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a) ĐKXD: x ≠ 2
\(\dfrac{1}{x-2}+3=\dfrac{3-x}{x-2}\)
\(\Leftrightarrow\dfrac{1}{x-2}-\dfrac{3-x}{x-2}=-3\)
\(\Leftrightarrow\dfrac{1-3+x}{x-2}=-3\)
\(\Leftrightarrow\dfrac{-2+x}{x-2}=-3\)
\(\Leftrightarrow-2+x=-3\left(x-2\right)\)
\(\Leftrightarrow-2+x=-3x+6\)
\(\Leftrightarrow x+3x=6+2\)
\(\Leftrightarrow4x=8\)
\(\Leftrightarrow x=2\) (loại vì không thỏa mãn điều kiện)
Vậy S = ∅
b) ĐKXĐ: x ≠ 7
\(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)
\(\Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{1}{x-7}=8\)
\(\Leftrightarrow\dfrac{7-x}{x-7}=8\)
\(\Leftrightarrow-1=8\left(vô-lý\right)\)
Vậy S = ∅
P/s: Ko chắc ạ!
c) ĐKXĐ: x ≠ 1
\(\dfrac{1}{x-1}+\dfrac{2x}{x^2+x+1}=\dfrac{3x^2}{x^3-1}\)
Quy đồng và khử mẫu ta được:
\(x^2+x+1+2x\left(x-1\right)=3x^2\)
\(\Leftrightarrow x^2+x+1+2x^2-2x-3x^2=0\)
\(\Leftrightarrow-x+1=0\)
\(\Leftrightarrow x=1\) (loại vì ko t/m đk)
Vậy S = ∅
a: =-1/5x^5y^2
b: =-9/7xy^3
c: =7/12xy^2z
d: =2x^4
e: =3/4x^5y
f: =11x^2y^5+x^6
f: \(=\dfrac{5x-3-x+3}{4x^2y}=\dfrac{4x}{4x^2y}=\dfrac{1}{xy}\)
g: \(=\dfrac{3x+10-x-4}{x+3}=\dfrac{2x+6}{x+3}=2\)
h: \(=\dfrac{4-2+x}{x-1}=\dfrac{x+2}{x-1}\)
n: \(=\dfrac{3x-x+6}{x\left(x+3\right)}=\dfrac{2\left(x+3\right)}{x\left(x+3\right)}=\dfrac{2}{x}\)
p: \(=\dfrac{x^2-9-x^2+9}{x\left(x-3\right)}=0\)
k: \(=\dfrac{x-2x-4+x-2}{\left(x+2\right)\left(x-2\right)}=\dfrac{-6}{x^2-4}\)
m: \(=\dfrac{3x-x+6}{x\left(2x+6\right)}=\dfrac{2x+6}{x\left(2x+6\right)}=\dfrac{1}{x}\)
a: \(\left(\dfrac{1}{3}x+2y\right)\left(\dfrac{1}{9}x^2-\dfrac{2}{3}xy+4y^2\right)=\dfrac{1}{27}x^3+8y^3\)
b: \(\left(x^2-\dfrac{1}{3}\right)\left(x^4+\dfrac{1}{3}x^2+\dfrac{1}{9}\right)=x^6-\dfrac{1}{27}\)
c: \(\left(y-5\right)\left(y^2+5y+25\right)=y^3-125\)
Lời giải:
ĐKXĐ: \(y\neq \pm 5; y\neq 0\)
Ta có:
\(\frac{y+1}{y^2-5y}-\frac{y-5}{2y^2+10y}=\frac{y+25}{2y^2-50}\)
\(\Leftrightarrow \frac{y+1}{y(y-5)}-\frac{y-5}{2y(y+5)}-\frac{y+25}{2(y^2-25)}=0\)
\(\Leftrightarrow \frac{2(y+1)(y+5)}{2y(y-5)(y+5)}-\frac{(y-5)(y-5)}{2y(y+5)(y-5)}-\frac{y(y+25)}{2y(y^2-25)}=0\)
\(\Leftrightarrow \frac{2(y^2+6y+5)}{2y(y^2-25)}-\frac{y^2-10y+25}{2y(y^2-25)}-\frac{y^2+25y}{2y(y^2-25)}=0\)
\(\Leftrightarrow \frac{2(y^2+6y+5)-(y^2-10y+25)-(y^2+25y)}{2y(y^2-25)}=0\)
\(\Leftrightarrow \frac{-3(y+5)}{2y(y^2-25)}=0\)
\(\Leftrightarrow -3(y+5)=0\Leftrightarrow y+5=0\Leftrightarrow y=-5\) (không t/m ĐKXĐ)
Vậy PT vô nghiệm.
\(\dfrac{y+1}{y^2-5y}-\dfrac{y-5}{2y^2+10y}=\dfrac{y+25}{2y^2-50}\left(ĐKXĐ:y\ne O;y\ne\pm5\right)\)
\(\Leftrightarrow\dfrac{y+1}{y\left(y-5\right)}-\dfrac{y-5}{2y\left(y+5\right)}=\dfrac{y+25}{2\left(y-5\right)\left(y+5\right)}\)
\(\Leftrightarrow\dfrac{2\left(y+1\right)\left(y+5\right)-\left(y-5\right)^2}{2y\left(y-5\right)\left(y+5\right)}=\dfrac{y\left(y+25\right)}{2y\left(y-5\right)\left(y+5\right)}\)
\(\Rightarrow2\left(y+1\right)\left(y+5\right)-\left(y-5\right)^2=y\left(y+25\right)\)
\(\Leftrightarrow\left(2y+2\right)\left(y+5\right)-\left(y^2-10y+25\right)=y^2+25y\)
\(\Leftrightarrow2y^2+10y+2y+10-y^2+10y-25=y^2+25y\)
\(\Leftrightarrow y^2+22y-15=y^2+25y\)
\(\Leftrightarrow y^2-y^2+22y-25y=15\)
\(\Leftrightarrow-3y=15\)
\(\Leftrightarrow y=-5\) (ko thỏa mãn ĐKXĐ)
Vậy ....................