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d, \(\frac{x+1}{9}+\frac{x+2}{8}=\frac{x+3}{7}+\frac{x+4}{6}\)
\(\Leftrightarrow\frac{x+1}{9}+1+\frac{x+2}{8}+1=\frac{x+3}{7}+1+\frac{x+4}{6}+1\)
\(\Leftrightarrow\frac{x+10}{9}+\frac{x+10}{8}-\frac{x+10}{7}-\frac{x+10}{6}=0\)
\(\Leftrightarrow\left(x+10\right)\left(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\right)=0\)
\(\Leftrightarrow x+10=0\) (Vì \(\frac{1}{9}+\frac{1}{8}-\frac{1}{7}-\frac{1}{6}\) ≠ 0)
\(\Leftrightarrow x=-10\)
Vậy x = -10 là nghiệm của phương trình.
Bài 3 :
Ta có : \(A=x^2+x+2012\)
=> \(A=x^2+x+\left(\frac{1}{2}\right)^2+\frac{8047}{4}\)
=> \(A=\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\)
- Ta thấy : \(\left(x+\frac{1}{2}\right)^2\ge0\forall x\)
=> \(\left(x+\frac{1}{2}\right)^2+\frac{8047}{4}\ge\frac{8047}{4}\forall x\)
- Dấu "=" xảy ra <=> \(x+\frac{1}{2}=0\)
<=> \(x=-\frac{1}{2}\)
Vậy MinA = \(\frac{8047}{4}\) <=> x = \(-\frac{1}{2}\) .
Bài 1 :
a, Ta có : \(\left(3x-2\right)\left(4+5x\right)=0\)
=> \(\left[{}\begin{matrix}3x-2=0\\4+5x=0\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}3x=2\\5x=-4\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{2}{3}\\x=-\frac{4}{5}\end{matrix}\right.\)
Vậy phương trình có nghiệm là x = \(\frac{2}{3}\), x = \(-\frac{4}{5}\) .
b,- ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) => \(\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)
=> \(x\ne\pm1\)
Ta có : \(\frac{x+1}{x-1}-\frac{4}{x+1}=\frac{3-x^2}{1-x^2}\)
=> \(\frac{\left(x+1\right)^2}{x^2-1}-\frac{4\left(x-1\right)}{x^2-1}=\frac{x^2-3}{x^2-1}\)
=> \(\left(x+1\right)^2-4\left(x-1\right)=x^2-3\)
=> \(x^2+2x+1-4x+4=x^2-3\)
=> \(-2x=-3-5\)
=> \(x=4\left(TM\right)\)
Vậy phương trình có nghiệm là x = 4 .
c, Ta có : \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}-\frac{2-10x}{2014}\)
=> \(\frac{10x+3}{2009}+\frac{10x-1}{2013}=\frac{10x+1}{2011}+\frac{10x-2}{2014}\)
=> \(\frac{10x+3}{2009}+1+\frac{10x-1}{2013}+1=\frac{10x+1}{2011}+1+\frac{10x-2}{2014}+1\)
=> \(\frac{10x+3}{2009}+\frac{2009}{2009}+\frac{10x-1}{2013}+\frac{2013}{2013}=\frac{10x+1}{2011}+\frac{2011}{2011}+\frac{10x-2}{2014}+\frac{2014}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}=\frac{10x+2012}{2011}+\frac{10x+2012}{2014}\)
=> \(\frac{10x+2012}{2009}+\frac{10x+2012}{2013}-\frac{10x+2012}{2011}-\frac{10x+2012}{2014}=0\)
=> \(\left(10x+2012\right)\left(\frac{1}{2009}+\frac{1}{2013}-\frac{1}{2011}-\frac{1}{2014}\right)=0\)
=> \(10x+2012=0\)
=> \(x=-\frac{2012}{10}\)
Vậy phương trình có nghiệm là x = \(-\frac{2012}{10}\) .
Bài 3:
Giải:
Ta có : A = x2 + x + 2012
= x2 + 2.\(\frac{1}{2}\).x + \(\frac{1}{4}\) + \(\frac{8047}{4}\)
= (x + \(\frac{1}{2}\))2 + \(\frac{8047}{4}\) ≥ \(\frac{8047}{4}\)
⇒ Amin = \(\frac{8047}{4}\) ⇔ (x + \(\frac{1}{2}\))2 = 0 ⇔ x = \(-\frac{1}{2}\)
Vậy Amin = \(\frac{8047}{4}\) tại x = \(-\frac{1}{2}\)
Chúc bạn học tốt@@
a) \(\frac{4x-8}{2x^2+1}=0\)
\(\Rightarrow4x-8=0\left(2x^2+1\ne0\right)\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\)
Vậy x=2
b)
\(\frac{x^2-x-6}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+2\right)}{x-3}=0\)
\(\Rightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
Vậy x=-2
\(a.\frac{x-6}{x-4}=\frac{x}{x-2}\\\Leftrightarrow \frac{\left(x-6\right)\left(x-2\right)}{\left(x-4\right)\left(x-2\right)}=\frac{x\left(x-4\right)}{\left(x-4\right)\left(x-2\right)}\\\Leftrightarrow \left(x-6\right)\left(x-2\right)=x\left(x-4\right)\\\Leftrightarrow \left(x-6\right)\left(x-2\right)-x\left(x-4\right)=0\\ \Leftrightarrow x^2-2x-6x+12-x^2+4x=0\\\Leftrightarrow -4x+12=0\\\Leftrightarrow -4x=-12\\ \Leftrightarrow x=3\)
\(b.1+\frac{2x-5}{x-2}-\frac{3x-5}{x-1}=0\\ \Leftrightarrow\frac{\left(x-2\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}+\frac{\left(2x-5\right)\left(x-1\right)}{\left(x-2\right)\left(x-1\right)}-\frac{\left(3x-5\right)\left(x-2\right)}{\left(x-2\right)\left(x-1\right)}=0\\ \Leftrightarrow\left(x-2\right)\left(x-1\right)+\left(2x-5\right)\left(x-1\right)-\left(3x-5\right)\left(x-2\right)=0\\ \Leftrightarrow x^2-x-2x+3+2x^2-2x-5x+5-3x^2+6x+5x-10=0\\ \Leftrightarrow x-2=0\\ \Leftrightarrow x=2\\ \)
a)2x-5/x+5=3=>2x-5=3(x+5)=3x+15
=>2x=3x+20=>x=-20
b)(x^2-6)/x=x+3/2
=>(x^2-6)/x - x=3/2
=>-6/x[quy đồng]=3/2
=>x=-4
c)Để (x^2+2x)−(3x+6)/x−3=0
thì (x^2+2x)−(3x+6)=0
=x(x+2)-3(x+2)=(x-3)(x+2)=0
=>x=3 hoặc x=-2
Mà ở mẫu có x-3 nếu x=3 thì mẫu =0=>loại
Vậy x=2
d)5/3x+2=2x−1
=>5=(3x+2)(2x-1)
Tìm ước của 5 rùi thay vào 3x+2 và 2x-1 rùi tìm x,cái đó dễ nên bn tự lm nhé
e)
(2x−1/x−1)+1=1/x−1
=>1/x-1-2x-1/x-1=1
=>-2x/x-1=1
=>-2x=x-1
=>x=1/3
g)(x+3/x+1)+(x−2/x)=2
=>quy đồng rùi tính và tìm x nhé bn,mk mỏi tay rùi
nhớ tick cho mk nha,mk siêng lắm ms ghi cho bn nhiều thế này nè,nhớ tick nha,thanks
a) \(\frac{2x-5}{x+5}=3\)
\(\Leftrightarrow2x-5=3\left(x+5\right)\)
\(\Leftrightarrow2x-5=3x+15\)
\(\Leftrightarrow2x-3x=15+5\)
\(\Leftrightarrow-x=20\\ \)
\(\Leftrightarrow x=-20\)
b) \(\frac{x^2-6}{x}=x+\frac{3}{2}\)
\(\Leftrightarrow\frac{x^2-6}{x}=\frac{2x+3}{2}\)
\(\Leftrightarrow2\left(x^2-6\right)=x\left(2x+3\right)\)
\(\Leftrightarrow2x^2-12=2x^2+3x\)
\(\Leftrightarrow3x=-12\)
\(\Leftrightarrow x=-4\)
c) \(\frac{\left(x^2+2x\right)-\left(3x+6\right)}{x-3}=0\)
\(\Leftrightarrow\frac{x\left(x+2\right)-3\left(x+2\right)}{x-3}=0\)
\(\Leftrightarrow\frac{\left(x+2\right)\left(x-3\right)}{x-3}=0\)
\(\Leftrightarrow x+2=0\)
\(\Leftrightarrow x=-2\)
d) \(\frac{5}{3x+2}=2x-1\)
\(\Leftrightarrow5=\left(2x-1\right)\left(3x+2\right)\)
\(\Leftrightarrow5=6x^2+x-2\)
\(\Leftrightarrow6x^2+x-7=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}1\\\frac{-7}{6}\end{array}\right.\)
e) \(\frac{2x-1}{x-1}+1=\frac{1}{x-1}\)
\(\Leftrightarrow2x-1+x-1=1\)
\(\Leftrightarrow3x=3\)
\(\Leftrightarrow x=1\)
g) \(\frac{x+3}{x+1}+\frac{x-2}{x}=2\)
\(\Leftrightarrow\frac{x\left(x+3\right)}{x\left(x+1\right)}+\frac{\left(x-2\right)\left(x+1\right)}{x\left(x+1\right)}=\frac{2x\left(x+1\right)}{x\left(x+1\right)}\)
\(\Leftrightarrow x\left(x+3\right)+\left(x-2\right)\left(x+1\right)=2x\left(x+1\right)\)
\(\Leftrightarrow x^2+3x+x^2-x-2=2x^2+2x\)
\(\Leftrightarrow2x-2x-2=0\)
\(\Leftrightarrow-2=0\) \(\Rightarrow\)Phương trình vô nghiệm
a) ĐK: \(x\ne2\).
\(\frac{x-3}{x-2}-2=\frac{x+1}{x-2}\)
\(\Leftrightarrow\frac{x-3}{x-2}-\frac{2\left(x-2\right)}{x-2}=\frac{x+1}{x-2}\)
\(\Rightarrow x-3-2\left(x-2\right)=x+1\)
\(\Leftrightarrow2x=0\)
\(\Leftrightarrow x=0\)(thỏa mãn)
2) ĐK: \(x\ne\pm1\).
\(\frac{x+2}{x^2-1}+\frac{1}{x-1}=0\)
\(\Leftrightarrow\frac{x+2}{x^2-1}+\frac{x+1}{\left(x-1\right)\left(x+1\right)}=0\)
\(\Rightarrow x+2+x+1=0\)
\(\Leftrightarrow x=-\frac{3}{2}\)(tm)
c) ĐK:\(x\ne0,x\ne-1\).
\(\frac{x}{x+1}-\frac{1}{x}=\frac{1}{x^2+x}\)
\(\Leftrightarrow\frac{x^2}{x\left(1+x\right)}-\frac{x+1}{x\left(x+1\right)}=\frac{1}{x\left(x+1\right)}\)
\(\Rightarrow x^2-\left(x+1\right)=1\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\x=-1\left(l\right)\end{cases}}\)
d) ĐK: \(x\ne1\).
\(\frac{3x-1}{x^3-1}=\frac{1}{x-1}-\frac{x}{x^2+x+1}\)
\(\Leftrightarrow\frac{3x-1}{x^3-1}=\frac{x^2+x+1}{\left(x-1\right)\left(x^2+x+1\right)}-\frac{x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\Rightarrow3x-1=x^2+x+1-x\left(x-1\right)\)
\(\Leftrightarrow3x-1=2x+1\)
\(\Leftrightarrow x=2\)(tm)