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ĐKXĐ: \(\left[{}\begin{matrix}x< -1\\x>2\end{matrix}\right.\)
\(\frac{\left|2x-1\right|}{\left(x+1\right)\left(x-2\right)}>\frac{1}{2}\) (*)
+) Nếu \(x>2\) thì (*) \(\Leftrightarrow\frac{2x-1}{x^2-x-2}>\frac{1}{2}\)
\(\Leftrightarrow4x-2>x^2-x-2\)
\(\Leftrightarrow x^2-5x< 0\)
\(\Leftrightarrow x\left(x-5\right)< 0\)
\(\Leftrightarrow0< x< 5\)
\(\Leftrightarrow2< x< 5\)
+) Nếu \(x< -1\) thì (*) \(\Leftrightarrow\frac{1-2x}{x^2-x-2}>\frac{1}{2}\)
\(\Leftrightarrow2-4x>x^2-x-2\)
\(\Leftrightarrow x^2+3x-4< 0\)
\(\Leftrightarrow\left(x+4\right)\left(x-1\right)< 0\)
\(\Leftrightarrow-4< x< 1\)
\(\Leftrightarrow-4< x< -1\)
Vậy...
Giải PT:
a/ \(\Leftrightarrow\left(x+2\right)^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x+2\right|=0\\\left|x+2\right|=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x+2=3\\x+2=-3\end{matrix}\right.\)
b/
\(\Leftrightarrow\left|x+2\right|^2-3\left|x+2\right|-4=0\)
\(\Leftrightarrow\left(\left|x+2\right|+1\right)\left(\left|x+2\right|-4\right)=0\)
\(\Leftrightarrow\left|x+2\right|-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=4\\x+2=-4\end{matrix}\right.\)
c/
\(\Leftrightarrow\left|x^2-3\right|^2-6\left|x^2-3\right|+5=0\)
\(\Leftrightarrow\left(\left|x^2-3\right|-1\right)\left(\left|x^2-3\right|-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x^2-3\right|=1\\\left|x^2-3\right|=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-3=1\\x^2-3=-1\\x^2-3=5\\x^2-3=-5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x^2=4\\x^2=2\\x^2=8\\x^2=-2\left(l\right)\end{matrix}\right.\)
d/ ĐKXĐ: ...
\(\Leftrightarrow\frac{\left|x-2\right|^2}{\left(x-1\right)^2}+\frac{2\left|x-4\right|}{x-1}=3\)
Đặt \(\frac{\left|x-2\right|}{x-1}=a\)
\(a^2+2a-3=0\Rightarrow\left[{}\begin{matrix}a=1\\a=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\\\left|x-2\right|=-3\left(x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x-2\right|=x-1\left(x\ge1\right)\\\left|x-2\right|=3-3x\left(x\le1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=x-1\left(vn\right)\\x-2=1-x\\x-2=3-3x\\x-2=3x-3\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\frac{3}{2}\\x=\frac{4}{5}\\x=\frac{1}{2}\end{matrix}\right.\)
e/ ĐKXĐ: ...
Đặt \(\left|\frac{2x-1}{x+2}\right|=a>0\)
\(a-\frac{2}{a}=1\Leftrightarrow a^2-a-2=0\)
\(\Rightarrow\left[{}\begin{matrix}a=-1\left(l\right)\\a=2\end{matrix}\right.\) \(\Rightarrow\left|\frac{2x-1}{x+2}\right|=2\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=2\left(x+2\right)\\2x-1=-2\left(x+2\right)\end{matrix}\right.\)
PT cho tđuong với: (x^2 +9). (x^2 + 9x) = 22 (x-1)^2
Đặt t = [x^2 + 9 + x^2 + 9x]/2 hay t= x^2 + (9x + 9)/2.
Khi đó: x^2 + 9 = t - 9(x-1)/2
x^2 + 9x = t + 9(x-1)/2
PT cho trở thành: [t - 9(x-1)/2]. [t + 9(x-1)/2] = 22(x-1)^2
<=> t^2 -(81/4)(x-1)^2 = 22(x-1)^2
<=> t^2 = (169/4)(x-1)^2
<=> t = 13/2. (x-1) hoặc t= -13/2. (x-1)
<=> 2t =13x -13 hoặc 2t =-13x + 13
hay 2x^2 + 9x+ 9 =13x -13 hoặc 2x^2 + 9x +9 = -13x +13
hay 2x^2 - 4x +22 =0 hoặc 2x^2 + 22x - 4 =0
PT bậc hai thứ nhất vô nghiệm, PT bậc hai thứ hai cho ta hai nghiệm là:
x= (-11 +căn(129))/2 , x= (-11 - căn(129))/2.
cách 2:đặt x-1=k
pt trở thành (k+1)(k2+2k+10)(k+10)=22k2
<=>(k2+2k+10)(k2+11k+10)=22k2
tự làm tiếp
Ta có:
Xét \(x=0;y=0\) không là nghiệm của hệ phương trình
Xét \(x\ne0;y\ne0\), ta có:\(\left\{{}\begin{matrix}y\left(x^2+1\right)=2x\left(y^2+1\right)\\\left(x^2+y^2\right)\left(1+\dfrac{1}{x^2y^2}\right)=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{x^2+1}{x}=2.\dfrac{y^2+1}{y}\\x^2+y^2+\dfrac{1}{x^2}+\dfrac{1}{y^2}=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=2\left(y+\dfrac{1}{y}\right)\\\left(x+\dfrac{1}{x}\right)^2+\left(y+\dfrac{1}{y}\right)^2=20\end{matrix}\right.\)
Đặt \(a=x+\dfrac{1}{x};b=y+\dfrac{1}{y}\)
Ta có: \(\left\{{}\begin{matrix}a=2b\\a^2+b^2=20\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=2b\\5b^2=20\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)hoặc\(\left\{{}\begin{matrix}a=-4\\b=-2\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}a=4\\b=2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=4\\y+\dfrac{1}{y}=2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2-4x+1=0\\y^2-2y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=2-\sqrt{3}\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=2+\sqrt{3}\\y=1\end{matrix}\right.\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}a=-4\\b=-2\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+\dfrac{1}{x}=-4\\y+\dfrac{1}{y}=-2\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x^2+4x+1=0\\y^2+2y+1=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=-2+\sqrt{3}\\y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-2-\sqrt{3}\\y=-1\end{matrix}\right.\end{matrix}\right.\)
Vậy tập nghiệm của hệ phương trình là \(\left(2-\sqrt{3};1\right),\left(2+\sqrt{3};1\right),\left(-2+\sqrt{3};-1\right),\left(-2-\sqrt{3};-1\right)\)
a/
\(\Leftrightarrow\left[{}\begin{matrix}x^2-5x-4=x^2-4\\x^2-5x-4=4-x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-5x=0\\2x^2-5x-8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{5\pm\sqrt{89}}{4}\\\end{matrix}\right.\)
b/ - Với \(x\ge3\) pt trở thành:
\(x-1+3\left(x-3\right)=6\Leftrightarrow4x=16\Rightarrow x=4\)
- Với \(x\le1\) pt trở thành:
\(1-x+3\left(3-x\right)=6\)
\(\Leftrightarrow x=1\)
- Với \(1< x< 3\) pt trở thành:
\(x-1+3\left(3-x\right)=6\)
\(\Leftrightarrow-2x=-2\Rightarrow x=1\) (loại)
c/ ĐKXĐ: \(x\ne\pm2\)
\(\left[{}\begin{matrix}\frac{x^2-6x-4}{x^2-4}=1\\\frac{x^2-6x-4}{x^2-4}=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-6x-4=x^2-4\\x^2-6x-4=4-x^2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-6x=0\\2x^2-6x-8=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=4\end{matrix}\right.\)
d/ - Với \(x\ge2\) pt trở thành:
\(x-1-2\left(x-2\right)=x^2-x-3\)
\(\Leftrightarrow x^2=6\Rightarrow\left[{}\begin{matrix}x=\sqrt{6}\\x=-\sqrt{6}\left(l\right)\end{matrix}\right.\)
- Với \(x\le1\) pt trở thành:
\(1-x-2\left(2-x\right)=x^2-x-3\) làm tương tự
- Với \(1< x< 2\):
\(x-1-2\left(2-x\right)=x^2-x-3\)
ĐKXĐ: \(x\ne\left\{0;2\right\}\)
- Với \(x>0\Leftrightarrow x^2-1+x+1=2x\left(x-2\right)\)
\(\Leftrightarrow x^2+x=2x^2-4x\Leftrightarrow x^2-5x=0\Rightarrow x=5\)
- Với \(x< -1\Leftrightarrow x^2-1-x-1=-2x\left(x-2\right)\)
\(\Leftrightarrow x^2-x-2=-2x^2+4x\)
\(\Leftrightarrow3x^2-5x-2=0\Rightarrow\left[{}\begin{matrix}x=2\\x=-\frac{1}{3}\end{matrix}\right.\) (đều loại)
- Với \(-1< x< 0\Leftrightarrow x^2-1+x+1=-2x\left(x-2\right)\)
\(\Leftrightarrow x^2+x=-2x^2+4x\Leftrightarrow3x^2-3x=0\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) (loại)
Vậy pt có nghiệm duy nhất \(x=5\)