Độ dài đường trung bình là 30cm nhé bạn

\(x^2-7x+8=0\)

\(\Leftrightarrow\)\(x^2-2.\frac{7}{2}x+\frac{49}{4}-\frac{17}{4}=0\)

\(\Leftrightarrow\)\(\left(x-\frac{7}{2}\right)^2-\frac{17}{4}=0\)

\(\Leftrightarrow\)\(\left(x-\frac{7}{2}-\frac{\sqrt{17}}{2}\right)\left(x-\frac{7}{2}+\frac{\sqrt{17}}{2}\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-\frac{7}{2}-\frac{\sqrt{17}}{2}=0\\x-\frac{7}{2}+\frac{\sqrt{17}}{2}=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=\frac{7+\sqrt{17}}{2}\\x=\frac{7-\sqrt{17}}{2}\end{cases}}\)

Vậy....

a, =[ x^2 - 2x. \(\left(\frac{1}{2}\right)^2\)+\(\left(\frac{1}{2}\right)^2]-\left(\frac{1}{2}\right)^2\)+ 5

= (x^2 - \(\frac{1}{2}\))^2 -\(\frac{1}{4}\)+5

= (x^2 - 1/2)^2 + 19/4 \(\ge\)19/4

Vậy GTNN là 19/4

\(A=6x-x^2-5\)

\(=-x^2-6x-5\)

\(=-\left(x^2+6x+5\right)\)

\(=-\left(x^2+2x.3+3^2-3^2+5\right)\)

\(=-\left(x^2+2x.3+3^2\right)-3^2+5\)

\(=-\left(x+3\right)^2-9+5\)

\(=-\left(x+3\right)^2-4\)

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

\(\Rightarrow-\left(x+3\right)^2-2^2\ge4\)

Vậy muốn A đạt Max thì<=>  \(A\ge4\)

= -(-6x + x^2 + 5)

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

= - ( x^2 - 2.x.3 + 3^2 - 3^2 + 5)

= - (x^2 - 2.x.3 + 3^2) -3^2 + 5

= - (x - 3)^2 - 9 + 5

=> - (x - 3)^2 - 4 \(\le\)-4

GTLN = -4

= (1,6 ^2) + 2.1,6.3,4 +(3,4)^2

= (1,6+3,4)^2 

= 5^2 =25

Chúc bạn học tốt.

\(\Rightarrow\left(1,6^2\right)+2.1,6.3,4+\left(3,4\right)^2\)

\(\Rightarrow\left(1,6+3,4\right)^2\)

\(\Rightarrow5^2=25\)