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

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

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

\(A=1-3x+1+9x^2-6x+1\)

\(A=9x^2-9x+3\)

\(A=\left(3x\right)^2-2.3x.\frac{9}{6}+\frac{81}{36}-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{27}{36}\)

\(A=\left(3x-\frac{9}{6}\right)^2-\frac{3}{4}\ge0\forall x\)

Dấu = xảy ra khi:

\(3x-\frac{9}{6}=0\Leftrightarrow3x=\frac{9}{6}\Leftrightarrow x=0,5\)

Vậy A_min = -3/4 tại x = 0,5

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

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

đặt t=/3x-1/ với t>=0

khi đó A=t^2-t+1

A=t^2-t+1/4+3/4

A=(t-1/2)^2+3/4

khi đó A>=3/4

dấu bằng xảy ra khi t=1/2 hay x=1/2

Chúc bạn học tốt!

Bài này em tham khảo nhé cu :

https://olm.vn/hoi-dap/detail/9206014220.html

Lời giải :

\(\frac{a^2+b^2}{ab}=\frac{\left(a+b\right)^2-2ab}{ab}\)

\(=\frac{1-2ab}{ab}=\frac{1}{ab}-\frac{2ab}{ab}=\frac{1}{ab}-2\)

Ta có : \(ab\le\left(\frac{a+b}{2}\right)^2=\left(\frac{1}{2}\right)^2=\frac{1}{4}\)

\(\Rightarrow\frac{1}{ab}-2\ge\frac{1}{\frac{1}{4}}-2=2\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=\frac{1}{2}\)

TL:

\(\frac{a^2+b^2}{ab}=\frac{a^2+2ab+b^2-2ab}{ab}\) 

\(=\frac{\left(a+b\right)^2-2ab}{ab}=\frac{1-2ab}{ab}=\frac{1}{ab}-\frac{2ab}{ab}=\frac{1}{ab}-2\)

mà \(ab\le(\frac{a+b}{2})^2=\frac{1}{4}\) 

\(\frac{1}{ab}-2\ge\frac{\frac{1}{1}}{4}-2=\frac{-7}{4}\)  

\(\Rightarrow ab\ge4\) Dấu "=" xảy ra <=>ab=4(bạn tự tìm a,b nha)

Vậy GTNN của BT=\(\frac{-7}{4}\)

\(B=\frac{x^2+17}{x^2+7}\)

\(\Leftrightarrow Bx^2+7B=x^2+17\)

\(\Leftrightarrow Bx^2+7B-x^2-17=0\)

\(\Leftrightarrow x^2\left(B-1\right)+7B-17=0\)

Để pt có nghiệm thì \(\Delta'\ge0\)

\(\Leftrightarrow0^2-\left(B-1\right)\left(7B-17\right)\ge0\)

\(\Leftrightarrow7B^2-24B+17\le0\)

\(\Leftrightarrow1\le B\le\frac{17}{7}\)

Vậy \(max_B=\frac{17}{7}\Leftrightarrow x=0\)

Phuongdeptrai274:e có cách khác a thử check nha!

\(B=\frac{x^2+17}{x^2+7}\)

\(B=\frac{x^2+7+10}{x^2+7}\)

\(B=1+\frac{10}{x^2+7}\)

\(\Rightarrow B\le1+\frac{10}{0+7}=\frac{17}{7}\)

Dấu "=" xảy ra khi x=0

Lời giải :

\(A=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

\(A=\sqrt{2}\cdot\sqrt{3-\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\left(\sqrt{5}-1\right)\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(A=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)\)

\(A=\left(5-2\sqrt{5}+1\right)\left(3+\sqrt{5}\right)\)

\(A=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(A=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(A=2\left(9-5\right)\)

\(A=8\)