Rut gon bieu thuc sau
3(2*2+1)(2*4+1)(2*8+1)(2*16+1)
Rut gon bieu thuc
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
=\(\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=\(\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
=...=2^32-1
nhân hết ra là xong:))
bài về nhà hs phải tự làm
Cái bước (22-1)(22 + 1)(24 +1)(216+1) làm như thế nào mà ra vậy
Cho a - b = 1. Rut gon bieu thuc
M = ( a + b)( a^2 + b^2 )( a^4 + b^4 )( a^8 + b^8)( a^16 + b^16)
Cho a - b = 1. Rut gon bieu thuc
M = ( a + b)( a^2 + b^2 )( a^4 + b^4 )( a^8 + b^8)( a^16 + b^16)
\(M=1.\left(a+b\right)\left(a^2+b^2\right).......\)
\(=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)....\)
\(=\left(a^2-b^2\right)\left(a^2+b^2\right)....\)
\(=\left(a^4-b^4\right)\left(a^4+b^4\right)......\)
\(=\left(a^8-b^8\right)\left(a^8+b^8\right)\left(a^{16}+b^{16}\right)\)
\(=\left(a^{16}-b^{16}\right)\left(a^{16}+b^{16}\right)\)
\(=a^{32}-b^{32}\)
rut gon bieu thuc
3(2^2+1).(2^4+1)...(2^64+1)+1
\(3\left(2^2+1\right).\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)+1\)
\(=\left(2^4-1\right)\left(2^4+1\right)....\left(2^{64}+1\right)+1\)
\(=\left(2^8-1\right).\left(2^8+1\right)\left(2^{16}+1\right)....\left(2^{64}+1\right)+1\)
\(=\left(2^{64}-1\right).\left(2^{64}+1\right)+1\)
\(=2^{64}-1+1=2^{64}\)
Vậy : \(3\left(2^2+1\right).\left(2^4+1\right)...\left(2^{64}+1\right)+1=2^{64}\)
B1: rut gon bieu thuc
a, (x+y)^2-4(x-y)^2
b, 2(x-y)(x+y)+(x+y)^2+(x-y)^2
B2: tim X
a, (2X-1)^2-4(X+2)^2=9
b, 3(X-1)^2-3X(X-5)=21
B3: Cho bieu thuc
M=(x+3)^3-(x-1)^3+12x(x-1)
a, Rut gon bieu thuc tren
b, Tinh gia tri M tai x=-2/3
c, Tim x de M=16
1)a)=>x2+y2+2xy-4(x2-y2-2xy)
=>x2+y2+2xy-4.x2+4y2+8xy
=>-3.x2+5y2+10xy
rut gon bieu thuc:(1/2+1)(1/3+1)(1/4+1).....(1/99+1)
Bài này có rắc rối đâu em?
Thực hiện phép tính trong ngoặc lại là ra dạng (n+1)/n.
1 dãy các số liên tục kéo dài nhân với nhau thì triệt tiêu là xong!
Chúc em học tốt!
Rut gon bieu thuc
A= 1+1/2+1/2^+1/2^+....+1/2^2017
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\)
\(\Rightarrow2A=2+1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\)
\(\Rightarrow2A-A=\left(2+1+\dfrac{1}{2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\right)\)
\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)
\(\Rightarrow A=2-\dfrac{1}{2^{2017}}=\dfrac{2^{2018}-1}{2^{2017}}\)
\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\)
\(2A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)\)
\(2A-A=\left(2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2016}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2017}}\right)\)
\(A=2-2^{2017}\)
rut gon bieu thuc (6x-1)^2+(6x+1)^2-2(1+6x)(6x+1)
rut gon bieu thuc
B=1/2+(1/2)^2+(1/2)^3+...+(1/2)^99