\(a.A=\left(x-2\right)^2+\left(y+1\right)^2+1\ge1\forall x;y\) . " = " \(\Leftrightarrow x=2;y=-1\) 

b.\(B=7-\left(x+3\right)^2\le7\forall x\)  " = " \(\Leftrightarrow x=-3\)

c.\(C=\left|2x-3\right|-13\ge-13\forall x\)  " = " \(\Leftrightarrow x=\dfrac{3}{2}\)

d.\(D=11-\left|2x-13\right|\le11\forall x\)  " = " \(\Leftrightarrow x=\dfrac{13}{2}\)

\(...=0\) \(\Leftrightarrow\left(2x-5\right)^2-\left(x+2\right)^2=0\Leftrightarrow\left(x-7\right)\left(3x-3\right)=0\)  \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=1\end{matrix}\right.\)

\(...=0,36\left(550+126+324\right)=0,36.1000=360\)

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

Đa thức \(⋮\left(x-1\right)^2\Leftrightarrow a-2⋮\left(x-1\right)^2\Leftrightarrow a-2=k\left(x-1\right)^2\Leftrightarrow a=k\left(x-1\right)^2+2\)  

\(P=\sqrt{a^2+3ab^2}+\sqrt{b^2+3ba^2}\le\sqrt{\left(a^2+b^2\right)+3ab\left(a+b\right)}\sqrt{2}\)  

\(\le\sqrt{2+3.\dfrac{a^2+b^2}{2}.\sqrt{2\left(a^2+b^2\right)}}.\sqrt{2}\le\sqrt{2+3.\dfrac{2}{2}.\sqrt{2.2}}\sqrt{2}=2\sqrt{2}.\sqrt{2}=4\)

" = " \(\Leftrightarrow a=b=1\)

\(\dfrac{4x+3}{6}-\dfrac{6x-2}{8}=\dfrac{5x+4}{3}\)

\(\Leftrightarrow x\left(\dfrac{4}{6}-\dfrac{6}{8}-\dfrac{5}{3}\right)+\dfrac{3}{6}+\dfrac{2}{8}-\dfrac{4}{3}=0\)

\(\Leftrightarrow\dfrac{-7}{4}x-\dfrac{7}{12}=0\)  \(\Leftrightarrow x=-\dfrac{1}{3}\) 

 Vậy ... 

Câu 15 : \(P=tan^3a+cot^3a+tan^2a+cot^2a=\left(tana+cota\right)^3-3tana.cota\left(tana+cota\right)+\left(tana+cota\right)^2-2tana.cota\)

\(=3^3-3.3+3^2-2=25\) . Chọn A 

Câu 16 : Đặt a = \(sin^2x;b=cos^2x\Rightarrow a+b=1\)

Ta có :  \(3a^2+b^2=\dfrac{3}{4}\) \(\Leftrightarrow3a^2+\left(1-a\right)^2=\dfrac{3}{4}\Leftrightarrow4a^2-2a+\dfrac{1}{4}=0\Leftrightarrow\left(a-\dfrac{1}{4}\right)^2=0\Leftrightarrow a=\dfrac{1}{4}\) 

Suy ra : \(a^2=\dfrac{1}{16}\Rightarrow b^2=\dfrac{9}{16}\)

\(P=sin^4x+3cos^4x=a^2+3b^2=\dfrac{1}{16}+3.\dfrac{9}{16}=\dfrac{7}{4}\)  . Chọn C 

Giả sử bán kính hình tròn là r ( r > 0 ) ( cm ) 

Diện tích hình tròn ban đầu là \(S_1\)

Ta có : \(S_1=\pi r^2\) ;  \(S_2=\pi\left(r-20\%r\right)^2=\pi r^2\dfrac{16}{25}=\dfrac{16}{25}S_1=S_1-452,16\)

\(\Rightarrow\dfrac{9}{25}S_1=452,16\Rightarrow S_1=1256\left(cm^2\right)\)

a. \(x^2-\left(2m-6\right)+m-13=0\)  (1) 

\(\left(1\right):\Delta'=\left(3-m\right)^2-\left(m-13\right)=m^2-6m+9-m+13=m^2-7m+22\)

\(=\left(m-\dfrac{7}{2}\right)^2+\dfrac{39}{4}>0\forall m\)  => P/t luôn có 2 no p/b

b. Theo viet ta có : \(\left\{{}\begin{matrix}x_1+x_2=2m-6\\x_1x_2=m-13\end{matrix}\right.\)

\(A=x_1x_2-\left(x_1^2+x_2^2\right)=3x_1x_2-\left(x_1+x_2\right)^2=3\left(m-13\right)-\left(2m-6\right)^2\)

\(=3m-39-4m^2+24m-36\)  \(=-4m^2+27m-75\)  

Đoạn này bn tự tìm 

c . P/t có 2 no đối nhau \(\Leftrightarrow x_1+x_2=0\Leftrightarrow2m-6=0\Leftrightarrow m=3\)

Chiều rộng thửa ruộng : \(216:\left(3+5\right).3=81\left(m\right)\)

Chiều dài thửa ruộng : \(216-81=135\left(m\right)\)

Diện tích thửa ruộng : \(135.81=10935\left(m^2\right)\)

Đ/s : ...