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a: Ta có: \(A=\left(x+2\right)\left(x-4\right)+\left(x+1\right)\left(x-6\right)\)
\(=x^2-4x+2x-8+x^2-6x+x-6\)
\(=2x^2-7x-14\)
b: \(B=\left(2a-b\right)\left(4a^2+2ab+b^2\right)=8a^3-b^3\)
c: \(C=\left(2+x\right)\left(2-x\right)\left(x+4\right)\)
\(=\left(4-x^2\right)\left(x+4\right)\)
\(=4x+16-x^3-4x^2\)
\(a,=x^2-4-x^2+2x+3=2x-1\\ b,=x^3+3x^2-5x-15+x^2-x^3+4x-4x^2=-x-15\\ c,=2x^2+3x-10x-15-2x^2+6x+x+7=-8\\ d,=\left(2x+1+3x-1\right)^2=25x^2\)
a) (x+2)2+x(x-4)
=x2+4x+4+x2-4x
=2x2+4
b)(x-3)2-(x+3)(x-4)
=x2-6x+9-x2+4x-3x+12
=-5x+12
c) (3x+1)2+3x(2-4x)
=9x2+6x+1+6x-12x2
=-3x2+12x+1
d) (2x-4y)2-(2x-3)(2x-3y)
=4x2-16xy+16y2-4x2+6xy+6x-9y
=16y2-10xy+6x-9y
Tìm GTNN của : \(x^2-4x+3\)
\(x^2-4x+3=x^2-4x+4-1=\left(x-2\right)^2-1\)
Vì \(\left(x-2\right)^2\ge0\) nên \(\left(x-2\right)^2-1\ge-1\)
Vậy GTNN của biểu thức là -1 . Dấu bằng xảy ra khi x = 2
2) \(\left(2x-1\right)\left(x+5\right)-3.\left(x-2\right)^2+\left(x+4\right)\left(x-4\right)\)
\(=2x^2+10x-x-5-3.\left(x^2-4x+4\right)+x^2-16\)
\(=2x^2+9x-5-3x^2+12x-12+x^2-16=21x-33\)
Khi x = -2 thì A = 21 . (-2) -33 = -75
\(A=x^2-6x+10\)
\(=x^2-6x+9+1\)
\(=\left(x-3\right)^2+1\)
\(\left(x-3\right)^2\ge0\)
\(\Rightarrow\left(x-3\right)^2+1\ge1>0\)
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) MTC = (x -2)(x + 2). Ta rút gọn được M = 1 x − 2
b) Gợi ý: x 2 + 5 x + 6 = ( x + 2 ) ( x + 3 ) ; x 2 + x − 12 = ( x − 3 ) ( x + 4 )
Ta có N = ( x + 2 ) ( x + 3 ) ( x − 3 ) ( x + 4 ) : ( x + 2 ) 2 x ( x − 3 ) = x ( x + 3 ) ( x + 2 ) ( x + 4 )