\(S=\sqrt{x-1}+\sqrt{2x^2-5x+7}\)

\(\Rightarrow S^2=2x^2-4x+6+2\sqrt{x-1.2x^2-5x+7}\)

\(=2.x-1^2+4+2\sqrt{x-1.2x^2+5x-7}\ge4\)

\(Min_A=4\Leftrightarrow x=1\)

Vậy: \(x=1\)

P/s: Đúng ko nhỉ?

bạn ơ\(\sqrt{x-1}+\sqrt{2x^2-5x+7}\)i  sao ra cai do vay

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Thông tin đến bạn!

Ta có :

\(A=\frac{x-2}{\left(x^3-1\right)-x^2-x-1}=\frac{x-2}{\left(x-1\right)\left(x^2+x+1\right)-\left(x^2+x+1\right)}=\frac{x-2}{\left(x-2\right)\left(x^2+x+1\right)}=\frac{1}{x^2+x+1}\)

Để \(A\) đạt GTLN \(\Leftrightarrow x^2+x+1\) đạt GTNN

Ta có : \(x^2+x+1=\left(x^2+2\cdot\frac{1}{2}\cdot x+\frac{1}{4}\right)+\frac{3}{4}=\left(x+\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\) có GTNN là 3/4

\(\Rightarrow A\le\frac{1}{\frac{3}{4}}=\frac{4}{3}\) có GTLN là \(\frac{4}{3}\)

Dấu "=" xảy ra \(\Leftrightarrow x=-\frac{1}{2}\)

Vậy \(A_{max}=\frac{4}{3}\) tại \(x=-\frac{1}{2}\)

a: ĐKXĐ: x<>-3

b: \(Q=\left(\dfrac{x}{x^2-3x+9}-\dfrac{11}{\left(x+3\right)\left(x^2-3x+9\right)}+\dfrac{1}{x+3}\right)\cdot\dfrac{x+3}{x^2-1}\)

\(=\dfrac{x^2+3x-11+x^2-3x+9}{\left(x+3\right)\left(x^2-3x+9\right)}\cdot\dfrac{x+3}{x^2-1}\)

\(=\dfrac{2x^2-2}{x^2-1}\cdot\dfrac{1}{x^2-3x+9}=\dfrac{2}{x^2-3x+9}\)