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Câu 1:
a: \(=a^2+2ab+b^2-a^2-2ab-b^2=0\)
b: \(=x^3+27-54-x^3=-27\)
Câu 4:
\(\Leftrightarrow3x^3+x^2+9x^2+3x-3x-1-4⋮3x+1\)
\(\Leftrightarrow3x+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(x\in\left\{0;1\right\}\)
Bài 2 :
a) (2x + 1)(1 - 2x) + (2x - 1)2 = 22
=> 1 - 4x2 + (4x2 - 4x + 1) = 22
=> 1 - 4x2 + 4x2 + 4x + 1 = 22
=> 4x + 2 = 22
=> 4x = 20
=> x = 5
Vậy x = 5
#)Giải :
a) \(ab-ac-b+c=a\left(b-c\right)-\left(b-c\right)=\left(a-1\right)\left(b-c\right)\)
b) \(5a^2-5=5\left(a^2-1\right)=5\left(a-1\right)\left(a+1\right)\)
c) \(x^2-2x+1-a^2-2ab-b^2=\left(x-1\right)^2-\left(a+b\right)^2\)
\(=\left(x-1-a-b\right)\left(x-1+a+b\right)\)
d) \(7x^2-14x+7=7\left(x^2-2x+1\right)=7\left(x-1\right)^2\)
e) \(81x^4+4=81x^4+36x^2+4-36x^2=\left(9x^2+6x+2\right)\left(9x^2-6x+2\right)\)
f) \(x^7+x^2+1=\left(x^7+x^6+x^5\right)-\left(x^6+x^5+x^4\right)+...+\left(x^2+x+1\right)\)
\(=x^5\left(x^2+x+1\right)-x^4\left(x^2+x+1\right)+...+\left(x^2+x+1\right)\)
\(=\left(x^5-x^4+x^2-x+1\right)\left(x^2+x+1\right)\)
g) \(\left(a+b\right)\left(a^2-b^2\right)+\left(b+c\right)\left(b^2-c^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a+b\right)\left(a^2-b^2\right)-\left(b+c\right)\left(a^2-b^2+c^2-a^2\right)+\left(c+a\right)\left(c^2-a^2\right)\)
\(=\left(a-b\right)\left(a-c\right)\left(a+b\right)-\left(a-b\right)\left(a-c\right)\left(a+c\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\)
Câu 1. D
Câu 4. A, C
Câu 5. Xem lại đề!
tui gõ chưa xong lỡ ấn enter á . bạn xem lại giúp mik vs