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a) 1x−1−3x2x3−1=2xx2+x+11x−1−3x2x3−1=2xx2+x+1
Ta có: x3−1=(x−1)(x2+x+1)x3−1=(x−1)(x2+x+1)
=(x−1)[(x+12)2+34]=(x−1)[(x+12)2+34] cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
x2+x+1−3x2=2x(x−1)⇔−2x2+x+1=2x2−2xx2+x+1−3x2=2x(x−1)⇔−2x2+x+1=2x2−2x
⇔4x2−3x−1=0⇔4x2−3x−1=0
⇔4x(x−1
a) \(\dfrac{2\left(x+7\right)}{x+1}-\dfrac{x+11}{x^2-1}=4-\dfrac{x-1}{x+1}\left(ĐKXĐ:x\ne1;x\ne-1\right)\)
\(\Leftrightarrow\dfrac{2\left(x+7\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}-\dfrac{x+11}{\left(x-1\right)\left(x+1\right)}=\dfrac{4\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}\)
\(\Rightarrow2x^2-2x+14x-14-x-11=4x^2-4-x^2+2x-1\)
\(\Leftrightarrow2x^2+11x-25=3x^2+2x-5\)
\(\Leftrightarrow2x^2+11x-25-3x^2-2x+5=0\)
\(\Leftrightarrow-x^2+9x-20=0\)
\(\Leftrightarrow-x^2-4x-5x-20=0\)
\(\Leftrightarrow-x\left(x+4\right)-5\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+\text{4}\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+4=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\end{matrix}\right.\)(thõa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm S = \(\left\{-4;-5\right\}\)
b) \(\dfrac{x+2}{x-3}+\dfrac{x+2}{x+3}=\dfrac{2\left(x+6\right)}{x^2-9}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x+2\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{2\left(x+6\right)}{x^2-9}\)
\(\Rightarrow x^2+3x+2x+6+x^2-3x+2x-6=2x+12\)
\(\Leftrightarrow2x^2+4x=2x+12\)
\(\Leftrightarrow2x^2+4x-2x-12=0\)
\(\Leftrightarrow2x^2+2x-12=0\)
\(\Leftrightarrow2x^2+6x-4x-12=0\)
\(\Leftrightarrow2x\left(x+3\right)-4\left(x+3\right)=0\)
\(\Leftrightarrow\left(2x-4\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-4=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)
\(\Leftrightarrow x=2\)(vì x=-3 ko thõa mãn ĐKXĐ)
Vậy phương trình có tập nghiệm S = \(\left\{2\right\}\)
d. ĐKXĐ: x khác 1, x khác 3
\(\dfrac{x+5}{x-1}=\dfrac{x+1}{\left(x-3\right)}-\dfrac{8}{x^2-4x+3}\)
\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+5\right)}{\left(x-1\right)\left(x-3\right)}=\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x-1\right)\left(x-3\right)}-\dfrac{8}{\left(x-1\right)\left(x-3\right)}\) \(\Leftrightarrow x^2+2x-15=x^2-1-8\)
\(\Leftrightarrow2x-15+1+8=0\)
\(\Leftrightarrow2x-6=0\)
\(\Leftrightarrow x=3\) (loại)
Vậy pt vô nghiệm
\(a.\Leftrightarrow\frac{3\left(x-2\right)-\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\frac{-9}{\left(x+1\right)\left(x-2\right)}.DKXD:x\ne-1;x\ne2\)
\(\Rightarrow3x-6-x-1=-9\)
\(\Leftrightarrow2x=-2\)
\(\Leftrightarrow x=-1\)
\(b.\frac{\left(x-4\right)\left(x+1\right)+\left(x+4\right)\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}=\frac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}.DKXDx\ne1;-1\)
\(\Rightarrow x^2+x-4x-4+x^2-x+4x-4=2x^2+2x-2x-2\)
\(\Leftrightarrow-6=0\left(voly\right)\)
vay \(S=\varnothing\)
a. (x + 2)(x2 – 3x + 5) = (x + 2)x2
⇔ (x + 2)(x2 – 3x + 5) – (x + 2)x2 = 0
⇔ (x + 2)[(x2 – 3x + 5) – x2] = 0
⇔ (x + 2)(\(x^2\) – 3x + 5 – \(x^2\)) = 0
⇔ (x + 2)(5 – 3x) = 0
⇔ x + 2 = 0 hoặc 5 – 3x = 0
x + 2 = 0 ⇔ x = -2
5 – 3x = 0 ⇔ x = \(\dfrac{5}{3}\)
Vậy phương trình có nghiệm x = -2 hoặc x =\(\dfrac{5}{3}\)
c.\(2x^2\) – x = 3 – 6x
⇔ \(2x^2\) – x + 6x – 3 = 0
⇔ (\(2x^2\) + 6x) – (x + 3) = 0
⇔ 2x(x + 3) – (x + 3) = 0
⇔ (2x – 1)(x + 3) = 0
⇔ 2x – 1 = 0 hoặc x + 3 = 0
2x – 1 = 0 ⇔ x = 1/2
x + 3 = 0 ⇔ x = -3
Vậy phương trình có nghiệm x = \(\dfrac{1}{2}\) hoặc x = -3