Question

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

\(P=\dfrac{a^4+b^4}{ab\left(a^3+b^3\right)}+\dfrac{b^4+c^4}{bc\left(b^3+c^3\right)}+\dfrac{c^4+a^4}{ac\left(c^3+a^3\right)}\)

Answer 1

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

Answer 2

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 \(\Leftrightarrow a=b=c=3\)

Answer 3

3/ a/ \(2x^6+y^2-2x^3y=320\)

\(\Leftrightarrow x^6-320=-\left(x^3-y\right)^2\le0\)

\(\Leftrightarrow x^6\le320\)

\(\Leftrightarrow-2\le x\le2\)

Thế vô làm tiếp