\(2003.2005\) và \(2004^2\)

b)

K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

AH
Akai Haruma
Giáo viên
30 tháng 9 2018

Lời giải:
Ta sử dụng các hằng đẳng thức đáng nhớ, cụ thể là công thức:
\((a-b)(a+b)=a^2-b^2\)

a)

\(2003.2005=(2004-1)(2004+1)=2004^2-1^2=2004^2-1< 2004^2\)

Vậy \(2003.2005< 2004^2\)

b)

\(8(7^8+1)(7^4+1)(7^2+1)=(7+1)(7^2+1)(7^4+1)(7^8+1)\)

\(=\frac{1}{6}.(7-1)(7+1)(7^2+1)(7^4+1)(7^8+1)\)

\(=\frac{1}{6}(7^2-1)(7^2+1)(7^4+1)(7^8+1)\)

\(=\frac{1}{6}(7^4-1)(7^4+1)(7^8+1)\)

\(=\frac{1}{6}(7^8-1)(7^8+1)=\frac{1}{6}(7^{16}-1)< 7^{16}-1\)

30 tháng 9 2018

Tks

18 tháng 7 2017

Câu 3 kiểm tra lại đề lại với , nếu đúng thì phức tạp lắm, còn sửa lại đề thì là :

\(y^2+2y+4^x-2^{x+1}+2=0\)

\(=>\left(y^2+2y+1\right)+2^{2x}-2^x.2+1=0\)

\(=>\left(y+1\right)^2+\left(\left(2^x\right)^2-2^x.2.1+1^2\right)=0\)

\(=>\left(y+1\right)^2+\left(2^x-1\right)^2=0\)

Dấu = xảy ra khi :

\(\hept{\begin{cases}y+1=0\\2^x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}y=-1\\x=0\end{cases}}}\)

CHÚC BẠN HỌC TỐT........... 

18 tháng 7 2017

mk chịu

23 tháng 7 2016

a) \(A=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)

\(=.............................................................\)

\(=\left(2^{16}-1\right)\left(2^{16}+1\right)=2^{32}-1=B-1\)

Suy ra A < B

b) \(A=2015.2017=\left(2016-1\right)\left(2016+1\right)=2016^2-1=B-1\)

Suy ra A < B

23 tháng 7 2016

Phần a bạn nhân thêm ở A là (2-1) là ra hằng đẳng thức, cứ thế mà triển. (Kết quả: A<B)

Phần b: phân tích A, ta có:

2015.2017= (2016-1).(2016+1)= 2016^2 -1 <2016^2

Suy ra: A<B

1: =>2x-5=4 hoặc 2x-5=-4

=>2x=9 hoặc 2x=1

=>x=9/2hoặc x=1/2

2: \(\Leftrightarrow\left|2x+1\right|=\dfrac{3}{4}-\dfrac{7}{8}=\dfrac{-1}{8}\)(vô lý)

3: \(\Leftrightarrow\left|5x-3\right|=x+5\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(5x-3-x-5\right)\left(5x-3+x+5\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>=-5\\\left(4x-8\right)\left(6x+2\right)=0\end{matrix}\right.\Leftrightarrow x\in\left\{2;-\dfrac{1}{3}\right\}\)

21 tháng 1 2018

câu b sai rồi bạn

\(x^8+4=\left(x^4+2\right)^2-4x^4\) mới đúng

26 tháng 9 2017

Bài1:

\(a,\left(-8\right)^9\)\(\left(-32\right)^5\)

Ta có:

\(\left(-8\right)^9=-2^{27}\)

\(\left(-32\right)^5=\left(-8.4\right)^5=-2^{27}.2^{10}\)

\(-2^{27}.10< -2^{27}\) nên \(\left(-8\right)^9>\left(-32\right)^5\)

Các câu sau tương tự

Bài2:

\(a,2\left|x-1\right|-3x=7\)

+)Xét \(x\ge1\Rightarrow\left|x-1\right|=x-1\)

Do đó:

\(2\left(x-1\right)-3x=7\\ \Leftrightarrow2x-2-3x=7\\ \Leftrightarrow-x=9\\ \Leftrightarrow x=-9\left(loại\right)\)

+)Xét \(x< 1\Rightarrow\left|x-1\right|=1-x\)

Do đó:

\(2\left(1-x\right)-3x=7\\ \Leftrightarrow2-2x-3x=7\\ \Leftrightarrow-5x=5\\ x=-1\left(chon\right)\)

Vậy x=-1

Câu b tương tự

26 tháng 9 2017

Bài 1:

\(a,\left(-8\right)^9\)\(\left(-32\right)^5\)

\(\left(-8\right)^9=\left[\left(-2\right)^3\right]^9=\left(-2\right)^{27}\)

\(\left(-32\right)^5=\left[\left(-2\right)^5\right]^5=\left(-2\right)^{25}\)

\(\left(-2\right)^{27}< \left(-2\right)^{25}\)

\(\Rightarrow\left(-8\right)^9< \left(-32\right)^5\)

\(b,2^{21}\)\(3^{14}\)

\(2^{21}=\left(2^3\right)^7\)

\(3^{14}=\left(3^2\right)^7\)

\(2^3< 3^2\)\(\Rightarrow2^{21}< 3^{14}\)

\(c,12^8\)\(8^{12}\)

\(12^8=\left(12^2\right)^4=144^4\)

\(8^{12}=\left(8^3\right)^4=512^4\)

\(144^4< 512^4\)\(\Rightarrow12^8< 8^{12}\)

\(d,\left(-5\right)^{39}\)\(\left(-2\right)^{91}\)

\(\left(-5\right)^{39}=\left[\left(-5\right)^3\right]^{13}\)

\(\left(-2\right)^{91}=\left[\left(-2\right)^7\right]^{13}\)

\(\left(-5\right)^3>\left(-2\right)^7\)\(\Rightarrow\left(-5\right)^{39}>\left(-2\right)^{91}\)

Bài 2:

\(a,2.\left|x-1\right|-3x=7\)

\(\left|x-1\right|=\dfrac{7+3x}{2}\)

Ta có 2 trường hợp:

Th1:\(x-1=\dfrac{7-3x}{2}\)

\(\dfrac{2x-2}{2}=\dfrac{7+3x}{2}\)

\(\Rightarrow2x-2=7+3x\)

\(2x-3x=7+2\)

\(-x=9\Rightarrow x=-9\)

Th2:\(x+1=-\dfrac{7+3x}{2}\)

\(\dfrac{2x-2}{2}=\dfrac{-7-3x}{2}\)

\(\Rightarrow2x-2=-7-3x\)

\(2x+3x=-7+2\)

\(5x=-5\Rightarrow x=-1\)

Vậy \(x\in\left\{-9;-1\right\}\)

\(b,\left|5x-3\right|=\left|7-x\right|\)

Ta có: Th1: \(\left|7-x\right|=7-x\) khi \(7-x\ge0\)\(\Rightarrow x\le7\)

\(5x-3=7-x\)

\(5x+x=7+3\)

\(6x=10\Rightarrow x=\dfrac{10}{6}=\dfrac{5}{3}\)( thoả mãn )

vì x thoả mãn \(x\le7\)\(\Rightarrow\) th1 thoả mãn x

Ta có: Th2: \(\left|7-x\right|=-\left(7-x\right)\) khi \(7-x< 0\Rightarrow x>7\)

\(5x-3=-\left(7-x\right)\)

\(5x-3=-7+x\)

\(5x-x=-7+3\)

\(4x=-4\Rightarrow x=-1\) ( loại )

Vì x thoả mãn \(x>7\)\(x=-1\Rightarrow\)th2 loại

AH
Akai Haruma
Giáo viên
22 tháng 9 2020

a)

$A=(1^2-2^2)+(3^2-4^2)+....+(2003^2-2004^2)+2005^2$

$=(1-2)(1+2)+(3-4)(3+4)+....+(2003-2004)(2003+2004)+2005^2$

$=-(1+2)-(3+4)-...-(2003+2004)+2005^2$

$=-(1+2+3+...+2004)+2005^2=-\frac{2004.2005}{2}+2005^2$

$=2005^2-1002.2005=2005(2005-1002)=2011015$

b)

$B=(2-1)(2+1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^8-1)(2^8+1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^{16}-1)(2^{16}+1)(2^{32}+1)-2^{64}$

$=(2^{32}-1)(2^{32}+1)-2^{64}$

$=2^{64}-1-2^{64}=-1$

AH
Akai Haruma
Giáo viên
22 tháng 9 2020

c) Do $x=16$ nên $x-16=0$

$R(x)=x^4-17x^3+17x^2-17x+20$

$=(x^4-16x^3)-(x^3-16x^2)+x^2-16x-x+20$

$=x^3(x-16)-x^2(x-16)+x(x-16)-x+20$

$=x^3.0-x^2.0+x.0-x+20=-x+20=-16+20=4$

d) Do $x=12$ nên $x-12=0$. Khi đó:

$S(x)=(x^{10}-12x^9)-(x^9-12x^8)+(x^8-12x^7)-....+(x^2-12x)-x+10$

$=x^9(x-12)-x^8(x-12)+x^7(x-12)-....+x(x-12)-x+10$

$=(x-12)(x^9-x^8+x^7-....+x)-x+10$

$=0-x+10=-x+10=-12+10=-2$

18 tháng 7 2016

a) \(3x\left(2x+1\right)=5\left(2x+1\right)\)

\(3x=5\)

\(x=\frac{5}{3}\)

b) \(\left(3x-8\right)^2=\left(2x-7\right)^2\)

\(3x-8=2x-7\)

\(x=1\)

c) \(\left(4x^2-3x-18\right)^2-\left(4x^2+3x\right)^2=0\)

\(\left(4x^2-3x-18\right)^2=\left(4x^2+3x\right)^2\)

\(4x^2-3x-18=4x^2+3x\)

\(6x=-18\)

\(x=-3\)

d) Sai đề

e) ko bt

10 tháng 8 2020

Ta có : \(\hept{\begin{cases}A=1999.2001\\B=2000^2\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+1999\\B=2000\cdot2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=1999.2000+2000+1\\B=1999.2000+2000\end{cases}}\)

\(< =>\hept{\begin{cases}A=2000.2000+1\\B=2000.2000\end{cases}}\)

\(< =>A>B\)

10 tháng 8 2020

a. Ta có : \(A=1999.2021=\left(2000-1\right)\left(2000+1\right)=2020^2-1< 2020\)

\(\Rightarrow A< B\)

b. Ta có : \(B=\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2-1\right)\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\)

...

\(=\left(2^8-1\right)\left(2^8+1\right)=2^{16}-1< 2^{16}\)

\(\Rightarrow A>B\)

c,d tương tự

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\) \(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\) \(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\) \(\Leftrightarrow x^2+5x-2x-10-5=1x+5\) \(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0 \(\Leftrightarrow\) \(x^2+2x-20=0\) \(\Leftrightarrow x^2+2x-10x-20=0\) \(\Leftrightarrow\) (x\(^2\) + 2x) - (10x +...
Đọc tiếp

3) \(\frac{x-2}{x-5}\) \(-\frac{5}{x^2-5x}=\frac{1}{x}\)

\(\Leftrightarrow\) \(\frac{x-2}{x-5}-\frac{5}{x.\left(x-5\right)}=\frac{1}{x}\)

\(\Leftrightarrow\frac{\left(x-2\right).\left(x+5\right)}{x.\left(x-5\right)}-\frac{5}{x.\left(x-5\right)}=\frac{1.\left(x+5\right)}{x.\left(x-5\right)}\)

\(\Leftrightarrow x^2+5x-2x-10-5=1x+5\)

\(\Leftrightarrow x^2+5x-2x-1x-10-5-5\) = 0

\(\Leftrightarrow\) \(x^2+2x-20=0\)

\(\Leftrightarrow x^2+2x-10x-20=0\)

\(\Leftrightarrow\) (x\(^2\) + 2x) - (10x + 20) = 0

\(\Leftrightarrow\) x.(x + 2) - 10.(x + 2) = 0

\(\Leftrightarrow\)

4) \(\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x^2+7x}\)

\(\Leftrightarrow\frac{x-4}{x+7}-\frac{1}{x}=\frac{-7}{x\left(x+7\right)}\)

\(\Leftrightarrow\frac{\left(x-4\right).\left(x+7\right)}{x.\left(x+7\right)}-\frac{1.\left(x+7\right)}{x.\left(x+7\right)}=\frac{-7}{x.\left(x+7\right)}\)

\(\Leftrightarrow\) \(x^2+7x-4x-28-x-7=-7\)

\(\Leftrightarrow x^2+7x-4x-x-28-7+7=0\)

\(\Leftrightarrow\) x\(^2\) + 2x - 28 = 0

\(\Leftrightarrow\) x\(^2\) + 2x - 14x - 28 = 0

\(\Leftrightarrow\) (x\(^2\) + 2x) - (14x + 28) = 0

\(\Leftrightarrow\) x.(x + 2) - 14.(x + 2) = 0

\(\Leftrightarrow\) (x - 14) = 0 hoặc (x + 2) = 0

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = -2 (Loại)

5) \(\frac{x+2}{x-2}+\frac{x-2}{x+2}=\frac{8x}{x^2-4}\)

\(\Leftrightarrow\) \(\frac{\left(x+2\right).\left(x+2\right)}{\left(x-2\right).\left(x+2\right)}+\frac{\left(x-2\right).\left(x-2\right)}{\left(x+2\right).\left(x-2\right)}=\frac{8x}{\left(x-2\right).\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x+2x+4+x^2-2x-2x+4=8x\)

\(\Leftrightarrow\) \(x^2+x^2+2x+2x-2x-2x-8x+4+4=0\)

\(\Leftrightarrow2x^2-8x+8=0\)

\(\Leftrightarrow\) 2x\(^2\) - 2x - 8x + 8 = 0

\(\Leftrightarrow\) 2x(x - 1) - 8(x - 1) = 0

\(\Leftrightarrow\) 2x - 8 = 0 hoặc x - 1 = 0

\(\Leftrightarrow\) 2x = 8 hoặc x = 1

\(\Leftrightarrow\) x = 4 (Nhận) hoặc x = 1 (Nhận)

Vậy S = {4; 1}

6) \(\frac{x+1}{x-1}-\frac{x-1}{x+1}=\frac{4}{x^2-1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}-\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x-1\right)}=\frac{4}{\left(x-1\right).\left(x+1\right)}\)

\(\Leftrightarrow\) x\(^2\) + x + x + 1 - x\(^2\) + x + x - 1 = 4

\(\Leftrightarrow\) 4x - 4 = 0

\(\Leftrightarrow\) 4 (x - 1) =0

\(\Leftrightarrow\) x - 1 = 0 / 4 = 0

\(\Leftrightarrow\) x = 1 (Nhận)

Vậy S = {1}

7) \(\frac{x+1}{x-1}+\frac{-4x}{x^2-1}=\frac{x-1}{x+1}\)

\(\Leftrightarrow\) \(\frac{\left(x+1\right).\left(x+1\right)}{\left(x-1\right).\left(x+1\right)}+\frac{-4x}{\left(x-1\right).\left(x+1\right)}=\frac{\left(x-1\right).\left(x-1\right)}{\left(x+1\right).\left(x+1\right)}\)

\(\Leftrightarrow x^2+x+x+1-4x=x^2-x-x+1\)

\(\Leftrightarrow\) 0

Vậy S ={\(\varnothing\)}

0