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a: \(A=\left(1+x+x^2-x\right):\dfrac{1-x^2}{x^3-x^2-x+1}\)
\(=\left(x^2+1\right)\cdot\dfrac{\left(x-1\right)\left(x^2-1\right)}{-\left(x^2-1\right)}=\left(1-x\right)\left(x^2+1\right)\)
b: Khi x=-5/3 thì \(A=\left(1+\dfrac{5}{3}\right)\left(\dfrac{25}{9}+1\right)=\dfrac{8}{3}\cdot\dfrac{34}{9}=\dfrac{272}{27}\)
c: Để A<0 thì 1-x<0
hay x>1
a: \(B=\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x+3\right)\left(x-3\right)}:\dfrac{x+3-1}{x+3}\)
\(=\dfrac{3x+6}{\left(x+3\right)\left(x-3\right)}\cdot\dfrac{x+3}{x+2}\)
\(=\dfrac{3}{x-3}\)
b: |2x+1|=5
=>2x+1=5 hoặc 2x+1=-5
=>x=-3(loại) hoặc x=2(nhận)
Khi x=2 thì \(B=\dfrac{3}{2-3}=-3\)
c: Để B=-3/5 thì x-3=-5
=>x=-2(loại)
d: Để B<0 thì x-3<0
=>x<3
\(M=\left(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}\right):\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
\(M=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{6}{3\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{x^2-4+10-x^2}{x+2}\right)\)
\(M=\left(\dfrac{x^2}{x\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right):\left(\dfrac{6}{x+2}\right)\)
a) dkxd : x khac {0;1;-2)
\(M=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{\left(x-2\right)}+\dfrac{1}{x+2}\right).\left(\dfrac{x+2}{6}\right)\)
\(M=\left(\dfrac{x-2\left(x+2\right)+\left(x-2\right)}{\left(x-1\right)\left(x+2\right)}\right).\left(\dfrac{x+2}{6}\right)=\dfrac{-6}{6\left(x-2\right)}=\dfrac{1}{2-x}\)
b)
GTLN M =1 khi x =1
a: \(A=\dfrac{x+2+2x+x-2}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{2-x}{x}\)
\(=\dfrac{4x}{\left(x+2\right)}\cdot\dfrac{-1}{x}=\dfrac{-4}{x+2}\)
b: 2x^2+x=0
=>x(2x+1)=0
=>x=0(loại) hoặc x=-1/2(nhận)
Khi x=-1/2 thì \(A=-4:\left(-\dfrac{1}{2}+2\right)=-4:\dfrac{3}{2}=-4\cdot\dfrac{2}{3}=-\dfrac{8}{3}\)
c: Để A=1/2 thì -4/x+2=1/2
=>x+2=-2
=>x=-4
a: \(A=\dfrac{x^2-3x+x+3+18}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2-2x+21}{\left(x-3\right)\left(x+3\right)}\)
b: Khi x=-1/2 thì \(A=\dfrac{\dfrac{1}{4}+1+21}{\dfrac{1}{4}-9}=-\dfrac{89}{35}\)
c: Để A=4 thì \(4x^2-36=x^2-2x+21\)
=>3x^2+2x-57=0
=>\(x=\dfrac{-1\pm2\sqrt{43}}{3}\)
a) \(x^3-\dfrac{1}{4}x=0\)
⇔ \(x.\left(x^2-\dfrac{1}{4}\right)=0\)
⇔ \(x\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{1}{2}\right)=0\)
⇔ x = 0 hoặc \(x=\dfrac{1}{2}\) hoặc \(x=\dfrac{-1}{2}\)
b) (2x - 1)2 - (x + 3)2 = 0
⇔ (2x - 1 - x - 3)(2x - 1 + x + 3) = 0
⇔ (x - 4)(3x +2) = 0
⇔ x = 4 hoặc \(x=\dfrac{-2}{3}\)
c) 2x2 - x - 6 = 0
⇔ 2x2 - 4x + 3x - 6 = 0
⇔ 2x(x - 2) + 3(x - 2) = 0
⇔ (x - 2) (2x + 3) = 0
⇔ x = 2 hoặc \(x=\dfrac{-3}{2}\)
2)a.
\(B=\left(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}\right):\dfrac{2x-6}{x^2+6x}\\ =\left(\dfrac{x\left(x^2+6x\right)-\left(x-6\right)\left(x^2-36\right)}{\left(x^2-36\right)\left(x^2+6x\right)}\right).\dfrac{x^2+6x}{2x-6}\\ =\dfrac{x^2\left(x+6\right)-\left(x-6\right)^2.\left(x+6\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x+6\right)\left(x^2-\left(x-6\right)^2\right)}{x^2-36}.\dfrac{1}{2x-6}\\ =\dfrac{\left(x-x+6\right)\left(x+x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6.\left(2x-6\right)}{x-6}.\dfrac{1}{2x-6}\\ =\dfrac{6}{x-6}\)
b)
\(x=2\Leftrightarrow B=\dfrac{6}{x-6}=\dfrac{6}{2-6}=\dfrac{6}{-4}=-\dfrac{3}{2}\)
Bài 1:
\(D=-3x^2+x+15x-5-3\left(2x^2-5x+2\right)\)
\(=-3x^2+16x-5-6x^2+15x-6\)
\(=-9x^2+31x-11\)
\(=-9\cdot\dfrac{1}{9}+\dfrac{31}{3}-11\)
=-11-1+31/3=-12+31/3=-5/3
b: \(E=x^2+x-56-x^2+7x-10=8x-66\)
\(=-\dfrac{8}{5}-66=-\dfrac{338}{5}\)
c: \(F=-3\left(2x^2+x-16x-8\right)-\left(-3x^2+2x-15x+10\right)-4x^2+24x\)
\(=-6x^2+45x+24+3x^2+13x-10-4x^2+24x\)
\(=-4x^2+82x+14\)
\(=-4\cdot9-82\cdot3+14=-268\)
a) ĐK \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\x\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\\x\ne0\end{matrix}\right.\)
b) \(A=\left(\dfrac{x}{x-3}-\dfrac{x}{x+3}\right).\dfrac{x^2+6x+9}{6x}\)
\(A=\dfrac{x\left(x+3\right)-x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(x-3\right)^2}{6x}\)
\(A=\dfrac{6x}{\left(x-3\right)\left(x+3\right)}.\dfrac{\left(x-3\right)^2}{6x}=\dfrac{x-3}{x+3}\)
c) \(A=\dfrac{x-3}{x+3}=\dfrac{x+3-6}{x+3}=1-\dfrac{6}{x+3}\)
Để A nguyên khi \(6⋮\left(x+3\right)\Rightarrow\left(x+3\right)\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
Để A là nguyên dương thì \(\dfrac{6}{x+3}< 1\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=-1\\x+3=-2\\x+3=-3\\x+3=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-4\\x=-5\\x=-6\\x=-9\end{matrix}\right.\)
a, ĐKXĐ:\(5x^2-45\ne0\Rightarrow x^2-9\ne0\Rightarrow x\ne\pm3\)
b, \(\dfrac{x^2-6x+9}{5x^2-45}=\dfrac{\left(x-3\right)^2}{5\left(x^2-9\right)}=\dfrac{\left(x-3\right)^2}{5\left(x-3\right)\left(x+3\right)}=\dfrac{x-3}{5\left(x+3\right)}=\dfrac{x-3}{5x+15}\)
\(c,A=-\dfrac{2}{3}\Rightarrow\dfrac{x-3}{5x+15}=\dfrac{-2}{3}\\ \Rightarrow-2\left(5x+15\right)=3\left(x-3\right)\\ \Rightarrow-10x-30=3x-9\\ \Rightarrow3x-9+10x+30=0\\ \Rightarrow13x+39=0\\ \Rightarrow13x=-39\\ \Rightarrow x=-3\)