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1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)
\(=x^3+27-x^3-54\)
=-27
2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
1. Đề bài sai, các biểu thức này chỉ có giá trị lớn nhất, không có giá trị nhỏ nhất
2.
\(A=\left(2x\right)^3-3^3-\left(8x^3+2\right)\)
\(=8x^3-27-8x^3-2\)
\(=-29\)
\(B=x^3+9x^2+27x+27-\left(x^3+9x^2+27x+243\right)\)
\(=27-243=-216\)
sửa đề lại thành tìm Max nhé1, vì mấy ý này ko có min
\(1,=>D=-\left(x^2-4x-3\right)=-\left(x^2-2.2x+4-7\right)\)
\(=-[\left(x-2\right)^2-7]=-\left(x-2\right)^2+7\le7\)
dấu"=" xảy ra<=>x=2
2, \(E=-2\left(x^2-x+\dfrac{5}{2}\right)=-2[x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+\dfrac{9}{4}]\)
\(=-2[\left(x-\dfrac{1}{2}\right)^2+\dfrac{9}{4}]\le-\dfrac{9}{2}\) dấu"=" xảy ra<=>x=1/2
3, \(F=-\left(x^2+4x-20\right)=-\left(x^2+2.2x+4-24\right)\)
\(=-[\left(x+2\right)^2-24]\le24\) dấu"=" xảy ra<=>x=-2
Bài 3:
a) Ta có: \(A=25x^2-20x+7\)
\(=\left(5x\right)^2-2\cdot5x\cdot2+4+3\)
\(=\left(5x-2\right)^2+3>0\forall x\)(đpcm)
d) Ta có: \(D=x^2-2x+2\)
\(=x^2-2x+1+1\)
\(=\left(x-1\right)^2+1>0\forall x\)(đpcm)
Bài 1:
a) Ta có: \(A=x^2-2x+5\)
\(=x^2-2x+1+4\)
\(=\left(x-1\right)^2+4\ge4\forall x\)
Dấu '=' xảy ra khi x=1
b) Ta có: \(B=x^2-x+1\)
\(=x^2-2\cdot x\cdot\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}\)
\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{2}\)
\(a,\Rightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1-3x^2=54\\ \Rightarrow26x=26\Rightarrow x=1\\ b,\Rightarrow x^3-9x^2+27x-27-x^3+27+6x^2+12x+6+3x^2=-33\\ \Rightarrow39x=-39\Rightarrow x=-1\)
\(a,=4x^2+3xy-y^2+4xy-4x^2=7xy-y^2\\ b,=x^2-9-x^3+3x+x^2-3=-x^3+2x^2+3x-12\\ c,=-2x^2+12x-18+5x^2+4x-1=3x^2+16x-19\\ d,=8x^3+1-3x^3+6x^2=5x^3+6x^2+1\\ e,=\left(3x^2+4x+15x+20\right):\left(3x+4\right)\\ =\left(3x+4\right)\left(x+5\right):\left(3x+4\right)\\ =x+5\\ f,=\left(x^3+4x^2-3x+3x^2+12x-9+3x+3\right):\left(x^2+4x-3\right)\\ =\left[\left(x^2+4x-3\right)\left(x+3\right)+3x+3\right]:\left(x^2+4x-3\right)\\ =x+3\left(dư.3x+3\right)\)
\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)
\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)
\(=4x^2-4x+5-8x^2+24x-18\)
\(=-4x^2+20x-13\)
e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)