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Ta có :
\(\frac{x+1}{65}+\frac{x+3}{63}< \frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{65}+1\right)+\left(\frac{x+3}{63}+1\right)< \left(\frac{x+5}{61}+1\right)+\left(\frac{x+7}{59}+1\right)\)
\(\Leftrightarrow\)\(\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+5}{61}-\frac{x+7}{59}< 0\)
\(\Leftrightarrow\)\(\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)
Vì \(\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)< 0\)
\(\Rightarrow\)\(x+66>0\)
\(\Rightarrow\)\(x>-66\)
Vậy \(x>-66\)
Bài 15:
Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)
\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)
\(\Leftrightarrow\frac{x+2+2008}{2008}+\frac{x+3+2007}{2007}+\frac{x+4+2006}{2006}+\frac{x+2028-18}{6}=0\)
\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)
Vì \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}>0\)
nên x+2010=0
hay x=-2010
Vậy: x=-2010
Bài 17:
Ta có: \(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+1+65}{65}+\frac{x+3+63}{63}=\frac{x+5+61}{61}+\frac{x+7+59}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}-\frac{x+66}{61}-\frac{x+66}{59}=0\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
Vì \(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\ne0\)
nên x+66=0
hay x=-66
Vậy: x=-66
a, <=> (59-x/41 + 1) + (57-x/43 + 1) + (55-x/45 + 1) + (53-x/47 + 1) + (51-x/49 + 1) = 0
<=> 100-x/41 + 100-x/43 + 100-x/45 + 100-x/47 + 100-x/49 = 0
<=> (100-x).(1/41+1/43+1/45+1/47+1/49) = 0
<=> 100-x=0 ( vì 1/41+1/43+1/45+1/47+1/49 > 0 )
<=> x=100
Vậy x = 100
b, <=> 2-x/2016 + 1 = (1-x/2017 + 1) + (1 - x/2018)
<=> 2018-x/2016 = 2018-x/2017 + 2018-x/2018
<=> 2018-x/2016 - 2018-x/2017 - 2018-x/2018 = 0
<=> (2018-x).(1/2016-1/2017-1/2018) = 0
<=> 2018-x=0 ( vì 1/2016-1/2017-1/2018 khác 0 )
<=> x=2018
Vậy x=2018
Tk mk nha
Bài 1:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
b) \(\frac{m^2\left(\left(x+2\right)^2-\left(x-2\right)^2\right)}{8}-4x=\left(m-1\right)^2+3\left(2m+1\right)\)
\(\Leftrightarrow m^2x-4x=m^2+4m+4\)
\(\Leftrightarrow\left(m^2-4\right)x=m^2+4m+4\)
Để phương trình vô nghiệm thì \(\hept{\begin{cases}m^2-4=0\\m^2+4m+4\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}m=2\vee m=-2\\\left(m+2\right)^2\ne0\end{cases}}\Leftrightarrow m=2\)
(x -1)/59 -1 +(x-2)/58 -1 +(x-3)/57 -1 = (x-3)/56 -1 +(x-4)/55 -1 +(x-5)/54 -1
<=> (x-60)/59 +(x-60)/58 + (X-60)/57 -(x-60)/56 - (X-60)/55 -(X-60)/54 =0
<=> (x-60).(1/59 +1/58 +1/57 -1/56 -1/55 - 1/54)=0
vì 1/59 +1/58 +1/57 -1/56 -1/55 -1/54 <0
nên x-60 =0 <=> x=60
đề bài của bạn bi sai vì vế trái không thể bằng vế phải nếu đề đúng thì phải là :
(x-1)/59 +(x-2)/58 +(x-3)/57 =(x-4)/56 +(x-5)/55 +(x-6)/54
khí đó bạn giải cách như trên ,chúc bạn học toán tốt
\(\left(8x^3-60x^2+150x-125\right)-\left(27x^3-108x^2+144x-64\right)+\left(x^3+3x^2+3x+1\right)=0\)
\(-18x^3+51x^2+9x-60=0\)
\(\left(2x-5\right)\left(x+1\right)\left(3x-4\right)=0\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-1\\x=\frac{4}{3}\end{array}\right.\)
\(\frac{59-x}{41}+\frac{57-x}{43}+\frac{55-x}{45}+\frac{53-x}{47}+\frac{51-x}{49}=-5\)
\(\Rightarrow\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\frac{51-x}{49}+1\)\(=-5+5\)
\(\Rightarrow\frac{59-x+49}{41}+\frac{57-x+43}{43}+\frac{55-x+45}{45}+\frac{53-x+47}{47}\)\(+\frac{51-x+49}{49}=0\)
\(\Rightarrow\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\frac{100-x}{49}=0\)
\(\Rightarrow\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
Vì \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\ne0\)
\(\Rightarrow100-x=0\)
\(\Rightarrow x=100\)
\(=\frac{59-x}{41}+1+\frac{57-x}{43}+1+\frac{55-x}{45}+1+\frac{53-x}{47}+1+\)
\(\frac{51-x}{49}+1=-5+5\)
đoạn này có 5 là do mik mượn 5 con 1 khi đó nha
\(=\frac{100-x}{41}+\frac{100-x}{43}+\frac{100-x}{45}+\frac{100-x}{47}+\)
\(\frac{100-x}{49}=0\)
\(=\left(100-x\right)\left(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}\right)=0\)
mà \(\frac{1}{41}+\frac{1}{43}+\frac{1}{45}+\frac{1}{47}+\frac{1}{49}< 0\)
nên 100-x=0
còn lại bn từ lm
trừ 1 vào mỗi phân thức ở hai vế
\(\left(x-2016\right)\left(\frac{1}{1953}+\frac{1}{1955}+\frac{1}{1957}+\frac{1}{1959}-\frac{1}{63}-\frac{1}{61}-\frac{1}{59}-\frac{1}{57}\right)=0\)
vì 1/1953 + 1/1955 + 1/1957 + 1/1959 -1/63 -1/61-1/59-1/57 khác0
=> x-2016=0 => x=2016
bạn có thể làm rỏ hơn được không.mình cảm ơn bạn nhiều