a)\(\dfrac{x+2}{(x)^{2}-4 }-\dfrac{2}{(x)^{2}-2x } \)
b)\((2x)^{3}-(x)^{2}+x+1:2x+1 \)
Những câu hỏi liên quan
a) dfrac{4x}{x^2+2x}+dfrac{8}{x^2+2x} b) dfrac{2x-3x}{x-2}-dfrac{2x-4}{x-2} c) dfrac{2x-1}{x+3}-dfrac{3x+2}{x+3}d) dfrac{11x}{2x-3}-dfrac{18-x}{2x-3}e) dfrac{3left(x-2right)}{2x+1}-dfrac{9x-3}{2x+1}
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a) \(\dfrac{4x}{x^2+2x}\)+\(\dfrac{8}{x^2+2x}\)
b) \(\dfrac{2x-3x}{x-2}\)-\(\dfrac{2x-4}{x-2}\)
c) \(\dfrac{2x-1}{x+3}\)-\(\dfrac{3x+2}{x+3}\)
d) \(\dfrac{11x}{2x-3}\)-\(\dfrac{18-x}{2x-3}\)
e) \(\dfrac{3\left(x-2\right)}{2x+1}\)-\(\dfrac{9x-3}{2x+1}\)
\(a,=\dfrac{4x+8}{x^2+2x}=\dfrac{4\left(x+2\right)}{x\left(x+2\right)}=\dfrac{4}{x}\\ b,=\dfrac{\left(2x-3\right)-\left(2x-4\right)}{x-2}=\dfrac{2x-3-2x+4}{x-2}=\dfrac{1}{x-2}\\ c,=\dfrac{2x-1-3x-2}{x+3}=\dfrac{-x-3}{x+3}=\dfrac{-\left(x+3\right)}{x+3}=-1\\ d,=\dfrac{11x-18+x}{2x-3}=\dfrac{12x-18}{2x-3}=\dfrac{6\left(2x-3\right)}{2x-3}=6\)
\(e,=\dfrac{3x-6-9x+3}{2x+1}=\dfrac{-6x-3}{2x+1}=\dfrac{-3\left(2x+1\right)}{2x+1}=-3\)
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giải phương trình
a, \(\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1}\)
b,\(\dfrac{3x-1}{x-1}-\dfrac{2x+5}{x+3}+\dfrac{4}{x^2+2x-3}=1\)
c,\(\dfrac{5}{x^2+x-6}-\dfrac{2}{x^2+4x+3}=\dfrac{-3}{2x-1}\)
d, \(\left(x^2-4\right)\left(2x+3\right)=\left(x^2-4\right)\left(x-1\right)\)
e, \(x^3+x^2+x+1=0\)
\(a,\dfrac{3}{2x-1}+1=\dfrac{2x-1}{2x+1};ĐKXĐ:x\ne\pm\dfrac{1}{2}\\ \Leftrightarrow\dfrac{3}{2x-1}-\dfrac{2x-1}{2x+1}+1=0\\ \Leftrightarrow\dfrac{3\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}-\dfrac{\left(2x-1\right)\left(2x-1\right)}{\left(2x+1\right)\left(2x-1\right)}+\dfrac{\left(2x-1\right)\left(2x+1\right)}{\left(2x-1\right)\left(2x+1\right)}=0\\ \Rightarrow3\left(2x+1\right)-\left(2x-1\right)^2+\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow6x+3-\left(4x^2-4x+1\right)+\left(4x^2-1\right)=0\\ \Leftrightarrow6x+3-4x^2+4x-1+4x^2-1=0\\ \Leftrightarrow10x+1=0\\ \Leftrightarrow10x=-1\\ \Leftrightarrow x=-\dfrac{1}{10}\)
Vậy \(x\in\left\{-\dfrac{1}{10}\right\}\)
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a,\(\dfrac{x+1}{x-3}+\dfrac{-2x^2+2x}{x^2-9}+\dfrac{x-1}{x+3}\)
b,\(\dfrac{1-2x}{6x^3y}+\dfrac{3+2y}{6x^3y}+\dfrac{2x-4}{6x^3y}\)
c,\(\dfrac{5}{2x^2y}+\dfrac{3}{5xy^2}+\dfrac{x}{3y^3}\)
d,\(\dfrac{5}{4\left(x+2\right)}+\dfrac{8-x}{4x^2+8x}\)
c,\(\dfrac{x^2+2}{x^3+1}+\dfrac{2}{x^2+x+1}+\dfrac{1}{1-x}\)
\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)
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Bài 2:a) (x+1)(2x-3)-3(x-2)2(x-1)^2b) (x+1)(x^2-x+1)-2xx(x-1)(x+1)c) dfrac{x}{3}-dfrac{5x}{6}-dfrac{15x}{12}dfrac{x}{4}-5d) dfrac{x-1}{2}-dfrac{x+1}{15}-dfrac{2x-13}{6}0e) dfrac{3left(5x-2right)}{4}-2dfrac{7x}{3}-5(x-7)g) dfrac{x-3}{11}+dfrac{x+1}{3}dfrac{x+7}{9}-1h) dfrac{3x-0,4}{2}+dfrac{1,5-2x}{3}dfrac{x+0,5}{5}
Đọc tiếp
Bài 2:
a) (x+1)(2x-3)-3(x-2)
=2(x-1)\(^2\)
b) (x+1)(x\(^2\)-x+1)-2x
=x(x-1)(x+1)
c) \(\dfrac{x}{3}\)-\(\dfrac{5x}{6}\)-\(\dfrac{15x}{12}\)=\(\dfrac{x}{4}\)-5
d) \(\dfrac{x-1}{2}\)-\(\dfrac{x+1}{15}\)-
\(\dfrac{2x-13}{6}\)=0
e) \(\dfrac{3\left(5x-2\right)}{4}\)-2
=\(\dfrac{7x}{3}\)-5(x-7)
g) \(\dfrac{x-3}{11}\)+\(\dfrac{x+1}{3}\)
=\(\dfrac{x+7}{9}\)-1
h) \(\dfrac{3x-0,4}{2}\)+\(\dfrac{1,5-2x}{3}\)
=\(\dfrac{x+0,5}{5}\)
a) Ta có: \(\left(x+1\right)\left(2x-3\right)-3\left(x-2\right)=2\left(x-1\right)^2\)
\(\Leftrightarrow2x^2-3x+2x-3-3x+6=2\left(x^2-2x+1\right)\)
\(\Leftrightarrow2x^2-4x+3-2x^2+4x-2=0\)
\(\Leftrightarrow1=0\)(vô lý)
Vậy: \(S=\varnothing\)
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a/\(\dfrac{1-x}{x+1}+3=\dfrac{2x+3}{x+1}\)
b/\(\dfrac{\left(x+2\right)^2}{2x-3}-1=\dfrac{x^2+10}{2x-3}\)
c/\(\dfrac{5x-2}{2-2x}+\dfrac{2x-1}{2}=1-\dfrac{x^2+x-3}{1-x}\)
đk: \(_{x+1\ne0\Leftrightarrow x\ne-1}\)\(\dfrac{1-x}{x+1}+3=\dfrac{2x-3}{x+1}\Leftrightarrow\dfrac{1-x}{x+1}+\dfrac{3\left(x+1\right)}{x+1}=\dfrac{2x+3}{x-1}\Leftrightarrow1-x+3x+3-2x-3=0\Leftrightarrow-2x+1=0\Leftrightarrow-2x=-1\Leftrightarrow x=0,5\)
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tìm điều kiện xác định của các phương trình saua,3x^2-2x0 b,dfrac{1}{x-1}3c,dfrac{2}{x-1}dfrac{x}{2x-4} d,dfrac{2x}{x^2-9}dfrac{1}{x+3}e,2xdfrac{1}{x^2-2x+1} f,dfrac{1}{x-2}dfrac{2x}{x^2-5x+6} giúp mik với , mik cần gấp
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tìm điều kiện xác định của các phương trình sau
\(a,3x^2-2x=0\) \(b,\dfrac{1}{x-1}=3\)
\(c,\dfrac{2}{x-1}=\dfrac{x}{2x-4}\) \(d,\dfrac{2x}{x^2-9}=\dfrac{1}{x+3}\)
\(e,2x=\dfrac{1}{x^2-2x+1}\) \(f,\dfrac{1}{x-2}=\dfrac{2x}{x^2-5x+6}\)
giúp mik với , mik cần gấp
a)\(x\in R\)
b)\(x\ne1\)
c) \(x\notin\left\{1;2\right\}\)
d) \(x\notin\left\{3;-3\right\}\)
e) \(x\ne1\)
f) \(x\notin\left\{2;3\right\}\)
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a) x∈R
b) x≠1
c) x∉{1;2}
d) x∉{3;−3}
e) x≠1
f) x∉{2;3}
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Thực hiện phép tính: (câu nào khó quá bỏ qua)a) dfrac{x^2+2}{x^3+1}-dfrac{1}{x+1}b)dfrac{x}{x^2-2x}-dfrac{x^2+4x}{x^3-4x}-dfrac{2}{x^2+2x}c)dfrac{1}{2-2x}-dfrac{3}{2+2x}+dfrac{2x}{x^2-1}d) dfrac{1}{left(a-bright)left(b-cright)}+dfrac{1}{left(b-cright)left(c-aright)}+dfrac{1}{left(c-aright)left(a-bright)}À mà nay sinh nhật tui á
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Thực hiện phép tính: (câu nào khó quá bỏ qua)
a) \(\dfrac{x^2+2}{x^3+1}\)-\(\dfrac{1}{x+1}\)
b)\(\dfrac{x}{x^2-2x}\)-\(\dfrac{x^2+4x}{x^3-4x}\)-\(\dfrac{2}{x^2+2x}\)
c)\(\dfrac{1}{2-2x}\)-\(\dfrac{3}{2+2x}\)+\(\dfrac{2x}{x^2-1}\)
d) \(\dfrac{1}{\left(a-b\right)\left(b-c\right)}\)+\(\dfrac{1}{\left(b-c\right)\left(c-a\right)}\)+\(\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
À mà nay sinh nhật tui á
a:
ĐKXĐ: x<>-1
\(\dfrac{x^2+2}{x^3+1}-\dfrac{1}{x+1}\)
\(=\dfrac{x^2+1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\)
\(=\dfrac{x^2+1-x^2+x-1}{\left(x+1\right)\left(x^2-x+1\right)}=\dfrac{x}{\left(x+1\right)\left(x^2-x+1\right)}\)
b: \(\dfrac{x}{x^2-2x}-\dfrac{x^2+4x}{x^3-4x}-\dfrac{2}{x^2+2x}\)
\(=\dfrac{x}{x\left(x-2\right)}-\dfrac{x\left(x+4\right)}{x\left(x^2-4\right)}-\dfrac{2}{x\left(x+2\right)}\)
\(=\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}-\left(\dfrac{1}{x}-\dfrac{1}{x+2}\right)\)
\(=\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}-\dfrac{1}{x}+\dfrac{1}{x+2}\)
\(=\left(\dfrac{1}{x-2}-\dfrac{x+4}{x^2-4}+\dfrac{1}{x+2}\right)-\dfrac{1}{x}\)
\(=\dfrac{x+2-x-4+x-2}{\left(x-2\right)\left(x+2\right)}-\dfrac{1}{x}\)
\(=\dfrac{x-4}{x^2-4}-\dfrac{1}{x}\)
\(=\dfrac{x^2-4x-x^2+4}{x\left(x^2-4\right)}=\dfrac{-4x+4}{x\left(x-2\right)\left(x+2\right)}\)
c: \(\dfrac{1}{2-2x}-\dfrac{3}{2+2x}+\dfrac{2x}{x^2-1}\)
\(=\dfrac{-1}{2\left(x-1\right)}-\dfrac{3}{2\left(x+1\right)}+\dfrac{2x}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x-1-3x+3+4x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{2}{2\left(x-1\right)\left(x+1\right)}=\dfrac{1}{x^2-1}\)
d:
\(\dfrac{1}{\left(a-b\right)\left(b-c\right)}+\dfrac{1}{\left(b-c\right)\left(c-a\right)}+\dfrac{1}{\left(c-a\right)\left(a-b\right)}\)
\(=\dfrac{c-a+a-b+b-c}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)
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Thực hiện phép tính:
a) \(\dfrac{x}{2x-y}-\dfrac{2x-y}{4x-2y}\)
b)\(\dfrac{3x+1}{x^2-1}-\dfrac{x}{2x-2}\)
c) \(\dfrac{x-2}{x^2-4}-\dfrac{-8-x}{3x^2+6x}\)
d) \(\dfrac{2}{2x-3}-\dfrac{x}{2x+3}-\dfrac{2x+1}{9-4x^2}\)
a: \(=\dfrac{2x-2x+y}{2\left(2x-y\right)}=\dfrac{y}{2\left(2x-y\right)}\)
b: \(=\dfrac{3x+1}{\left(x-1\right)\left(x+1\right)}-\dfrac{x}{2\left(x-1\right)}\)
\(=\dfrac{6x+2-x^2-x}{2\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{-x^2+5x+2}{2\left(x-1\right)\left(x+1\right)}\)
c: \(=\dfrac{1}{x+2}+\dfrac{x+8}{3x\left(x+2\right)}\)
\(=\dfrac{3x+x+8}{3x\left(x+2\right)}=\dfrac{4x+8}{3x\left(x+2\right)}=\dfrac{4}{3x}\)
d: \(=\dfrac{4x+6-2x^2+3x+2x+1}{\left(2x-3\right)\left(2x+3\right)}\)
\(=\dfrac{-2x^2+9x+7}{\left(2x-3\right)\left(2x+3\right)}\)
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a\(8\left(x+\dfrac{1}{x}\right)^{2^{ }}+4\left(x^{2^{ }}+\dfrac{1}{x^2}\right)-4\left(x^2+\dfrac{1}{x^2}\right)\left(x+\dfrac{1}{x}\right)=\left(x+4\right)^2\)giải các phương trình\(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
b)
ĐKXĐ: \(x\notin\left\{2;3;\dfrac{1}{2}\right\}\)
Ta có: \(\dfrac{x+4}{2x^2-5x+2}+\dfrac{x+1}{2x^2-7x+3}=\dfrac{2x+5}{2x^2-7x+3}\)
\(\Leftrightarrow\dfrac{x+4}{\left(x-2\right)\left(2x-1\right)}+\dfrac{x+1}{\left(x-3\right)\left(2x-1\right)}=\dfrac{2x+5}{\left(2x-1\right)\left(x-3\right)}\)
\(\Leftrightarrow\dfrac{\left(x+4\right)\left(x-3\right)}{\left(x-2\right)\left(2x-1\right)\left(x-3\right)}+\dfrac{\left(x+1\right)\left(x-2\right)}{\left(x-2\right)\left(x-3\right)\left(2x-1\right)}=\dfrac{\left(2x+5\right)\left(x-2\right)}{\left(2x-1\right)\left(x-3\right)\left(x-2\right)}\)
Suy ra: \(x^2-3x+4x-12+x^2-2x+x-2=2x^2-4x+5x-10\)
\(\Leftrightarrow2x^2-14=2x^2+x-10\)
\(\Leftrightarrow2x^2-14-2x^2-x+10=0\)
\(\Leftrightarrow-x-4=0\)
\(\Leftrightarrow-x=4\)
hay x=-4(nhận)
Vậy: S={-4}
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