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=> 1.\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)\) \(+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}-b^{64}\)
=> \(\left(a-b\right)\left(a+b\right)+...+\left(a^{32}+b^{32}\right)\)= \(a^{64}+b^{64}\)
=> \(\left(a^2-b^2\right)\left(a^2+b^2\right)+...+\left(a^{32}+b^{32}\right)\)= a^64 + b^64
tương tự sẽ ra kết quả cuối là \(\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)=a^{64}-b^{64}\left(đpcm\right)\)
ta có \(a^2-b^2=\left(a+b\right)\left(a-b\right)\) => \(\frac{a^2-b^2}{a-b}=a+b\)
\(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)\)=> \(\frac{a^4-b^4}{a^2-b^2}=a^2+b^2\)
\(a^8-b^8=\left(a^4-b^4\right)\left(a^4+b^4\right)\) => \(\frac{a^8-b^8}{a^4-b^4}=a^4+b^4\)
...............................................................................................
\(a^{64}-b^{64}=\left(a^{32}-b^{32}\right)\left(a^{32}+b^{32}\right)\) => \(\frac{a^{64}-b^{64}}{a^{32}-b^{32}}=a^{32}+b^{32}\)
thay vào ta được
\(\left(a+b\right)\left(a^2+b^2\right)\left(a^4+b^4\right)......\left(a^{32}+b^{32}\right)\)
\(=\frac{a^2-b^2}{a-b}.\frac{a^4-b^4}{a^2-b^2}.\frac{a^8-b^8}{a^4-b^4}.............\frac{a ^{64}-b^{64}}{a^{32}-b^{32}}\)
\(=\frac{a^{64}-b^{64}}{a-b}\)
mà a-b= 1 nên \(\frac{a^{64}-b^{64}}{a-b}=a^{64}-b^{64}\)
\(VT=1.\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)
\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)...\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)...\left(2^{16}+1\right)\)
\(=...=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1\)
a) Ta có: \(\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2bc-2ab-2ac+a^2+b^2+c^2-2ab-2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ca\)
\(=a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2\)
\(=4a^2+4b^2+4c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
b) Đặt x = b + c - a
y = c + a - b
z = a + b - c
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Rightarrow a+b+c=x+y+z\)
Ta có: \(\left(a+b+c\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left[xy+\left(x+y\right)z+z^2\right]\)
\(=3\left(x+y\right)\left[z^2+xy+xz+yz\right]\)
\(=3\left(x+y\right)\left[z\left(x+y\right)+y\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=3.2a.2b.2c\)
\(=24abc\) (đpcm)
\(a.\)
Ta sẽ biến đổi biểu thức \(B\) quy về dạng có thể dùng được hằng đẳng thức \(\left(x-y\right)\left(x+y\right)=x^2-y^2\), khi đó:
\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1\)
Vì \(2^{16}>2^{26}-1\) nên \(2^{16}>\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)
Vậy, \(A>B\)
Tương tự với câu \(b\) kết hợp với phương pháp tách hạng tử, khi đó xuất hiện hằng đẳng thức mới và dễ dàng đơn giản hóa biểu thức \(A\). Ta có:
\(A=4\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)=\frac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^4-1\right)\left(3^4+1\right)...\left(3^{64}+1\right)\)
\(=\frac{1}{2}\left(3^{64}-1\right)\left(3^{64}+1\right)=\frac{1}{2}\left(3^{128}-1\right)\)
Mặt khác, do \(\frac{1}{2}<1\) nên \(\frac{1}{2}\left(3^{128}-1\right)<3^{128}-1\)
Vậy, \(B>A\)
Có a = b+1
=> a - b =1
=> (a-b)(a+b)(a^2+b^2)(a^4+b^4)...(a^32+b^32) = (a-b)(a^64-b^64)
=> (a^2-b^2)(a^2+b^2)(a^4+b^4)...(a^32+b^32) = 1 . (a^64 - b^64)
=> (a^4-b^4)(a^4+b^4)(a^8+b^8)(a^16+b^16)(a^32+b^32) = a^64 - b^64
=> (a^8-b^8)(a^8+b^8)(a^16+b^16)(a^32+b^32) = a^64 - b^64
=> (a^16-b^16)(a^16+b^16)(a^32+b^32) = a^64 - b^64
=> (a^32-b^32)(a^32+b^32) = a^64 - b^64
=> a^64-b^64 = a^64 - b^64
=> đpcm