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Lời giải:
ĐKXĐ:.....
Ta có: \(\frac{2x}{3x^2-x+2}-\frac{7x}{3x^2+5x+2}=1\)
\(\Leftrightarrow \frac{1}{6}+\frac{2x}{3x^2-x+2}-7\left(\frac{x}{3x^2+5x+2}+\frac{1}{6}\right)=0\)
\(\Leftrightarrow \frac{3x^2+11x+2}{6(3x^2-x+2)}-\frac{7(3x^2+11x+2)}{6(3x^2+5x+2)}=0\)
\(\Leftrightarrow \frac{1}{6}(3x^2+11x+2)\left(\frac{1}{3x^2-x+2}-\frac{7}{3x^2+5x+2}\right)=0\)
TH1: \(3x^2+11x+2=0\)
\(\Leftrightarrow x=\frac{-11\pm \sqrt{97}}{6}\) (thỏa mãn)
TH2: \(\frac{1}{3x^2-x+2}-\frac{7}{3x^2+5x+2}=0\)
\(\Leftrightarrow \frac{2}{3x^2-x+2}-\frac{7}{3x^2+5x+2}=\frac{1}{3x^2-x+2}\)
\(\Leftrightarrow \frac{1}{x}=\frac{1}{3x^2-x+2}\)
\(\Leftrightarrow x=3x^2-x+2\)
\(\Leftrightarrow 3x^2-2x+2=0\)
\(\Leftrightarrow 2x^2+(x-1)^2+1=0\) (vô lý)
Do đó PT có nghiệm \(x=\frac{-11\pm \sqrt{97}}{6}\)
Edogawa Conan: đúng hay không bạn cứ thử giá trị của x đã tính vào là được :)
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Ta có: 6x4 -x3-7x2+x+1=0
<=> 6x4-6x3+5x3-5x2-2x2+2x-x+1=0
<=> 6x3(x-1)+5x2(x-1)-2x(x-1)-(x-1)=0
<=> (x-1) ( 6x3+5x2-2x-1)=0
<=> ( x-1) ( 6x3-3x2+8x2-4x+2x-1)=0
<=> (x-1)\(\left[3x^2\left(2x-1\right)+4x\left(2x-1\right)+\left(2x-1\right)\right]\)=0
<=> (x-1) ( 2x-1) ( 3x2+4x+1)=0
<=> (x-1) ( 2x-1) (3x2+3x+x+1)=0
<=> (x-1) (2x-1) \(\left[3x\left(x+1\right)+\left(x+1\right)\right]\)=0
<=> (x-1)(2x-1)(x+1)(3x+1)=0
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-1=0\\x+1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=1\\x=-1\\3x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2}\\x=-1\\x=\dfrac{-1}{3}\end{matrix}\right.\)
vậy \(S=\left\{\pm1;\dfrac{1}{2};\dfrac{-1}{3}\right\}\)
\(6x^4-x^3-7x^2+x+1=0\)
\(\Leftrightarrow6x^4-6x^3+5x^3-5x^2-2x^2+2x-x+1=0\)
\(\Leftrightarrow6x^3\left(x-1\right)+5x^2\left(x-1\right)-2x\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+5x^2-2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(6x^3+6x^2-x^2-x-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[6x^2\left(x+1\right)-x\left(x+1\right)-\left(x+1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(6x^2-3x+2x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(2x-1\right)\left(3x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\2x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=\dfrac{1}{2}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\dfrac{2x-1}{3x^2+7x+2}+\dfrac{3}{9x^2+15x+4}-\dfrac{2x+7}{3x^2-5x-12}=\dfrac{5}{x+2}\)
\(\Leftrightarrow\dfrac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\dfrac{3}{\left(3x+1\right)\left(3x+4\right)}-\dfrac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\dfrac{5}{x+2}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{3x+1}+\dfrac{1}{3x+1}-\dfrac{1}{3x+4}+\dfrac{1}{3x+4}-\dfrac{1}{x-3}=\dfrac{5}{x+2}\)
\(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x-3}=\dfrac{5}{x+2}\)
\(\Leftrightarrow\dfrac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\dfrac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow5x-3=-5\)
\(\Leftrightarrow x=-\dfrac{2}{5}\)
Vậy...
a: \(\Leftrightarrow20x^2-12x+15x+5< 10x\left(2x+1\right)-30\)
\(\Leftrightarrow20x^2+3x+5< 20x^2+10x-30\)
=>3x+5<10x-30
=>-7x<-35
hay x>5
b: \(\Leftrightarrow4\left(5x-20\right)-6\left(2x^2+x\right)>4x\left(1-3x\right)-15x\)
\(\Leftrightarrow20x-80-12x^2-6x>4x-12x^2-15x\)
=>14x-80>-11x
=>25x>80
hay x>16/5
`(2x)/(3x^2-x+2)-(7x)/(3x^2+5x+2)=1(x ne -1,-2/3)`
Đặt `a=3x^2+2x+2(a>=5/3)`
`pt<=>(2x)/(a-3x)-(7x)/(a+3x)=1`
`=>2x(a+3x)-7x(a-3x)=a^2-9x^2`
`<=>2ax+6x^2-7ax+21x^2=a^2-9x^2`
`<=>-5ax+27x^2=a^2-9x^2`
`<=>a^2-36x^2+5ax=0`
`<=>a^2-4ax+9ax-36x^2=0`
`<=>a(a-4x)+9x(a-4x)=0`
`<=>(a-4x)(a+9x)=0`
`+)a=4x`
`=>3x^2+2x+2=4x`
`=>3x^2-2x+2=0`
`=>x^2-2/3x+2/3=0`
`=>(x-1/3)^2+5/9=0` vô lý
`+)a+9x=0`
`=>3x^2+2x+2+9x=0`
`=>3x^2+11x+2=0`
`=>x^2+11/3x+2/3=0`
`=>x=(-11+-\sqrt{97})/6`
ĐKXĐ: \(x\ne-1;x\ne-\dfrac{2}{3}\)
Ta có: \(\dfrac{2x}{3x^2-x+2}-\dfrac{7x}{3x^2+5x+2}=1\)(1)
\(\Leftrightarrow\dfrac{2}{3x-1+\dfrac{2}{x}}-\dfrac{7}{3x+5+\dfrac{2}{x}}=1\)
Đặt: \(3x+\dfrac{2}{x}=a\) (x khác 0) thì pt(1) trở thành:
\(\dfrac{2}{a-1}-\dfrac{7}{a+5}=1\)
\(\Leftrightarrow\dfrac{2\left(a+5\right)-7\left(a-1\right)}{\left(a-1\right)\left(a+5\right)}=1\)
\(\Leftrightarrow2\left(a+5\right)-7\left(a-1\right)=\left(a-1\right)\left(a+5\right)\)
\(\Leftrightarrow-5a+17=a^2+4a-5\)
\(\Leftrightarrow a^2+4a+5-5-17=0\)
\(\Leftrightarrow a^2+9a-22=0\)
\(\Leftrightarrow\left(a-2\right)\left(a+11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\a=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x+\dfrac{2}{x}=2\\3x+\dfrac{2}{x}=-11\end{matrix}\right.\)
Vì \(\left\{{}\begin{matrix}3x^2+2-2x\ne0\\3x^2+11x+2\ne0\end{matrix}\right.\)
=> PT vô nghiệm
Ủa hình như sai:vvv
Câu 1:
\(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{\left(x-7\right)\left(x-3\right)}{\left(x-7\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
\(\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}=\dfrac{2x^2-6x+5x-15}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{\left(2x+5\right)\left(x-3\right)}{\left(2x+5\right)\left(x^2+1\right)}=\dfrac{x-3}{x^2+1}\)
Do đó: \(\dfrac{x^2-10x+21}{x^3-7x^2+x-7}=\dfrac{2x^2-x-15}{2x^3+5x^2+2x+5}\)
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 \(S=\left\{-1\right\}\)
