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a) A = (-1.2axy^2)^3
A = (-1/2)^3.a^3x^3y^6
A = -1/8.a^3.x^3.y^6
- Hệ số: -1/8.a^3
các bn còn lại bn lm tương tự nha!
- Bậc: 9
\(\text{a) }ax-bx+ab-x^2\\ \\=\left(ax+ab\right)-\left(x^2+bx\right)\\ \\=a\left(x+b\right)-x\left(x+b\right)\\ \\=\left(a-x\right)\left(x+b\right)\\ \)
\(\text{b) }x^2-y^2+4x+4\\ \\ =\left(x^2+4x+4\right)-y^2\\ \\ =\left(x+2\right)^2-y^2\\ \\ =\left(x+2+y\right)\left(x+2-y\right)\\ \)
\(\text{c) }ax+ay-3x-3y\\ \\=\left(ax+ay\right)-\left(3x+3y\right)\\ \\ =a\left(x+y\right)-3\left(x+y\right)\\ \\=\left(a-3\right)\left(x+y\right)\\ \)
\(\text{d) }x^3+x^2+x+1\\ \\=\left(x^3+x^2\right)+\left(x+1\right)\\ \\=x^2\left(x+1\right)+\left(x+1\right)\\ \\=\left(x^2+1\right)\left(x+1\right)\\ \)
\(\text{e) }x^3-3x^2+3x-9\\ \\=\left(x^3-3x^2\right)+\left(3x-9\right)\\ \\ =x^2\left(x-3\right)+3\left(x-3\right)\\ \\=\left(x^2+3\right)\left(x-3\right)\\ \)
\(\text{f) }x^2+ab+ax+bx\\ \\=\left(x^2+ax\right)+\left(bx+ab\right)\\ \\ =x\left(x+a\right)+b\left(x+a\right)\\ \\=\left(x+b\right)\left(x+a\right)\\ \)
\(\text{g) }xy+1+x+y\\ \\=\left(xy+x\right)+\left(y+1\right)\\ \\=x\left(y+1\right)+\left(y+1\right)\\ \\=\left(x+1\right)\left(y+1\right)\)
\(\text{h) }9-x^2-2xy-y^2\\ \\=9-\left(x^2+2xy+y^2\right)\\ \\=3^2-\left(x+y\right)^2\\ \\=\left(3-x-y\right)\left(3+x+y\right)\\ \)
\(\text{i) }x^2-2xy+y^2-1\\ \\=\left(x^2-2xy+y^2\right)-1\\ \\=\left(x-y\right)^2-1^2\\ \\=\left(x-y-1\right)\left(x-y+1\right)\\ \)
d) x3 + x2 + x + 1
= x2(x + 1) + (x + 1)
= (x + 1)(x2 + 1)
e) x3 - 3x2 + 3x - 9
= x2(x - 3) + 3(x - 3)
= (x - 3)(x2 + 3)
f) x2 + ab + ax + bx
= x2 + bx + ab + ax
= x(x + b) + a(b + x)
= (b + x)(x + a)
g) xy + 1 + x + y
= xy + x + y + 1
= x(y + 1) + (y + 1)
= (y + 1)(x + 1)
h) 9 - x2 - 2xy - y2
= 9 - (x2 + 2xy + y2)
= 32 - (x + y)2
= (3 - x - y)(3 + x + y)
i) x2 - 2xy + y2 - 1
= (x - y)2 - 1
= (x - y - 1)(x - y + 1)
\(\left(ax^2+bx+c\right)\left(x+1\right)=ax^3+\left(a+b\right)x^2+\left(b+c\right)x+c\)
đồng nhất đa thức trên với đa thức đã cho ta được
\(\left\{{}\begin{matrix}a=1\\a+b=8\\b+c=19\\c=12\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=1\\b=7\\c=12\end{matrix}\right.\)
3 phần kia làm tương tự
b: \(\left(ax^2+bx+c\right)\left(x+3\right)\)
\(=ax^3+3ax^2+bx^2+3bx+cx+3c\)
\(=ax^3+x^2\left(3a+b\right)+x\left(3b+c\right)+3c\)
Theo đề, ta có:
\(\left\{{}\begin{matrix}3c=0\\3b+c=-3\\3a+b=2\\a=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}c=0\\b=-1\\a=1\end{matrix}\right.\)
c: \(\left(x^2+cx+2\right)\left(ax+b\right)\)
\(=a\cdot x^3+bx^2+ac\cdot x^2+bc\cdot x+2a\cdot x+2b\)
\(=a\cdot x^3+x^2\left(b+ac\right)+x\left(bc+2a\right)+2b\)
Theo đề, ta có: 2b=-2; bc+2a=0; b+ac=1; a=1
=>b=-1; a=1; c=2
d: \(\left(x^2+cx+1\right)\left(ax+b\right)\)
\(=a\cdot x^3+bx^2+ac\cdot x^2+bc\cdot x+a\cdot x+b\)
\(=a\cdot x^3+x^2\left(b+ac\right)+x\left(bc+a\right)+b\)
Theo đề, ta có:
b=2; bc+a=-3; b+ac=0; a=1
=>b=2; a=1; bc=-3-a=-3-1=-4
=>b=2; a=1; 2c=-4
=>b=2; a=1; c=-2
1,Thực hiện phép tính :
a, (x + 2)9 : (x + 2)6
=(x+2)9-6
=(x+2)3
b, (x - y) 4 : (x - 2)3
=(x-y)4-3
=x-y
c, ( x2+ 2x + 4)5 : (x2 + 2x + 4)
=(x2+2x+4)5-1
=(x2+2x+4)4
d, 2(x2 + 1)3 : 1/3(x2 + 1)
=(2÷1/3).[(x2+1)3÷(x2+1)]
=6(x2+1)2
e, 5 (x - y)5 : 5/6 (x - y)2
=(5÷5/6).[(x-y)5÷(x-y)2]
=6(x-y))3
1) \(D=\left|x^2+x+3\right|+\left|x^2+x-6\right|\)
\(D=\left|x^2+x+3\right|+\left|6-x^2-x\right|\)
Áp dụng bđt \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\) ta có :
\(D\ge\left|x^2+x+3+6-x^2-x\right|=\left|9\right|=9\)
Dấu "=" xảy ra \(\Leftrightarrow\left(x^2+x+3\right)\left(6-x^2-x\right)\ge0\Leftrightarrow-3\le x\le2\)
2) \(C=x^2+xy+y^2-3x-3y\)
\(C=\left(x^2-2x+1\right)+\left(y^2-2y+1\right)+\left(xy-x-y+1\right)-3\)
\(C=\left(x-1\right)^2+\left(y-1\right)^2+\left(x-1\right)\left(y-1\right)-3\)
\(C=\left(x-1\right)^2+2\cdot\left(x-1\right)\cdot\frac{\left(y-1\right)}{2}+\frac{\left(y-1\right)^2}{4}+\frac{3\left(y-1\right)^2}{4}-3\)
\(C=\left(x-1-\frac{y-1}{2}\right)^2+\frac{3\left(y-1\right)^2}{4}-3\ge-3\forall x;y\)
Dấu "=" xảy ra \(\Leftrightarrow\left\{{}\begin{matrix}x-1-\frac{y-1}{2}=0\\y-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=1\end{matrix}\right.\)
3) \(B=x^4-2x^3+3x^2-2x+1\)
\(B=x^2\left(x^2-2x+3-\frac{2}{x}+\frac{1}{x^2}\right)\)
\(B=x^2\left[\left(x^2+2+\frac{1}{x^2}\right)-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left[\left(x+\frac{1}{x}\right)^2-2\left(x+\frac{1}{x}\right)+1\right]\)
\(B=x^2\left(x+\frac{1}{x}-1\right)^2\)
\(B=\left[x\left(x+\frac{1}{x}-1\right)\right]^2\)
\(B=\left(x^2-x+1\right)^2\)
Xét \(x^2-x+1=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}\ge\frac{3}{4}\forall x\)
\(\Rightarrow B=\left(x^2-x+1\right)^2\ge\left(\frac{3}{4}\right)^2=\frac{9}{16}\forall x\)
Dấu "=" xảy ra \(\Leftrightarrow x=\frac{1}{2}\)