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17 tháng 12 2017

Ta có :

\(\dfrac{a}{x-1}+\dfrac{bx+x}{x^2-x+1}=\dfrac{ax^2-ax+a+bx^2-bx+x^2-x}{\left(x-1\right)\left(x^2-x+1\right)}\)

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

Đồng nhất hai vế , ta có :

\(x^2\left(a+b+1\right)-x\left(a+b+1\right)+a=1\)

Suy ra :

* a + b +1 = 0 => 2 + b = 0 => b = - 2

* a = 1

Vậy,....

17 tháng 12 2017

nếu chuyển thành bx+c thì lm thế nào

NV
20 tháng 12 2020

Quy đồng vế phải:

\(VP=\dfrac{a\left(x+1\right)\left(x+2\right)+b\left(x+2\right)+c\left(x+1\right)^2}{\left(x+1\right)^2\left(x+2\right)}\)

\(=\dfrac{ax^2+3ax+2a+bx+2b+cx^2+2cx+c}{\left(x+1\right)^2\left(x+2\right)}\)

\(=\dfrac{\left(a+c\right)x^2+\left(3a+b+2c\right)x+2a+2b+c}{\left(x+1\right)^2\left(x+2\right)}\)

Đồng nhất hệ số với tử số vế trái ta được:

\(\left\{{}\begin{matrix}a+c=0\\3a+b+2c=0\\2a+2b+c=1\end{matrix}\right.\)  \(\Rightarrow\left\{{}\begin{matrix}a=-1\\b=1\\c=1\end{matrix}\right.\)

8 tháng 8 2017

ngonhuminh

Y
13 tháng 2 2019

a) PT \(\Leftrightarrow\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A+B\left(x-1\right)+C\left(x-1\right)^2}{\left(x-1\right)^3}\)

\(\Leftrightarrow x^2-x+2=A+Bx-B+Cx^2-2Cx+C\)

\(\Leftrightarrow x^2-x+2=Cx^2+x\left(B-2C\right)+\left(A+C-B\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}C=1\\B-2C=-1\\A+C-B=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=1\\C=1\end{matrix}\right.\)

27 tháng 11 2022

b: \(\Leftrightarrow\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{A\cdot x^2+A+\left(Bx+C\right)\left(x-1\right)}{\left(x^2+1\right)\left(x-1\right)}\)

\(\Leftrightarrow x^2\cdot A+A+x^2\cdot B-x\cdot B+x\cdot C-C=x^2+2x-1\)

\(\Leftrightarrow x^2\left(A+B\right)+x\left(-B+C\right)+A-C=x^2+2x-1\)

=>A+B=1; -B+C=2; A-C=-1

=>A+C=3; A-C=-1; A+B=1

=>A=1; C=2; B=1-A=0

a: =>a(x+1)(x+2)+bx(x+2)+cx(x+1)=1

=>a(x^2+3x+2)+bx^2+2bx+cx^2+cx=1

=>ax^2+3ax+2a+bx^2+2bx+cx^2+cx=1

=>x^2(a+b+c)+x(3a+2b+c)+2a=1

=>a+b+c=0 và 3a+2b+c=0 và a=1/2

=>a=1/2; b+c=-1/2; 2b+c=-3/2

=>b=-1; c=1/2; a=1/2

b: =>1=(ax+b)(x-1)+c(x^2+1)

=>x^2*a-a*x+bx-b+cx^2+c=1

=>x^2(a+c)+x(-a+b)-b+c=1

=>a+c=0 và -a+b=0 và -b+c=1

=>a+b=-1 và -a+b=0 và a+c=0

=>a=-1/2; b=-1/2; c=-a=1/2

10 tháng 12 2017

Bài 1:

\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)

\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)

\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)

\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)

10 tháng 12 2017

đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "

bn xem lại đề đi

12 tháng 5 2022

-ĐKXĐ: \(x\ne\pm1\)

\(\dfrac{x\left(1-x^2\right)^2}{1+x^2}:\left[\left(\dfrac{1-x^3}{1-x}+x\right)\left(\dfrac{1+x^3}{1+x}-x\right)\right]\)

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

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

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

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

\(=\dfrac{x}{x^2+1}\)

\(\Rightarrow m=\dfrac{x}{x^2+1}\)

-Khi \(x< 0\), mà \(x^2+1>0\forall x\)

\(\Rightarrow m=\dfrac{x}{x^2+1}< 0\).

\(\Rightarrow m< 0\)

-Vậy khi \(m< 0\) và \(m\ne\dfrac{-1}{2}\) thì \(x< 0\)

17 tháng 3 2017

Ta có: \(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a}{x-1}\)+\(\dfrac{bx+c}{x^2+1}\)

<=>\(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}\)=\(\dfrac{a\left(x^2+1\right)+\left(bx+c\right)\left(x-1\right)}{\left(x-1\right)\left(x^2+1\right)}\)

=>a(x2+1)+(bx+c)(x-1)=x2+2x-1

<=>ax2+a+bx2-bx+cx-c=x2+2x-1

<=>(a+b)x2+(c-b)x-(c-a)=x2+2x-1

=>\(\left\{{}\begin{matrix}a+b=1\\c-b=2\\c-a=1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}a+b=1\\b=c-2\\a=c-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}c-1+c-2=1\\b=c-2\\a=c-1\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}c=2\\b=0\\a=1\end{matrix}\right.\)

Vậy a=1,b=0,c=2