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\(P=\frac{12x^2-6x+4}{x^2+1}=\frac{\left(9x^2-6x+1\right)+3\left(x^2+1\right)}{x^2+1}=\frac{\left(3x-1\right)^2}{x^2+1}+3\ge3\forall x\)
Dấu "=" xảy ra khi: \(3x-1=0\Rightarrow x=\frac{1}{3}\)
Vậy \(P_{min}=3\Leftrightarrow x=\frac{1}{3}\)
a: \(\dfrac{2x-3}{35}+\dfrac{x\left(x-2\right)}{7}\le\dfrac{x^2}{7}-\dfrac{2x-3}{5}\)
\(\Leftrightarrow2x-3+5x\left(x-2\right)\le5x^2-7\left(2x-3\right)\)
\(\Leftrightarrow2x-3+5x^2-10x< =5x^2-14x+21\)
=>-8x-3<=-14x+21
=>6x<=24
hay x<=4
b: \(\dfrac{6x+1}{18}+\dfrac{x+3}{12}>=\dfrac{5x+3}{6}+\dfrac{12-5x}{9}\)
=>2(6x+1)+3(x+3)>=6(5x+3)+4(12-5x)
=>12x+2+3x+9>=30x+18+48-20x
=>15x+11>=10x+66
=>5x>=55
hay x>=11
a/ \(\left(a^2+b^2\right)+\left(a^2+1\right)+\left(b^2+1\right)\ge2ab+2a+2b\)
\(\Leftrightarrow a^2+b^2+1\ge ab+a+b\)
b/ \(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\)
\(\Leftrightarrow a^2-2ab+b^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2\ge0\) đúng
c/ \(M=x^4-6x^3+13x^2-12x-5\)
Đặt \(x^2-3x=a\)thì ta có:
\(M=a^2+4a-5=\left(a+2\right)^2-9\ge-9\)
Dấu = xảy ra khi:
\(x^2-3x+2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
2. \(Q=\left(x-3\right)\left(4x+5\right)+2019\)
\(Q=4x^2+5x-12x-15+2019\)
\(Q=4x^2-7x+2004\)
\(Q=\left(2x\right)^2-2.2x.\frac{7}{4}+\frac{49}{16}+2019-\frac{49}{16}\)
\(Q=\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\)
\(Do\) \(\left(2x-\frac{7}{4}\right)^2\ge0\forall x\) \(Nên\) \(\left(2x-\frac{7}{4}\right)^2+\frac{32255}{16}\ge\frac{32255}{16}\)
\(\Rightarrow Q\ge\frac{32255}{16}\)
\(Vậy\) \(MinQ=\frac{32255}{16}\Leftrightarrow x=\frac{7}{8}\)
3. \(T=4\left(a^3+b^3\right)-6\left(a^2+b^2\right)\)
\(T=4\left(a+b\right)\left(a^2-ab+b^2\right)-6a^2-6b^2\)
\(T=4\left(a^2-ab+b^2\right)-6a^2-6b^2\) (do a+b=1)
\(T=4a^2-4ab+4a^2-6a^2-6b^2\)
\(T=-2a^2-4ab-2b^2\)
\(T=-2\left(a^2+2ab+b^2\right)\)
\(T=-2\left(a+b\right)^2\)
\(T=-2.1^2=-2.1=-2\) (do a+b=1)
\(1,a,A=x^2-6x+25\)
\(=x^2-2.x.3+9-9+25\)
\(=\left(x-3\right)^2+16\)
Ta có :
\(\left(x-3\right)^2\ge0\)Với mọi x
\(\Rightarrow\left(x-3\right)^2+16\ge16\)
Hay \(A\ge16\)
\(\Rightarrow A_{min}=16\)
\(\Leftrightarrow x=3\)
Ta có :
\(P=\dfrac{12x^2-6x+4}{x^2+1}\)
\(=\dfrac{3x^2+3+9x^2-6x+1}{x^2+1}\)
\(=\dfrac{3\left(x^2+1\right)+\left(3x-1\right)^2}{x^2+1}\)
\(=3+\dfrac{\left(3x-1\right)^2}{x^2+1}\)
Do : \(\left\{{}\begin{matrix}\left(3x-1\right)^2\ge0\\x^2+1>0\end{matrix}\right.\Rightarrow3+\dfrac{\left(3x-1\right)^2}{x^2+1}\ge3\)
Vậy GTNN của P là 3 . Dấu \("="\) xảy ra khi \(\left(3x-1\right)^2=0\Leftrightarrow x=\dfrac{1}{3}\)