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a) \(3\left(4x-1\right)-2x\left(5x+2\right)>8x-2\)
\(\Leftrightarrow12x-3-10x^2-4x>8x-2\)
\(\Leftrightarrow-10x^2>5\)
\(\Leftrightarrow x^2< \dfrac{-1}{2}\)(vô lí)
Vậy bất phương trình đã cho vô nghiệm.
h)
\(\dfrac{x+5}{x+7}-1>0\)
\(\Leftrightarrow\dfrac{x+5}{x+7}-\dfrac{x+7}{x+7}>0\)
\(\Leftrightarrow\dfrac{x+5-x-7}{x+7}>0\)
\(\Leftrightarrow\dfrac{-2}{x+7}>0\)
\(\Leftrightarrow x+7< 0\)
\(\Leftrightarrow x< -7\)
g)
\(\dfrac{4-x}{3x+5}\ge0\)
* TH1:
\(4-x\ge0\) và \(3x+5>0\)
\(\Leftrightarrow x\le4\) và \(x>\dfrac{-5}{3}\)
* TH2:
\(4-x\le0\) và \(3x+5< 0\)
\(\Leftrightarrow x\ge4\) và \(x< \dfrac{-5}{3}\) ( loại)
Vậy: \(-\dfrac{5}{3}< x\le4\)
1. ĐKXĐ: $x\neq 1$
Sửa lại đề 1 chút:
$\frac{1}{x-1}-\frac{3x^2}{x^3-1}=\frac{2x}{x^2+x+1}$
$\Leftrightarrow \frac{x^2+x+1}{(x-1)(x^2+x+1)}-\frac{3x^2}{(x-1)(x^2+x+1)}=\frac{2x(x-1)}{(x-1)(x^2+x+1)}$
$\Leftrightarrow x^2+x+1-3x^2=2x(x-1)$
$\Leftrightarrow 4x^2-3x-1=0$
$\Leftrightarrow (4x+1)(x-1)=0$
Vì $x\neq 1$ nên $x=-\frac{1}{4}$
2. ĐKXĐ: $x\neq 0;2$
PT \(\Leftrightarrow \frac{7}{8x}+\frac{5-x}{4x(x-2)}=\frac{x-1}{2x(x-2)}+\frac{1}{8(x-2)}\)
\(\Leftrightarrow \frac{7(x-2)}{8x(x-2)}+\frac{2(5-x)}{8x(x-2)}=\frac{4(x-1)}{8x(x-2)}+\frac{x}{8x(x-2)}\)
\(\Leftrightarrow 7(x-2)+2(5-x)=4(x-1)+x\)
\(\Leftrightarrow 5x-4=5x-4\) (luôn đúng)
Vậy pt có nghiệm $x\in\mathbb{R}$ với $x\neq 0;2$
a: =>5-x+6=12-8x
=>-x+11=12-8x
=>7x=1
hay x=1/7
b: \(\dfrac{3x+2}{2}-\dfrac{3x+1}{6}=2x+\dfrac{5}{3}\)
\(\Leftrightarrow9x+6-3x-1=12x+10\)
=>12x+10=6x+5
=>6x=-5
hay x=-5/6
d: =>(x-2)(x-3)=0
=>x=2 hoặc x=3
a.x-\(\dfrac{5x+2}{6}=\dfrac{7-3x}{4}\)
⇔\(x=\dfrac{7-3x}{4}+\dfrac{5x+2}{6}\)
⇔\(x=\dfrac{21-9x+10x+4}{12}\)
⇔x=\(\dfrac{x+25}{12}\)
⇔12x=x+25
⇔x=\(\dfrac{25}{11}\)
Vậy pt đã cho có n0 là S=\(\left\{\dfrac{25}{11}\right\}\)
b.ĐKXĐ:x≠-2;x≠2
\(\dfrac{x-2}{x+2}-\dfrac{3}{x-2}=\dfrac{2\left(x-11\right)}{x^2-4}\)
⇔\(\dfrac{\left(x-2\right)\cdot\left(x-2\right)-3\cdot\left(x+2\right)}{\left(x-2\right)\cdot\left(x+2\right)}\)=\(\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇔\(\dfrac{x^2-7x-2}{\left(x-2\right)\cdot\left(x+2\right)}=\dfrac{2x-22}{\left(x-2\right)\cdot\left(x+2\right)}\)
⇒\(\left(x^2-7x-2\right)\cdot\left(x-2\right)\cdot\left(x+2\right)=\left(2x-22\right)\cdot\left(x-2\right)\cdot\left(x+2\right)\)
⇔x2-7x-2=2x-22
⇔x2-9x+20=0
⇔(x-4)(x-5)=0
⇔\(\left\{{}\begin{matrix}x-4=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)
Vậy pt đã cho có n0 là S={4;5}
bài 1:
b,\(\dfrac{x+2}{x}=\dfrac{x^2+5x+4}{x^2+2x}+\dfrac{x}{x+2}\)(ĐKXĐ:x ≠0,x≠-2)
<=>\(\dfrac{\left(x+2\right)^2}{x\left(x+2\right)}=\dfrac{x^2+5x+4}{x\left(x+2\right)}+\dfrac{x^2}{x\left(x+2\right)}\)
=>\(x^2+4x+4=x^2+5x+4+x^2\)
<=>\(x^2-x^2-x^2+4x-5x+4-4=0\)
<=>\(-x^2-x=0< =>-x\left(x+1\right)=0< =>\left[{}\begin{matrix}x=0\left(loại\right)\\x+1=0< =>x=-1\left(nhận\right)\end{matrix}\right.\)
vậy...............
d,\(\left(x+3\right)^2-25=0< =>\left(x+3-5\right)\left(x+3+5\right)=0< =>\left(x-2\right)\left(x+8\right)=0< =>\left[{}\begin{matrix}x-2=0\\x+8=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=2\\x=-8\end{matrix}\right.\)
vậy............
bài 3:
g,\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-x-2}\)(ĐKXĐ:x khác -1,x khác 2)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x^2-2x+x-2}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{x\left(x-2\right)+\left(x-2\right)}\)
<=>\(\dfrac{4}{x+1}-\dfrac{2}{x-2}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
<=>\(\dfrac{4\left(x-2\right)}{\left(x+1\right)\left(x-2\right)}-\dfrac{2\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{x+3}{\left(x+1\right)\left(x-2\right)}\)
=>\(4x-8-2x-2=x+3\)
<=>\(x=13\)
vậy..............
mấy ý khác bạn làm tương tụ nhé
chúc bạn học tốt ^ ^
ĐKXĐ: \(x\ne0\)
\(\Leftrightarrow x^2+\dfrac{1}{x^2}-\dfrac{5}{2}\left(x+\dfrac{1}{x}\right)+3=0\)
Đặt \(x+\dfrac{1}{x}=a\Rightarrow x^2+2+\dfrac{1}{x^2}=a^2\Rightarrow x^2+\dfrac{1}{x^2}=a^2-2\)
Phương trình trở thành:
\(a^2-2-\dfrac{5}{2}a+3=0\Leftrightarrow2a^2-5a+2=0\Rightarrow\left[{}\begin{matrix}a=2\\a=\dfrac{1}{2}\end{matrix}\right.\)
- Với \(a=2\Rightarrow x+\dfrac{1}{x}=2\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Rightarrow x=1\)
- Với \(a=\dfrac{1}{2}\Rightarrow x+\dfrac{1}{x}=\dfrac{1}{2}\Leftrightarrow2x^2-x+2=0\)
\(\Leftrightarrow2\left(x-\dfrac{1}{4}\right)^2+\dfrac{15}{8}=0\) (vô nghiệm)
Vậy pt có nghiệm duy nhất \(x=1\)
1: =>x^3-5x^2+x^2-5x+3x-15=0
=>(x-5)(x^2+x+3)=0
=>x-5=0
=>x=5
2: =>x^3+6x^2+12x+35=0
=>x^3+5x^2+x^2+5x+7x+35=0
=>(x+5)(x^2+x+7)=0
=>x+5=0
=>x=-5
3: \(\Leftrightarrow\left(\dfrac{x+43}{57}+1\right)+\left(\dfrac{x+46}{54}+1\right)=\left(\dfrac{x+49}{51}+1\right)+\left(\dfrac{x+52}{48}+1\right)\)
=>x+100=0
=>x=-100
ĐKXĐ : \(\left\{{}\begin{matrix}x\ne-1\\x\ne-\dfrac{1}{2}\end{matrix}\right.\)
\(\dfrac{x^2-4x+1}{x+1}+2=-\dfrac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\) \(\dfrac{x^2-2x+3}{x+1}=-\dfrac{x^2-5x+1}{2x+1}\)
\(\Leftrightarrow\) \(\dfrac{\left(x^2-2x+3\right)\left(2x+1\right)}{\left(x+1\right)\left(2x+1\right)}=\dfrac{-\left(x^2-5x+1\right)\left(x+1\right)}{\left(2x+1\right)\left(x+1\right)}\)
\(\Leftrightarrow\left(x^2-2x+3\right)\left(2x+1\right)=-\left(x^2-5x+1\right)\left(x+1\right)\)
\(\Leftrightarrow2x^3-3x^2+4x+3=-x^3+4x^2+4x-1\)
\(\Leftrightarrow2x^3-3x^2+4x+3+x^3-4x^2-4x+1=0\)
\(\Leftrightarrow3x^3-7x^2+4=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-\dfrac{2}{3}\end{matrix}\right.\)
Vậy \(S=\left\{-\dfrac{2}{3};1;2\right\}\)