de ot tich cho minh di roi minh lam cho minh the day

dễ ~ tích cho mik đi mình làm x= -1/2;1/2;-2;-1/2

\(\frac{6}{x^2+2}+\frac{12}{x^2+8}=3-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-1+\frac{12}{x^2+8}-1=1-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{6}{x^2+2}-\frac{x^2+2}{x^2+2}+\frac{12}{x^2+8}-\frac{x^2+8}{x^2+8}=\frac{x^2+3}{x^2+3}-\frac{7}{x^2+3}\)

\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}=\frac{x^2-4}{x^2+3}\)

\(\Leftrightarrow\frac{-x^2+4}{x^2+2}+\frac{-x^2+4}{x^2+8}+\frac{-x^2+4}{x^2+3}=0\)

\(\Leftrightarrow\left(-x^2+4\right)\left(\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\right)=0\)

\(\Leftrightarrow-x^2+4=0\left(\text{vì : }\frac{1}{x^2+2}+\frac{1}{x^2+8}+\frac{1}{x^2+3}\ne0\right)\)

<=>(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}

( x² - 2x +4 )( x² +3x + 4 ) = 14x²

Đặt t=x²-2x+4 ta được:

t.(t+5x)=14x²

<=>t²+5tx=14x²

<=>t²+5tx-14x²=0

<=>t²-2tx+7tx-14x²=0

<=>t.(t-2x)+7x.(t-2x)=0

<=>(t-2x)(t+7x)=0

<=>t-2x=0 hoặc t+7x=0

<=>x²-2x+4-2x=0 hoặc x²-2x+4+7x=0

<=>x²-4x+4=0 hoặc x²+5x+4=0

<=>(x-2)²=0 hoặc x²+4x+x+4=0

<=>x-2=0 hoặc x.(x+4)+(x+4)=0

<=>x=2 hoặc (x+4)(x+1)=0

<=>x=2 hoặc x=-4 hoặc x=-1

ccamr ơn rất rất nhìu 

\(\frac{x^2+2x+1}{x^2+2x+1}+\frac{x^2+2x+2}{x^2+2x+3}=\frac{7}{6}\)

\(\Leftrightarrow\frac{x^2+2x+2-1}{x^2+2x+2}+\frac{x^2+2x+3-1}{x^2+3x+3}=\frac{7}{6}\)

\(\Leftrightarrow1-\frac{1}{x^2+2x+2}+1-\frac{1}{x^2+2x+3}=\frac{7}{6}\)

Đặt \(y=x^2+2x+1\), ta được:

\(2-\left(\frac{1}{y+1}+\frac{1}{y+2}\right)=\frac{7}{6}\)

\(\Leftrightarrow\frac{1}{y+1}+\frac{1}{y+2}=2-\frac{7}{6}=\frac{5}{6}\)

\(\Leftrightarrow\frac{1}{y+1}+\frac{1}{y+2}-\frac{5}{6}=0\)

\(\Leftrightarrow\frac{6\left(y+2\right)+6\left(y+1\right)-5\left(y+1\right)\left(y+2\right)}{6\left(y+1\right)\left(y+2\right)}=0\)

\(\Leftrightarrow6y+12+6y+6-\left(5y+5\right)\left(y+2\right)=0\)

\(\Leftrightarrow6y+12+6y+6-5y^2-10y-5y-10=0\)

\(\Leftrightarrow-5y^2-3y+8=0\)

\(\Leftrightarrow-5y^2+5y-8y+8=0\)

\(\Leftrightarrow-5y\left(y-1\right)-8\left(y-1\right)=0\)

\(\Leftrightarrow-\left(y-1\right)\left(5y+8\right)=0\)

Th1  \(y-1=0\Leftrightarrow y=1\) 

               \(\Leftrightarrow x^2+2x+1=1\)

                \(\Leftrightarrow\left(x+1\right)^2=1\Leftrightarrow x+1=1;x=1=-1\)

               \(\Leftrightarrow x=0\)   hoặc   \(x=-2\)

Th2  \(5y+8=0\Leftrightarrow5y=-8\Leftrightarrow y=\frac{-8}{5}\) 

       \(\Leftrightarrow x^2+2x+1=\frac{-8}{5}\)

        \(\Leftrightarrow\left(x+1\right)^2=-\frac{8}{5}\)

        Vì \(\left(x+1\right)^2\ge0\) mà   \(\left(x+1\right)^2=\frac{-8}{5}\)  ( vô lý) nên k có giá trị của x

Vậy   \(S=\left\{0;-2\right\}\)