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Do \(0\le a,b,c\le1\) nên
\(\left(a^2-1\right)\left(b-1\right)\ge0\Leftrightarrow a^2b+1\ge a^2+b\)
Tương tự
\(3+a^2b+b^2c+c^2a\ge a^2+b^2+c^2+a+b+c\ge2\left(a^3+b^3+c^3\right)\)...
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ĐKXĐ \(x\ge0;x\ne4\)
1. Với x = 25 :
\(A=\dfrac{\sqrt{25}+1}{25-4}=\dfrac{2}{7}\)
2. \(B=\dfrac{18-\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x+2}\right)}+\dfrac{4\left(\sqrt{x}+2\right)}{\left(2-\sqrt{x}\right)\left(\sqrt{x+2}\right)}+\dfrac{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)
\(=\dfrac{18-\sqrt{x}-4\left(\sqrt{x}+2\right)+\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4\sqrt{x}+4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
3.\(P=A.B=\dfrac{\sqrt{x}+1}{x-4}.\dfrac{\sqrt{x}-2}{\sqrt{x}+2}=\dfrac{\sqrt{x}+1}{\left(\sqrt{x}+2\right)^2}\)
<=> 4P = \(\dfrac{4\sqrt{x}+4}{\left(\sqrt{x}+2\right)^2}=\dfrac{x+4\sqrt{x}+4-x}{\left(\sqrt{x}+2\right)^2}=1-\dfrac{x}{\left(\sqrt{x}+2\right)^2}\le1\)(Do \(x\ge0\))
<=> \(P\le\dfrac{1}{4}\)("Dấu "=" xảy ra <=> x = 0)
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Ta có: \(\dfrac{a}{1+9b^2}=a-\dfrac{9ab^2}{1+9b^2}\ge a-\dfrac{3ab}{2}\)
\(\Rightarrow\)\(\text{Σ}\dfrac{a}{1+9b^2}\ge a+b+c-\dfrac{3\left(ab+bc+ca\right)}{2}\ge a+b+c-\dfrac{\left(a+b+c\right)^2}{2}=\dfrac{1}{2}\)
(Áp dụng BĐT Cô Si cho 2 số dương, ta có:
\(\text{ }ab+bc+ca\le a^2+b^2+c^2\Rightarrow3\left(\text{ }ab+bc+ca\right)\le\left(a+b+c\right)^2\))
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{3}\)
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\(\left\{{}\begin{matrix}x+y-xy=0\left(1\right)\\\dfrac{2y-xy}{3-2x}=1\end{matrix}\right.\)
\(\left\{{}\begin{matrix}y-xy=-x\\y-xy=3-2x-y\end{matrix}\right.\)
\(\Rightarrow x=3-y\)
Thay vào (1)
\(\Rightarrow\left(3-y\right)+y-\left(3-y\right)y=0\)
\(\Leftrightarrow y^2-3y+3=0\)
Đến đây dùng delta tìm y, xong thay y vào chỗ x=3-y để tìm x