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a, \(\dfrac{4\left(x-3\right)^2-\left(2x-1\right)^2-12x}{12}< 0\)
\(\Rightarrow4\left(x^2-6x+9\right)-4x^2+4x-1-12x< 0\)
\(\Leftrightarrow-32x+35< 0\Leftrightarrow x>\dfrac{35}{32}\)
b, \(\dfrac{24+12\left(x+1\right)-36+3\left(x-1\right)}{12}< 0\)
\(\Rightarrow-12x+15x+9< 0\Leftrightarrow3x< -9\Leftrightarrow x>-3\)
1: \(\Leftrightarrow x^2+6x+9-6x+3>x^2-4x\)
=>-4x<12
hay x>-3
2: \(\Leftrightarrow6+2x+2>2x-1-12\)
=>8>-13(đúng)
4: \(\dfrac{2x+1}{x-3}\le2\)
\(\Leftrightarrow\dfrac{2x+1-2x+6}{x-3}< =0\)
=>x-3<0
hay x<3
6: =>(x+4)(x-1)<=0
=>-4<=x<=1
a ) \(\dfrac{\left(x-3\right)^2}{3}-\dfrac{\left(2x-1\right)^2}{12}\le x\)
\(\Leftrightarrow4\left(x-3\right)^2-\left(2x-1\right)^2\le12x\)
\(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)-12x\le0\)
\(\Leftrightarrow4x^2-24x+36-4x^2+4x-1-12x\le0\)
\(\Leftrightarrow-36x\le-35\)
\(\Leftrightarrow x\ge\dfrac{35}{36}\)
Vậy bất phương trình có nghiệm \(x\ge\dfrac{35}{36}\).
b ) \(2+\dfrac{3\left(x+1\right)}{3}< 3-\dfrac{x-1}{4}\)
\(\Leftrightarrow2+x+1< 3-\dfrac{x-1}{4}\)
\(\Leftrightarrow x+3< 3-\dfrac{x-1}{4}\)
\(\Leftrightarrow4\left(x+3\right)< 12-x+1\)
\(\Leftrightarrow4x+12+x< 13\)
\(\Leftrightarrow5x< 13-12\)
\(\Leftrightarrow5x< 1\)
\(\Leftrightarrow x< \dfrac{1}{5}\)
Vậy bất phương trình có nghiệm \(x< \dfrac{1}{5}\)
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
a)Để biểu thức vô nghĩa thì \(\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\Leftrightarrow x\in\left\{-2;1\right\}\)
ĐKXĐ: \(\left\{{}\begin{matrix}x+2\ne0\\x-1\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne1\end{matrix}\right.\Leftrightarrow x\notin\left\{-2;1\right\}\)
b) Ta có: \(\dfrac{5x-2}{12}-\dfrac{2x^2+1}{8}=\dfrac{x-3}{6}+\dfrac{1-x^2}{4}\)
\(\Leftrightarrow\dfrac{2\left(5x-2\right)}{24}-\dfrac{3\left(2x^2+1\right)}{24}=\dfrac{4\left(x-3\right)}{24}+\dfrac{6\left(1-x^2\right)}{24}\)
\(\Leftrightarrow10x-4-6x^2-3=4x-12+6-6x^2\)
\(\Leftrightarrow-6x^2+10x-7+6x^2-4x+6=0\)
\(\Leftrightarrow6x-1=0\)
\(\Leftrightarrow6x=1\)
\(\Leftrightarrow x=\dfrac{1}{6}\)
Vậy: \(S=\left\{\dfrac{1}{6}\right\}\)
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
+ Pt thứ nhất :
Ta có mẫu thức chung là : \(2\left(x-3\right)\left(x+1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x\ne2\\x-3\ne0\\x+1\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne3\\x\ne-1\end{matrix}\right.\)
Vậy \(ĐKXĐ\) là :\(x\ne2;3;-1\)
+ Pt thứ hai :
Ta có mẫu thức chung là : \(\left(x-2\right)\left(x+3\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2\ne0\\x+3\ne0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x\ne2\\x\ne-3\end{matrix}\right.\)
Vậy \(DKXD:\) \(\) \(x\ne2;-3\)
a)
Ta có:
cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
(1): \(\Leftrightarrow4\left(x^2-6x+9\right)-\left(4x^2-4x+1\right)< =12x\)
\(\Leftrightarrow4x^2-24x+36-4x^2+4x-1< =12x\)
=>-20x+35<=12x
=>-32x<=-35
hay x>=35/32(3)
(2): \(\Leftrightarrow24+4\left(x+1\right)< 36-3\left(x-1\right)\)
=>24+4x+4<36-3x+3
=>4x+28<-3x+39
=>7x<=11
hay x<=11/7(4)
Từ (3) và (4) suy ra 35/32<=x<=11/7