a) \(A=-\left(\dfrac{x+18}{1203}\right)^2-\dfrac{183}{121}\le-\dfrac{183}{121}\)

Dấu bằng xảy ra 

\(\Leftrightarrow x+18=0\Leftrightarrow x=-18\)

b) \(B=\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\)

Ta có : \(\left(x+\dfrac{1}{3}\right)^2+5\ge5\)

\(\Leftrightarrow\dfrac{4}{\left(x+\dfrac{1}{3}\right)^2+5}\le\dfrac{4}{5}\)

Dấu bằng xảy ra 

\(\Leftrightarrow x+\dfrac{1}{3}=0\Leftrightarrow x=-\dfrac{1}{3}\)

\(\sqrt{\dfrac{9}{7-4\sqrt{3}}}-\sqrt{\dfrac{4}{7+4\sqrt{3}}}\)

\(=\sqrt{\dfrac{9\left(7+4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}-\sqrt{\dfrac{4\left(7-4\sqrt{3}\right)}{7^2-\left(4\sqrt{3}\right)^2}}\)

\(=\dfrac{3\sqrt{7+4\sqrt{3}}-2\sqrt{7-4\sqrt{3}}}{\sqrt{49-48}}\)

\(=3\sqrt{4+2.2.\sqrt{3}+3}-2\sqrt{4-2.2.\sqrt{3}+3}\)

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

\(=3.\left|2+\sqrt{3}\right|-2.\left|2-\sqrt{3}\right|\)

\(=6+3\sqrt{3}-2+2\sqrt{3}\)

\(=4+5\sqrt{3}\)

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 

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

Sơ đồ : 

Tuổi bố : |.....|.....|.....|

Tuổi con : |.....|

Tổng số phần bằng nhau là : 3 + 1 = 4 ( phần )

Tuổi con hiện nay là : 48 : 4 x 1 = 12 ( tuổi )

Số cây tháng này đội A trồng được là :

1400 + 1400 x 12 % = 1568 ( cây ) 

Đáp số : .........

Phần trăm số sảm phẩm đạt chuẩn là : 

\(\dfrac{80}{120}\times100\approx66,7\%\)

a) \(x^{15}=x\)

\(\Leftrightarrow x\left(x^{14}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^{14}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)

b) \(\left(x-5\right)^6=\left(x-5\right)^4\)

\(\Leftrightarrow\left(x-5\right)^6-\left(x-5\right)^4=0\)

\(\Leftrightarrow\left(x-5\right)^4\left[\left(x-5\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=0\\x-5=1\\x-5=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=6\\x=4\end{matrix}\right.\)

c) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)

\(\Leftrightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)

\(\Leftrightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-15=0\\2x-15=1\\2x-15=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{15}{2}\\x=8\\x=7\end{matrix}\right.\)

Điều kiện : \(x\ge-\dfrac{3}{2}\)

\(x^2+4x+5=2\sqrt{2x+3}\)

\(\Leftrightarrow x^2+4x+4+1=2\sqrt{2x+4-1}\)

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

Đặt \(x+2=t\left(t\ge\dfrac{1}{2}\right)\)

Ta có phương trình :

\(t^2+1=2\sqrt{2t-1}\)

\(\Leftrightarrow\left(t^2+1\right)^2=4\left(2t-1\right)\)

\(\Leftrightarrow t^4+2t^2+1-8t+4=0\)

\(\Leftrightarrow t^4+2t^2-8t+5=0\)

\(\Leftrightarrow t=1\)

\(\Leftrightarrow x+2=1\)

\(\Leftrightarrow x=-1\left(tm\right)\)

Vậy \(\Leftrightarrow x=-1\left(tm\right)\)

Điều kiện : \(x\le5\)

+) Với \(2x-7< 0\Leftrightarrow x< \dfrac{7}{2}\Rightarrow VP< 0\)

=> Phương trình vô nghiệm 

+) Với \(2x-7>0\Leftrightarrow x>\dfrac{7}{2}\Rightarrow VP>0\)

Ta có phương trình :

\(5-x=\left(2x-7\right)^2\)

\(\Leftrightarrow5-x=4x^2-28x+49\)

\(\Leftrightarrow4x^2-27x+44=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\left(tm\right)\\x=\dfrac{11}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy x = 4

Điều kiện : \(x\ne1\)

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

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

\(\Leftrightarrow\sqrt{x^2+2x+3}=x^2-x+3x-3\)

\(\Leftrightarrow\sqrt{x^2+2x+3}=x^2+2x-3\)

+) Với \(-3< x< 1\) thì VP < 0 => Phương trình vô nghiệm

+) Với \(\left[{}\begin{matrix}x\le-3\\x>1\end{matrix}\right.\) thì VP > 0 , lúc này ta có phương trình : 

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

\(\Leftrightarrow x^2+2x+3=\left(x^2+2x+1-4\right)^2\)

\(\Leftrightarrow x^2+2x+3=\left(x+1\right)^4-8\left(x+1\right)^2+16\)

\(\Leftrightarrow\left(x+1\right)^2+2=\left(x+1\right)^4-8\left(x+1\right)^2+16\)

\(\Leftrightarrow\left(x+1\right)^4-9\left(x+1\right)^2+14=0\)

Đặt \(\left(x+1\right)^2=t\left(t>0\right)\) , ta có : 

\(t^2-9t+14=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=7\\t=2\end{matrix}\right.\)

+) \(t=7\Rightarrow\left(x+1\right)^2=7\Leftrightarrow x=\sqrt{7}-1\left(tm\right)\)

+) \(t=2\Leftrightarrow\left(x+1\right)^2=2\Leftrightarrow x=\sqrt{2}-1\left(ktm\right)\)

Vậy \(x=\sqrt{7}-1\)