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\(A=x\left(x+1\right)\left(x^2+x-4\right)=\left(x^2+x\right)\left(x^2+x-4\right)\)
Đặt \(x^2+x=a\) nên \(A=a\left(a-4\right)=a^2-4a+4-4=\left(a-2\right)^2-4\ge-4\)
Dấu "=" xảy ra \(\Leftrightarrow a-2=0\Leftrightarrow x^2+x-2=0\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Rightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
Vậy \(A_{min}=-4\) tại \(\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)
B;C tương tự
Tìm GTNN
a/ \(A=4x^2+7x+13=\left(4x^2+7x+\frac{49}{16}\right)+\frac{159}{16}=\left(2x+\frac{7}{4}\right)^2+\frac{159}{16}\ge\frac{159}{16}\)
b/ \(B=5-8x+x^2=\left(x^2-8x+16\right)-11=\left(x-4\right)^2-11\ge-11\)
c/ \(C=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)
\(=\left(x-1\right)\left(x+6\right)\left(x+2\right)\left(x+3\right)\)
\(=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)
\(=\left(x^2+5x\right)^2-36\ge-36\)
a, \(A=x^4-2x^3+2x^2-2x+3\)
\(=\left(x^4+2x^2+1\right)-\left(2x^3+2x\right)+2\)
\(=\left(x^2+1\right)^2-2x\left(x^2+1\right)+2\)
\(=\left(x^2+1\right)\left(x^2-2x+1\right)+2\)
\(=\left(x^2+1\right)\left(x-1\right)^2+2\)
Vì \(\hept{\begin{cases}x^2\ge0\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow\hept{\begin{cases}x^2+1\ge1\\\left(x-1\right)^2\ge0\end{cases}\Rightarrow}\left(x^2+1\right)\left(x-1\right)^2\ge0}\)
\(\Rightarrow A=\left(x^2+1\right)\left(x-1\right)^2+2\ge2\)
Dấu "=" xảy ra khi x = 1
Vậy Amin = 2 khi x = 1
b, \(B=4x^2-2\left|2x-1\right|-4x+5=\left(4x^2-4x+1\right)-2\left|2x-1\right|+4=\left(2x-1\right)^2-2\left|2x-1\right|+4\)
đề sai ko
c, \(C=4-x^2+2x=-\left(x^2-2x+1\right)+5=-\left(x-1\right)^2+5\)
Vì \(-\left(x-1\right)^2\le0\Rightarrow C=-\left(x-1\right)^2+5\le5\)
Dấu "=" xảy ra khi x=1
Vậy Cmin = 5 khi x = 1
2/
+) \(D=-x^2-y^2+x+y+3=-\left(x^2-x+\frac{1}{4}\right)-\left(y^2-y+\frac{1}{4}\right)+\frac{7}{2}=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\)
Vì \(\hept{\begin{cases}-\left(x-\frac{1}{2}\right)^2\le0\\-\left(y-\frac{1}{2}\right)^2\le0\end{cases}\Rightarrow-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2\le0}\Rightarrow D=-\left(x-\frac{1}{2}\right)^2-\left(y-\frac{1}{2}\right)^2+\frac{7}{2}\le\frac{7}{2}\)
Dấu "=" xảy ra khi x=y=1/2
Vậy Dmax=7/2 khi x=y=1/2
+) Đề sai
+)bài này là tìm min
\(G=x^2-3x+5=\left(x^2-3x+\frac{9}{4}\right)+\frac{11}{4}=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\ge\frac{11}{4}\)
Dấu "=" xảy ra khi x=3/2
Vậy Gmin=11/4 khi x=3//2
a, =[ x^2 - 2x. \(\left(\frac{1}{2}\right)^2\)+\(\left(\frac{1}{2}\right)^2]-\left(\frac{1}{2}\right)^2\)+ 5
= (x^2 - \(\frac{1}{2}\))^2 -\(\frac{1}{4}\)+5
= (x^2 - 1/2)^2 + 19/4 \(\ge\)19/4
Vậy GTNN là 19/4
a)
\(A=\left(x^2-4x+4\right)+1=\left(x-2\right)^2+1\)
CÓ: \(\left(x-2\right)^2\ge0\forall x\Rightarrow\left(x-2\right)^2+1\ge1\)
=> \(A\ge1\)
DẤU "=" XẢY RA <=> \(x=2\)
b)
\(2B=4x^2+6x+2=\left(2x+\frac{3}{2}\right)^2-0,25\)
CÓ: \(\left(2x+\frac{3}{2}\right)^2\ge0\forall x\Rightarrow\left(2x+\frac{3}{2}\right)^2-0,25\ge-0,25\)
DẤU "=" XẢY RA <=> \(2x+\frac{3}{2}=0\Leftrightarrow x=-\frac{3}{4}\)
c)
\(C=\left(2x+\frac{5}{4}\right)^2-\frac{73}{16}\ge-\frac{73}{16}\)
DẤU "=" XẢY RA <=> \(2x+\frac{5}{4}=0\Leftrightarrow x=-\frac{5}{8}\)
a. Ta có :
\(A=x^2-4x+5=x^2-4x+4+1=\left(x-2\right)^2+1\)
Vì \(\left(x-2\right)^2\ge0\forall x\)\(\Rightarrow\left(x-2\right)^2+1\ge1\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x-2\right)^2=0\Leftrightarrow x-2=0\Leftrightarrow x=2\)
b. \(B=2x^2+3x+1=2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\)
Vì \(\left(x+\frac{3}{4}\right)^2\ge0\forall x\)\(\Rightarrow2\left(x+\frac{3}{4}\right)^2-\frac{1}{8}\ge-\frac{1}{8}\)
Dấu "=" xảy ra \(\Leftrightarrow2\left(x+\frac{3}{4}\right)^2=0\Leftrightarrow x+\frac{3}{4}=0\Leftrightarrow x=-\frac{3}{4}\)
Vậy Bmin = - 1/8 <=> x = - 3/4
c. \(C=5x-3+4x^2=4\left(x+\frac{5}{8}\right)^2-\frac{73}{16}\)
Vì \(\left(x+\frac{5}{8}\right)^2\ge0\forall x\)\(\Rightarrow4\left(x+\frac{5}{8}\right)^2-\frac{73}{16}\ge-\frac{73}{16}\)
Dấu "=" xảy ra \(\Leftrightarrow4\left(x+\frac{5}{8}\right)^2=0\Leftrightarrow x+\frac{5}{8}=0\Leftrightarrow x=-\frac{5}{8}\)
Vậy Cmin = - 73/16 <=> x = - 5/8