b)  (2x-6)(x+4)=0

c)  (x-3)(x+4)<0

d)  (x+2)(X-5)>0

bạn đăg tách ra cho m.n cùng giúp nhé

Bài 2 : 

a, \(A=\left|2x-4\right|+2\ge2\)

Dấu ''='' xảy ra khi x = 2 

Vậy GTNN A là 2 khi x = 2 

b, \(B=\left|x+2\right|-3\ge-3\)

Dấu ''='' xảy ra khi x = -2 

Vậy GTNN B là -3 khi x = -2 

Giups mik vs

Bài 4:

\(A=2x^2-15\ge-15\\ A_{min}=-15\Leftrightarrow x=0\\ B=2\left(x+1\right)^2-17\ge-17\\ B_{min}=-17\Leftrightarrow x=-1\)

Bài 5:

\(A=-x^2+14\le14\\ A_{max}=14\Leftrightarrow x=0\\ B=25-\left(x-2\right)^2\le25\\ B_{max}=25\Leftrightarrow x=2\)

\(A=-x^2+2xy-4y^2+2x+10y-3\)

\(=-x^2+2xy-y^2+2x-2y-1-3y^2+12y-12+10\)

\(=-\left(x^2-2xy+y^2-2x+2y+1\right)-3\left(y^2-4y+4\right)+10\)

\(=-\left(x-y-1\right)^2-3\left(y-2\right)^2+10< =10\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y-1=0\\y-2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=y+1=3\end{matrix}\right.\)

\(B=-4x^2-5y^2+8xy+10y+12\)

\(=-4x^2+8xy-4y^2-y^2+10y-25+37\)

\(=-4\left(x^2-2xy+y^2\right)-\left(y^2-10y+25\right)+37\)

\(=-4\left(x-y\right)^2-\left(y-5\right)^2+37< =37\)

Dấu '=' xảy ra khi \(\left\{{}\begin{matrix}x-y=0\\y-5=0\end{matrix}\right.\)

=>x=y=5

Bắt quả tang dũng nhá!~

\(I=-\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)+2021\)

\(=-\left(x^2+5x-6\right)\left(x^2+5x+6\right)+2021\)

\(=-\left[\left(x^2+5x\right)^2-6^2\right]+2021\)

\(=-\left(x^2+5x\right)^2+2057\le2057\)

\(I_{max}=2057\) khi \(x^2+5x=0\)

\(K=-\left(x-2\right)\left(x-7\right)\left(x-5\right)\left(x-4\right)+102\)

\(=-\left(x^2-9x+14\right)\left(x^2-9x+20\right)+102\)

\(=-\left(x^2-9x+14\right)\left(x^2+9x+14+6\right)+102\)

\(=-\left[\left(x^2-9x+14\right)^2+6\left(x^2-9x+14\right)\right]+102\)

\(=-\left[\left(x^2-9x+14\right)+6\left(x^2-9x+14\right)+9-9\right]+102\)

\(=-\left(x^2-9x+17\right)^2+111\le111\)

\(K_{max}=111\) khi \(x^2-9x+17=0\)

\(M=-\left(4x^2+4x+1\right)\left(16x^2+16x+3\right)-11\)

Đặt \(4x^2+4x+1=t\Rightarrow16x^2+16x=4t-4\)

\(\Rightarrow M=-t\left(4t-4+3\right)-11\)

\(M=-4t^2+t-11\)

\(M=-4\left(t-\dfrac{1}{8}\right)^2-\dfrac{175}{16}\le-\dfrac{175}{16}\)

\(M_{max}=-\dfrac{175}{16}\) khi \(t=\dfrac{1}{8}\)

đang cần gấp ạ

a) \(A=-x^2+2x=-\left(x^2-2x+1\right)+1=-\left(x-1\right)^2+1\le1\)

\(maxA=1\Leftrightarrow x=1\)

b) \(B=\left(2-3x\right)\left(3+2x\right)=-6x^2-5x+6=-6\left(x^2+\dfrac{5}{6}x+\dfrac{25}{144}\right)+\dfrac{169}{24}=-6\left(x+\dfrac{5}{12}\right)^2+\dfrac{169}{24}\le\dfrac{169}{24}\)

\(minB=\dfrac{169}{24}\Leftrightarrow x=-\dfrac{5}{12}\)

c) \(C=4xy-4x-2y-4x^2-2y^2-3=-\left[4x^2-4x\left(y-1\right)+\left(y-1\right)^2\right]+\left(y^2-4y+4\right)-6=\left(2x-y+1\right)^2+\left(y-2\right)^2-6\le-6\)

\(minC=-6\Leftrightarrow\)\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=2\end{matrix}\right.\)