1) \(\left|1,4+x\right|\ge0\Leftrightarrow-\left|1,4+x\right|\le0\Rightarrow\left|1,4+x\right|-2\le-2\Leftrightarrow A\le-2\Rightarrow MaxA=-2\Leftrightarrow x=-1,4\)

\(\left|5x-2\right|\ge0\Leftrightarrow-\left|5x-2\right|\le0;\left|3y+12\right|\ge0\Leftrightarrow-\left|3y+12\right|\le0\Rightarrow4-\left|5x-2\right|-\left|3y+12\right|\le4\Rightarrow B\le4\Rightarrow MaxB=4\)

<=> x=2/5 và y=-4

Bài 1 :A có GTLN <=> -|1,4 + x| có GTLN

=> x không tồn tại.

Bài 2 : B có GTLN <=>  | 5x - 2 | - | 3y + 12 | có GTNN

<=>  | 5x - 2 | - | 3y + 12 | = 0

Vậy GTLN của B = 4 - 0 = 4

\(=4\)

1) \(P=-2x^2-12x=-2\left(x^2+6x+9\right)+18=-2\left(x+3\right)^2+18\le18\)

\(maxP=18\Leftrightarrow x=-3\)

2) \(Q=-5x^2+10x=-5\left(x^2-2x+1\right)+5=-5\left(x-1\right)^2+5\le5\)

\(maxQ=5\Leftrightarrow x=1\)

3) \(A=-3x^2+12x-6=-3\left(x^2-4x+4\right)+6=-3\left(x-2\right)^2+6\le6\)

\(maxA=6\Leftrightarrow x=2\)

4) \(B=-2x^2-24x+12=-2\left(x^2+12x+36\right)+84=-2\left(x+6\right)^2+84\le84\)

\(maxB=84\Leftrightarrow x=-6\)

Ta có:

\(M=3x\left(x-5y\right)+\left(y-5x\right)\left(-3y\right)-3\left(x^2-y^2\right)-1\)

\(M=3x^2-15xy-3y^2+15xy-3x^2+3y^2\)

\(M=0\left(đpcm\right)\)

M=3x²-15xy-3y²+15xy-3x²+3y²-1

M=-1

Ta có 10 - 5x = 0

=> -5x = 0 - 10

=> -5x = -10

=> x = -10 : -5

=> x = 2

Vậy đa thức B (x) co nghiệm là 2

\(\Leftrightarrow B\left(x\right)=10-5x=0\\ \Leftrightarrow B\left(x\right)=-5x=-10\\ \Leftrightarrow B\left(x\right)=x=2\)

vậy...

2) \(P=\frac{4}{2x^2+2xy+y^2+5x+20}=\frac{4}{\left(x^2+2xy+y^2\right)+\left(x^2+5x+\frac{25}{4}\right)+\frac{75}{4}}\)

\(=\frac{4}{\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}}\)

Để P đạt GTLN 

=> Mẫu thức đạt GTNN

mà \(\left(x+y\right)^2+\left(x+\frac{5}{2}\right)^2+\frac{75}{4}\ge\frac{75}{4}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\x+\frac{5}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{5}{2}\\y=\frac{5}{2}\end{cases}}\)

Thay x = -5/2 và y = 5/2 vào P 

Khi đó P = \(\frac{4}{\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\left(-\frac{5}{2}+\frac{5}{2}\right)^2+\frac{75}{4}}=\frac{4}{\frac{75}{4}}=\frac{16}{75}\)

Vậy Max P = 16/75 <=> x = -5/2 ; y = 5/2

1) Ta có P = x² + 2xy + 3y² + 5y + 10

= (x² + 2xy + y²) + (2y² + 5y + 10) 

= \(\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+5\right)=\left(x+y\right)^2+2\left(y^2+\frac{5}{2}y+\frac{25}{16}+\frac{55}{16}\right)\)

= \(\left(x+y\right)^2+2\left(y+\frac{5}{4}\right)^2+\frac{55}{8}\ge\frac{55}{8}\)

Dấu "=" xảy ra <=> \(\hept{\begin{cases}x+y=0\\y+\frac{5}{4}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{5}{4}\\y=-\frac{5}{4}\end{cases}}\)

Vạy Min P = 55/8 <=> x = 5/4 ; y = -5/4 

Mình làm lại nhé !

D= 4 - |5x - 2| - |3y + 12|

Ta có : 5x - 2 > 0 => -|5x -2|<0

3x + 12 >0 => -|3x + 12 |< 0

=> 4 - |5x -2 | - |3y + 12 |>4 hay D >4

\(\left[{}\begin{matrix}5x-2=0\\3y+12=0\end{matrix}\right.\)

Sau đó tính ra nhé !

Chúc bạn học tốt !

D= 4 - |5x - 2| - |3y + 12|

Ta có : 5x - 2 > 0 => -|5x -2|<0

3x + 12 >0 => -|3x + 12 |< 0

=> 4 - |5x -2 | - |3y + 12 |>4 hay D >4

=>\(\left[{}\begin{matrix}5x-2=0\\3y+12=0\end{matrix}\right.\)

=>

\(A=-\left|x\right|\le0\)

Dấu "=" xảy ra khi:

\(x=0\)

\(B=4-\left|5x-2\right|\le4\)

Dấu "=" xảy ra khi:

\(\left|5x-2\right|=0\Rightarrow x=\dfrac{2}{5}\)

\(C=\left|x-3\right|-\left|5-x\right|=\left|x-3\right|-\left|x-5\right|\)

\(C\le\left|x-3-x-5\right|=8\)

Dấu "=" xảy ra khi:

\(3\le x\le5\)

\(D=4-\left|5x-2\right|-\left|3y+12\right|\le4\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|5x-2\right|=0\\\left|3y+12\right|=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{2}{5}\\y=-4\end{matrix}\right.\)

\(E=5-3\left(2x-1\right)\le5\)

Dấu "=" xảy ra khi:

\(2x-1=0\Leftrightarrow x=\dfrac{1}{2}\)