\(C=\sqrt{2x^2-6x+5}=\sqrt{2\left(x^2-3x+\frac{9}{4}\right)+\frac{1}{2}}\)

\(C=\sqrt{2\left(x^2-2.\frac{3}{2}x+\left(\frac{3}{2}\right)^2\right)+\frac{1}{2}}=\sqrt{2\left(x-\frac{3}{2}\right)^2+\frac{1}{2}}\ge\frac{1}{2}\)

Vậy GTNN của C là \(\frac{1}{2}\) \(\Leftrightarrow x=\frac{3}{2}\)

mình nhầm. thay GTNN \(\frac{1}{2}\)thành \(\sqrt{\frac{1}{2}}\)

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

A=\(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}=\sqrt{\left(x-1\right)^2}+\sqrt{\left(x+3\right)^2}\)=|x-1|+|x+3|=|1-x|+|x+3|

Áp dụng bđt |a|+|b|\(\ge\)|a+b| ta được: A=|1-x|+|x+3|\(\ge\)|1-x+x+3|=4

Dấu "=" xảy ra khi (1-x)(x+3)\(\ge\)0 <=> \(-3\le x\le1\)

Vậy A_min=4 khi \(-3\le x\le1\)

A = \(\sqrt{x^2-2x+1}+\sqrt{x^2+6x+9}\)

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

 = 1 - x + x + 3

  = 4 

kết quả là 

 =4

    đs...

Mình nghĩ ra câu C rồi bạn nào giúp mình nghĩ nốt câu A,B hộ mình nhé mình cảm ơn!

a:6x-5-9x^2

=-(9x^2-6x+5)

=-(9x^2-6x+1+4)

=-(3x-1)^2-4<=-4

=>A>=2/-4=-1/2

Dấu = xảy ra khi x=1/3

b: \(B=\dfrac{4x^2-6x+4-1}{2x^2-3x+2}=2-\dfrac{1}{2x^2-3x+2}\)

2x^2-3x+2=2(x^2-3/2x+1)

=2(x^2-2*x*3/4+9/16+7/16)

=2(x-3/4)^2+7/8>=7/8

=>-1/2x^2-3x+2<=-1:7/8=-8/7

=>B<=-8/7+2=6/7

Dâu = xảy ra khi x=3/4

Ta có \(A=\sqrt{2x^2+6x+5}\)

\(=\sqrt{2\left(x^2+3x+\dfrac{5}{2}\right)}\)

\(=\sqrt{2\left(x^2+3x+\dfrac{9}{4}+\dfrac{1}{4}\right)}\)

\(=\sqrt{2\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{1}{4}\right]}\ge\sqrt{2.\dfrac{1}{4}}=\dfrac{\sqrt{2}}{2}\)

Vậy GTNN của A là \(\dfrac{\sqrt{2}}{2}\) khi \(\left(x+\dfrac{3}{2}\right)^2=0\Leftrightarrow x=-\dfrac{3}{2}\)

Học tốt nhé

\(A=x^2-6x+10\)

\(\Leftrightarrow A=x^2-2\cdot x\cdot3+3^2-9+10\)

\(\Leftrightarrow A=\left(x-3\right)^2+1\ge1\)     \(\forall x\in z\)

\(\Leftrightarrow A_{min}=1khix=3\)

\(B=3x^2-12x+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x\right)^2-2\cdot\sqrt{3}x\cdot2\sqrt{3}+\left(2\sqrt{3}\right)^2-12+1\)

\(\Leftrightarrow B=\left(\sqrt{3}x-2\sqrt{3}\right)^2-11\ge-11\)    \(\forall x\in z\)

\(\Leftrightarrow B_{min}=-11khix=2\)

\(A=\left(x^2+2\cdot\dfrac{3}{2}x+\dfrac{9}{4}\right)-\dfrac{5}{4}=\left(x+\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ge-\dfrac{5}{4}\\ A_{min}=-\dfrac{5}{4}\Leftrightarrow x=-\dfrac{3}{2}\\ B=\left(x^2+2xy+y^2\right)+\left(x^2+6x+9\right)+3\\ B=\left(x+y\right)^2+\left(x+3\right)^2+3\ge3\\ B_{min}=3\Leftrightarrow\left\{{}\begin{matrix}x+y=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=-3\end{matrix}\right.\\ C=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\\ C_{max}=1\Leftrightarrow x=1\)