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13 tháng 1 2018

minh giai phan d, nha bn :

x-a/b+c + x-b/c+a + x-c/a+b=3

=> (x-a/b+c - 1)+(x-b/a+c - 1 )+(x-c/a+b - 1) = 3-3=0

=>x-a-b-c/b+c + x-a-b-c/a+c + x-a-b-c/a+b =0

=>(x-a-b-c)(1/b+c + 1/a+c + 1/a+b )=0

Vi 1/b+c + 1/a+c + 1/a+b luon lon hon 0=>x-a-b-c=0

=>x=a+b+c

13 tháng 1 2018

g, x - a / b + c + x - b/ c+a + x - c/ a+b = 3x / a+b+c

19 tháng 11 2018

a, \(\dfrac{x^2-x}{x-2}+\dfrac{4-3x}{x-2}\)

\(=\dfrac{x^2-x+4-3x}{x-2}=\dfrac{x^2-4x+4}{x-2}\)

19 tháng 11 2018

c) \(\dfrac{2}{x^2-9}+\dfrac{1}{x+3}\)

Ta có: \(\dfrac{1}{x+3}=\dfrac{1\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}=\dfrac{x-3}{x^2-9}\)

\(\Rightarrow\dfrac{2}{x^2-9}+\dfrac{1}{x+3}=\dfrac{2}{x^2-9}+\dfrac{x-3}{x^2-9}=\dfrac{2+x-3}{x^2-9}=\dfrac{x-1}{x^2-9}\)

NV
18 tháng 3 2021

1a.

ĐKXĐ: \(x\ne\left\{1;3\right\}\)

\(\Leftrightarrow\dfrac{6}{x-1}=\dfrac{4}{x-3}+\dfrac{4}{x-3}\)

\(\Leftrightarrow\dfrac{3}{x-1}=\dfrac{4}{x-3}\Leftrightarrow3\left(x-3\right)=4\left(x-1\right)\)

\(\Leftrightarrow3x-9=4x-4\Rightarrow x=-5\)

b.

ĐKXĐ: \(x\ne\left\{-1;2\right\}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{3}{2-x}+\dfrac{1}{2-x}\)

\(\Leftrightarrow\dfrac{5}{x+1}=\dfrac{4}{2-x}\Leftrightarrow5\left(2-x\right)=4\left(x+1\right)\)

\(\Leftrightarrow10-2x=4x+4\Leftrightarrow6x=6\Rightarrow x=1\)

NV
18 tháng 3 2021

1c.

ĐKXĐ: \(x\ne\left\{2;5\right\}\)

\(\Leftrightarrow\dfrac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\dfrac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}=\dfrac{-3x}{\left(x-2\right)\left(x-5\right)}\)

\(\Leftrightarrow3x\left(x-5\right)-x\left(x-2\right)=-3x\)

\(\Leftrightarrow2x^2-10x=0\Leftrightarrow2x\left(x-5\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x=5\left(loại\right)\end{matrix}\right.\)

2a.

\(\Leftrightarrow-4x^2-5x+6=x^2+4x+4\)

\(\Leftrightarrow5x^2+9x-2=0\Rightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{5}\end{matrix}\right.\)

2b.

\(2x^2-6x+1=0\Rightarrow x=\dfrac{3\pm\sqrt{7}}{2}\)

\(A=\left[\dfrac{\left(a-1\right)^2}{a^2+a+1}+\dfrac{2a^2-4a-1}{a^3-1}+\dfrac{1}{a-1}\right]\cdot\dfrac{a\left(a^2+1\right)}{2a}\)

\(=\dfrac{a^3-3a^2+3a-1+2a^2-4a-1+a^2+a+1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}\)

\(=\dfrac{a^3-1}{\left(a-1\right)\left(a^2+a+1\right)}\cdot\dfrac{a^2+1}{2}=\dfrac{a^2+1}{2}\)

2: \(\left(\dfrac{7}{a+7}+\dfrac{a^2+49}{a^2-49}-\dfrac{7}{a-7}\right):\dfrac{a+1}{2}\)

\(=\dfrac{7a-49+a^2+49-7a-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}\)

\(=\dfrac{a^2-49}{\left(a-7\right)\left(a+7\right)}\cdot\dfrac{2}{a+1}=\dfrac{2}{a+1}\)

3: \(=\dfrac{x^4-4x^2+4x^2}{x^2-4}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x^2-4\right)}\cdot\dfrac{x^2-4}{x-2}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\left(\dfrac{x+2}{x-4}+\dfrac{2-3x}{x\left(x-2\right)}\right)\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x\left(x^2-4\right)+\left(2-3x\right)\left(x-4\right)}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-4x+2x-8-3x^2+12x}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-3x^2+10x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x^3-x^2-2x^2+2x+8x-8}{x\left(x-2\right)\left(x-4\right)}\)

\(=\dfrac{x^3\left(x-1\right)\left(x^2-2x+8\right)}{\left(x-2\right)^2\cdot\left(x+2\right)\left(x-4\right)}\)