1-x/(x+1)+3=2x+3/x+1giải pt sau mong mọi người giúp đỡ.Mình đang cần gấp
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\(a,\Leftrightarrow-\dfrac{1}{2}x=\dfrac{1}{4}\Leftrightarrow x=-\dfrac{1}{2}\\ b,\Leftrightarrow\dfrac{1}{6}:x=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\Leftrightarrow x=\dfrac{1}{6}:\dfrac{5}{6}=\dfrac{1}{5}\\ c,\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{5}=3\\x+\dfrac{1}{5}=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{14}{5}\\x=-\dfrac{16}{5}\end{matrix}\right.\)
\(d,\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2=\dfrac{22}{9}-\dfrac{7}{3}=\dfrac{1}{9}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{1}{3}\\x+\dfrac{1}{2}=-\dfrac{1}{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{6}\\x=-\dfrac{5}{6}\end{matrix}\right.\\ e,\Leftrightarrow2\left|x\right|=2-\dfrac{1}{2}=\dfrac{3}{2}\\ \Leftrightarrow\left[{}\begin{matrix}2x=\dfrac{3}{2}\\2x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{4}\\x=-\dfrac{3}{4}\end{matrix}\right.\)
\(f,\Leftrightarrow\left|x+\dfrac{1}{2}\right|=1+\dfrac{1}{6}=\dfrac{7}{6}\\ \Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=\dfrac{7}{6}\\x+\dfrac{1}{2}=-\dfrac{7}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-\dfrac{5}{3}\end{matrix}\right.\)
e: ta có: \(2\left|x\right|+\dfrac{1}{2}=2\)
\(\Leftrightarrow2\left|x\right|=\dfrac{3}{2}\)
\(\Leftrightarrow\left|x\right|=\dfrac{3}{4}\)
hay \(x\in\left\{\dfrac{3}{4};-\dfrac{3}{4}\right\}\)
`2)x^4+2x^3-x^2-2x+1=0`
`<=>x^4+2x^3+x^2-2x^2-2x+1=0`
`<=>(x^2+x)^2-2(x^2+x)+1=0`
`<=>(x^2+x-1)^2=0`
`<=>x^2+x-1=0`
`\Delta=1+4=5`
`=>x_{1,2}=(-1+-sqrt5)/2`
Vậy `S={(-1+sqrt5)/2,(-1+sqrt5)/2`
`3)x^4-4x^3-9x^2+8x+4=0`
`<=>x^4-x^3-3x^3+3x^2-12x^2+12x-4x+4=0`
`<=>(x-1)(x^3-3x^2-12x-4)=0`
`<=>(x-1)(x^3+2x^2-5x^2-10x-2x-4)=0`
`<=>(x-1)(x+2)(x^2-5x-10)=0`
`+)x=1`
`+)x=-2`
`+)x^2-5x-10=0`
`Delta=25+40=65`
`=>x_{12}=(5+sqrt{65})/2`
\(\left(x-\frac{1}{5}\right)^2=\frac{1}{25}\)
\(\Leftrightarrow x-\frac{1}{5}=\frac{1}{5}\)
\(\Rightarrow x=\frac{2}{5}\)
P/s tham khảo nha
\(\left(\frac{3}{2}-x\right)^3=-8\)
\(\Leftrightarrow\frac{3}{2}-x=-2\)
\(\Rightarrow x=\frac{7}{2}\)
P/s tham khảo nha
x^3-3x^2+5x+2007=0
nên \(x\simeq-11,57\)
y^3-3y^2+5y-2013=0
nên \(y\simeq13,57\)
=>x+y=2
1) \(25x^4-10x^2y+y^2\)
\(\Leftrightarrow\left(5x^2\right)^2+2\cdot\left(5x^2\right)\cdot y+y^2\)
\(\Leftrightarrow\left(5x^2+y\right)^2\)
2) \(x^4+2x^3-4x-4\)
\(\Leftrightarrow\left(x^4-4\right)+\left(2x^3-4x\right)\Leftrightarrow\left(x^2-2\right)\left(x^2+2\right)+2x\left(x^2-2\right)\)
\(\Leftrightarrow\left(x^2-2\right)\left(x^2+2+2x\right)\)
3) \(x^4+x^2+1\)
\(\Leftrightarrow x^4+x^2-x+x+1\)
\(\Leftrightarrow\left(x^4-x\right)+\left(x^2+x+1\right)\)
\(\Leftrightarrow x\left(x-1\right)\left(x^2+x+1\right)+\left(x^2+x+1\right)\)\(\Leftrightarrow\left(x^2+x+1\right)\left(x^2-x+1\right)\)
4) \(x^3-5x^2-14x\)\(\Leftrightarrow x^3-7x^2+2x^2-14x\)
\(\Leftrightarrow x^2\left(x-7\right)+2x\left(x-7\right)\)\(\Leftrightarrow x\left(x+2\right)\left(x-7\right)\)
5) \(x^2yz+5xyz-14yz\)\(\Leftrightarrow yz\left(x^2+5x-14\right)\)
\(\Leftrightarrow yz\left(x^2+7x-2x-14\right)\)
\(\Leftrightarrow yz\left[x\left(x+7\right)-2\left(x+7\right)\right]\)
\(\Leftrightarrow yz\left(x+7\right)\left(x-2\right)\)
DKXD : x khac -1
\(\frac{-x}{x+1}\)+ 3 =\(\frac{2x+3}{x+1}\)
<=> \(\frac{-x}{x+1}\)+\(\frac{3\left(x+1\right)}{x+1}\)= \(\frac{2x+3}{x+1}\)
=> -x + 3x +3 = 2x +3
<=> 2x -2x =3-3
<=> 0x=0
<=> x=0(TMDK)