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a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)
\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)
c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)
Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)
\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)
câu d
\(D=\dfrac{\left(1-x^2\right)}{x}\left(\dfrac{x^2}{x+3}-1\right)+\dfrac{3x^2-14x+3}{x^2+3x}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{\left(1-x^2\right)\left(x^2-x-3\right)+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{x^2-x-3-x^4+x^3-3x^2+3x^2-14x+3}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x^4+x^3+x^2-15x}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-x\left(x^3-x^2-x+15\right)}{x\left(x+3\right)}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{-3;0\right\}\\D=\dfrac{-\left(x^3-x^2-x+15\right)}{\left(x+3\right)}\end{matrix}\right.\)
d) \(\dfrac{x+1}{x-1}-\dfrac{x+2}{x+3}+\dfrac{4}{x^2+2x-3}=0\) (ĐKXĐ: \(x\ne1;-3\))
\(\Leftrightarrow\dfrac{\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)+4}{\left(x+3\right)\left(x-1\right)}=0\)
\(\Rightarrow\left(x+1\right)\left(x+3\right)-\left(x+2\right)\left(x-1\right)+4=0\)
\(\Leftrightarrow x^2+4x+3-x^2-x+2+4=0\)
\(\Leftrightarrow3x+9=0\Leftrightarrow x=-3\left(loại\right)\)
vậy phương trình đã cho vô nghiệm
c)\(\dfrac{2}{x-1}-\dfrac{3x^2}{x^3-1}=\dfrac{x}{x^2+x+1}\) (ĐKXĐ: \(x\ne1\))
\(\Leftrightarrow\dfrac{2\left(x^2+x+1\right)-3x^2}{x^3-1}=\dfrac{x\left(x-1\right)}{x^3-1}\)
\(\Rightarrow2x^2+2x+2-3x^2=x^2-x\)
\(-2x^2+3x+2=0\)
\(\left(x-2\right)\left(-2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\Leftrightarrow x=2\\-2x-1=0\Leftrightarrow x=-\dfrac{1}{2}\end{matrix}\right.\)
vậy tập nghiệm của phương trình là S={2;-0,5)
a: \(A=\left(\dfrac{\sqrt{3}\left(x-\sqrt{3}\right)+3}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\right)\cdot\dfrac{x^2+3+x\sqrt{3}}{x\sqrt{3}}\)
\(=\dfrac{x\sqrt{3}}{\left(x-\sqrt{3}\right)\left(x^2+x\sqrt{3}+3\right)}\cdot\dfrac{x^2+x\sqrt{3}+3}{x\sqrt{3}}\)
\(=\dfrac{1}{x-\sqrt{3}}\)
b: \(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}+x+1\)
\(=x-\sqrt{x}-x-\sqrt{x}+x+1\)
\(=x-2\sqrt{x}+1\)
c: \(C=\left(\dfrac{\sqrt{x}+2}{\left(\sqrt{x}+1\right)^2}-\dfrac{\sqrt{x}-2}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\right)\cdot\dfrac{x\left(\sqrt{x}+1\right)-\left(\sqrt{x}+1\right)}{\sqrt{x}}\)
\(=\dfrac{x+\sqrt{x}-2-\left(x-\sqrt{x}-2\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)^2}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}}=2\)
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: \(=\dfrac{3b+4a}{6ab}\)
b: \(=\dfrac{x^2-2x+1-x^2-2x-1}{x^2-1}=\dfrac{-4x}{x^2-1}\)
c: \(=\dfrac{xz+yz-xy-xz}{xyz}=\dfrac{yz-xy}{xyz}=\dfrac{z-x}{xz}\)
d: \(=\dfrac{2x+6-12}{\left(x-3\right)\left(x+3\right)}=\dfrac{2x-6}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x+3}\)
e: \(=\dfrac{x-2+2}{\left(x-2\right)^2}=\dfrac{x}{\left(x-2\right)^2}\)
a: \(=\dfrac{4a^2-4a+1-4a^2-2a+6a+3}{\left(2a-1\right)\left(2a+1\right)}\)
\(=\dfrac{4}{\left(2a-1\right)\left(2a+1\right)}\)
b: \(=\dfrac{x-1-x-1+2x^2}{\left(x-1\right)\left(x+1\right)}=2\)
d: \(=\dfrac{x-5+6x}{x\left(x+3\right)}=\dfrac{7x-5}{x\left(x+3\right)}\)
e: \(=\dfrac{x^2-4+3}{x-2}=\dfrac{x^2-1}{x-2}\)
i: \(=\dfrac{x}{x\left(x-4\right)}-\dfrac{3}{5x}=\dfrac{1}{x-4}-\dfrac{3}{5x}\)
\(=\dfrac{5x-3x+12}{5x\left(x-4\right)}=\dfrac{2x+12}{5x\left(x-4\right)}\)
\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\)
\(=\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x+1\right)\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\left(x+1\right)^2\left(x-1\right)}-\dfrac{x+1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{2\left(x-1\right)-x-1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{2x-2-x-1}{\left(x+1\right)^2\left(x-1\right)}\)
\(=\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\)
⇒Chọn B
\(\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{x^2-1}\\ =\dfrac{2}{\left(x+1\right)^2}-\dfrac{1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{2.\left(x-1\right)-\left(x+1\right)}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{2x-2-x-1}{\left(x+1\right)^2.\left(x-1\right)}\\ =\dfrac{x-3}{\left(x+1\right)^2\left(x-1\right)}\\ =>B\)