Sai đề === khi a=b=1 thì
VT=(1+1)(1+1)(1+1)(1+1)=24=16
VP=1-1=0
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Các câu hỏi tương tự
Cho a=b+1
Chứng minh: \(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\left(a^8+b^8\right)=a^{16}-b^{16}\)
BT7: Tính
\(1,A=8\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
\(2,B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{16}+1\right)\)
chứng minh:\(\frac{b}{\left(a-b\right).\left(b-c\right)}+\frac{c}{\left(b-c\right).\left(c-a\right)}+\frac{ca}{\left(c-b\right)\left(a-b\right)}=0\)
b,\(\frac{1}{1-x}+\frac{1}{1+x}+\frac{2}{1+x^2}+\frac{4}{1+x^4}+\frac{8}{1+x^8}=\frac{6}{1-x^{16}}\)
mọi người giải gấp giúp mk nha .giải kĩ giúp mk
cảm ơn nhìu
1) CMR \(\frac{X^{32}+X^{16}+1}{X^2+X+1}\)= \(\left(X^2-X+1\right)\left(X^4-X^2+1\right)\left(X^8-X^4+1\right)\left(X^{16}-X^8+1\right)\)
2)\(Tinh\left(a-b\right)^{2017}Biet\left(a+b\right)=7;a.b=12\)(a<b)
Chứng minh bất đẳng thức
a)\(8\left(a^4+b^4\right)\ge\left(a+b\right)^4\)
b)\(\left(a^2+b^2\right)^2\ge ab\left(a+b\right)^2\)
Cho x=y+1. Chứng minh rằng:
a)\(x^3-y^3-3xy=1\)
b)\(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)=x^{16}-y^{16}\)
chung minh rang\(\left(a^{10}+b^{10}\right)\left(a^2+b^2\right)>=\left(a^8+b^8\right)\left(a^4+b^4\right)\)voi moi a,b
Bài 1 : rút gọn các biểu thức sau
A = \(\left(3x+1\right)^2-2\left(3x+1\right)\left(5x+5\right)+\left(5x+5\right)^2\)
B = \(\left(a+b+c\right)^2+\left(a-b-c\right)^2+\left(b-c-a\right)^2+\left(c-b-a\right)^2\)
C = \(\left(3x+1\right)\left(3x^2+1\right)\left(3x^4+1\right)\left(3x^8+1\right)\left(3x^{16}+1\right)\left(3x^{32}+1\right)\)
Tính
a) \(A=1+\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\)
b) \(B=10^2+8^2+.....+2^2-\left(9^2+7^2+5^2+3^2+1^2\right)\)