\left(x+y\right)\left(1+xy\right)\)

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

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

\(=\left(x+y+1+xy\right)^2\) là SCP

(1+x²)(1+y²)+4xy+2(x+y)(1+xy)

 = 1+y²+x²+x²y²+2xy+2xy+2(x+y)(1+xy)

 =(x²+2xy+y²)+(x²y²+2xy+1)+2(x+y)(1+xy)

 =(x+y)²+(xy+1)²+2(x+y)(1+xy)

 =(x+y+xy+1)²

\(x\left(x+2\right)-\left(x-2\right)^2=8\)

\(x^2+2x-x^2+4x-4=8\)

\(6x-4=8\)

\(x=2\)