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B = x x 3 + 1 + 1 − x x 2 − x + 1 + 1 x + 1 = x (x + 1)(x 2 − x + 1) + (1 − x)(x + 1) (x + 1)(x 2 − x + 1) + 1.(x 2 − x + 1) (x + 1)(x 2 − x + 1) = x + 1 − x 2 + x 2 − x + 1 (x + 1)(x 2 − x + 1) = 2 x 3 + 1
\(A=x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)
Vậy A > 0 với mọi x.
\(B=x^2-2xy+y^2+1\)
\(=\left(x-y\right)^2+1\)
\(\left(x-y\right)^2\ge0\)
\(\Rightarrow\left(x-y\right)^2+1\ge1>0\)
Vậy B > 0 với mọi x, y.
\(M=x^2-6x+12\)
\(=x^2-6x+9+3\)
\(=\left(x-3\right)^2+3\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+3\ge3\)
\(MinB=3\Leftrightarrow x=3\)
\(\left(x+3\right)^2+\left(x-2\right)\left(x+2\right)-2\left(x-1\right)^2=7\)
\(x^2+6x+9+x^2-4-2\left(x^2-2x+1\right)=7\)
\(2x^2+6x+5-2x^2+4x-2=7\)
\(10x=7+3\)
\(10x=10\)
\(x=1\)
\(x^2+x=0\)
\(x\left(x+1\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x+1=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=-1\end{array}\right.\)
\(x^3-\frac{1}{4}x=0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\left[\begin{array}{nghiempt}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{array}\right.\)
\(\left(x+10\right)^2-\left(x^2+2x\right)\)
\(=x^2+20x+100-x^2-2x\)
\(=18x+100\)
\(\left(x+2\right)\left(x-2\right)+\left(x-1\right)\left(x^2+x+1\right)-x\left(x^2+x\right)\)
\(=x^2-4+x^3-1-x^3-x^2\)
\(=-5\)
a, Do \(x=-3\)\(=>A=\frac{x+3}{x+2}=\frac{-3+3}{-3+2}=\frac{0}{-1}=0\)
Vậy A = 0 khi x = -3
b, Ta có : \(B=\frac{x}{x+1}+\frac{2}{x-1}-\frac{4}{x^2-1}=\frac{x\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}+\frac{2\left(x+1\right)}{\left(x+1\right)\left(x-1\right)}-\frac{4}{x^2-1}\)
\(=\frac{x^2-x+2x-2}{x^2-1}=\frac{x\left(x-1\right)+2\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}=\frac{\left(x+2\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\)
\(=\frac{x+2}{x+1}\)(đpcm)
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Bài 2:
(1 + x)3 + (1 - x)3 - 6x(x + 1) = 6
<=> x3 + 3x2 + 3x + 1 - x3 + 3x2 - 3x + 1 - 6x2 - 6x = 6
<=> -6x + 2 = 6
<=> -6x = 6 - 2
<=> -6x = 4
<=> x = -4/6 = -2/3
Bài 3:
a) (7x - 2x)(2x - 1)(x + 3) = 0
<=> 10x3 + 25x2 - 15x = 0
<=> 5x(2x - 1)(x + 3) = 0
<=> 5x = 0 hoặc 2x - 1 = 0 hoặc x + 3 = 0
<=> x = 0 hoặc x = 1/2 hoặc x = -3
b) (4x - 1)(x - 3) - (x - 3)(5x + 2) = 0
<=> 4x2 - 13x + 3 - 5x2 + 13x + 6 = 0
<=> -x2 + 9 = 0
<=> -x2 = -9
<=> x2 = 9
<=> x = +-3
c) (x + 4)(5x + 9) - x2 + 16 = 0
<=> 5x2 + 9x + 20x + 36 - x2 + 16 = 0
<=> 4x2 + 29x + 52 = 0
<=> 4x2 + 13x + 16x + 52 = 0
<=> 4x(x + 4) + 13(x + 4) = 0
<=> (4x + 13)(x + 4) = 0
<=> 4x + 13 = 0 hoặc x + 4 = 0
<=> x = -13/4 hoặc x = -4
Đáp án cần chọn là: D