.A) x/x+1=1/2 , B) x/2=x/3 , C) x+1/2=x+1/2017, i)3/x=x/2017 , O) x:3=x+1/2 , m) x+1/2=x+2/3 ,P) x+1/2=x , M) 2/x=x/8 , Q) x2/2=8/x2 , r) x3/2=32/x
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N
Giải pt : a) 2/-x2+6x-8 - x-1/x-2 = x+3/x-4
b) 2/x3-x2-x+1 = 3/1-x2 - 1/x+1
c) x+2/x-2 - 2/x2-2x = 1/x
3
PN
10 tháng 5 2021
a,\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\left(đkxđ:x\ne2;4\right)\)
\(< =>\frac{-2}{\left(x-2\right)\left(x-4\right)}-\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}=\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}\)
\(< =>-2-\left(x^2-5x+4\right)=x^2+x-5\)
\(< =>-x^2+5x-6-x^2-x+5=0\)
\(< =>-2x^2+4x-1=0\)
\(< =>2x^2-4x+1=0\)
đến đây thì pt bậc 2 dể rồi
PN
10 tháng 5 2021
\(\frac{2}{x^3-x^2-x+1}=\frac{3}{1-x^2}-\frac{1}{x+1}\left(đkxđ:x\ne\pm1\right)\)
\(< =>\frac{2}{x^2\left(x-1\right)-\left(x-1\right)}=\frac{3}{1-x^2}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x^2-1\right)\left(x-1\right)}=-\frac{3}{x^2-1}-\frac{1}{x+1}\)
\(< =>\frac{2}{\left(x+1\right)\left(x-1\right)^2}=\frac{-3\left(x-1\right)}{\left(x-1\right)^2\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x-1\right)^2\left(x+1\right)}\)
\(< =>2+3x-3+x^2-2x+1=0\)
\(< =>x^2+x=0< =>x\left(x+1\right)=0< =>\orbr{\begin{cases}x=-1\left(loai\right)\\x=0\left(tm\right)\end{cases}}\)
20 tháng 7 2021
a) Đặt A(x)=0
\(\Leftrightarrow-4x-5=0\)
\(\Leftrightarrow-4x=5\)
hay \(x=-\dfrac{5}{4}\)
b) Đặt B(x)=0
\(\Leftrightarrow3\left(2x-1\right)-2\left(x+1\right)=0\)
\(\Leftrightarrow6x-3-2x-2=0\)
\(\Leftrightarrow4x=5\)
hay \(x=\dfrac{5}{4}\)
29 tháng 10 2021
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
CN
31 tháng 12 2017
Bài 12:
1) A = x2 - 6x + 11
= (x2 - 6x + 9) + 2
= (x - 3)2 + 2
Ta có: (x - 3)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 3 = 0 ⇔ x = 3
Do đó: (x - 3)2 + 2 ≥ 2
Hay A ≥ 2
Dấu ''='' xảy ra khi x = 3
Vậy Min A = 2 tại x = 3
2) B = x2 - 20x + 101
= (x2 - 20x + 100) + 1
= (x - 10)2 + 1
Ta có: (x - 10)2 ≥ 0 ∀ x
Dấu ''='' xảy ra khi x - 10 = 0 ⇔ x = 10
Do đó: (x - 10)2 + 1 ≥ 1
Hay B ≥ 1
Dấu ''='' xảy ra khi x = 10
Vậy Min B = 1 tại x = 10
a) \(\frac{x}{x+1}=\frac{1}{2}\)
=> 2x = x + 1
=> 2x - x = 1
=> x = 1
b) \(\frac{x}{2}=\frac{x}{3}\)
=> 3x = 2x
=> 3x - 2x = 0
=> x = 0
c) \(\frac{x+1}{2}=\frac{x+1}{2017}\)
=> \(2017\left(x+1\right)=2\left(x+1\right)\)
=> 2017x + 2017 = 2x + 2
=> 2017x - 2x = 2 - 2017
=> 2015x = -2015
=> x = -2015 : 2015
=> x = -1
i) \(\frac{3}{x}=\frac{x}{2017}\)
=> x2 = 2017.3
=> x2 = 6051
=> \(\orbr{\begin{cases}x=\sqrt{6051}\\x=-\sqrt{6051}\end{cases}}\)
còn lại tự lm
\(a,\frac{x}{x+1}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}.\left(x+1\right)\)
\(\Rightarrow x=\frac{1}{2}x+\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}x=\frac{1}{2}\)
\(\Rightarrow x=1\)
\(b,\frac{x}{2}=\frac{x}{3}\)
\(\Rightarrow x=\frac{x}{3}.2\)
\(\Rightarrow x=\frac{2x}{3}\)
\(\Rightarrow3x=2x\)
\(\Rightarrow x=0\)
\(c,\frac{x+1}{2}=\frac{x+1}{2017}\)
\(\Rightarrow x+1=\frac{x+1}{2017}.2\)
\(\Rightarrow x+1=\frac{2x+2}{2017}\)
\(\Rightarrow2017x+2017=2x+2\)
\(\Rightarrow2017x-2x=2-2017\)
\(\Rightarrow2015x=-2015\)
\(\Rightarrow x=-1\)
\(i,\frac{3}{x}=\frac{x}{2017}\)
\(\Rightarrow x=3:\frac{x}{2017}\)
\(\Rightarrow x=\frac{6051}{x}\)
\(\Rightarrow x^2=6051\)
\(\Rightarrow x=\sqrt{6051}\)
\(o,\frac{x}{3}=\frac{x+1}{2}\)
\(\Rightarrow x=\frac{x+1}{2}.3\)
\(\Rightarrow x=\frac{3x+3}{2}\)
\(\Rightarrow2x=3x+3\)
\(\Rightarrow-x=3\)
\(\Rightarrow x=-3\)
\(m,\frac{x+1}{2}=\frac{x+2}{3}\)
\(\Rightarrow x+1=\frac{x+2}{3}.2\)
\(\Rightarrow x+1=\frac{2x+4}{3}\)
\(\Rightarrow3x+3=2x+4\)
\(\Rightarrow x=1\)
\(p,\frac{x+1}{2}=x\)
\(\Rightarrow2x=x+1\)
\(\Rightarrow x=1\)
\(m,\frac{2}{x}=\frac{x}{8}\)
\(\Rightarrow x=2:\frac{x}{8}\)
\(\Rightarrow x=\frac{16}{x}\)
\(\Rightarrow x^2=16\)
\(\Rightarrow x=4\)
\(Q,\frac{x^2}{2}=\frac{8}{x^2}\)
\(\Rightarrow x^2=\frac{8}{x^2}.2\)
\(\Rightarrow x^2=\frac{16}{x^2}\)
\(\Rightarrow x^4=16\)
\(\Rightarrow x=2\)
\(r,\frac{x^3}{2}=\frac{32}{x}\)
\(\Rightarrow x^3=\frac{32}{x}.2\)
\(\Rightarrow x^3=\frac{64}{x}\)
\(\Rightarrow x^4=64\)
\(\Rightarrow x=\sqrt[4]{64}\)