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\(\Leftrightarrow x^3-\left(m-1\right)x^2-\left(m-1\right)x-2x^2+2\left(m-1\right)x+2m-2=0\)
\(\Leftrightarrow x\left(x^2-\left(m-1\right)x-m+1\right)-2\left(x^2-\left(m-1\right)x-m+1\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-\left(m-1\right)x-m+1\right)=0\)
Ta có: \(\text{Δ}=\left(1-4m\right)^2-4\left(3-2m\right)\left(1-2m\right)\)
\(=16m^2-8m+4-4\left(2m-3\right)\left(2m-1\right)\)
\(=16m^2-8m+4-4\left(4m^2-2m-6m+3\right)\)
\(=16m^2-8m+4-4\left(4m^2-8m+3\right)\)
\(=16m^2-8m+4-16m^2+32m-12\)
\(=24m-8\)
Để phương trình có hai nghiệm phân biệt thì
\(\left\{{}\begin{matrix}3-2m\ne0\\24m-8>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}2m\ne3\\24m>8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m\ne\dfrac{3}{2}\\m>\dfrac{1}{3}\end{matrix}\right.\)
1:
\(=\left(\dfrac{1}{x-2\sqrt{x}}+\dfrac{2}{3\sqrt{x}-6}\right):\dfrac{2\sqrt{x}+3}{3\sqrt{x}}\)
\(=\dfrac{3+2\sqrt{x}}{3\sqrt{x}\left(\sqrt{x}-2\right)}\cdot\dfrac{3\sqrt{x}}{2\sqrt{x}+3}=\dfrac{1}{\sqrt{x}-2}\)
`B=(x-x/(x+1))-(1-x/(x+1))`
`đkxđ:x ne +-1`
`=((x^2+x-x)/(x+1))-(x+1-x)/(x+1)`
`=x^2/(x+1)-1/(x+1)`
`=(x^2-1)/(x+1)`
`=((x-1)(x+1))/(x+1)`
`=x-1`
`2)(x-1)^2-25`
`=(x-1)^2-5^2`
`=(x-1-5)(x-1+5)`
`=(x-6)(x+4)`
Bài 1:
Ta có: \(B=\left(x-\dfrac{x}{x+1}\right)-\left(1-\dfrac{x}{x+1}\right)\)
\(=\left(\dfrac{x\left(x+1\right)-x}{x+1}\right)-\left(\dfrac{x+1-x}{x+1}\right)\)
\(=\dfrac{x^2+x-x-\left(x+1-x\right)}{x+1}\)
\(=\dfrac{x^2-1}{x+1}=x-1\)
câu a
Gọi x0 là nghiệm chung của PT(1) và (2)
\(\Rightarrow\left\{{}\begin{matrix}2x^2_0+\left(3m-1\right)x_0-3=0\left(\times3\right)\\6.x^2_0-\left(2m-1\right)x_0-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x^2_0+3\left(3m-1\right)x_0-9=0\left(1\right)\\6x^2_0-\left(2m-1\right)x_0-1=0\left(2\right)\end{matrix}\right.\) Lấy (1)-(2) ,ta được
PT\(\Leftrightarrow3\left(3m-1\right)-9+\left(2m-1\right)+1\)=0
\(\Leftrightarrow9m-3-9+2m-1+1=0\Leftrightarrow11m-12=0\)
\(\Leftrightarrow m=\dfrac{12}{11}\)
\(x^3+\left(2m+5\right)x^2+\left(2m+6\right)x-4m-12=\left(x^3-x^2\right)+\left[\left(2m+6\right)x^2-\left(2m+6\right)x\right]+\left[\left(4m+12\right)x-\left(4m+12\right)\right]=\left[x^2+\left(2m+6\right)x+\left(4m+12\right)\right]\left(x-1\right)\)