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8 tháng 8 2017

sau khi rút gọn ta được \(P=\frac{x-4}{x-2}\left(x\ne-3;x\ne2;x\ne-2\right)\)

d,ta có \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\left(x\ne-2;x\ne-3;x\ne2\right)\)

để P nguyên mà x nguyên \(\Leftrightarrow x-2\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

ta có bảng:

x-21-12-2
x3(tm)1(tm)4(tm)0(tm)

vậy \(P\in Z\Leftrightarrow x\in\left\{3;1;4;0\right\}\)

e,x2-9=0

\(\Leftrightarrow x^2=9\Leftrightarrow\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(kotm\right)\end{cases}}\)

thay x=3 vào P đã rút gọn ta có \(P=\frac{3-4}{3-2}=-1\)

vậy với x=3 thì p có giá trị bằng -1

tích mình đi

làm ơn

rùi mình

tích lại

thanks

31 tháng 12 2018

M xác định

\(\Leftrightarrow\hept{\begin{cases}x-1\ne0\\x^2-x\ne0\end{cases}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\left(x-1\right)\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne1\\x\ne0;x\ne1\end{cases}}\Leftrightarrow}\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)

Vậy ĐKXĐ của M là \(\hept{\begin{cases}x\ne1\\x\ne0\end{cases}}\)

\(M=\frac{3}{x-1}+\frac{1}{x^2-x}=\frac{3}{x-1}+\frac{1}{x\left(x-1\right)}=\frac{3x}{x\left(x-1\right)}+\frac{1}{x\left(x-1\right)}=\frac{3x+1}{x\left(x-1\right)}\)

Thay x=5 ta có: 

\(M=\frac{3.5+1}{5\left(5-1\right)}=\frac{15+1}{5.4}=\frac{16}{20}=\frac{4}{5}\)

Vậy \(M=5\)tại  x=5

31 tháng 12 2018

\(M=0\)

\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=0\Leftrightarrow3x+1=0\Leftrightarrow x=-\frac{1}{3}\)( thỏa mãn đkxđ)

Vậy với \(x=-\frac{1}{3}\)thì \(M=0\)

\(M=-1\)

\(\Leftrightarrow\frac{3x+1}{x\left(x-1\right)}=-1\Leftrightarrow3x+1=-x^2+x\Leftrightarrow x^2+2x+1=0\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)

Vậy với \(x=-1\)thì \(M=-1\)

6 tháng 12 2020

Bài làm

\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

\(=\frac{x+2}{x+3}-\frac{5}{x^2+3x-2x-6}-\frac{1}{x-2}\)

\(=\frac{x+2}{x+3}-\frac{5}{x\left(x+3\right)-2\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x^2-4x+3x-12}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{x\left(x-4\right)+3\left(x-4\right)}{\left(x+3\right)\left(x-2\right)}\)

\(=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

b) x2 - 9 = 0 <=> ( x - 3 )( x + 3 ) = 0

<=> \(\orbr{\begin{cases}x=3\left(nhan\right)\\x=-3\left(loai\right)\end{cases}}\)

x = 3 => \(P=\frac{3-4}{3-2}=-1\)

c) \(P=\frac{x-4}{x-2}=\frac{x-2-2}{x-2}=1-\frac{2}{x-2}\)

Để P đạt giá trị nguyên => \(\frac{2}{x-2}\)nguyên

=> \(2⋮x-2\)

=> \(x-2\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

x-21-12-2
x3140

Vậy ...

24 tháng 6 2019

\(P=\frac{1}{a^2-a}+\frac{1}{a^2-3a+2}+\frac{1}{a^2-5a+6}+\frac{1}{a^2-7a+12}+\frac{1}{a^2-9a+20}\)

\(=\frac{1}{a.\left(a-1\right)}+\frac{1}{\left(a-1\right).\left(a-2\right)}+\frac{1}{\left(a-2\right).\left(a-3\right)}+\frac{1}{\left(a-3\right).\left(a-4\right)}+\frac{1}{\left(a-4\right).\left(a-5\right)}\)

a) ĐKXĐ: \(a\ne0;1;2;3;4;5;6\)

b) \(P=\frac{1}{a-1}-\frac{1}{a}+\frac{1}{a-2}-\frac{1}{a-1}+\frac{1}{a-3}-\frac{1}{a-2}+\frac{1}{a-4}-\frac{1}{a-3}+\frac{1}{a-5}-\frac{1}{a-4}\)

\(A=\frac{1}{a-5}-\frac{1}{a}=\frac{a-\left(a-5\right)}{a.\left(a-5\right)}=\frac{5}{a.\left(a-5\right)}\)

c) \(a^3-a^2+2=0\)

\(\Leftrightarrow a^3+a^2-2a^2-2a+2a+2=0\)

\(\Leftrightarrow a^2.\left(a+1\right)-2a.\left(a+1\right)+2.\left(a+1\right)=0\)

\(\Leftrightarrow\left(a+1\right).\left(a^2-2a+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}a+1=0\\a^2-2a+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}a=-1\\\left(a-1\right)^2=-1\left(loai\right)\end{cases}}}\)

Thay a=-1 vào P

\(P=\frac{5}{a.\left(a-5\right)}=\frac{5}{-1.\left(-1-5\right)}=\frac{5}{6}\)

9 tháng 1 2019

a, ĐKXĐ: \(x\ne0;x\ne\pm1\)

\(P=\left(\frac{2x}{x^2-1}+\frac{x-1}{2x+2}\right):\frac{x+1}{2x}=\left(\frac{2x}{\left(x-1\right)\left(x+1\right)}+\frac{x-1}{2\left(x+1\right)}\right):\frac{x+1}{2x}\)

\(=\left(\frac{2x.2}{2\left(x-1\right)\left(x+1\right)}+\frac{\left(x-1\right)^2}{2\left(x-1\right)\left(x+1\right)}\right):\frac{x+1}{2x}\)

\(=\frac{4x+x^2-2x+1}{2\left(x-1\right)\left(x+1\right)}:\frac{x+1}{2x}=\frac{x^2+2x+1}{2\left(x-1\right)\left(x+1\right)}=\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}\cdot\frac{2x}{x+1}=\frac{x}{x-1}\)

b,Để \(P=2\Leftrightarrow\frac{x}{x-1}=2\Leftrightarrow2\left(x-1\right)=x\Leftrightarrow2x-2-x=0\Leftrightarrow x-2=0\Leftrightarrow x=2\left(tmđk\right)\)

Vậy để P=2 <=> x=2

26 tháng 12 2017

Bài 4:

\(b,\dfrac{\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}}{1+\dfrac{x^3}{1-x^3}}\)

\(=\dfrac{\dfrac{\left(x+1\right)^2}{\left(x-1\right)\left(x+1\right)}-\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}}{\dfrac{1-x^3}{1-x^3}+\dfrac{x^3}{1-x^3}}\)

\(=\dfrac{\dfrac{x^2+2x+1-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}}{\dfrac{1-x^3+x^3}{1-x^3}}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{1}{\left(1-x\right)\left(1+x+x^2\right)}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}:\dfrac{-1}{\left(x-1\right)\left(1+x+x^2\right)}\)

\(=\dfrac{4x}{\left(x-1\right)\left(x+1\right)}.\left[-\left(x-1\right)\left(x^2+x+1\right)\right]\)

\(=\dfrac{-4x\left(x^2+x+1\right)}{x+1}\)

15 tháng 12 2017

Hỏi đáp Toán

8 tháng 12 2018

\(B=\frac{x^2-2}{x^2+1}=\frac{x^2+1-3}{x^2+1}=1-\frac{3}{x^2+1}\)

 \(B_{min}\Rightarrow\left(\frac{3}{x^2+1}\right)_{max}\Rightarrow\left(x^2+1\right)_{min}\)

\(x^2+1\ge1\). dấu = xảy ra khi x2=0

=> x=0

Vậy \(B_{min}\Leftrightarrow x=0\)

ta có: \(x^2+2x-2=x^2+2x+1^2-3=\left(x+1\right)^2-3\ge-3\)

dấu = xảy ra khi \(x+1=0\)

\(\Rightarrow x=-1\)

Vậy\(\left(x^2+2x-2\right)_{min}\Leftrightarrow x=-1\)

8 tháng 12 2018

Để A xác định 

\(\Rightarrow\hept{\begin{cases}x-1\ne0\\x^2-1\ne0\\x^2-2x+1\ne0\end{cases}}\)

\(\Rightarrow x^2-1\ne0\)

\(\Rightarrow\hept{\begin{cases}x\ne1\\x\ne-1\end{cases}}\)

b,