Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a)(x+1)(x+2)(x+3)(x+4)+1
=(x+1)(x+4)(x+2)(x+3)+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4) thì (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
Thay: =(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16) thì (x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
Thay: =(x2+10x+16+4)2
=(x2+5x+20)2
a)(x+1)(x+2)(x+3)(x+4)+1
=[(x+1)(x+4][(x+2)(x+3)]+1
=(x2+5x+4)(x2+5x+6)+1
Đặt a=(x2+5x+4)
Ta có: (x2+5x+4)(x2+5x+6)+1
= a.(a+2)+1
=a2+2a+1
=(a+1)2
=(x2+5x+4+1)2
=(x2+5x+5)2
b)(x+2)(x+4)(x+6)(x+8)+16
=(x+2)(x+8)(x+4)(x+6)+16
=(x2+10x+16)(x2+10x+24)+16
Đặt a=(x2+10x+16)
Ta có:(x2+10x+16)(x+5x+24)+1
= a.(a+8)+16
=a2+8x+16
=(a+4)2
=(x2+10x+16+4)2
=(x2+5x+20)2
Mk yêu bé Shin-Conan lém
A= \(\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{2}{x+3}-...+\frac{8}{x+5}-\frac{8}{x+6}\)
A=\(\frac{1}{x+1}+\frac{1}{x+3}+\frac{2}{x+4}+\frac{4}{x+5}-\frac{8}{x+6}\)
Rồi tiếp tục làm nhé bạn.
-x^61+5*x^60+x^59-5*x^58-x^55+5*x^54+x^53-5*x^52-x^49+5*x^48+x^47-5*x^46x^43+5*x^42+x^41-5*x^40-x^37+5*x^36+x^35-5*x^34-x^49+5*x^48+x^47-5*x^46x^43+5*x^42+x^41-5*x^40-x^37+5*x^36+x^35-5*x^34-x^31+5*x^30+x^27-5*x^26-x^25+5*x^24+x^21-5*x^20-x^19+5*x^18+x^15-5*x^14-x^13+5*x^12+x^9-5*x^8-x^7+5*x^6+x^3-5*x^2-x+5
Đặt \(A=3\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^{16}-1\right)\left(x^{16}+1\right)\)
\(\Leftrightarrow\left(x^2-1\right)A=3\left(x^{32}-1\right)\)
\(\Leftrightarrow A=\frac{3\left(x^{32}-1\right)}{x^2-1}\)
a: Ta có: \(\left(7x+4\right)^2-\left(7x-4\right)\left(7x+4\right)\)
\(=\left(7x+4\right)\left(7x+4-7x+4\right)\)
\(=8\left(7x+4\right)\)
=56x+32
b: Ta có: \(8\left(x-2\right)^2-3\left(x^2-4x-5\right)-5x^2\)
\(=8x^2-32x+32-3x^2+12x+15-5x^2\)
\(=-20x+47\)
c: Ta có: \(\left(x+1\right)^3-\left(x-1\right)\left(x^2+x+1\right)-3x\left(x+1\right)\)
\(=x^3+3x^2+3x+1-x^3+1-3x^2-3x\)
=2
\(\left(x+2\right)\left(x-2\right)-\left(x-3\right)\left(x+1\right)\)
\(=x^2-4-\left(x^2-2x-3\right)\)
\(=2x-1\)
\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
a, \(\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow x^2+8x+16-\left(x^2-x+x-1\right)=16\)
\(\Leftrightarrow8x+1=0\Leftrightarrow x=-\frac{1}{8}\)
b, \(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow2x+255=0\Leftrightarrow x=-\frac{225}{2}\)
c, \(\left(x+2\right)\left(x-2\right)-x^3-2x=15\)
\(\Leftrightarrow x^2-4-x^3-2x=15\)( vô nghiệm )
d, \(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3+6x^2-x+8x^3+1=28\)
\(\Leftrightarrow15x^2+26=0\Leftrightarrow x^2\ne-\frac{26}{15}\)( vô nghiệm )
Tính nhẩm hết á, sai bỏ quá nhá, sắp đi hc ... nên chất lượng hơi kém xíu ~~~
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
Ta có: \(\left(x-1\right)\left(x+1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^2-1\right)\left(x^2+1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^4-1\right)\left(x^4+1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^8-1\right)\left(x^8+1\right)\left(x^{16}+1\right)\)
\(=\left(x^{16}-1\right)\left(x^{16}+1\right)\)
\(=x^{32}-1\)
Bạn tham khảo nhé!