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\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)
\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}=\frac{x^2-4}{x^2+3}\)
\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)
\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)
\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)
<=>(2-x)(2+x)=0
<=>x=2 hoặc x=-2
Vậy S={-2;2}
\(\frac{148-x}{25}+\frac{169-x}{23}+\frac{186-x}{21}+\frac{199-x}{19}=10\)
\(\Leftrightarrow\frac{148-x}{25}-1+\frac{169-x}{23}-2+\frac{186-x}{21}-3+\frac{199-x}{19}-4=0\)
\(\Leftrightarrow\frac{148-x}{25}-\frac{25}{25}+\frac{169-x}{23}-\frac{46}{23}+\frac{186-x}{21}-\frac{63}{21}+\frac{199-x}{19}-\frac{76}{19}=0\)
\(\Leftrightarrow\frac{123-x}{25}+\frac{123-x}{23}+\frac{123-x}{21}+\frac{123-x}{19}=0\)
\(\Leftrightarrow\left(123-x\right).\left(\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\right)=0\)
\(\Leftrightarrow123-x=0\left(\text{vì }\frac{1}{25}+\frac{1}{23}+\frac{1}{21}+\frac{1}{19}\ne0\right)\)
<=>x=123
Vậy S={123}
\(\text{ĐKXĐ: }x+2\ne0\Leftrightarrow x\ne-2\)
\(x^2+\frac{4x^2}{\left(x+2\right)^2}=5\Leftrightarrow\frac{x^2\left(x+2\right)^2}{\left(x+2\right)^2}+\frac{4x^2}{\left(x+2\right)^2}=\frac{5\left(x+2\right)^2}{\left(x+2\right)^2}\)
\(\Leftrightarrow x^2\left(x+2\right)^2+4x^2=5\left(x+2\right)^2\)
<=>x2.(x2+4x+4)+4x2=5.(x2+4x+4)
<=>x4+4x3+4x2+4x2=5x2+20x+20
<=>x4+4x3+3x2-20x-20=0
<=>x4-2x3+6x3-12x2+15x2-30x+10x+20
<=>x3.(x-2)+6x2.(x-2)+15x.(x-2)+10.(x-2)=0
<=>(x-2)(x3+6x2+15x+10)=0
<=>(x-2)(x3+x2+5x2+5x+10x+10)=0
<=>(x-2).[x2(x+1)+5x.(x+1)+10.(x+1)]=0
<=>(x-2)(x+1)(x2+5x+10)=0
<=>x=2 hoặc x=-1 (vì x2+5x+10 = (x+5/2)2+15/4 >0)
Vậy S={-1;2}
phương trình hệ x hay phương trinhg hệ f vậy?
mấy bạn giúp mink vs