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 = \(\dfrac{1}{6}\)

\(A=\dfrac{1}{x-2}+\dfrac{2}{x-3}=\dfrac{x-3+2x-4}{\left(x-2\right)\left(x+3\right)}=\dfrac{3x-7}{\left(x-2\right)\left(x-3\right)}\)

\(M=A+B=\dfrac{3x-7}{\left(x-2\right)\left(x-3\right)}+\dfrac{x^2-5x+7}{x^2-5x+6}=\dfrac{3x-7+x^2-5x+7}{x^2-5x+6}=\dfrac{x^2-2x}{x^2-5x+6}\)

\(H=\dfrac{6-8x}{x^2+1}\)

<=> Hx² + H = 6 - 8x 

<=> Hx² + 8x + H - 6 = 0 (1) 

Phương trình (1) có nghiệm khi 

\(\Delta=8^2-4H\left(H-6\right)\ge0\)

<=> \(H^2-6H-16\le0\)

<=> \(\left(H-8\right)\left(H+2\right)\le0\)

\(\Leftrightarrow-2\le H\le8\) 

=> Min H = -2 

Dấu "=" xảy ra khi x = 2 

Mới xong câu a , b thui nhé

Thanks bạn nka

hi

Kẻ đường cao AK ứng với cạnh BC ( K thuộc BC )

+) Xét \(\Delta ABC\) cân tại A có : AK là đường cao ứng với BC

=> AK đồng thời là đường trung trực của BC 

\(\Rightarrow BK=KC=\dfrac{1}{2}BC=\dfrac{1}{2}.12=6\left(cm\right)\)

+) Áp dụng định lí Py - ta - go vào \(\Delta AKB\left(\widehat{AKB}=90^o\right)\) , có :

\(AK^2=AB^2-BK^2\)

\(\Leftrightarrow AK=\sqrt{10^2-6^2}=8\left(cm\right)\)

+) Ta có : \(S_{\Delta ABC}=S_{\Delta ABH}+S_{\Delta BHC}\)

\(\Leftrightarrow\dfrac{1}{2}AH.BC=\dfrac{1}{2}BH.AH+\dfrac{1}{2}BH.HC\)

\(\Leftrightarrow8.12=BH\left(AH+HC\right)\)

\(\Leftrightarrow BH.AC=96\left(cm\right)\)

\(\Leftrightarrow BH=\dfrac{96}{10}=9,6\left(cm\right)\)

4-x^2

4-x^2