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<=> \(\dfrac{x+2}{x-2}\)-\(\dfrac{1}{x}\)=\(\dfrac{2}{x\left(x-2\right)}\)
<=> \(\dfrac{x\left(x+2\right)}{x\left(x-2\right)}-\dfrac{x-2}{x\left(x-2\right)}=\dfrac{2}{x\left(x-2\right)}\)
ok, ở đây đã có mẫu chung rồi, em cứ vậy làm tiếp thôi :D
\(\dfrac{x+2}{x-2}-\dfrac{1}{x}=\dfrac{2}{x^2-2x}\) (ĐKXĐ: \(x\ne0;x\ne2\))
\(\Leftrightarrow x\left(x+2\right)-\left(x-2\right)=2\)
\(\Leftrightarrow x^2+2x-x+2=2\)
\(\Leftrightarrow x^2+x+2-2=0\Leftrightarrow x^2+x=0\)
\(\Leftrightarrow x\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(ktm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)
\(\Rightarrow S=\left\{-1\right\}\)
ta có: x4-4x3-2x2+12x+9 < x4-4x3-2x2+15x-3
=> x4-4x3-2x2+15x-3 - (x4-4x3-2x2+12x+9) > 0
=> 3x+6>0
(đề bài có cho điều kiện của x thì chứng minh 3x+6>0 là xong ạ)
Ta có: \(\left(x^2-2x-3\right)^2< x^2\left(x^2-4x-2\right)+3\left(5x-1\right)\)
\(\Leftrightarrow x^4+4x^2+9-4x^3-6x^2+12x< x^4-4x^3-2x^2+15x-3\)
\(\Leftrightarrow3x-12>0\)
\(\Leftrightarrow x-4>0\Rightarrow x>4\)
Vậy x > 4
a) \(\frac{2-x}{3}< \frac{3-2x}{5}\)
<=> \(10-5x< 9-6x\)
<=> x < - 1
Vậy S = { x| x < -1 }
b)
0 -1
2.a)\(\dfrac{3\text{x}-2}{2}\)=\(\dfrac{1-2\text{x}}{3}\)
<=>\(\dfrac{9\text{x}-6}{6}\)=\(\dfrac{2-4\text{x}}{6}\)
<=>9x-6=2-4x
<=>9x+4x=2+6
<=>13x=8
<=>x=\(\dfrac{8}{13}\)
1.a)2(x-0,5)+3=0,25(4x-1)
<=>2x-1+3=x-1phần4
<=>2x-x=-1/4+1-3
<=>x=-3/4
\(x^5+y^5-\left(x+y\right)^5\)
\(=x^5+y^5-\left(x^5+5x^4y+10x^3y^2+10x^2y^3+8xy^4+y^5\right)\)
\(=-5xy\left(x^3+2x^2y+2xy^2+y^3\right)\)
\(=-5xy\left[\left(x+y\right)\left(x^2-xy+y^2\right)+2xy\left(x+y\right)\right]\)
\(=-5xy\left(x+y\right)\left(x^2+xy+y^2\right)\)
Bài 4:
Đặt P =\(\dfrac{ab}{a+b}+\dfrac{bc}{b+c}+\dfrac{ca}{c+a}\)
\(P=a-\dfrac{a^2}{a+b}+b-\dfrac{b^2}{b+c}+c-\dfrac{c^2}{c+a}\)
\(P=a+b+c-\left(\dfrac{a^2}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}\right)\le a+b+c-\dfrac{\left(a+b+c\right)^2}{2\left(a+b+c\right)}\)
\(P\le a+b+c-\dfrac{a+b+c}{2}=\dfrac{a+b+c}{2}\left(đpcm\right)\)
Dấu "=" xảy ra khi a = b = c
Bài 3 :
\(Ta.có:2x^2+3y^2-2z^2=0\)
\(\Leftrightarrow3y^2=2\left(z^2-x^2\right)=2\left(z-x\right)\left(z+x\right)\)
\(y>0=>3y^2>0;z+x>0\left(x,z>0\right)\)
\(=>z-x>0=>z>x\left(1\right)\)
\(2x^2+3y^2-2z^2=2x^2+y^2=2\left(z^2-y^2\right)\)
\(=>z>y\left(2\right)\)
\(\left(1\right),\left(2\right)=>z>x,y\)
Vậy............................
\(x^3+2x^2+3x-6=0\\ \Leftrightarrow x^3-x^2+3x^2-3x+6x-6=0\\ \Leftrightarrow x^2\left(x-1\right)+3x\left(x-1\right)+6\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x^2+3x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2+3x+6=0\left(Vn\right)\end{matrix}\right.\\ \Leftrightarrow x=1\)
Đặt \(x^2+2x+3=a\ge2\)
\(\left(a+1\right)a=a+4\)
\(\Leftrightarrow a^2=4\)
\(\Rightarrow\left[{}\begin{matrix}a=2\\a=-2\left(l\right)\end{matrix}\right.\)
\(\Rightarrow x^2+2x+3=2\Rightarrow\left(x+1\right)^2=0\Rightarrow x=-1\)
\(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\Leftrightarrow\dfrac{x^2-2x-4}{x^2-2x-3}-1>0\)
\(\Leftrightarrow\dfrac{x^2-2x-4-x^2+2x+3}{x^2-3x+x-3}>0\)
\(\Leftrightarrow\dfrac{-1}{\left(x-3\right)\left(x+1\right)}>0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)< 0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x+1< 0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x+1>0\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x< -1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x>-1\end{matrix}\right.\end{matrix}\right.\)
TH1 : vô lý
Vậy \(-1< x< 3\) thì \(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\dfrac{x^2-2x-4}{x^2-2x-3}>1\)
\(\Leftrightarrow x^2-2x-4>x^2-2x-3\)
\(\Leftrightarrow x^2-x^2-2x+2x>-3+4\)
\(\Leftrightarrow0x>1\) (vô lí)
Vậy bpt vô nghiệm