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a ) \(A=\dfrac{3x+15}{x^2-9}+\dfrac{1}{x+3}-\dfrac{2}{x-3}\)
\(A=3x+15+x-3-2\left(x+3\right)\)
\(A=4x+10-2x-6\)
\(A=2x+4\)
b ) Để \(A=\dfrac{1}{2}\) thì \(2x+4=\dfrac{1}{2}\), ta có :
\(2x+4=\dfrac{1}{2}\)
\(\Leftrightarrow2x=\dfrac{1}{2}-4\)
\(\Leftrightarrow2x=\dfrac{-7}{2}\)
\(\Leftrightarrow x=\dfrac{-7}{4}\)
Vậy để \(A=\dfrac{1}{2}\) thì \(x=\dfrac{-7}{4}\)
Lời giải:
ĐKXĐ: \(x\neq \left\{2;\pm 3\right\}\)
a) Ta có:
\(P=\left(\frac{x^2-3x}{x^2-9}-1\right):\left(\frac{9-x^2}{x^2+x-6}-\frac{x-3}{2-x}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x(x-3)}{(x-3)(x+3)}-1\right):\left(\frac{(3-x)(3+x)}{(x-2)(x+3)}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\left(\frac{x}{x+3}-1\right):\left(\frac{3-x}{x-2}-\frac{3-x}{x-2}-\frac{x-2}{x+3}\right)\)
\(P=\frac{x-(x+3)}{x+3}:\left(-\frac{x-2}{x+3}\right)=\frac{-3}{x+3}.\frac{x+3}{-(x-2)}=\frac{3}{x-2}\)
b) \(x^3-3x+2=0\)
\(\Leftrightarrow (x^3-x)-2(x-1)=0\)
\(\Leftrightarrow x(x-1)(x+1)-2(x-1)=0\)
\(\Leftrightarrow (x-1)(x^2+x-2)=0\)
\(\Leftrightarrow (x-1)[(x^2-1)+(x-1)]=0\)
\(\Leftrightarrow (x-1)^2(x+2)=0\) \(\Leftrightarrow \left[\begin{matrix} x=1\\ x=-2\end{matrix}\right.\)
Với \(x=1\Rightarrow P=\frac{3}{1-2}=-3\)
Với \(x=-2\Rightarrow P=\frac{3}{-2-2}=\frac{-3}{4}\)
c)
\(P=\frac{3}{x-2}\in\mathbb{Z}\Leftrightarrow 3\vdots x-2\)
\(\Leftrightarrow x-2\in \text{Ư}(3)\Rightarrow x-2\in\left\{\pm 1; \pm 3\right\}\)
\(\Leftrightarrow x\in \left\{3,1,5,-1\right\}\)
Do \(x\neq 3\Rightarrow x\in \left\{-1,1,5\right\}\)
a) \(\dfrac{x+1}{2}+\dfrac{3x-2}{3}=\dfrac{x-7}{12}\)
\(\Leftrightarrow\dfrac{6\left(x+1\right)+4\left(3x-2\right)}{12}=\dfrac{x-7}{12}\)
\(\Leftrightarrow6\left(x+1\right)+4\left(3x-2\right)=x-7\)
\(\Leftrightarrow6x+6+12x-8=x-7\)
\(\Leftrightarrow6x+12x-x=-7-6+8\)
\(\Leftrightarrow17x=-5\)
\(\Leftrightarrow x=\dfrac{-5}{17}\)
Vậy .........................
b) \(\dfrac{2x}{x-3}-\dfrac{5}{x+3}=\dfrac{x^2+21}{x^2-9}\left(ĐKXĐ:x\ne\pm3\right)\)
\(\Leftrightarrow\dfrac{2x\left(x+3\right)-5\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{x^2+21}{\left(x-3\right)\left(x+3\right)}\)
\(\Rightarrow2x\left(x+3\right)-5\left(x-3\right)=x^2+21\)
\(\Leftrightarrow2x^2+6x-5x+15=x^2+21\)
\(\Leftrightarrow2x^2-x^2+x+15-21=0\)
\(\Leftrightarrow x^2+x-6=0\)
\(\Leftrightarrow x^2-2x+3x-6=0\)
\(\Leftrightarrow x\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(n\right)\\x=-3\left(l\right)\end{matrix}\right.\)
Vậy \(S=\left\{2\right\}\)
d) \(\left(x-4\right)\left(7x-3\right)-x^2+16=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x^2-16\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3\right)-\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(7x-3-x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(6x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\6x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{6}\end{matrix}\right.\)
Vậy .........................
P/s: các câu còn lại tương tự, bn tự giải nha
Sửa đề:
\(Q=1+\left(\dfrac{x+1}{x^3+1}-\dfrac{1}{x^2-x+1}-\dfrac{2}{x+1}\right):\dfrac{x^3-2x^2}{x^3-x^2+x}\)
\(=1+\left(\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{x+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\right):\dfrac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\dfrac{x+1-x-1-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x^2\left(x-2\right)}{x\left(x^2-x+1\right)}\)
\(=1+\dfrac{-2x^2+2x-2}{\left(x+1\right)\left(x^2-x+1\right)}:\dfrac{x\left(x-2\right)}{x^2-x+1}\)
\(=1+\dfrac{-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{x^2-x+1}{x\left(x-2\right)}\)
\(=1+\dfrac{-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-2x\right)}=\dfrac{\left(x+1\right)\left(x^2-2x\right)-2\left(x^2-x+1\right)}{\left(x+1\right)\left(x^2-2x\right)}\)
\(=\dfrac{x^3-2x^2+x^2-2x-2x^2+2x-2}{\left(x+1\right)\left(x^2-2x\right)}\)
\(=\dfrac{x^3-3x^2-2}{\left(x+1\right)\left(x^2-2x\right)}\)
a) Rút gọn :
P = \(\left(\dfrac{2x}{x+3}+\dfrac{10}{x-3}-\dfrac{2x^2+14}{x^2-9}\right):\dfrac{4}{x+3}\)
\(ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\x-3\ne0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ne-3\\x\ne3\end{matrix}\right.\)
Ta có : \(P=\left[\dfrac{2x\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{10\left(x+3\right)}{\left(x+3\right)\left(x-3\right)}-\dfrac{2x^2+14}{\left(x+3\right)\left(x-3\right)}\right].\dfrac{x+3}{4}\)
\(P=\dfrac{2x^2-6x+10x+30-2x^2-14}{\left(x+3\right)\left(x-3\right)}.\dfrac{x+3}{4}\)
\(P=\dfrac{4x+16}{4x-13}=\dfrac{x+4}{x-3}\)
b) |x| = 3 => \(\left\{{}\begin{matrix}\left|x\right|=3khix\ge0\\\left|x\right|=-3khix< 0\end{matrix}\right.\)
* TH1 : x \(\ge0\)
\(P=\dfrac{x+4}{x-3}=\dfrac{3+4}{3-3}\left(koTMvìmẫu\ne0\right)\)
* TH2 : x < 0
\(P=\dfrac{x+4}{x-3}=\dfrac{-3+4}{-3-3}=\dfrac{-1}{6}\left(Tm\right)\)
c) Để P = \(\dfrac{-1}{2}\) thì :
\(\dfrac{x+4}{x-3}=\dfrac{-1}{2}\)
\(\Leftrightarrow2x+8=3-x\)
\(\Leftrightarrow2x+x=-8+3\)
\(\Leftrightarrow3x=-5\Rightarrow x=\dfrac{-5}{3}\)
d) P \(\le\) 2
<=> \(\dfrac{x+4}{x-3}\le2\)
\(\Leftrightarrow\dfrac{x+4}{x-3}-\dfrac{2x-6}{x-3}\le0\)
\(\Leftrightarrow\dfrac{10-x}{x-3}\le0\)
Lập bang xét dấu và tìm x nhé!!
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
a: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
b: |x|=1/3 thì x=1/3 hoặc x=-1/3
Khi x=1/3 thì \(Q=\left(\dfrac{1}{3}\right)^2:\left(\dfrac{1}{3}-1\right)=-\dfrac{1}{6}\)
Khi x=-1/3 thì \(Q=\left(-\dfrac{1}{3}\right)^2:\left(-\dfrac{1}{3}-1\right)=-\dfrac{1}{12}\)
c: Để Q là số nguyên thì \(x^2-1+1⋮x-1\)
=>\(x-1\in\left\{1;-1\right\}\)
=>x=2
d: Để Q=4 thì x^2=4x-4
=>x=2
đkxđ: x\(\ne\pm3\)
a/ \(P=\left(\dfrac{x}{x+3}-\dfrac{x^2+5}{x^2-9}+\dfrac{7}{x-3}\right)\cdot\dfrac{x+3}{4}=\left(\dfrac{x\left(x-3\right)-x^2-5+7\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\right)\cdot\dfrac{x+3}{4}=\dfrac{x^2-3x-x^2-5+7x+21}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{4}=\dfrac{4x+16}{x-3}\cdot\dfrac{1}{4}=\dfrac{4\left(x+4\right)}{4\left(x-3\right)}=\dfrac{x+4}{x-3}\)
b/ tại x = 5 thì:
\(P=\dfrac{5+4}{5-3}=\dfrac{9}{2}\)
c/ Ta có: \(\dfrac{x+4}{x-3}=\dfrac{x-3+7}{x-3}=\dfrac{x-3}{x-3}+\dfrac{7}{x-3}=1+\dfrac{7}{x-3}\)
để P ∈ Z thì \(\dfrac{7}{x-3}\in Z\Leftrightarrow x-3\inƯ\left(7\right)\)
=> x - 3 = {-7;-1;1;7}
=> x = {-4;2;4;10}
Vậy.............
minh giai phan d, nha bn :
x-a/b+c + x-b/c+a + x-c/a+b=3
=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0
=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0
=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0
Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0
=>x=a+b+c