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\(A=x^2-3x+5\)
\(=x^2-3x+\frac{9}{4}+\frac{11}{4}\)
\(=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)
\(\left(x-\frac{3}{2}\right)^2\ge0\Rightarrow A\ge\frac{11}{4}\)
Dấu "=" xảy ra khi \(x-\frac{3}{2}=0\Rightarrow x=\frac{3}{2}\)
Vậy Min A = \(\frac{11}{4}\Leftrightarrow x=\frac{3}{2}\)
a) \(A=x^2-3x+5\)
\("="\Leftrightarrow x=\frac{11}{4}\Rightarrow x=\frac{3}{2};\frac{11}{4}\)
b) \(B=\left(2x-1\right)^2+\left(x+2\right)^2\)
\("="\Leftrightarrow x=5\Rightarrow x=0;5\)
c) \(C=4x-x^2+3\)
\("="\Leftrightarrow x=7\Rightarrow x=2;7\)
d) \(D=x^4+x^2+2\)
\("="\Leftrightarrow x=2\Rightarrow x=0;2\)
1.\(A=\frac{2x^2-16x+41}{x^2-8x+22}\) \(=\frac{2\left(x^2-8x+22\right)-3}{x^2-8x+22}=2-\frac{3}{\left(x-4\right)^2+6}\ge\frac{1}{2}\)
Dấu '' = '' xảy ra khi x = 4.
Vậy MinA= \(\frac{1}{2}\) tại x = 4.
a/ \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)=\left[\left(x+1\right)\left(x-6\right)\right].\left[\left(x-2\right)\left(x-3\right)\right]\)
\(=\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\ge-36\)
Suy ra Min A = -36 <=> \(x^2-5x=0\Leftrightarrow x\left(x-5\right)=0\) \(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b/ \(B=19-6x-9x^2=-9\left(x-\frac{1}{3}\right)^2+20\le20\)
Suy ra Min B = 20 <=> x = 1/3
a) \(A=\left(x+1\right)\left(x-2\right)\left(x-3\right)\left(x-6\right)\)
\(=\left[\left(x+1\right)\left(x-6\right)\right]\left[\left(x-2\right)\left(x-3\right)\right]\)
\(\left(x^2-5x-6\right)\left(x^2-5x+6\right)=\left(x^2-5x\right)^2-36\)
Vì \(\left(x^2-5x\right)^2\ge0\)
=> \(\left(x^2-5x\right)^2-36\ge-36\)
Vậy GTNN của A là -36 khi \(x^2-5x=0\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b) \(B=19-6x-9x^2=-\left(9x^2+6x+1\right)+20=-\left(3x+1\right)^2+20\)
Vì \(-\left(3x+1\right)^2\le0\)
=> \(-\left(3x+1\right)+20\le20\)
Vậy GTLN của B là 20 khi \(x=-\frac{1}{3}\)
a) \(27x^3+27x^2+9x+1=64\)
\(\Rightarrow27x^3+27x^2+9x-63=0\)
\(\Rightarrow27x^3-27x^2+54x^2-54x+63x-63=0\)
\(\Rightarrow27x^2\left(x-1\right)+54x\left(x-1\right)+63\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(27x^2+54x+63\right)=0\)
\(\Rightarrow\left(x-1\right).9\left(3x^2+6x+7\right)=0\)
\(\Rightarrow\left(x-1\right)\left(3x^2+6x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\3x^2+6x+7=0\end{matrix}\right.\)
Mà ta có:
\(3x^2+6x+7\)
\(=3\left(x^2+2x+\dfrac{7}{3}\right)\)
\(=3\left(x^2+2x+1-1+\dfrac{7}{3}\right)\)
\(=3\left(x+1\right)^2+4\)
Vì \(3\left(x+1\right)^2\ge0\) với mọi x
\(\Rightarrow3\left(x+1\right)^2+4\ge4\)
\(\Rightarrow3x^2+6x+7\) vô nghiệm
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
b) \(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(\Rightarrow x^3-6x^2+12x-8-x^3+6x^2=4\)
\(\Rightarrow12x-8=4\)
\(\Rightarrow12x=12\)
\(\Rightarrow x=1\)
c) \(\left(x-1\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+3\left(x-2\right)\left(x+2\right)=2\)
\(\Rightarrow x^3-3x^2+3x-1-\left(x^3+3^3\right)+3\left(x^2-2^2\right)=2\)
\(\Rightarrow x^3-3x^2+3x-1-x^3-9+3x^2-12=2\)
\(\Rightarrow3x-22=2\)
\(\Rightarrow3x=24\)
\(\Rightarrow x=8\)
a) (x + 2)(x + 3) - (x - 2)(x + 5) = 6
x2 + 3x + 2x + 6 - (x2 + 5x - 2x - 10) = 6
x2 + 5x + 6 - x2 - 3x + 10 = 6
2x +16 = 6
\(\Rightarrow\) 2x = -10
\(\Rightarrow\) x = -5
b) (3x + 2)(2x + 9) - (x + 2)(6x + 1) = (x + 1) - (x - 6)
6x2 + 27x + 4x + 18 - (6x2 + x + 12x + 2) = x + 1 - x + 6
6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
18x + 16 = 7
\(\Rightarrow\) 18x = -9
\(\Rightarrow\) x = -0.5
c) 3(2x - 1)(3x - 1) - (2x - 3)(9x - 1) = 0
3(6x2 - 2x - 3x + 1) - (18x2 - 2x - 27x + 3) = 0
3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
18x2 - 15x + 3 - 18x2 + 29x - 3 = 0
14x = 0
\(\Rightarrow\) x = 0