Tìm GTNN
a, A = \(\frac{5}{-x^2-2x-3}\)
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a: \(=\sqrt{x-3-2\sqrt{x-3}+3}\)
\(=\sqrt{x-3-2\sqrt{x-3}+1+2}=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}>=\sqrt{2}\)
Dấu = xảy ra khi x-3=1
=>x=4
\(A=\left|3-x\right|+8\ge8\)
\(minA=8\Leftrightarrow x=3\)
\(B=\left|x+2\right|-4\ge-4\)
\(minB=-4\Leftrightarrow x=-2\)
\(A=\dfrac{x^2-4x+1}{x^2}=\dfrac{1}{x^2}-\dfrac{4}{x}+1=\left(\dfrac{1}{x^2}-\dfrac{4}{x}+4\right)-3=\left(\dfrac{1}{x}-2\right)^2-3\ge-3\)
\(A_{min}=-3\) khi \(x=\dfrac{1}{2}\)
2) \(A=-x^2-y^2+2x-6y+9=-\left(x^2-2x+1\right)-\left(y^2+6y+9\right)+19=-\left(x-1\right)^2-\left(y+3\right)^2+19\)
\(maxA=19\Leftrightarrow\)\(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
a) Đặt \(x-1=a\)
\(pt\Leftrightarrow\frac{13}{a}+\frac{5}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2a}=\frac{6}{3a}\)
\(\Leftrightarrow\frac{31}{2}=2\)(vô lí)
Vậy pt vô nghiệm
a) \(\frac{13}{x-1}+\frac{5}{2x-2}=\frac{6}{3x-3}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{6}{3\left(x-1\right)}\)
\(\frac{13}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2\left(x-1\right)}=\frac{2}{x-1}\)
\(\frac{31}{2}=2\)
=> không có x thỏa mãn đề bài.
b) \(\frac{1}{x-1}+\frac{-2}{3}\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\frac{1}{x-1}+\frac{-2}{3}.\frac{-9}{20}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}-\frac{-18}{60}=\frac{5}{2\left(1-x\right)}\)
\(\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2\left(1-x\right)}\)
\(10\left(1-x\right)+3\left(x-1\right)\left(1-x\right)=25\left(x-1\right)\)
\(7-4x-3x^2=25x-25\)
\(7-4x-3x^2-25x+25=0\)
\(32-29x-3x^2=0\)
\(3x^2+29x-30=0\)
\(3x^2+32x-3x-32=0\)
\(x\left(3x+32\right)-\left(3x+32\right)=0\)
\(\left(3x+32\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}3x+32=0\\x-1=0\end{cases}}\)
\(\orbr{\begin{cases}x=-\frac{32}{3}\\x=1\end{cases}}\)
\(A=\left|x-201\right|+\left|x-204\right|=\left|x-201\right|+\left|204-x\right|\ge\left|x-201+204-x\right|=\left|3\right|=3\)
\(minA=3\Leftrightarrow\left(x-201\right)\left(204-x\right)\ge0\Leftrightarrow204\ge x\ge201\)
\(a,\frac{5}{x-2}=\frac{3}{2x+1}\)
=>\(5\left(2x+1\right)=3\left(x-2\right)\)
=>\(10x+5=3x-6\)
=>\(10x-3x=-6-5\)
=>\(7x=-11\)
=> \(x=-\frac{11}{7}\)
b,\(\frac{2x-3}{5}=\frac{x+2}{2}\)
=>\(2\left(2x-3\right)=5\left(x+2\right)\)
=>\(4x-6=5x+10\)
=>\(4x-5x=10+6\)
=>\(-x=16\)
=>\(x=-16\)
Chúc Bạn May Mắn
A = \(\frac{5}{-x^2-2x-3}=\frac{5}{-\left(x+1\right)^2-2}\)
Do: \(-\left(x+1\right)^2-2\le-2\)=> \(\frac{5}{-\left(x+1\right)^2-2}\ge-\frac{5}{2}\)
Dấu "=" xảy ra <=> x + 1 = 0 <=> x = -1
Vậy MinA = -5/2 <=> x = -1
Bài làm
Ta có : -x2 - 2x - 3
= -x2 - 2x - 1 - 2
= -( x2 + 2x + 1 ) - 2
= -( x - 1 )2 - 2 ≤ -2 ∀ x
=> \(\frac{1}{-\left(x-1\right)^2-2}\ge-\frac{1}{2}\)
=> \(\frac{5}{-\left(x-1\right)^2-2}\ge-\frac{5}{2}\)
hay A ≥ -5/2
Đẳng thức xảy ra khi x = 1
=> MinA = -5/2 <=> x = 1