BT1: Cho \(x=\sqrt{2}+1\). Tính \(P=\left(x^4-4x^3+4x^2-2\right)^5+\left(x^3-3x^2-x-1\right)^6\)
BT2: Cho \(x,y>0\), \(x+y=1\). Tìm min
\(P=\dfrac{x+2y}{\sqrt{1-x}}+\dfrac{y+2x}{\sqrt{1-y}}\)
BT3: Tìm nghiệm nguyên:
a) \(2x^6+y^2-2x^3y=320\)
b) \(y^2-1=x\left(1+x\right)\left(1+x^2\right)\)
BT4: Cho \(f\left(x^2-1\right)=x^4-3x^2+3\) đúng vs mọi \(x\). Tìm \(f\left(x^2+1\right)\)
BT5: Cho \(ab+bc+ca=abc\). Tìm GTNN
\(P=\dfrac{a^4+b^4}{ab\left(a^3+b^3\right)}+\dfrac{b^4+c^4}{bc\left(b^3+c^3\right)}+\dfrac{c^4+a^4}{ac\left(c^3+a^3\right)}\)
1/ Ta có: \(x^2-2x-1=\left(\sqrt{2}+1\right)^2-2\left(\sqrt{2}+1\right)-1=0\)
\(\Rightarrow P=\left(x^4-4x^3+4x^2-2\right)^5+\left(x^3-3x^2-x-1\right)^6\)
\(=\left[\left(x^4-2x^3-x^2\right)+\left(-2x^3+4x^2+2x\right)+\left(x^2-2x-1\right)-1\right]^5+\left[\left(x^3-2x^2-x\right)+\left(-x^2+2x+1\right)-2x-2\right]^6\)
\(=\left(-1\right)^5+\left(-2x-2\right)^6\)
Xong
5) Lợi dụng AM-GM :v
\(a^4+a^4+a^4+b^4\ge4a^3b\)
\(b^4+b^4+b^4+a^4\ge4b^3a\)
\(\Rightarrow2a^4+2b^4\ge a^4+a^4+ab^3+a^3b=\left(a^3+b^3\right)\left(a+b\right)\)
\(\Rightarrow P\ge\dfrac{a+b}{2ab}+\dfrac{b+c}{2bc}+\dfrac{c+a}{2ac}=\dfrac{\left(a+b\right)c}{2abc}+\dfrac{\left(b+c\right)a}{2abc}+\dfrac{\left(c+a\right)b}{2abc}=\dfrac{2\left(ab+bc+ca\right)}{2abc}=1\)
Đẳng thức xảy ra \(\Leftrightarrow a=b=c=3\)