Tìm các số A, B, C để có:
a) \(\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A}{\left(x-1\right)^3}+\dfrac{B}{\left(x-1\right)^2}+\dfrac{C}{x-1}\)
b) \(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}\)
Tìm các số A,B,C để có:
a)\(\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A}{\left(x-1\right)^3}+\dfrac{B}{\left(x-1\right)^2}+\dfrac{C}{x-1}\)
b)\(\dfrac{x^2+2x-1}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{A}{x-1}+\dfrac{Bx+C}{x^2+1}\)
a) PT \(\Leftrightarrow\dfrac{x^2-x+2}{\left(x-1\right)^3}=\dfrac{A+B\left(x-1\right)+C\left(x-1\right)^2}{\left(x-1\right)^3}\)
\(\Leftrightarrow x^2-x+2=A+Bx-B+Cx^2-2Cx+C\)
\(\Leftrightarrow x^2-x+2=Cx^2+x\left(B-2C\right)+\left(A+C-B\right)\)
\(\Leftrightarrow\left\{{}\begin{matrix}C=1\\B-2C=-1\\A+C-B=2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}A=2\\B=1\\C=1\end{matrix}\right.\)
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
Thực hiên phép tính:
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
b) \(\dfrac{\left[a^2-\left(b+c\right)^2\right]\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
d) \(\left[\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right]:\dfrac{x-y}{x}\)
a) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
b) \(\dfrac{\left(a^2-\left(b+c\right)^2\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(a^2+c^2-2ac-b^2\right)}\)
\(=\dfrac{\left(a-b-c\right)\left(a+b+c\right)\left(a+b-c\right)}{\left(a+b+c\right)\left(\left(a-c\right)^2-b^2\right)}\)
\(=\dfrac{\left(a-c-b\right)\left(a-c+b\right)}{\left(a-c-b\right)\left(a-c+b\right)}=1\)
c) \(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
\(=\dfrac{x-1}{x^3}-\dfrac{x+1}{x^2\left(x-1\right)}+\dfrac{3}{x\left(x-1\right)^2}\)
\(=\dfrac{\left(x-1\right)^3-x\left(x+1\right)\left(x-1\right)+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{x^3-3x^2+3x-1-x^3+x+3x^2}{x^3\left(x-1\right)^2}\)
\(=\dfrac{4x-1}{x^3\left(x-1\right)^2}\)
d) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
tìm các hệ số a,b,c sao cho
a) \(\dfrac{1}{x\left(x+1\right)\left(x+2\right)}\)= \(\dfrac{a}{x}\)+\(\dfrac{b}{x+1}\)+\(\dfrac{c}{x+2}\)
b) \(\dfrac{1}{\left(x^2+1\right)\left(x-1\right)}\)=\(\dfrac{ax+b}{x^2+1}\)+\(\dfrac{c}{x-1}\)
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
Mina ơi~~~Ai giải giùm em vài bài này với a~~Em làm rùi nhưng cứ thấy hoang mang quá nên hỏi mina cho chắc a~Em cảm ơn mina nhiều a~
Bài 1:
b,\(\left(\dfrac{3x}{1-3x}+\dfrac{2x}{3x+1}\right):\dfrac{6x^2+10x}{1-6x+9x^2}\)
c,\(\left(\dfrac{9}{x^3-9x}+\dfrac{1}{x+3}\right):\left(\dfrac{x-3}{x^2+3x}-\dfrac{x}{3x+9}\right)\)
d,\(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
c,\(\dfrac{x-1}{x^3}-\dfrac{x+1}{x^3-x^2}+\dfrac{3}{x^3-2x^2+x}\)
d,\(\dfrac{xy}{ab}+\dfrac{\left(x-a\right)\left(y-a\right)}{a\left(a-b\right)}-\dfrac{\left(x-b\right)\left(y-b\right)}{b\left(a-b\right)}\)
e,\(\dfrac{x^3}{x-1}-\dfrac{x^2}{x+1}-\dfrac{1}{x-1}+\dfrac{1}{x+1}\)
g,\(\left(\dfrac{x-y}{x+y}+\dfrac{x+y}{x-y}\right).\left(\dfrac{x^2+y^2}{2xy}+1\right).\dfrac{xy}{x^2+y^2}\)
h,\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
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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 = \(\dfrac{\sqrt{x}}{\sqrt{x}-5}-\dfrac{10\sqrt{x}}{x-25}-\dfrac{5}{\sqrt{x}+5}\)
1, tính
a,\(\dfrac{x+1}{x+2}:\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\)
b,\(\dfrac{x+1}{x+2}:\left(\dfrac{x+2}{x+3}:\dfrac{x+3}{x+1}\right)\)
c,\(\dfrac{x+3}{x+1}-\dfrac{2x-1}{x-1}-\dfrac{x-3}{x^2-1}\)
d,\(\dfrac{bc}{\left(a-b\right).\left(a-c\right)}+\dfrac{ac}{\left(b-a\right).\left(b-c\right)}+\dfrac{ab}{\left(c-a\right).\left(a-b\right)}\)
a: \(=\dfrac{x+1}{x+2}\cdot\dfrac{x+3}{x+2}\cdot\dfrac{x+1}{x+3}=\dfrac{\left(x+1\right)^2}{\left(x+2\right)^2}\)
b: \(=\dfrac{x+1}{x+2}:\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x+3\right)^2}\)
\(=\dfrac{x+1}{x+2}\cdot\dfrac{\left(x+3\right)^2}{\left(x+1\right)\left(x+2\right)}=\dfrac{\left(x+3\right)^2}{\left(x+2\right)^2}\)
c: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x-3-2x^2-2x+x+1-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+1}{\left(x-1\right)\left(x+1\right)}=-1\)
1.Cho B=\(\left(\dfrac{x-2}{x^2-5x+6}-\dfrac{x+3}{2-x}-\dfrac{x+2}{x-3}\right):\left(2-\dfrac{x}{x+1}\right)\)
a) Tìm đkxđ của B
b) Rút gọn B
c) Tìm x để B = 0
2. Cho C = \(\left(\dfrac{1}{x-1}-\dfrac{2x}{x^3-x^2+x-1}\right):\left(\dfrac{x^2+x}{x^3+x^2+x+1}+\dfrac{1}{x+1}\right)\)
a) Tìm đkxđ của C
b) Rút gọn C
c) Tìm x để C = \(\dfrac{2}{5}\)
d) Tìm x thuộc Z để giá trị C là số nguyên
a)\(\dfrac{2}{x+2}-\dfrac{1}{x+3}+\dfrac{2x+5}{\left(x+2\right)\left(x+3\right)}\)
b)\(\dfrac{2}{x+1}-\dfrac{1}{x+5}+\dfrac{2x+6}{\left(x+5\right)\left(x+1\right)}\)
c)\(\dfrac{-6}{x^2-9}-\dfrac{1}{x+3}+\dfrac{3}{x-3}\)
d)\(\dfrac{x}{x-2}-\dfrac{x}{x+2}+\dfrac{8}{x^2-4}\)