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a) \(x.\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x.\left(x^2-16\right)-\left(x^4-1\right)=x^3-16x-x^4+1\)
ý này ko rút gọn được hết đâu.
b) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4=-77\)
c) \(\left(a+b-c\right)^2-\left(a-c\right)^2-2ab+2bc=a^2+b^2+c^2+2ab-2bc-2ac-a^2+2ac-c^2-2ab+2bc=b^2\)
1/ \(P=\left(a+b\right)^2-4ab=a^2+2ab+b^2-4ab=a^2-2ab+b^2=\left(a-b\right)^2=5^2=25\)
2/\(M=a^3+b^3+3ab=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab=a^2-ab+b^2+3ab=a^2+2ab+b^2=\left(a+b\right)^2=1\)
3/
\(\left(2n+1\right)^2-\left(2n-1\right)^2=\left(2n+1-2n+1\right)\left(2n+1+2n-1\right)=2.4n=8n⋮8\)
\(a-b=5\)=> \(a=5+b\)
thay vào biểu thức P ta có
\(\left(5+b+b\right)^2-4.\left(5+b\right).b\)
=\(\left(5+2b\right)^2-\left(20+4b\right).b\)
= \(25+20b+4b^2-20b-4b^2\)
\(=25\)
ta có \(a+b=1\)
=> \(\left(a+b\right)^3=1\)
<=> \(a^3+3a^2b+3ab^2+b^3=1\)
<=> \(a^3+b^3+3ab.\left(a+b\right)=1\)
mà \(a+b=1\)
<=> \(a^3+b^3+3ab=1\)
hay M =1
\(\left(2n+1\right)^2-\left(2n-1\right)^2\)
\(=4n^2+4n+1-\) \(\left(4n^2-4n+1\right)\)
\(=4n^2+4n+4-\) \(4n^2+4n-1\)
\(=8n+3\)
câu cuối mk làm được thế thôi
sorry nha
Bài 1: \(P=\left(a+b\right)^2-4ab\)
\(=a^2+2ab+b^2-4ab\)
\(=a^2+\left(2ab-4ab\right)+b^2\)
\(=a^2-2ab+b^2\)
\(=\left(a-b\right)^2\)
\(=5^2\)
\(=25\)
Bài 2: \(M=a^3+b^3+3ab\)
\(=\left(a^3+b^3\right)+3ab\)
\(=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\)
\(=1.\left(a^2-ab+b^2\right)+3ab\)
\(=a^2-ab+b^2+3ab\)
\(=a^2+\left(3ab-ab\right)+b^2\)
\(=a^2+2ab+b^2\)
\(=\left(a+b\right)^2\)
\(=1^2=1\)
Bài 3 : Ta có : \(\left(2n+1\right)^2-\left(2n-1\right)^2\)
\(=\left(2n\right)^2+2.2n.1+1^2-\left(2n\right)^2+2.2n.1-1^2\)
\(=4.n+4.n\)
\(=8n\)Chia hết cho 8
Nhận xét: \(b^3c-cb^3=0;b^2c-cb^2=0.\).Nên phân thức trở thành:
\(\frac{a^3b-ab^3+c^3a-ca^3}{a^2b-ab^2+c^2a-ca^2}=\frac{a^3\left(b-c\right)-a\left(b^3-c^3\right)}{a^2\left(b-c\right)-a\left(b^2-c^2\right)}\)
\(=\frac{a\left(b-c\right)\left\{a^2-\left(b^2-bc+c^2\right)\right\}}{a\left(b-c\right)\left\{a-\left(b+c\right)\right\}}\)
\(=\frac{a^2-\left(b^2-bc+c^2\right)}{a-\left(b+c\right)}=\frac{a^2-\left(b+c\right)^2+3bc}{a-\left(b+c\right)}\)
\(=a+b+c+\frac{3bc}{a-b-c}\).
\(a,5\left(x-2\right)\left(x+2\right)-5.x^{28}\)
\(=\left(5x-10\right)\left(x+2\right)-5x^{28}\)
\(=5x^2+10x-10x-20-5x^{28}\)
\(=5x^2-2x-5x^{28}\)
\(b,\left(a-b\right)^2+4ab\)
\(=a^2-2ab+b^2+4ab\)
\(=a^2+2ab+b^2\)
a) 5(x - 2)(x + 2) - 5.x28
= 5(x2 - 4) - 5.x28
= 5.x2 - 5.4 - 5.x2
= 5x2 - 20 - 5x28
= -5x28 + 5x2 - 20
b) (a - b)2 + 4ab
= a2 - 2ab + b2 + 4ab
= a2 - 2ab + b2
\(\left(a+b\right)^2-\left(a-b\right)^2\)
\(=\left(a+b-a+b\right)\left(a+b-a-b\right)\)
\(=2a.2b=4ab\)
=> câu trả lời là a . 4ab