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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
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) \(\left(\dfrac{x^2-y^2}{xy}-\dfrac{1}{x+y}\left(\dfrac{x^2}{y}-\dfrac{y^2}{x}\right)\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{1}{x+y}.\dfrac{x^3-y^3}{xy}\right):\dfrac{x-y}{x}\)
\(=\left(\dfrac{\left(x-y\right)\left(x+y\right)}{xy}-\dfrac{\left(x-y\right)\left(x^2+xy+y^2\right)}{xy\left(x+y\right)}\right):\dfrac{x-y}{x}\)
\(=\dfrac{\left(x-y\right)\left(x^2+2xy+y^2-x^2-xy-y^2\right)}{xy\left(x+y\right)}.\dfrac{x}{x-y}\)
\(=\dfrac{x}{x+y}\)
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: \(=\dfrac{\left(x+3\right)\left(x-1\right)-\left(2x-1\right)\left(x+1\right)-\left(x-3\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2+2x-3-2x^2-2x+x+1-x+3}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+1}{\left(x-1\right)\left(x+1\right)}=-1\)
a) Đk : \(x\ne0;\ne1\)
\(\dfrac{x+3}{x+1}+\dfrac{x-2}{x}=\dfrac{2\left(x^2+x-1\right)}{x\left(x+1\right)}\)
\(\Rightarrow\dfrac{x^2+3x}{x\left(x+1\right)}+\dfrac{x^2-x-2}{x\left(x+1\right)}-\dfrac{2x^2+2x-2}{x\left(x+1\right)}=0\)
\(\Rightarrow\dfrac{x^2+3x+x^2-x-2-2x^2-2x+2}{x\left(x-1\right)}=0\)
\(\Rightarrow\dfrac{0}{x-1}=0\)
=> Phương trình có vô số nghiệm x
b) Đk : \(x\ne2;x\ne3\)
\(\dfrac{2}{x-2}-\dfrac{x}{x+3}=\dfrac{5x}{\left(x-2\right)\left(x+3\right)}-1\)
\(\Rightarrow\dfrac{2x+6}{\left(x-2\right)\left(x+3\right)}-\dfrac{x^2-2x}{\left(x-2\right)\left(x+3\right)}-\dfrac{5x}{\left(x-2\right)\left(x+3\right)}+\dfrac{x^2+x-6}{\left(x-2\right)\left(x+3\right)}\)
=0
\(\Rightarrow\dfrac{2x+6-x^2+2x-5x+x^2+x+6}{\left(x-2\right)\left(x+3\right)}=0\)
\(\Rightarrow\dfrac{12}{\left(x-2\right)\left(x+3\right)}=0\)
=> Phương trình vô nghiệm
c)
\(\Leftrightarrow\dfrac{x^2-x+1}{x^4+x^2+1}-\dfrac{x^2+x+1}{x^4+x^2+1}-\dfrac{1-2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{x^2-x+1-x^2-x-1-1+2x}{x^4+x^2+1}=0\)
\(\Rightarrow\dfrac{-1}{x^4+x^2+1}=0\)
=> PTVN
d) Thôi tự làm đi, câu này dễ :Vvv
e)
\(\left(x+1\right)\left(x+2\right)\left(x+4\right)\left(x+5\right)\)=40
\(\Rightarrow\left[\left(x+1\right)\left(x+5\right)\right]\cdot\left[\left(x+2\right)\left(x+4\right)\right]=40\)
\(\Rightarrow\left(x^2+6x+5\right)\left(x^2+6x+8\right)=40\)
Đặt
\(x^2+6x+7=t\)
Phương trình tương đương
\(\left(t-1\right)\left(t+1\right)=40\)
\(t^2=41\)
\(\)\(t=\pm\sqrt{41}\)
Thay vào tìm x.
a: \(=\dfrac{1}{x-y}-\dfrac{3xy}{\left(x-y\right)\left(x^2+xy+y^2\right)}+\dfrac{x-y}{x^2+xy+y^2}\)
\(=\dfrac{x^2+xy+y^2-3xy+x^2-2xy+y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}\)
\(=\dfrac{2x^2-4xy+2y^2}{\left(x-y\right)\left(x^2+xy+y^2\right)}=\dfrac{2\left(x-y\right)}{x^2+xy+y^2}\)
d: \(=\dfrac{x^3-1}{x-1}-\dfrac{x^2-1}{x+1}\)
\(=x^2+x+1-x+1=x^2+2\)
b: \(=\left[\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{x+1-3x^2-3x}{3x}\right]\cdot\dfrac{x}{x+1}\)
\(=\left(\dfrac{2}{3x}-\dfrac{2}{x+1}\cdot\dfrac{-3x^2-2x+1}{3x}\right)\cdot\dfrac{x}{x+1}\)
\(=\dfrac{2x+2+6x^2+4x-2}{3x\left(x+1\right)}\cdot\dfrac{x}{x+1}\)
\(=\dfrac{6x^2+6x}{3\left(x+1\right)}\cdot\dfrac{1}{x+1}\)
\(=\dfrac{6x\left(x+1\right)}{3\left(x+1\right)^2}=\dfrac{2x}{x+1}\)
c: \(VT=\left[\dfrac{2}{\left(x+1\right)^3}\cdot\dfrac{x+1}{x}+\dfrac{1}{\left(x+1\right)^2}\cdot\dfrac{1+x^2}{x^2}\right]\cdot\dfrac{x^3}{x-1}\)
\(=\left(\dfrac{2}{x\left(x+1\right)^2}+\dfrac{x^2+1}{x^2\cdot\left(x+1\right)^2}\right)\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{2x+x^2+1}{x^2\cdot\left(x+1\right)^2}\cdot\dfrac{x^3}{x-1}\)
\(=\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2}\cdot\dfrac{x}{x-1}=\dfrac{x}{x-1}\)
a) 1x−1−3x2x3−1=2xx2+x+11x−1−3x2x3−1=2xx2+x+1
Ta có: x3−1=(x−1)(x2+x+1)x3−1=(x−1)(x2+x+1)
=(x−1)[(x+12)2+34]=(x−1)[(x+12)2+34] cho nên x3 – 1 ≠ 0 khi x – 1 ≠ 0⇔ x ≠ 1
Vậy ĐKXĐ: x ≠ 1
Khử mẫu ta được:
x2+x+1−3x2=2x(x−1)⇔−2x2+x+1=2x2−2xx2+x+1−3x2=2x(x−1)⇔−2x2+x+1=2x2−2x
⇔4x2−3x−1=0⇔4x2−3x−1=0
⇔4x(x−1
Bài 1:
\(B=\dfrac{4\left(x+3\right)^2}{\left(3x+5\right)^2-4x^2}-\dfrac{\left(x^2-25\right)}{9x^2-\left(2x+5\right)^2}-\dfrac{\left(2x+3\right)^2-x^2}{\left(4x+15\right)^2-x^2}\)
\(=\dfrac{4\left(x+3\right)^2}{\left(3x+5-2x\right)\left(3x+5+2x\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{\left(3x-2x-5\right)\left(3x+2x+5\right)}-\dfrac{\left(2x+3-x\right)\left(2x+3+x\right)}{\left(4x+15-x\right)\left(4x+15+x\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x-5\right)\left(x+5\right)}{5\left(x-5\right)\left(x+1\right)}-\dfrac{3\left(x+3\right)\left(x+1\right)}{15\left(x+5\right)\left(x+3\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{x+5}{5\left(x+1\right)}-\dfrac{x+1}{5\left(x+5\right)}\)
\(=\dfrac{4\left(x+3\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+5\right)^2}{5\left(x+5\right)\left(x+1\right)}-\dfrac{\left(x+1\right)^2}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4\left(x^2+6x+9\right)-\left(x^2+10x+25\right)-\left(x^2+2x+1\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{4x^2+24x+36-x^2-10x-25-x^2-2x-1}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2x^2+12x+10}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+6x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x^2+5x+x+5\right)}{5\left(x+5\right)\left(x+1\right)}\)
\(=\dfrac{2\left(x+5\right)\left(x+1\right)}{5\left(x+5\right)\left(x+1\right)}=\dfrac{2}{5}\)
đc bn , nhg mà đề bài câu a b2 sao tự nhiên lại có " n "
bn xem lại đề đi
ngonhuminh