Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(b,\)\(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=1.\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{16}-1\right)\left(2^{16}+1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=\left(2^{32}-1\right)\left(2^{32}+1\right)-2^{64}\)
\(\Rightarrow B=2^{64}-1-2^{64}=-1\)
a) Đặt \(A=\left(\frac{1}{2}+1\right).\left(\frac{1}{4}+1\right).\left(\frac{1}{16}+1\right)...\left(1+\frac{1}{2^{2n}}\right)\)
Rút gọn: \(A=\frac{2+1}{2}.\frac{4+1}{4}.\frac{16+1}{16}...\frac{2^{2.n}+1}{2^{2.n}}=\frac{2^{2.0}+1}{2^{2.0}}.\frac{2^{2.1}+1}{2^{2.1}}.\frac{2^{2.2}+1}{2^{2.2}}...\frac{2^{2.n}+1}{2^{2.n}}\)
\(\Rightarrow A=\frac{\left(2^{2.0}+1\right).\left(2^{2.1}+1\right).\left(2^{2.2}+1\right)...\left(2^{2.n}+1\right)}{2^{2.0}.2^{2.1}.2^{2.2}...2^{2.n}}.\)
b) Đặt \(B=\left(2+1\right).\left(2^2+1\right).\left(2^4+1\right).\left(2^8+1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2-1\right).\left(2+1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^2-1\right).\left(2^2+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^4-1\right).\left(2^4+1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}=\left(2^8-1\right).\left(2^8+1\right)...\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=\left(2^{16}-1\right).\left(2^{16}+1\right).\left(2^{32}+1\right)-2^{64}=\left(2^{32}-1\right).\left(2^{32}+1\right)-2^{64}\)
\(\Leftrightarrow B=2^{64}-1-2^{64}=-1\)Vậy B =-1.
Đặt $A=\dfrac1{(x+y)^3}\left(\dfrac1{x^3}+\dfrac1{y^3}\right)+\dfrac3{(x+y)^4}\left(\dfrac1{x^2}+\dfrac1{y^2}\right)+\dfrac6{(x+y)^5}\left(\dfrac1x+\dfrac1y\right).$
$=\dfrac{(x+y)^2(x^3+y^3)+3xy(x+y)(x^2+y^2)+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)^2(x+y)(x^2-xy+y^2)+3xy(x+y)(x^2+y^2)+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)\left[(x+y)^2(x^2-xy+y^2)+3xy(x^2+y^2)\right]+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)\left[x^4+x^3y+x^2y^2+xy^3+y^4+3x^3y+3xy^3\right]+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)\left(x^4+4x^3y+x^2y^2+4xy^3+y^4\right)+6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)^4-6x^2y^2}{x^3y^3(x+y)^4}+\dfrac{6x^2y^2}{x^3y^3(x+y)^5}.$
$=\dfrac{(x+y)^5}{x^3y^3(x+y)^5}.$
$=\dfrac1{x^3y^3}.$
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
a
(x+1)-(x-1)-3(x+1)(x-1)
=(x+1)-(x-1)-3x+1.(x-1)
=(x+1)-(x-1)-3x+x-1
=x+1-x+1-3x+x-1
=x-x-3x+x+1+1-1
=-2x
b,
5(x+2)(x-2)-1/2(6-8x)^2+17
=5x+10(x-2)-1/2(36-64x2)+17
=5x+10x-20-18+32x2+17
=5x+10x-20-18+17+32x2
=15x-21+32x2
\(1+\frac{2}{n\left(n+3\right)}=\frac{n^2+3n+2}{n\left(n+3\right)}=\frac{\left(n+1\right)\left(n+2\right)}{n\left(n+3\right)}\)
\(\Rightarrow A=\frac{2.3}{1.4}.\frac{3.4}{2.5}.\frac{4.5}{3.6}...\frac{2020.2021}{2019.2022}\)
\(\Rightarrow A=\frac{2.3.4...2020}{1.2.3...2019}.\frac{3.4.5...2021}{4.5.6...2022}=\frac{2020}{1}.\frac{3}{2022}=\frac{1010}{337}\)
Bài 1 :
a) Ta có : \(\left(1-a\right)\left(1-b\right)\left(1-c\right)=\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
Áp dụng bđt Cauchy : \(a+b\ge2\sqrt{ab}\) , \(b+c\ge2\sqrt{bc}\) , \(c+a\ge2\sqrt{ca}\)
\(\Rightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)\ge8abc\) hay \(\left(1-a\right)\left(1-b\right)\left(1-c\right)\ge8abc\)
MTC: (x+y)(x+1)(1-y)
\(=\frac{x^2\left(1+x\right)-y^2\left(1-y\right)-x^2y^2\left(x+y\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}=\frac{\left(x+y\right)\left(1+x\right)\left(1-y\right)\left(x-y+xy\right)}{\left(x+y\right)\left(1+x\right)\left(1-y\right)}\)
\(=x-y+xy\)
Với \(x\ne-1;x\ne-y;y\ne1\)thì giá trị biểu thức được xác định
\(S=\frac{yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(x-z\right)}\)
+ \(yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left(x-z\right)+xy\left(z+1\right)\left(x-y\right)\)
\(=yz\left(x+1\right)\left(y-z\right)-zx\left(y+1\right)\left[\left(y-z\right)+\left(x-y\right)\right]\)
\(+xy\left(z+1\right)\left(x-y\right)\)
\(=\left(y-z\right)\left[yz\left(x+1\right)-zx\left(y+1\right)\right]+\left(x-y\right)\left[xy\left(z+1\right)-zx\left(y+1\right)\right]\)
\(=\left(y-z\right)\left[z\left(y-x\right)\right]+\left(x-y\right)\cdot x\cdot\left(y-z\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
\(\Rightarrow S=\frac{1}{xyz}\)
B =\(\frac{1}{1.5}\) + \(\frac{1}{5.9}\) + ...+ \(\frac{1}{\left(4n-3\right).\left(4n+1\right)}\)
B = \(\frac14\).(\(\frac{4}{1.5}+\frac{4}{5.9}+\cdots+\frac{4}{\left(4n-3\right).\left(4n+1\right)}\)
B = \(\frac14\).(\(\frac11\) - \(\frac15\) + \(\frac15\) - \(\frac19\) + ... + \(\frac{1}{4n-3}-\frac{1}{4n+1}\))
B = \(\frac14\).(\(\frac11\) - \(\frac{1}{4n+1}\))
B = \(\frac14\).\(\frac{4n}{4n+1}\)
B = \(\frac{n}{4n+1}\)