B
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
30 tháng 1 2019
\(b,\)\(\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Rightarrow\left(\frac{x+1}{2008}+1\right)+\left(\frac{x+2}{2007}+1\right)+\left(\frac{x+3}{2006}+1\right)=\left(\frac{x+4}{2005}+1\right)+\left(\frac{x+5}{2004}+1\right)+\left(\frac{x+6}{2003}+1\right)\)
\(\Rightarrow\frac{x+2009}{2008}+\frac{x+2009}{2007}+\frac{x+2009}{2006}=\frac{x+2009}{2005}+\frac{x+2009}{2004}+\frac{x+2009}{2003}\)
\(\Rightarrow\left(x+9\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=\left(x+9\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)\)
\(\Rightarrow\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}=\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\left(KTM\right)\)
S
30 tháng 1 2019
\(\text{Giải}\)
\(b,\frac{x+1}{2008}+\frac{x+2}{2007}+\frac{x+3}{2006}=\frac{x+4}{2005}+\frac{x+5}{2004}+\frac{x+6}{2003}\)
\(\Leftrightarrow\left(x+2009\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}-\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2009=0\Leftrightarrow x=-2009\)
16 tháng 3 2018
\(2x^4-9x^3+14x^2-9x+2=0\)
\(\Leftrightarrow2x^4-4x^3-5x^3+10x^2+4x^2-8x-x+2=0\)
\(\Leftrightarrow2x^3\left(x-2\right)-5x^2\left(x-2\right)+4x\left(x-2\right)-\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-5x^2+4x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^3-2x^2-3x^2+3x+x-1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[2x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(2x^2-3x+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-1\right)\left(2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=\dfrac{1}{2}\end{matrix}\right.\)
23 tháng 1 2020
1) 2x4 - 9x3 + 14x2 - 9x + 2 = 0
<=> (2x4 - 4x3) - (5x3 - 10x2) + (4x2 - 8x) - (x - 2) = 0
<=> 2x3(x - 2) - 5x2(x - 2) + 4x(x - 2) - (x - 2) = 0
<=> (2x3 - 5x2 + 4x - 1)(x - 2) = 0
<=> [(2x3 - 2x2) - (3x2 - 3x) + (x - 1)](x - 2) = 0
<=> [2x2(x - 1) - 3x(x - 1) + (x - 1)](x - 2) = 0
<=> (2x2 - 2x - x + 1)(x - 1)(x - 2) = 0
<=> (2x - 1)(x - 1)2(x - 2) = 0
<=> 2x - 1=0
hoặc x - 1 = 0
hoặc x - 2 = 0
<=> x = 1/2
hoặc x = 1
hoặc x = 2
Vậy S = {1/2; 1; 2}
27 tháng 11 2018
1/ Ta có
\(x^2+9x+20=x^2+4x+5x+20=x\left(x+4\right)+5\left(x+4\right)=\left(x+4\right)\left(x+5\right)\)
Tương tự
\(x^2+11x+30=\left(x+5\right)\left(x+6\right)\)
\(x^2+13x+42=\left(x+6\right)\left(x+7\right)\)
Đk: x khác 4, 5, 6, 7
\(\frac{1}{\left(x+4\right)\left(x+5\right)}+\frac{1}{\left(x+5\right)\left(x+6\right)}+\frac{1}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{\left(x+5\right)-\left(x+4\right)}{\left(x+4\right)\left(x+5\right)}+\frac{\left(x+6\right)-\left(x+5\right)}{\left(x+5\right)\left(x+6\right)}+\frac{\left(x+7\right)-\left(x+6\right)}{\left(x+6\right)\left(x+7\right)}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+5}+\frac{1}{x+5}-\frac{1}{x+6}+\frac{1}{x+6}-\frac{1}{x+7}=\frac{1}{18}\)
\(\Leftrightarrow\frac{1}{x+4}-\frac{1}{x+7}=\frac{1}{18}\) EM tự làm tiếp nhé
AH
Akai Haruma
Giáo viên
29 tháng 1 2020
Lời giải:
$2x^4-9x^3+14x^2-9x+2=0$
$\Leftrightarrow 2x^4-2x^3-7x^3+7x^2+7x^2-7x-2x+2=0$
$\Leftrightarrow 2x^3(x-1)-7x^2(x-1)+7x(x-1)-2(x-1)=0$
$\Leftrightarrow (x-1)(2x^3-7x^2+7x-2)=0$
$\Leftrightarrow (x-1)[2(x^3-1)-7x(x-1)]=0$
$\Leftrightarrow (x-1)(x-1)(2x^2+2x+2-7x)=0$
$\Leftrightarrow (x-1)^2(2x^2-5x+2)=0$
$\Leftrightarrow (x-1)^2(2x^2-4x-x+2)=0$
$\Leftrightarrow (x-1)^2[2x(x-2)-(x-2)]=0$
$\Leftrightarrow (x-1)^2(2x-1)(x-2)=0$
\(\Rightarrow \left[\begin{matrix} x=1\\ x=\frac{1}{2}\\ x=2\end{matrix}\right.\)
BN
2
18 tháng 4 2020
Ta có: \(2-x+2005=1-x+2006=-x+2007\)
\(\frac{2-x}{2005}-1=\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\frac{2-x}{2005}+1-2=\frac{1-x}{2006}+1+\left(\frac{-x}{2007}+1\right)-2\)
\(\Leftrightarrow\frac{2007-x}{2005}=\frac{2007-x}{2006}+\frac{2007-x}{2007}\)
\(\Leftrightarrow\left(2007-x\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Rightarrow2007-x=0\)
\(\Rightarrow x=2007\)
18 tháng 4 2020
\(\frac{2-x}{2005}-1=\frac{1-x}{2006}-\frac{x}{2007}\)
\(\Leftrightarrow\frac{2-x}{2005}-\frac{1-x}{2006}+\frac{x}{2007}-1=0\)
\(\Leftrightarrow\frac{2-x}{2005}+1-\frac{1-x}{2006}-1+\frac{x}{2007}-1=0\)
\(\Leftrightarrow\left(\frac{2-x}{2005}+1\right)-\left(\frac{1-x}{2006}+1\right)-\left(1-\frac{x}{2007}\right)=0\)
\(\Leftrightarrow\frac{2-x+2005}{2005}-\frac{1-x+2006}{2006}-\frac{2007-x}{2007}=0\)
\(\Leftrightarrow\frac{2007-x}{2005}-\frac{2007-x}{2006}-\frac{2007-x}{2007}=0\)
\(\Leftrightarrow\left(2007-x\right)\left(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\right)=0\)
\(\Leftrightarrow2007-x=0\) < Vì \(\frac{1}{2005}-\frac{1}{2006}-\frac{1}{2007}\ne0\)>
\(\Leftrightarrow x=2007\)
VẬY \(x=2007\)
12 tháng 3 2021
a/ \(9x^2+y^2=18x+6y-18\)
\(\Leftrightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)=0\)
\(\Leftrightarrow9\left(x-1\right)^2+\left(y-3\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}\)
LD
12 tháng 3 2021
a) \(9x^2+y^2=18x+6y-18\)
\(\Rightarrow9x^2+y^2-18x-6y+9=0\)
\(\Rightarrow\left(9x^2-18x+9\right)+\left(y^2-6y+9\right)=0\)
\(\Rightarrow9\left(x-1\right)^2+\left(y-3\right)^2=0\)
Mà \(\hept{\begin{cases}9\left(x-1\right)^2\ge0\\\left(y-3\right)^2\ge0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}9\left(x-1\right)^2=0\\\left(y-3\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=1\\y=3\end{cases}}}\)
Vậy ....................
Câu b để mik nghĩ tiếp
HB
1
2 tháng 12 2018
\(9x+2=y^2+y\Rightarrow9x+2=y\left(y+1\right)\)
\(\Rightarrow9x+2⋮2\Rightarrow9x⋮2\Rightarrow x⋮2\)
Vậy x chia hết cho 2 (cứ thay 1 số x chia hết cho 2 thì tìm được 1 số y)
Vậy có vô số x,y thỏa mãn đề.
CC
1
HN
26 tháng 1 2019
\(\frac{x-3}{2011}+\frac{x-5}{2009}+\frac{x-7}{2007}+\frac{x-9}{2005}=4\)
\(\Leftrightarrow\left(\frac{x-3}{2011}-1\right)+\left(\frac{x-5}{2009}-1\right)+\left(\frac{x-7}{2007}-1\right)+\left(\frac{x-9}{2005}-1\right)=0\)
\(\Leftrightarrow\frac{x-2014}{2011}+\frac{x-2014}{2009}+\frac{x-2014}{2007}+\frac{x-2014}{2005}=0\)
\(\Leftrightarrow\left(x-2014\right)\left(\frac{1}{2011}+\frac{1}{2009}+\frac{1}{2007}+\frac{1}{2005}\right)=0\)
|________________A________________|
Do A > 0
nên x - 2014 = 0
<=> x = 2014
\(x^{2007}-9x^{2005}+5x^2-14x-3=0\)
\(\Leftrightarrow x^{2005}(x^{2}-9)+5x^{2}-15x+x-3=0\)
\(\Leftrightarrow x^{2005}(x-3)(x+3)+5x(x-3)+x-3=0\)
\(\Leftrightarrow (x^{2006}+3x^{2005}+5x+1)(x-3)=0\)
Xét đa thức : \(P(x)=x^{2006}+3x^{2005}+5x+1\)
\(P(x)<0\) với \(x \in \{-1;-2;-3 \}\)
\(P(x)>0\) với \(x \ge 0\) hoặc \(x \le -4\)
Vậy \(P(x) \ne 0\) \(\forall x\inℤ\)nên x = 3