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\(A^3+B^3+A^2C+B^2C-ABC\)
\(=\left(A+B\right)\left(A^2-AB+B^2\right)+C\left(A^2-AB+B^2\right)\)
\(=\left(A^2-AB+B^2\right)\left(A+B+C\right)\)
\(=\left(A^2-AB+B^2\right).0\)
\(=o\)
ta có : M=2.(a^3 +b^3) -3.(a^2 + b^2)
<=>M=2.(a+b)(a^2 -ab +b^2) - 3(a^2 +3b^2)
<=>M=2(a^2 -ab +b^2) -3(a^2 +b^2) vì a+b=1(gt)
<=>M=-(a^2 +b^2 +2ab)
<=>M=-(a+b)^2
<=>M=-1 (vì a+b=1)
Bài1: Phân tích các đa thức sau thành nhân tử
a)36-4x2+4xy-y2
\(=6^2-\left(4x^2-4xy+y^2\right)\)
\(=6^2-\left(2x-y\right)^2\)
\(=\left(6+2x-y\right)\left(6-2x+y\right)\)
b)2x4+3x2-5
\(=2x^4-2x^2+5x^2-5\)
\(=2x^2\left(x^2-1\right)+5\left(x^2-1\right)\)
\(=\left(2x^2+5\right)\left(x^2-1\right)\)
\(=\left(2x^2+5\right)\left(x-1\right)\left(x+1\right)\)
B1:a)\(36-4x^2+4xy-y^2=36-\left(4x^2-4xy+y^2\right)=6^2-\left(2x-y\right)^2\)
\(=\left(6-2x+y\right)\left(6+2x-y\right)\)
c)\(a^3-ab^2+a^2+b^2-2ab=a\left(a^2-b^2\right)+\left(a-b\right)^2\)\(=a\left(a-b\right)\left(a+b\right)+\left(a-b\right)^2=\left(a-b\right)\left(a^2+ab+a-b\right)\)
d)\(x^2-\left(a^2+b^2\right)x+a^2b^2=x^2-a^2x-b^2x+a^2b^2\)\(=x\left(x-a^2\right)-b^2\left(x-a^2\right)=\left(x-a^2\right)\left(x-b^2\right)\)
e)\(x\left(x-y\right)+x^2-y^2=x\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)\(=\left(x-y\right)\left(x+x+y\right)=\left(x-y\right)\left(2x+y\right)\)
Bài 2:
a) \(VP=\left(a+b\right)^3-3ab\left(a+b\right)\)
\(=a^3+b^3+3ab\left(a+b\right)-3ab\left(a+b\right)\)
\(=a^3+b^3=VT\) (đpcm)
b) \(\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
\(=a^3+ab^2+ac^2-a^2b-abc-a^2c+a^2b+b^3+bc^2-ab^2-b^2c-abc\)\(+a^2c+b^2c+c^3-abc-bc^2-ac^2\)
\(=a^3+b^3+c^3-3abc\)
Bài 1:
\(N=\frac{x\left|x-2\right|}{x^2+8x-20}+12x-3\)
\(=\frac{x\left|x-2\right|}{\left(x-2\right)\left(x+10\right)}+12x-3\)
Nếu \(x\ge2\)thì: \(N=\frac{x\left(x-2\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)
\(=\frac{x}{x+10}+12x+3\) (lm tiếp nhé)
Nếu \(x< 2\) thì: \(N=\frac{x\left(2-x\right)}{\left(x-2\right)\left(x+10\right)}+12x-3\)
\(=\frac{-x}{x+10}+12x-3\) (lm tiếp nhé)
Ta có:M=a3+b3+c(a2+b2)-abc
=(a+b)(a2-ab+b2)-(a+b)(a2+b2)+(a+b).ab
=(a+b)(a2-ab+b2-a2-b2+ab)
=(a+b).0=0
Vậy GT của M là:0
1, \(a^3+b^3+3ab\left(a^2+b^2\right)+6a^2b^2\left(a+b\right)\)
\(=a^3+b^3+3a^3b+3ab^3+6a^2b^2\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left(a^2+2ab+b^2\right)\)
\(=a^2-ab+b^2+3ab\left(a+b\right)^2\)
\(=a^2-ab+b^2+3ab\)
\(=a^2+2ab+b^2=\left(a+b\right)^2\)
\(=1\)
Vậy A = 1
Bài 2: ( đặt đề bài là A )
Đặt \(b+c-a=x,a+c-b=y,a+b-c=z\)
\(\Rightarrow a+b+c=x+y+z\)
\(\Leftrightarrow A=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(x+z\right)-x^3-y^3-z^3\)
\(=3\left(x+y\right)\left(y+z\right)\left(x+z\right)\)
\(=3.2c.2a.2b=24abc\)
Vậy...
Bài 3:
+) Xét p = 3 có: \(p^2+2=11\in P\) ( t/m )
+) Xét \(p\ne3\) thì:
+ \(p=3k+1\Rightarrow p^2+2=\left(3k+1\right)^2+2=9k^2+6k+3⋮3\notin P\)
+ \(p=3k+2\Rightarrow p^2+2=\left(3k+2\right)^2+2=9k^2+12k+6⋮3\notin P\)
Vậy p = 3
Bài 4:
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=2\)
\(\Leftrightarrow\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)^2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{ab}+\dfrac{2}{bc}+\dfrac{2}{ac}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2c}{abc}+\dfrac{2a}{abc}+\dfrac{2b}{abc}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2\left(a+b+c\right)}{abc}=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+2=4\)
\(\Leftrightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}=2\)
\(\Rightarrowđpcm\)
Áp dụng hằng đẳng thức mở rộng phân tích a3+b3+c3 sau đó phân tích tiếp a2-b2