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`B4:`
`a)` Thay `x=3` vào ptr:
`3^3-3^2-9.3-9m=0<=>m=-1`
`b)` Thay `m=-1` vào ptr có: `x^3-x^2-9x+9=0`
`<=>x^2(x-1)-9(x-1)=0`
`<=>(x-1)(x-3)(x+3)=0<=>[(x=1),(x=+-3):}`
`B5:`
`a)` Thay `x=-2` vào có: `(-2)^3-(m^2-m+7).(-2)-3(m^2-m-2)=0`
`<=>-8+2m^2-2m+14-3m^2+3m+6=0`
`<=>-m^2+m+12=0<=>(m-4)(m+3)=0<=>[(m=4),(m=-3):}`
`b)`
`@` Với `m=4` có: `x^3-(4^2-4+7)x-3(4^2-4-2)=0`
`<=>x^3-19x-30=0`
`<=>x^3-5x^2+5x^2-25x+6x-30=0`
`<=>(x-5)(x^2+5x+6)=0`
`<=>(x-5)(x+2)(x+3)=0<=>[(x=5),(x=-2),(x=-3):}`
`@` Với `m=-3` có: `x^3-[(-3)^2-(-3)+7]x-3[(-3)^2-(-3)-2]=0`
`<=>x^3-19x-30=0<=>[(x=5),(x=-2),(x=-3):}`
Bài 1:
c) ĐKXĐ: \(x\notin\left\{\dfrac{1}{4};-\dfrac{1}{4}\right\}\)
Ta có: \(\dfrac{3}{1-4x}=\dfrac{2}{4x+1}-\dfrac{8+6x}{16x^2-1}\)
\(\Leftrightarrow\dfrac{-3\left(4x+1\right)}{\left(4x-1\right)\left(4x+1\right)}=\dfrac{2\left(4x-1\right)}{\left(4x+1\right)\left(4x-1\right)}-\dfrac{6x+8}{\left(4x-1\right)\left(4x+1\right)}\)
Suy ra: \(-12x-3=8x-2-6x-8\)
\(\Leftrightarrow-12x-3-2x+10=0\)
\(\Leftrightarrow-14x+7=0\)
\(\Leftrightarrow-14x=-7\)
\(\Leftrightarrow x=\dfrac{1}{2}\)(nhận)
Vậy: \(S=\left\{\dfrac{1}{2}\right\}\)
ĐKXĐ:\(x\ne\pm2;x\ne-3;x\ne0\)
\(P=1+\frac{x-3}{x^2+5x+6}\left(\frac{8x^2}{4x^3-8x^2}-\frac{3x}{3x^2-12}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{8x^2}{4x^2\left(x-2\right)}-\frac{3x}{3\left(x^2-4\right)}-\frac{1}{x+2}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left(\frac{2}{x-2}-\frac{x}{x^2-4}-\frac{1}{x+2}\right)\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\left[\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}-\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right]\)
\(=1+\frac{x-3}{\left(x+2\right)\left(x+3\right)}\cdot\frac{2x+4-x-x+4}{\left(x-2\right)\left(x+2\right)}\)
\(=1+\frac{8\left(x-3\right)}{\left(x+2\right)^2\left(x+3\right)\left(x-2\right)}\)
Đề sai à ??
Bài 1:
\(\frac{x+1}{65}+\frac{x+3}{63}=\frac{x+5}{61}+\frac{x+7}{59}\)
\(\Leftrightarrow\frac{x+1}{65}+1+\frac{x+3}{63}+1=\frac{x+5}{61}+1+\frac{x+7}{59}+1\)
\(\Leftrightarrow\frac{x+66}{65}+\frac{x+66}{63}=\frac{x+66}{61}+\frac{x+66}{59}\)
\(\Leftrightarrow\left(x+66\right)\left(\frac{1}{65}+\frac{1}{63}-\frac{1}{61}-\frac{1}{59}\right)=0\)
\(\Leftrightarrow x+66=0\)
\(\Leftrightarrow x=-66\)
b) \(\frac{m^2\left(\left(x+2\right)^2-\left(x-2\right)^2\right)}{8}-4x=\left(m-1\right)^2+3\left(2m+1\right)\)
\(\Leftrightarrow m^2x-4x=m^2+4m+4\)
\(\Leftrightarrow\left(m^2-4\right)x=m^2+4m+4\)
Để phương trình vô nghiệm thì \(\hept{\begin{cases}m^2-4=0\\m^2+4m+4\ne0\end{cases}}\Leftrightarrow\hept{\begin{cases}m=2\vee m=-2\\\left(m+2\right)^2\ne0\end{cases}}\Leftrightarrow m=2\)
Bài 1:
a) \(\left(m+2\right).3-5=4\)
\(\Leftrightarrow3m+6-5=4\)
\(\Leftrightarrow3m+1=4\)
\(\Leftrightarrow3m=4-1\)
\(\Leftrightarrow3m=3\)
\(\Leftrightarrow m=1\)
Vậy: m = 1
b) \(\left(m-3\right).\left(-2\right)+8=-10\)
\(\Leftrightarrow-2m+6+8=-10\)
\(\Leftrightarrow-2m+14=-10\)
\(\Leftrightarrow-2m=-10-14\)
\(\Leftrightarrow-2m=-24\)
\(\Leftrightarrow m=12\)
Vậy: m = 12
Bài 2:
a) \(\left(x-2\right)^2=9\)
\(\Leftrightarrow\left(x-2\right)^2=3^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=3\\x-2=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
b) \(\left(x+3\right)^2-0,16=0\)
\(\Leftrightarrow\left(x+3\right)^2=0,16\)
\(\Leftrightarrow\left(x+3\right)^2=\left(0,4\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0,4\\x+3=-0,4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2,6\\x=-3,4\end{matrix}\right.\)
c) \(x^3=25x\)
\(\Leftrightarrow x^3-25x=0\)
\(\Leftrightarrow x\left(x^2-25\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x^2-25=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\pm5\end{matrix}\right.\)