a) \(x^2+10x+26+y^2+2y\)

\(=x^2+10x+25+y^2+2y+1\)

\(=\left(x+5\right)^2+\left(y+1\right)^2\)

b) Xem lại đề 

c) \(x^2-2xy+2y^2+2y+1\)

\(=x^2-2xy+y^2+y^2+2y+1\)

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

ra 900

   \(\dfrac{3}{5}\)a³b x (10ab³ - \(\dfrac{5}{3}\)b² + \(\dfrac{5}{6}\)ab)

=\(\dfrac{3}{5}\)a³b . 10ab³ - \(\dfrac{3}{5}\)a³b. \(\dfrac{5}{3}\)b² + \(\dfrac{3}{5}\)a³b . \(\dfrac{5}{6}\)ab

= 6a⁴b⁴ - a³b³ + \(\dfrac{1}{2}\) a⁴b²

Mn giúp với

thk mn

Lời giải:

$C=(m+n)^3-m^2-2mn-n^2=(m+n)^3-(m^2+2mn+n^2)$

$=(m+n)^3-(m+n)^2=2^3-2^2=8-4=4$

\(C=\left(m+n\right)^3-\left(m^2+2mn+n^2\right)=\left(m+n\right)^3-\left(m+n\right)^2\)

Thay vào ta được 8 - 4 = 4 

đặt:

\(\left\{{}\begin{matrix}S_{ABC}=S^2\\S_{BED}=S_1^2\\S_{CFD}=S_2^2\end{matrix}\right.\)

ta có:

\(ED\) // \(AC\) \(\Rightarrow\Delta BED\sim\Delta BAC\)

\(\Rightarrow\dfrac{S_1^2}{S^2}=\dfrac{BD^2}{BC^2}\Leftrightarrow\dfrac{S_1}{S}=\dfrac{BD}{BC}\)

C/M tương tự:

\(\Rightarrow\dfrac{S_2^2}{S^2}=\dfrac{CD^2}{BC^2}\Leftrightarrow\dfrac{S_2}{S}=\dfrac{CD}{BC}\)

\(\Leftrightarrow\dfrac{BD}{BC}+\dfrac{DC}{BC}=\dfrac{S_1+S_2}{S}\)

\(\Leftrightarrow\dfrac{S_1+S_2}{S}=1\)

\(\Leftrightarrow\left(S_1+S_2\right)^2=S^2\)

\(\Leftrightarrow S^2=\left(\sqrt{100}+\sqrt{16}\right)^2=16^2=256\left(cm^2\right)\)

\(\Leftrightarrow S_{ABC}=256cm^2\)

a) \(A\left(x\right)=-5x^2-4x+1\)

\(\Leftrightarrow A\left(x\right)=-\left(5x^2+4x-1\right)\)

\(\Leftrightarrow A\left(x\right)=-\left[\left(\sqrt{5}x\right)^2+2.\sqrt{5}x.\dfrac{2\sqrt{5}}{5}+\dfrac{4}{5}-\dfrac{9}{5}\right]\)

\(\Leftrightarrow A\left(x\right)=-\left(\sqrt{5}x+\dfrac{4}{5}\right)^2+\dfrac{9}{5}\le\dfrac{9}{5}\)

Dấu bằng xảy ra

 \(\Leftrightarrow\sqrt{5}x+\dfrac{4}{5}=0\Leftrightarrow x==-\dfrac{4\sqrt{5}}{25}\)

b) \(B\left(x\right)=-3x^2+x+1\)

\(\Leftrightarrow B\left(x\right)=-\left[\left(\sqrt{3}x\right)^2-2.\sqrt{3}x.\dfrac{2\sqrt{3}}{3}+\left(\dfrac{2\sqrt{3}}{3}\right)^2-\dfrac{1}{3}\right]\)

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

Dấu bằng xảy ra 

\(\Leftrightarrow\sqrt{3}x-\dfrac{2\sqrt{3}}{3}=0\Leftrightarrow x=\dfrac{2}{3}\)

A = -5x² -4x + 1

A = - ( 5x² + 2.\(\sqrt{5}\).\(\dfrac{2}{\sqrt{5}}\)x +\(\dfrac{4}{5}\) ) +\(\dfrac{29}{25}\)

A = -( \(\sqrt{5}\) x+ \(\dfrac{2}{\sqrt{5}}\))²  + \(\dfrac{29}{25}\)

 -( \(\sqrt{5}\) x+ \(\dfrac{2}{\sqrt{5}}\))²  ≤ 0 ⇔ A(max) =\(\dfrac{29}{25}\) ⇔ x = -2/5

B = -3x² + x + 1

B = -(3x² - 2.\(\sqrt{3}\).\(\dfrac{1}{2\sqrt{3}}\).x + \(\dfrac{1}{12}\)) + \(\dfrac{13}{12}\)

B = -(\(\sqrt{3}\)x - \(\dfrac{1}{2\sqrt{3}}\))² + \(\dfrac{13}{12}\)

vì  - (\(\sqrt{3}\)x - \(\dfrac{1}{2\sqrt{3}}\))² ≤ 0 ⇔ B(max) =\(\dfrac{13}{12}\) ⇔ x = \(\dfrac{1}{6}\)