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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.
Bài 1:
a: \(A=\dfrac{x^4+x^3+x+1}{x^4-x^3+2x^2-x+1}=\dfrac{x^3\left(x+1\right)+\left(x+1\right)}{x^4-x^3+x^2+x^2-x+1}\)
\(=\dfrac{\left(x+1\right)\left(x^3+1\right)}{\left(x^2-x+1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)^2}{x^2+1}\)
Để A=0 thì x+1=0
hay x=-1
b: \(B=\dfrac{x^4-5x^2+4}{x^4-10x^2+9}=\dfrac{\left(x^2-1\right)\left(x^2-4\right)}{\left(x^2-1\right)\left(x^2-9\right)}=\dfrac{x^2-4}{x^2-9}\)
Để B=0 thi (x-2)(x+2)=0
=>x=2 hoặc x=-2
Bài 1:
a: ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)
\(A+\left(\dfrac{4x}{x+2}-\dfrac{8x^2}{\left(x+2\right)\left(x-2\right)}\right):\left(\dfrac{x-1}{x\left(x-2\right)}-\dfrac{2}{x}\right)\)
\(=\dfrac{4x^2-8x-8x^2}{\left(x+2\right)\left(x-2\right)}:\dfrac{x-1-2x+4}{x\left(x-2\right)}\)
\(=\dfrac{-4x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x\left(x-2\right)}{-x+3}\)
\(=\dfrac{-4x}{-x+3}=\dfrac{4x}{x-3}\)
b: Để A<0 thi x/x-3<0
=>0<x<3
Lời giải:
ĐK: $x\neq \pm 2; x\neq 0$
a)
\(A=\left[\frac{x+2}{(x+2)(x-2)}+\frac{2x}{(x-2)(x+2)}+\frac{x-2}{(x-2)(x+2)}\right].\frac{2-x}{x}=\frac{x+2+2x+x-2}{(x-2)(x+2)}.\frac{-(x-2)}{x}\)
\(=\frac{4x}{(x-2)(x+2)}.\frac{-(x-2)}{x}=\frac{-4}{x+2}\)
b) Để $A=1\Leftrightarrow \frac{-4}{x+2}=1$
$\Leftrightarrow x+2=-4$
$\Leftrightarrow x=-6$ (thỏa ĐKXĐ)
Vậy $x=-6$
Lời giải:
a)
\(\frac{x-2}{6x^2-6x}-\frac{1}{4x^2-4}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x^2-1)}=\frac{x-2}{6x(x-1)}-\frac{1}{4(x-1)(x+1)}\)
\(=\frac{2(x+1)(x-2)}{12x(x-1)(x+1)}-\frac{3x}{12x(x-1)(x+1)}=\frac{2(x+1)(x-2)-3x}{12x(x-1)(x+1)}\)
\(=\frac{2x^2-5x-4}{12x(x-1)(x+1)}=\frac{2x^2-5x-4}{12x^3-12x}\)
b) ĐK: \(x\neq \pm 1\)
\(\frac{(x+1)(x^2-2x+1)}{6x^3+6}:\frac{x^2-1}{4x^2-4x+4}\)
\(=\frac{(x+1)(x-1)^2}{6(x^3+1)}.\frac{4x^2-4x+4}{x^2-1}\)
\(=\frac{4(x+1)(x-1)^2(x^2-x+1)}{6(x+1)(x^2-x+1)(x^2-1)}\)
\(=\frac{2(x-1)}{3(x+1)}\)
a)Ta có: \(\left(\dfrac{1}{x-2}+\dfrac{2x}{x^2-4}+\dfrac{1}{x+2}\right)\left(\dfrac{2}{x}-1\right)\)
\(=\left(\dfrac{1}{x-2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x+2}\right)\left(\dfrac{2-x}{x}\right)\)
\(=\left(\dfrac{-\left(x-2\right)}{x\left(x-2\right)}+\dfrac{-2x\left(x-2\right)}{x\left(x-2\right)\left(x+2\right)}+\dfrac{2-x}{x\left(x+2\right)}\right)\)
\(=\left(-\dfrac{1}{x}+\dfrac{-2}{x+2}+\dfrac{2-x}{x\left(x+2\right)}\right)\)
\(=\left(\dfrac{-x-2-2x+2-x}{x\left(x+2\right)}\right)=\dfrac{-4}{x+2}\)
b) Ta có ĐKXĐ của A là \(x\ne\pm2\)
Lại có \(A=-\dfrac{4}{x+2}=1\)
\(\Rightarrow-4=x+2\Rightarrow x=-6\)
Vậy x=-6 thì A=1