Rút gọn biểu thức
P = 12 ( 52 + 1 ) ( 54 + 1 ) ( 58 + 1 ) ( 516 + 1 )
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
PT
1
CM
8 tháng 11 2018
Ta có:
( 5 2 - 1).P = ( 5 2 – 1).12.( 5 2 + 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 2 – 1).( 5 2 + 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 4 - 1)( 5 4 + 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 8 - 1)( 5 8 + 1)( 5 16 + 1)
= 12.( 5 16 - 1)( 5 16 + 1)
= 12.( 5 32 - 1)
28 tháng 12 2017
Ta có: \(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(\Rightarrow P=\dfrac{24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{2}\)
\(\Rightarrow P=\dfrac{5^{32}-1}{2}\)
31 tháng 8 2021
\(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)=2\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\left(5^{128}-1\right)=2.5^{128}-2\)
31 tháng 8 2021
c: Ta có: \(C=48\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{16}-1\right)\cdot\left(5^{16}+1\right)\cdot\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{64}-1\right)\left(5^{64}+1\right)\)
\(=2\cdot\left(5^{128}-1\right)\)
\(=2\cdot5^{128}-2\)
A
3 tháng 8 2023
\(\text{ P = (2x-1).4x^2+2x+1+(x+1)x^2-x+1}\)
\(\text{P =}\) \(\text{[(2x-1) . 4x^2 ]}\)\(\text{[(x+1) .x^2]}\)
\(\text{P = }\) \(\text{8x^3 - 4x^2 + 2x^3 + 2x^2 + 2x + 1 + x^3 - x + 1}\)
\(\text{P =}\) \(\text{(8x^3 + 2x^3 + x^3) + (-4x^2 + 2x^2) + (2x - x) + (1 + 1)}\)
\(\text{P =}\) \(\text{11x^3 - 2x^2 + x + 2}\)
AH
Akai Haruma
Giáo viên
2 tháng 3 2021
Lời giải:
ĐK: $a\geq 0; a\neq 1$
\(P=\left[\frac{\sqrt{a}(\sqrt{a}+1)}{(a+1)(\sqrt{a}+1)}+\frac{1}{a+1}\right].\frac{a+1}{\sqrt{a}-1}\)
\(=\left(\frac{\sqrt{a}}{a+1}+\frac{1}{a+1}\right).\frac{a+1}{\sqrt{a}-1}=\frac{\sqrt{a}+1}{a+1}.\frac{a+1}{\sqrt{a}-1}=\frac{\sqrt{a}+1}{\sqrt{a}-1}\)
18 tháng 2 2021
P = \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)DKXD: \(\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)
= \(\sqrt{x}+\sqrt{x}\)
= \(2\sqrt{x}\)
Vậy tại x ∈ ĐKXĐ thì P = \(2\sqrt{x}\)
18 tháng 2 2021
Ta có: \(P=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}+1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\)
\(=\sqrt{x}+\sqrt{x}\)
\(=2\sqrt{x}\)
20 tháng 10 2022
Bài4:
=>x(x^2+1)=0
>x=0
Bài 5:
=>\(3n^3+n^2+9n^2+3n-3n-1-4⋮3n+1\)
=>\(3n+1\in\left\{1;-1;2;-2;4;-4\right\}\)
hay \(n\in\left\{0;-\dfrac{2}{3};\dfrac{1}{3};-1;1;-\dfrac{5}{3}\right\}\)
DH
3 tháng 3 2021
\(P=\left(\dfrac{x}{x-\sqrt{x}}+\dfrac{1}{\sqrt{x}+1}-\dfrac{2}{1-x}\right):\left(\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\right)\)
\(P=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(x-1\right)}+\dfrac{\left(\sqrt{x}-1\right)}{\left(x-1\right)}+\dfrac{2}{x-1}\right):\left(\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\right)\)
\(P=\dfrac{x+\sqrt{x}+\sqrt{x}-1+2}{x-1}:\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=\dfrac{x+2\sqrt{x}+1}{x-1}:\dfrac{\sqrt{x}-2}{1-\sqrt{x}}\)
\(P=\dfrac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}:\dfrac{\sqrt{x}-2}{1-\sqrt{x}}=-\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\)
3 tháng 3 2021
Ta có: \(P=\left(\dfrac{2\sqrt{x}+x+1}{\sqrt{x}+1}\right)\cdot\left(1-\dfrac{x-\sqrt{x}}{\sqrt{x}-1}\right):\left(1-\sqrt{x}\right)\)
\(=\left(\dfrac{\left(\sqrt{x}+1\right)^2}{\sqrt{x}+1}\right)\cdot\left(1-\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}-1}\right)\cdot\dfrac{1}{1-\sqrt{x}}\)
\(=\left(\sqrt{x}+1\right)\left(1-\sqrt{x}\right)\cdot\dfrac{1}{1-\sqrt{x}}\)
\(=\sqrt{x}+1\)
DH
2 tháng 3 2021
\(P=\left(\dfrac{2\sqrt{x}+2}{x\sqrt{x}-\sqrt{x}+x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{1}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{2}{\left(x-1\right)}-\dfrac{1}{\sqrt{x}-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{2}{\left(x-1\right)}-\dfrac{\left(\sqrt{x}+1\right)}{\left(x-1\right)}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(P=\left(\dfrac{1-\sqrt{x}}{x-1}\right):\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)=-\left(\sqrt{x}+2\right)\)
NM
2 tháng 3 2021
ĐKXĐ x\(\ge0;x\ne1\)
P=\(\left(\dfrac{2\sqrt{x}-2}{x\sqrt{x}-\sqrt{x}+x-1}-\dfrac{1}{\sqrt{x}-1}\right):\left(1+\dfrac{1}{\sqrt{x}+1}\right)\)
P=\(\left(\dfrac{2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)^2}-\dfrac{1}{\sqrt{x}-1}\right):\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
P=\(\dfrac{2-\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
P=\(\dfrac{1-\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}.\dfrac{\sqrt{x}+1}{\sqrt{x}+2}=\dfrac{-1}{\sqrt{x}+2}\)
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)
2P=(5-1)(5+1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
2P=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
2P=(5^4-1)(5^4+1)(5^8+1)(5^16+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