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\(\dfrac{x}{x+4}+\dfrac{56-x^2}{x^2-16}=\dfrac{5}{x-4}\) (*)
ĐKXĐ: \(x\ne4\)
(*) \(\Leftrightarrow\dfrac{x^2-4x}{\left(x-4\right)\left(x+4\right)}+\dfrac{56-x^2}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x+20}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow\dfrac{56-4x}{\left(x-4\right)\left(x+4\right)}=\dfrac{5x+20}{\left(x-4\right)\left(x+4\right)}\)
\(\Leftrightarrow56-4x=5x+20\)
\(\Leftrightarrow9x=36\)
\(\Leftrightarrow x=4\) ( không thỏa mãn đk )
Vậy phương trình vô nghiệm
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a,\(5x+10y=5\left(x+2y\right)\)
b,\(=3xy\left(x+3yz\right)\)
c,\(=\left(x+1\right)\left(5-3x-3\right)=\left(x+1\right)\left(2-3x\right)\)
d,\(=\left(x-y\right)\left(2+3x\right)\)
e,\(=y\left(xy+3x^2+y^2\right)\)
f,\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\)
g,\(=x^2-x+\dfrac{1}{4}-\left(\dfrac{5}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2-\left(\dfrac{5}{2}\right)^2=\left(x-3\right)\left(x+2\right)\)
h,\(=x^2+9x+\dfrac{81}{4}-\left(\dfrac{7}{2}\right)^2=\left(x+\dfrac{9}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x+1\right)\left(x+8\right)\)
i,\(=x^2-10x+25-4^2=\left(x-5\right)^2-4^2=\left(x-1\right)\left(x-9\right)\)
k,\(=x^2+x+\dfrac{1}{4}-\left(\dfrac{7}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2=\left(x-3\right)\left(x+4\right)\)
l.\(=3\left(x^2+\dfrac{8}{3}x+\dfrac{4}{3}\right)=3\left(x^2+\dfrac{8}{3}x+\dfrac{16}{9}-\dfrac{4}{9}\right)\)
\(=3[\left(x+\dfrac{4}{3}\right)^2-\left(\dfrac{2}{3}\right)^2]=3\left(x+\dfrac{2}{3}\right)\left(x+2\right)\)
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TH1: x>=-5
=>3x+1=x+5
=>2x=4
=>x=2(nhận)
TH2: x<-5
=>-3x-1=x+5
=>-4x=6
=>x=-3/2(loại)
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a: Xét tứ giác HMKN có
I là trung điểm của HK
I là trung điểm của MN
Do đó: HMKN là hình bình hành
thi thì k giúp
Này làm luyện tập bạn ơi