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\(a.\)
\(P=\left[\left(\dfrac{1}{x^2}+1\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+1\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\left(\dfrac{1}{x^2}+\dfrac{x^2}{x^2}\right).\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{1}{x}+\dfrac{x}{x}\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^2}.\dfrac{1}{x^2+2x+1}+\dfrac{2}{\left(x+1\right)^3}.\left(\dfrac{x+1}{x}\right)\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^2\left(x^2+2x+1\right)}+\dfrac{2}{x\left(x+1\right)^2}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2}{x^3+2x^2+x}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x\left(x^3+2x^2+x\right)}\right].\dfrac{x-1}{x^3}\)
\(P=\left[\dfrac{x^2+1}{x^4+2x^3+x^2}+\dfrac{2x}{x^4+2x^3+x^2}\right].\dfrac{x-1}{x^3}\)
\(P=\dfrac{x^2+1+2x}{x^4+2x^3+x^2}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{x^2+2x+1}{x^2\left(x^2+2x+1\right)}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{1}{x^2}.\dfrac{x-1}{x^3}\)
\(P=\dfrac{x-1}{x^5}\)
a: A=[(3x^2+3-x^2+2x-1-x^2-x-1)/(x-1)(x^2+x+1)]*(x-2)/2x^2-5x+5
=(x^2+x+1)/(x-1)(x^2+x+1)*(x-2)/2x^2-5x+5
=(x-2)/(2x^2-5x+5)(x-1)
\(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\) ( Chữa đề nhé.)
a) \(ĐKXĐ:x\ne-3;x\ne2\)
\(\text{Với }x\ne-3;x\ne2,\text{ ta có: }A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\\ =\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\\ =\dfrac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{x+3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-4-5-x-3}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{\left(x+3\right)\left(x-4\right)}{\left(x-2\right)\left(x+3\right)}\\ =\dfrac{x-4}{x-2}\\ \text{Vậy }A=\dfrac{x-4}{x-2}\text{ với }x\ne-3;x\ne2\)
b) Lập bảng xét dấu:
x x-4 x-2 x-4 2 4 0 0 x-2 _ _ + _ + + 0 + _ +
\(\Rightarrow\left[{}\begin{matrix}x< 2\\x>4\end{matrix}\right.\)
Vậy để \(A>0\) thì \(x< 2\) hoặc \(x>4\)
c) \(\text{Với }x\ne-3;x\ne2\)
\(\text{Ta có : }A=\dfrac{x-4}{x-2}=\dfrac{x-2-2}{x-2}\\ =\dfrac{x-2}{x-2}-\dfrac{2}{x-2}=1-\dfrac{2}{x-2}\)
\(\Rightarrow\) Để A nhận giá trị nguyên
thì \(\Rightarrow\dfrac{2}{x-2}\in Z\)
\(\Rightarrow2⋮x-2\\ \Rightarrow x-2\inƯ_{\left(2\right)}\)
Mà \(Ư_{\left(2\right)}=\left\{\pm1;\pm2\right\}\)
Lập bảng giá trị:
| \(x-2\) | \(-2\) | \(-1\) | \(1\) | \(2\) |
| \(x\) | \(0\left(TM\right)\) | \(1\left(TM\right)\) | \(3\left(TM\right)\) | \(4\left(TM\right)\) |
\(\Rightarrow x\in\left\{-2;-1;1;2\right\}\)
Vậy với \(x\in\left\{-2;-1;1;2\right\}\)
thì \(A\in Z\)
Câu 2:
a) \(ĐKXĐ:x\ne\dfrac{3}{2};x\ne1\)
\(\text{Với }x\ne\dfrac{3}{2};x\ne1,\text{ ta có : }B=\left(\dfrac{2x}{2x^2-5x+3}-\dfrac{5}{2x-3}\right):\left(3+\dfrac{2}{1-x}\right)\\ =\left[\dfrac{2x}{\left(2x-3\right)\left(x-1\right)}-\dfrac{5\left(x-1\right)}{\left(2x-3\right)\left(x-1\right)}\right]:\left(\dfrac{3\left(1-x\right)}{1-x}+\dfrac{2}{1-x}\right)\\ =\dfrac{2x-5x+5}{\left(2x-3\right)\left(x-1\right)}:\dfrac{3-3x+2}{\left(1-x\right)}\\ =\dfrac{\left(-3x+5\right)\cdot\left(1-x\right)}{\left(2x-3\right)\left(x-1\right)\cdot\left(-3x+5\right)}\\ =-\dfrac{1}{2x-3}\)
Vậy \(B=-\dfrac{1}{2x-3}\) với \(x\ne\dfrac{3}{2};x\ne1\)
b) \(\text{Với }x\ne\dfrac{3}{2};x\ne1\)
Để \(B=\dfrac{1}{x^2}\)
\(\text{thì }\Rightarrow\dfrac{-1}{2x-3}=\dfrac{1}{x^2}\\ \Rightarrow2x-3=-x^2\\ \Leftrightarrow2x-3+x^2=0\\ \Leftrightarrow x^2-3x+x-3=0\\ \Leftrightarrow\left(x^2-3x\right)+\left(x-3\right)=0\\ \Leftrightarrow x\left(x-3\right)+\left(x-3\right)=0\\ \Leftrightarrow\left(x+1\right)\left(x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\left(TM\right)\)
Vậy với \(x=-1;x=3\) thì \(B=\dfrac{1}{x^2}\)
a) \(M=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{x^2-1}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-1}{x-1}+\dfrac{2}{x+1}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\left(\dfrac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{2\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\dfrac{5-x}{\left(x-1\right)\left(x+1\right)}\right):\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{-x-1+2x-2+5-x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}:\dfrac{1-2x}{x^2-1}\)
\(\Leftrightarrow M=\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{x^2-1}{1-2x}\)
\(\Leftrightarrow M=\dfrac{2\left(x^2-1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)
\(\Leftrightarrow M=\dfrac{2\left(x-1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)\left(1-2x\right)}\)
\(\Leftrightarrow M=\dfrac{2}{1-2x}\)
b) \(M=\dfrac{2}{1-2x}=\dfrac{-2}{3}\)
\(\Rightarrow2.3=\left(1-2x\right).\left(-2\right)\)
\(\Rightarrow6=-2+4x\)
\(\Rightarrow4x=6-\left(-2\right)\)
\(\Rightarrow4x=6+2\)
\(\Rightarrow4x=8\)
\(\Rightarrow x=8:4\)
\(\Rightarrow x=2\)
Vậy \(M=\dfrac{-2}{3}\) thì \(x=2\)
c) Để \(M=\dfrac{2}{1-2x}\in Z\) \(\Leftrightarrow2⋮1-2x\)
\(\Rightarrow1-2x\in U\left(2\right)=\left\{-1;1;-2;2\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}1-2x=-1\Rightarrow x=1\\1-2x=1\Rightarrow x=0\\1-2x=-2\Rightarrow x=1,5\\1-2x=2\Rightarrow x=-0,5\end{matrix}\right.\)
Mà \(x\in Z\)
\(\Rightarrow x\in\left\{1;0\right\}\)
Vậy \(x=1\) hoặc \(x=0\) thì \(M\in Z\)
a) M = \(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
= \(\left(\dfrac{1}{1-x}+\dfrac{2}{1+x}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{x^2-1}{1-2x}\)
= \(\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1-x\right)\left(1+x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
= \(\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)\(=\dfrac{-2}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
= \(\dfrac{2}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
=\(\dfrac{2}{1-2x}\)
b) M = \(\dfrac{-2}{3}\Leftrightarrow\dfrac{2}{1-2x}=\dfrac{-2}{3}\)
=> 2 . 3 = -2 (1 - 2x) (tích chéo)
=> 6 = -2 + 4x
=> 6 + 2 - 4x = 0
=> 8 - 4x = 0
=> 4x = 8
=> x = 2 (thỏa mãn đkxđ)
Vậy để M = \(\dfrac{-2}{3}\) thì x = 2
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: \(Q=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}:\dfrac{x^2-1+x+2-x^2}{x\left(x-1\right)}\)
\(=\dfrac{x\left(x+1\right)}{\left(x-1\right)^2}\cdot\dfrac{x\left(x-1\right)}{x+1}=\dfrac{x^2}{x-1}\)
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
Giải:
a) ĐKXĐ: \(x\ne0\)
\(H=1-\dfrac{1}{x^2+4}\left(\left(1-\dfrac{x^2+4}{4x}\right):\left(\dfrac{1}{x}-\dfrac{1}{2}\right)\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\left(\dfrac{4x}{4x}-\dfrac{x^2+4}{4x}\right):\left(\dfrac{4}{4x}-\dfrac{2x}{4x}\right)\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\dfrac{4x-x^2+4}{4x}:\dfrac{4-2x}{4x}\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}\left(\dfrac{4x-x^2+4}{4x}.\dfrac{4x}{4-2x}\right)\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{4x-x^2+4}{4-2x}\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{\left(2-x\right)^2}{2\left(2-x\right)}\)
\(\Leftrightarrow H=1-\dfrac{1}{x^2+4}.\dfrac{2-x}{2}\)
\(\Leftrightarrow H=1-\dfrac{2-x}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2\left(x^2+4\right)}{2\left(x^2+4\right)}-\dfrac{2-x}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2\left(x^2+4\right)-\left(2-x\right)}{2\left(x^2+4\right)}\)
\(\Leftrightarrow H=\dfrac{2x^2+8-2+x}{2x^2+8}\)
\(\Leftrightarrow H=\dfrac{2x^2+6+x}{2x^2+8}\)
b) Để \(H=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{2x^2+6+x}{2x^2+8}=\dfrac{1}{4}\)
\(\Rightarrow4\left(2x^2+6+x\right)=2x^2+8\)
\(\Leftrightarrow8x^2+24+4x=2x^2+8\)
\(\Leftrightarrow8x^2+24+4x-2x^2-8=0\)
\(\Leftrightarrow6x^2+4x+16=0\)
Đến đây phân tích đa thức thành nhân tử rồi tìm nghiệm.
TXĐ : \(x\ne\pm2\)
\(M=\left[\dfrac{1}{x+2}-\dfrac{2}{x-2}+\dfrac{x}{\left(x-2\right)\left(x+2\right)}\right]:\dfrac{10-x^2+\left(x-2\right)\left(x+2\right)}{x+2}\)
\(=\dfrac{x-2-2\left(x+2\right)+x}{\left(x-2\right)\left(x+2\right)}.\dfrac{x+2}{10-x^2+x^2-1}\)
\(=\dfrac{x-2-2x-4+x}{x-2}.\dfrac{1}{6}\)
\(=\dfrac{-6}{x-2}.\dfrac{1}{6}=\dfrac{1}{2-x}\)