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b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)
\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)
=>(7x+10)(x-3)=0
hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)
d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)
\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)
\(\Leftrightarrow x^2+14x+68=0\)
hay \(x\in\varnothing\)
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
a) \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)ĐKXĐ : \(x\ne\pm3\)
\(\Leftrightarrow\dfrac{x\left(x-1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{x^2\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{3x-7x^2}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow x^3-4x^2+3x-x^3-3x^2=3x-7x^2\)
\(\Leftrightarrow-7x^2+3x-3x+7x^2=0\)
\(\Leftrightarrow0x=0\)( luôn đúng )
Vậy \(x\in R\left(x\ne\pm3\right)\)
b) \(\dfrac{2x-1}{x^3+1}=\dfrac{2}{x^2-x+1}-\dfrac{1}{x+1}\)ĐKXĐ : \(x\ne-1\)
\(\Leftrightarrow\dfrac{2x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x^2-x+1}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(\Rightarrow2x-1=2x+2-x^2+x-1\)
\(\Leftrightarrow2x+2-x^2+x-1-2x+1=0\)
\(\Leftrightarrow x^2-x-2=0\)
\(\Leftrightarrow x^2-2x+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\left(\text{t/m ĐKXĐ}\right)\\x=-1\left(\text{không t/m ĐKXĐ}\right)\end{matrix}\right.\)
Vậy....
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
\(a,\dfrac{3\left(5x-2\right)}{4}-2=\dfrac{7x}{3}-5\left(x-7\right)\)
\(\Leftrightarrow\dfrac{15x-6-8}{4}=\dfrac{7x-15\left(x-7\right)}{3}\)
\(\Leftrightarrow\dfrac{15x-14}{4}=\dfrac{7x-15x+105}{3}\)
\(\Leftrightarrow\dfrac{45x-42}{12}=\dfrac{-32x+420}{12}\)
\(\Leftrightarrow45x+32x=420+42\)
\(\Leftrightarrow77x=462\)
\(\Leftrightarrow x=6\)
\(b,\dfrac{x+5}{2}+\dfrac{3-2x}{4}=x-\dfrac{7+x}{6}\)
\(\Leftrightarrow\dfrac{2x+10+3-2x}{4}=\dfrac{6x-7-x}{6}\)
\(\Leftrightarrow\dfrac{13}{4}=\dfrac{5x-7}{6}\)
\(\Leftrightarrow2\left(5x-7\right)=3.13\)
\(\Leftrightarrow10x-14=39\)
\(\Leftrightarrow10x=53\)
\(\Leftrightarrow x=5,3\)
\(c,\dfrac{x-3}{11}+\dfrac{x+1}{3}=\dfrac{x+7}{9}-1\)
\(\Leftrightarrow\dfrac{3x-9+11x+11}{33}=\dfrac{x+7-9}{9}\)
\(\Leftrightarrow\dfrac{14x+2}{33}=\dfrac{x-2}{9}\)
\(\Leftrightarrow33\left(x-2\right)=9\left(14x+2\right)\)
\(\Leftrightarrow33x-66=126x+18\)
\(\Leftrightarrow-93x=84\)
\(\Leftrightarrow x=-\dfrac{28}{31}\)
\(d,\dfrac{3x-0,4}{2}+\dfrac{1,5-2x}{3}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{3\left(3x-0,4\right)+2\left(1,5-2x\right)}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{9x-1,2+3-4x}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow\dfrac{5x+1,8}{6}=\dfrac{x+0,5}{5}\)
\(\Leftrightarrow5\left(5x+1,8\right)=6\left(x+0,5\right)\)
\(\Leftrightarrow25x+9=6x+3\)
\(\Leftrightarrow19x=-6\)
\(\Leftrightarrow x=-\dfrac{6}{19}\)
\(\Leftrightarrow77x=378\)
\(\Leftrightarrow x=\dfrac{54}{11}\)
\(\text{a) }\left(x^2-9\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x+3\right)^2\left(x-3\right)^2-9\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x+9-9\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left(x^2+6x\right)\left(x-3\right)^2=0\\ \Leftrightarrow x\left(x+6\right)\left(x-3\right)^2=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\\left(x-3\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x+6=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-6\\x=3\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{0;3;-6\right\}\)
\(\text{b) }\dfrac{3x^2+7x-10}{x}=0\\ ĐKXĐ:x\ne0\\ \Rightarrow3x^2+7x-10=0\\ \Leftrightarrow3x^2-3x+10x-10=0\\ \Leftrightarrow\left(3x^2-3x\right)+\left(10x-10\right)=0\\ \Leftrightarrow3x\left(x-1\right)+10\left(x-1\right)=0\\ \Leftrightarrow\left(3x+10\right)\left(x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3x+10=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-10\\x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{10}{3}\\x=1\end{matrix}\right.\left(T/m\right)\)
Vậy phương trình có tập nghiệm \(S=\left\{-\dfrac{10}{3};1\right\}\)
\(\text{c) }x+\dfrac{2x+\dfrac{x-1}{5}}{3}=1-\dfrac{3x+\dfrac{1-2x}{3}}{5}\left(\text{Chữa đề}\right)\\ \Leftrightarrow15x+5\left(2x+\dfrac{x-1}{5}\right)=15-3\left(3x+\dfrac{1-2x}{3}\right)\\ \Leftrightarrow15x+10x+\left(x-1\right)=15-9x+\left(1-2x\right)\\ \Leftrightarrow15x+10x+x-1=15-9x+1-2x\\ \Leftrightarrow26x+11x=16+1\\ \Leftrightarrow37x=17\\ \Leftrightarrow x=\dfrac{17}{37}\\ \)
Vậy phương trình có nghiệm \(x=\dfrac{17}{37}\)
Câu 2:
ĐKXĐ: \(\left[{}\begin{matrix}1-9x^2\ne0\\1+3x\ne0\\1-3x\ne0\end{matrix}\right.\Rightarrow \left[{}\begin{matrix}x\ne\dfrac{-1}{3}\\x\ne\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{12}{1-9x^2}=\dfrac{1-3x}{1+3x}-\dfrac{1+3x}{1-3x}\left(1\right)\)
\(\left(1\right):\dfrac{12}{\left(1-3x\right)\left(1+3x\right)}-\dfrac{\left(1-3x\right)\left(1-3x\right)}{\left(1-3x\right)\left(1+3x\right)}+\dfrac{\left(1+3x\right)\left(1+3x\right)}{\left(1-3x\right)\left(1+3x\right)}=0\)
\(\Leftrightarrow 12-\left(1-3x-3x+9x^2\right)+\left(1+3x+3x+9x^2\right)=0\)
\(\Leftrightarrow 12-1+3x+3x-9x^2+1+3x+3x+9x^2=0\)
\(\Leftrightarrow12x+12=0\\ \Leftrightarrow12x=-12\\ \Leftrightarrow x=-1\left(TM\right)\)
Vậy \(S=\left\{-1\right\}\)
ĐK: ` x \ne \pm 3`
`(x^2-x)/(x+3)-(x^2)/(x-3)=(7x^2-3x)/(9-x^2)`
`<=> (x^2-x)(x-3)-x^2 (x+3) = -(7x^2-3x)`
`<=> −7x^2+3x=-7x^2+3x`
`<=> 0x=0 forall x`
Vậy `S=RR \\ {+-3}`.