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a: \(A=\left(x+2y\right)^3=\left(-5\right)^3=-125\)
b: \(B=\left(2x-y\right)^3=\dfrac{1}{125}\)
c: \(=x^3+3x^2+3x+1-x^3+3x^2-3x+1-3x\left(x^2-2x+1+x+1\right)\)
\(=6x^2+2-3x\left(2x^2-x+2\right)\)
\(=6x^2+2-6x^3+3x^2-6x\)
\(=-6x^3+9x^2-6x+2\)
1. \(A=x^6-x^4+x^3-x-x^4+x^2=x^3\left(x^3-x\right)+\left(x^3-x\right)-x\left(x^3-x\right)=\left(x^3-x\right)^2+\left(x^3-x\right)\)
Thay \(x^3-x=6\) vào A, ta được:
\(A=36+6=42\)
KL : A=42
2.
a) đa thức đã cho \(=ab^2+ac^2+abc+bc^2+ba^2+abc+ca^2+cb^2+abc\)
\(=\left(ab^2+ba^2+abc\right)+\left(ac^2+ca^2+abc\right)+\left(bc^2+cb^2+abc\right)\)
\(=ab\left(a+b+c\right)+ac\left(a+b+c\right)+bc\left(a+b+c\right)\)
\(=\left(ab+bc+ac\right)\left(a+b+c\right)\)
b) đa thức đã cho \(=\left(a^2c+abc\right)+\left(abc+b^2c\right)+\left(ac^2+bc^2\right)+\left(a^2b+ab^2\right)\)
\(=ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)+ab\left(a+b\right)\)
\(=\left(ac+bc+c^2+ab\right)\left(a+b\right)\)
\(=\left[\left(ac+ab\right)+\left(bc+c^2\right)\right]\left(a+b\right)\)
\(=\left[a\left(c+b\right)+c\left(b+c\right)\right]\left(a+b\right)\)
\(=\left(a+c\right)\left(b+c\right)\left(a+b\right)\)
a) \(\left(x+a\right)\left(x+b\right)\left(x+c\right)\)
\(=\left[x^2+\left(a+b\right)x+ab\right]\left(x+c\right)\)
\(=x^3+\left(a+b+c\right)x^2+\left(ab+bc+ca\right)x+abc\)
b) \(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3-3ab\left(a+b\right)+c^3-3abc\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+2ab+b^2-ca-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
c) \(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)
\(=a^2\left(b-c\right)+b^2c-ab^2+c^2a-bc^2\)
\(=a^2\left(b-c\right)+bc\left(b-c\right)-a\left(b-c\right)\left(b+c\right)\)
\(=\left(b-c\right)\left(a^2+bc-ab-ca\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Nhầm đoạn cuối là \(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Cho a+x2=2006, b+x2=2007, c+x2= 2008 và abc=3
Tính a/bc+b/ca+c/ab-1/a-1/b-1/c
.
a) mk sửa lại đề chút nhé, bn tham khảo
TH1: \(\left(x^3-2x^2\right)-\left(x-2\right)=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-1\right)=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
TH2: \(\left(x^3+2x^2\right)-\left(x+2\right)=x^2\left(x+2\right)-\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-1\right)=\left(x+2\right)\left(x+1\right)\left(x-1\right)\)
đề bài là phân tích đa thức thành nhân tử nha! các bạn giúp mình vs