giúp mình giải bài này
rut gon bieu thuc 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
thanks
Bai 1: Rut gon bieu thuc sau
B=(2/3)^3×(-3/4)^4×(-1)^5/(2/5)^2 × (-5/12)^3
Bai 2: Tim ia tri bieu thuc sau
M= 512- 512/2- 512/2^2- 512/2^2 - 512/2^3...512/2^10
rut gon bieu thuc
\(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(P=12\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
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
Rut gon bieu thuc sau
3(2*2+1)(2*4+1)(2*8+1)(2*16+1)
Rut gon bt P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
P=2.(5^2-1).(5^2+1).(5^4+1).(5^8+1).(5^16+1)
=2.(5^4-1).(5^4+1).(5^8+1).(5^16+1)
= 2.(5^8-1).(5^8+1).(5^16+1)
= 2.(5^16-1).(5^16+1)
= 2.(5^32-1)
1)P= 12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=> 2P = 24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
=> P = (5^32-1)/2
rut gon bieu thuc M=1/x - 2/(5-x) - x+5/(x^2-5x) voi x khac 0 ; x khac 5
a, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 thì
P = [x/(x^2 - 25) - (x - 5)/(x^2 + 5x)] : (2x - 5)/(x^2 + 5x) + x/(x - 5)
<=>P = [x/(x - 5)(x + 5) - (x - 5)/x(x+5)] . x(x + 5)/(2x - 5) + x/(x - 5)
=> P = [x^2 - (x - 5)^2]/x(x - 5)(x + 5) . x(x + 5)/(2x - 5) + x/(x - 5)
<=> P = (x - x + 5)(x + x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5(2x - 5)/(x - 5)(2x - 5) + x/(x - 5)
<=> P = 5/(x - 5) + x/(x - 5)
<=> P = (5 + x)/(x - 5)
b, Với x ≠ 0,x ≠ ± 5 và x ≠ 5/2 (x ∈ Z) thì P ∈ Z <=> (5 + x)/(x - 5) ∈ Z
<=> (x - 5 + 10)/(x - 5) ∈ Z
<=> 1 + 10/(x - 5) ∈ Z
<=> 10/(x - 5) ∈ Z
<=> (x - 5) ∈ Ư(10)
<=> x - 5 = 10 <=> x = 15 (TM)
hoặc x - 5 = -10 <=> x = -5 (TM)
hoặc x - 5 = 5 <=> x = 10 (TM)
hoặc x - 5 = -5 <=> x = 0 (TM)
hoặc x - 5 = 2 <=> x = 7 (TM)
hoặc x - 5 = -2 <=> x = 3 (TM)
hoặc x - 5 = -1 <=> x = 4 (TM)
hoặc x - 5 = 1 <=> x = 6 (TM)
Vậy x ∈ {-5,0,3,4,6,7,10,15} thì P ∈ Z
(4x-3)(3x+2)-(6x+1)(2x-5)+1
rut gon bieu thuc
12x2 + 8x - 9x - 6 - 12x2 + 30x + 2x -5 + 1= 31x - 10
Rut gon :P=12.(52+1).(54+1).(58+1).(516+1)
2P = 24.(5^2 + 1 )(5^4 + 1) ... (5^16 + 1)
2P = (5^2 - 1) (5^2 + 1) (5^4 + 1) .. (5^16+1)
2P = (5^4 - 1 )(5^4 + 1 ) (5^8 + 1)
2P = (5^8 - 1 ) (5^8 + 1) (5^16 + 1)
2P = ( 5^ 16 - 1 ) 5^ 16 + 1)
2P = 5^32 - 1
P = (5^32 - 1) : 2
\(P=12.\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow2P=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(\Leftrightarrow2P=5^{32}-1\)
\(\Leftrightarrow P=\frac{5^{32}-1}{2}\)
2P = 24.(5^2 + 1 )(5^4 + 1) ... (5^16 + 1)
2P = (5^2 - 1) (5^2 + 1) (5^4 + 1) .. (5^16+1)
2P = (5^4 - 1 )(5^4 + 1 ) (5^8 + 1)
2P = (5^8 - 1 ) (5^8 + 1) (5^16 + 1)
2P = ( 5^ 16 - 1 ) 5^ 16 + 1)
2P = 5^32 - 1
P = (5^32 - 1) : 2
đúng nha
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