Bài 1:
\(B=\left(x-2\right)\left(x-5\right)\left(x^2-7x-10\right)\)
\(B=\left(x^2-7x+10\right)\left(x^2-7x-10\right)\)
\(B=\left(x^2-7x\right)^2-10^2\)
\(B=\left(x^2-7x\right)^2-100\)
Vì \(\left(x^2-7x\right)^2\ge0\) với mọi x
=> \(\left(x^2-7x\right)-100\ge-100\)
=> Bmin = -100 <=> \(x^2-7x=0\)
\(\Rightarrow x\left(x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-7=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=7\end{matrix}\right.\)
Vậy Bmin = -100 <=> x = 0; x = 7
Bài 2:
\(A=11-10x-x^2\)
\(A=-\left(x^2+10+11\right)\)
\(A=-\left(x^2+2.x.5+5^2-14\right)\)
\(A=-\left(x^2+2.x.5+5^2\right)+14\)
\(A=-\left(x+5\right)^2+14\)
Vì \(\left(x+5\right)^2\ge0\) với mọi x
\(\Rightarrow-\left(x+5\right)^2\le0\) với mọi x
\(\Rightarrow-\left(x+5\right)^2+14\le0\)
\(\Rightarrow\) Amax = 14 <=> x + 5 = 0
=> x = -5
Vậy Amax = 14 <=> x = -5
\(B=|x-4|\left(2-|x-4|\right)\)
Ta có hai trường hợp
TH1: \(x\ge4\)
\(\Rightarrow B=\left(x-4\right)\left(2-x+4\right)\)
\(\Rightarrow B=\left(x-4\right)\left(6-x\right)\)
\(\Rightarrow B=-x^2+10x-24\)
\(\Rightarrow B=-\left(x^2-2.x.5+5^2-1\right)\)
\(\Rightarrow-\left(x-5\right)^2+1\)
Vì \(\left(x-5\right)^2\ge0\) với mọi x
\(\Rightarrow-\left(x-5\right)^2\le0\)
\(\Rightarrow-\left(x-5\right)^2+1\le1\)
=> Bmax = 1 <=> x - 5 = 0
=> x = 5
Vậy Bmax = 1 <=> x = 5
Bài 2 câu b mình ghi thiếu một trường hợp rồi, bổ sung đây nhé
TH2: x < 4
\(\Rightarrow B=\left(4-x\right)\left(2-4+x\right)\)
\(\Rightarrow B=\left(4-x\right)\left(x-2\right)=-x^2+6x-8\)
\(\Rightarrow B=-\left(x^2-6x+9\right)+1\)
\(\Rightarrow B=-\left(x-3\right)^2+1\)
Vì \(\left(x-3\right)^2\ge0\) với mọi x
\(\Rightarrow-\left(x-3\right)^2\le0\)
\(\Rightarrow-\left(x-3\right)^2+1\le1\)
=> Bmax = 1 <=> x - 3 = 0
=> x = 3
Vậy Bmax = 1 <=> x = 3