\(P=\left(\dfrac{x^3+2x}{x^3+1}+\dfrac{2x}{x^2-x+1}\right)\text{/}\dfrac{x^2+2x+1}{x-1}\)
a)rút gọn P
b)tại P=-5
P=\(\left(\dfrac{3\left(x+2\right)}{2x^2+8}-\dfrac{2x^2-x-10}{\left(x+1\right)\left[\left(x+1\right)^2-2x\right]}\right):\left(\dfrac{5}{x^2+1}+\dfrac{3}{2\left(x+1\right)}-\dfrac{3}{x-1}\right)\cdot\dfrac{2}{x-1}\)
a) rút gọn P
b)tìm tất cả các giá trị nguyên của x để P có giá trị là bội của 4
a: \(P=\left(\dfrac{3x+6}{2\left(x^2+4\right)}-\dfrac{2x^2-x-10}{\left(x+1\right)\left(x^2+1\right)}\right):\left(\dfrac{10\left(x^2-1\right)+3\left(x^2+1\right)\left(x-1\right)-6\left(x+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x+1\right)\left(x-1\right)\cdot2}\right)\cdot\dfrac{2}{x-1}\)
\(=\left(\dfrac{\left(3x+6\right)\left(x^3+x^2+x+1\right)-\left(2x^2+8\right)\left(2x^2-x-10\right)}{2\left(x^2+4\right)\left(x+1\right)\left(x^2+1\right)}\right)\cdot\dfrac{\left(x^2+1\right)\left(x-1\right)\left(x+1\right)\cdot2}{-3x^3+x^2-3x-13}\cdot\dfrac{2}{x-1}\)
\(=\dfrac{-x^4+11x^3+13x^2+17x+16}{\left(x^2+4\right)}\cdot\dfrac{2}{-3x^3+x^2-3x-13}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
Rút gọn các biểu thức sau :
A = \(2x^2\left(-3x^3+2x^2+x-1\right)+2x\left(x^2-3x+1\right)\)
B = \(2x:\dfrac{1}{2}x+x^2\)
C = \(\left[1:\left(1+x\right)+2x:\left(1-x^2\right)\right]:\left(\dfrac{1}{x}-1\right)\)
D = \(\dfrac{x^2-y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}+\dfrac{y^2}{x+y}.\dfrac{\left(x+y\right)^2}{x}\)
E = \(\dfrac{\left|x-3\right|}{x^2-9}.\left(x^2+6x+9\right)\)
F = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
C=\(\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{\text{5-x}}{\text{1-x}^{\text{2}}}\right)\):\(\dfrac{1-2x}{\text{x}^{\text{2}}-1}\)
a) Rút gọn
\(C=\dfrac{-\left(x+1\right)+2\left(x-1\right)+5-x}{\left(x-1\right)\left(x+1\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(=\dfrac{2}{1-2x}\)
\(C=\left(\dfrac{1}{1-x}+\dfrac{2}{x+1}-\dfrac{5-x}{1-x^2}\right):\dfrac{1-2x}{x^2-1}\)
\(\Rightarrow C=\left(\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}+\dfrac{2\left(1-x\right)}{\left(1+x\right)\left(1-x\right)}-\dfrac{5-x}{\left(1-x\right)\left(1+x\right)}\right).\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2\left(1-x\right)-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{\left(x-1\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=\dfrac{1+x+2-2x-5+x}{\left(1-x\right)\left(1+x\right)}.\dfrac{-\left(1-x\right)\left(x+1\right)}{1-2x}\)
\(\Rightarrow C=-2.\dfrac{-1}{1-2x}\)
\(\Rightarrow C=\dfrac{2}{1-2x}\)
M= \(\left(\dfrac{x-1}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x-1}\right)\)
Điều kiện: x khác -1, +1
a, Rút gọn M
b, Tính giá trị của M tại x=\(\dfrac{-1}{2}\)
a: \(M=\left(\dfrac{\left(x-1\right)\left(x+1\right)}{2\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x+3\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right):\dfrac{x-1-x+3}{x-1}\)
\(=\dfrac{x^2-1-x^2-2x+3}{2\left(x-1\right)\left(x+1\right)}\cdot\dfrac{x-1}{2}\)
\(=\dfrac{-2x+2}{2\left(x+1\right)}\cdot\dfrac{1}{2}=\dfrac{-x+1}{2}\)
b: Thay x=-1/2 vào M, ta được:
\(M=\dfrac{\dfrac{1}{2}+1}{2}=\dfrac{3}{2}:2=\dfrac{3}{4}\)
a, \(M=\left(\dfrac{x^2-1-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-1-x+3}{x-1}\right)\)
\(=\left(\dfrac{-1+x-3x-3}{2\left(x-1\right)\left(x+1\right)}\right):\dfrac{2}{x-1}=\dfrac{-2x-4}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-\left(x+2\right)}{2\left(x+1\right)}\)
b, Thay x =-1/2 vào ta đc
\(-\dfrac{\left(\dfrac{-1}{2}+2\right)}{2\left(-\dfrac{1}{2}+1\right)}=\dfrac{-\dfrac{3}{2}}{2\left(\dfrac{1}{2}\right)}=\dfrac{-3}{2}\)
a, \(M=\left(\dfrac{x^2-1-\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right):\left(\dfrac{x-1-x+3}{x-1}\right)\)
\(=\dfrac{-2x+2}{2\left(x-1\right)\left(x+1\right)}:\dfrac{2}{x-1}=\dfrac{-2\left(x-1\right)^2}{4\left(x-1\right)\left(x+1\right)}=\dfrac{-\left(x-1\right)}{2\left(x+1\right)}\)
b, Thay x = -1/2 vào ta đc
\(M=\dfrac{\dfrac{1}{2}+1}{2\left(\dfrac{-1}{2}+1\right)}=\dfrac{\dfrac{3}{2}}{\dfrac{2.1}{2}}=\dfrac{3}{2}\)
Cho A = \(\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+3}{2x+2}\right).\dfrac{2x^2-2}{5}\)
a) Tìm đkxđ của A
b) Rút gọn A
a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)
b) Ta có: \(A=\left(\dfrac{x+1}{2x-2}+\dfrac{3}{x^2-1}-\dfrac{x+2}{2x+2}\right)\cdot\dfrac{2x^2-2}{5}\)
\(=\left(\dfrac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\dfrac{6}{2\left(x+1\right)\left(x-1\right)}-\dfrac{\left(x+2\right)\left(x-1\right)}{2\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{2x^2-2}{5}\)
\(=\left(\dfrac{x^2+2x+1+6-\left(x^2-x+2x-2\right)}{2\left(x+1\right)\left(x-1\right)}\right)\cdot\dfrac{2x^2-2}{5}\)
\(=\dfrac{x^2+2x+7-x^2-x+2}{2\left(x+1\right)\left(x-1\right)}\cdot\dfrac{2\left(x-1\right)\left(x+1\right)}{5}\)
\(=\dfrac{x+9}{5}\)
2 a. rút gọn biểu C = \(\dfrac{2x^{\text{2}}-x}{\text{x }-1}+\dfrac{x+1}{1-x}+\dfrac{2-x^2}{x-1}\)
b. Rút gọn biểu thức D = \(\left(\dfrac{1}{a-\sqrt{a}}+\dfrac{1}{\sqrt{\text{a}}-1}\right):\dfrac{\sqrt{\text{a}}+1}{a-2\sqrt{a}+1}\)
Vậy khi rút gọn một biểu thức hửu tỉ và một biểu thức chứa căn có tìm điều kiện xác định không?
\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)
Có
Cho \(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\)
a ) Rút gọn B
b ) Tính B tại x thỏa mãn |2x+1|=5
c ) Tìm x để \(B=-\dfrac{3}{5}\)
d ) Tìm x để B < 0
`đk:x ne +-3,x ne -2`
`B=(21/(x^2-9)-(x-4)/(3-x)-(x-1)/(3+x)):(1-1/(x+3))`
`=(21/(x^2-9)+(x-4)/(x-3)-(x-1)/(x+3)):((x+3-1)/(x+3))`
`=((21+x^2-x-12-x^2+4x-3)/((x-3)(x+3))):(x+2)/(x+3)`
`=(3x+6)/((x-3)(x+3))*(x+3)/(x+2)`
`=(3x+6)/((x-3)(x+2))`
`=3/(x-3)`
`b)|2x+1|=5`
`<=>` \(\left[ \begin{array}{l}2x=4\\2x=-6\end{array} \right.\)
`<=>` \(\left[ \begin{array}{l}x=2(tm)\\x=-3(l)\end{array} \right.\)
`=>B=3/(2-3)=-3`
`c)B=-3/5`
`<=>3/(x-3)=3/(-5)`
`<=>x-3=-5`
`<=>x=-2(l)`
`d)B<0`
`<=>3/(x-3)<0`
Mà `3>0`
`=>x-3<0<=>x<3`
a) đk: \(x\ne\pm3\)
\(B=\left[\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{x-4}{x-3}-\dfrac{x-1}{x+3}\right]:\left(\dfrac{x+3-1}{x+3}\right)\)
= \(\left[\dfrac{21+\left(x-4\right)\left(x+3\right)-\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right]:\dfrac{x+2}{x+3}\)
= \(\dfrac{21+x^2-x-12-x^2+4x-3}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}\)
= \(\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b) Để \(\left|2x+1\right|=5\)
<=> \(\left[{}\begin{matrix}2x+1=5< =>x=2\left(c\right)\\2x+1=-5< =>x=-3\left(l\right)\end{matrix}\right.\)
Thay x = 2, ta có;
B = \(\dfrac{3}{2-3}=-3\)
c) Để B = \(\dfrac{-3}{5}\)
<=> \(\dfrac{3}{x-3}=\dfrac{-3}{5}\)
<=> x - 3 = -5
<=> x = -2
d) Để B < 0
<=> \(\dfrac{3}{x-3}< 0\)
<=> x - 3 < 0
<=> x < 3
a)\(B=\left(\dfrac{21}{x^2-9}-\dfrac{x-4}{3-x}-\dfrac{x-1}{3+x}\right):\left(1-\dfrac{1}{x+3}\right)\\ =\left(\dfrac{21}{\left(x-3\right)\left(x+3\right)}+\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{\left(x-1\right)\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}\right):\dfrac{x+2}{x+3}\)
\(=\dfrac{3x+6}{\left(x-3\right)\left(x+3\right)}.\dfrac{x+3}{x+2}=\dfrac{3}{x-3}\)
b)\(\left|2x+1\right|=5\\ \Leftrightarrow\left[{}\begin{matrix}2x+1=5\\2x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\left(loại\right)\end{matrix}\right.\)
với x=2 gt của B là
\(B=\dfrac{3}{2-3}=-3\)
c)\(B=\dfrac{3}{x-3}=-\dfrac{3}{5}\Leftrightarrow x-3=-5\Leftrightarrow x=-2\)
d) \(B=\dfrac{3}{x-3}< 0\Leftrightarrow x-3< 0\Leftrightarrow x< 3\)
tự kết luận mỗi câu
rút gọn
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\)
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\)
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\)
` @ \color{Red}{m}`
` \color{lightblue}{Answer}`
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\\ =\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x}{x+1}\)
__
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)
__
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\\ =\dfrac{1}{1-x}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{1-x}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x}{\left(1-x\right)\left(1+x\right)}-\dfrac{2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1+x-2x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1-x}{\left(1-x\right)\left(1+x\right)}\\ =\dfrac{1}{1+x}\)
\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)
\(=\dfrac{x^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x}{x+1}\)
========================
\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)
\(=\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\)
\(=\dfrac{3x-2\left(x-3\right)}{2x\left(x+3\right)}\)
\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)
\(=\dfrac{x+6}{2x^2+6x}\)
==========================
\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)
\(=-\dfrac{1}{x-1}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-\left(x+1\right)+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1+2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x+1}\)