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16 tháng 6 2022
1: \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
\(\Leftrightarrow x^2+12x+36+x^2-10x+25=2x^2+23x+61\)
=>23x+61=2x+61
hay x=0
2: \(\dfrac{6}{x-5}+\dfrac{x+2}{x-8}=\dfrac{18}{\left(x-5\right)\left(8-x\right)}-1\)
\(\Leftrightarrow6x-48+x^2-3x-10=-18-x^2+13x-40\)
\(\Leftrightarrow x^2+3x-58+x^2-13x+58=0\)
\(\Leftrightarrow2x^2-10x=0\)
=>2x(x-5)=0
=>x=0
c: \(\dfrac{x^2-x}{x+3}-\dfrac{x^2}{x-3}=\dfrac{7x^2-3x}{9-x^2}\)
\(\Leftrightarrow\left(x^2-x\right)\left(x-3\right)-x^2\left(x+3\right)=-7x^2+3x\)
\(\Leftrightarrow x^3-3x^2-x^2+3x-x^3-3x^2+7x^2-3x=0\)
\(\Leftrightarrow x^2=0\)
hay x=0
LT
1
S
20 tháng 10 2018
a, \(x^3+5x^2+8x+4=x^3+x^2+4x^2+4x+4x+4\)
\(=x^2\left(x+1\right)+4\left(x^2+2x+1\right)=x^2\left(x+1\right)+4\left(x+1\right)^2=\left(x^2+4\right)\left(x+1\right)\)
b, \(x^3-6x^2-x+30=x^3+2x^2-8x^2-16x+15x+30\)
\(=x^2\left(x+2\right)-8x\left(x+2\right)+15\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2-8x+15\right)=\left(x+2\right)\left(x^2-8x+16-1\right)\)
\(=\left(x+2\right)\left[\left(x-4\right)^2-1\right]=\left(x+2\right)\left(x-3\right)\left(x-5\right)\)
1 tháng 9 2018
a) điều kiện xác định : \(x\ne2;x\ne-1\)
ta có : \(\dfrac{x+2}{x+1}+\dfrac{3}{x-2}=\dfrac{3}{x^2-x-2}+1\)
\(\Leftrightarrow\dfrac{\left(x+2\right)\left(x-2\right)+3\left(x+1\right)}{\left(x+1\right)\left(x-2\right)}=\dfrac{3+x^2-x-2}{\left(x+1\right)\left(x-2\right)}\)
\(\Rightarrow x^2-4+3x+3=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tmđk\right)\)
vậy \(x=\dfrac{1}{2}\)
b) điều kiện xác định : \(x\ne5;x\ne-6\)
ta có : \(\dfrac{x+6}{x-5}+\dfrac{x-5}{x+6}=\dfrac{2x^2+23x+61}{x^2+x-30}\)
\(\Leftrightarrow\dfrac{\left(x+6\right)^2+\left(x-5\right)^2}{\left(x-5\right)\left(x+6\right)}=\dfrac{2x^2+23x+61}{\left(x-5\right)\left(x+6\right)}\)
\(\Rightarrow x^2+12x+36+x^2-25x+25=2x^2+23x+61\)
\(\Leftrightarrow-13x=23x\Leftrightarrow x=0\left(tmđk\right)\)
vậy \(x=0\)
25 tháng 5 2022
c: \(x^3-8x^2+x+42\)
\(=x^3+2x^2-10x^2-20x+21x+42\)
\(=\left(x+2\right)\left(x^2-10x+21\right)\)
\(=\left(x+2\right)\left(x-3\right)\left(x-7\right)\)
a: \(x^3+6x^2+11x+6\)
\(=x^3+3x^2+3x^2+9x+2x+6\)
\(=\left(x+3\right)\left(x^2+3x+2\right)\)
\(=\left(x+3\right)\left(x+1\right)\left(x+2\right)\)
HL
0
N
1
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 8 2020
a/
\(\Leftrightarrow x^3-27+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+9\right)+x-3=0\)
\(\Leftrightarrow\left(x-3\right)\left(x^2+3x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x^2+3x+10=0\left(vn\right)\end{matrix}\right.\)
b/ Do \(x^2+x+6=\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0;\forall x\) nên pt tương đương:
\(x^2+x+6=x+7\)
\(\Leftrightarrow x^2=1\Rightarrow x=\pm1\)
c/ TH1: \(x\ge-11\)
\(\Leftrightarrow x+11=x^2+x+10\Leftrightarrow x^2=1\Rightarrow x=\pm1\)
TH2: \(x< -11\)
\(\Leftrightarrow-x-11=x^2+x+10\)
\(\Leftrightarrow x^2+2x+21=0\Leftrightarrow\left(x+1\right)^2+20=0\left(vn\right)\)
x3 - 6x2 - x + 30
= x3 - 5x2 - x2 + 5x - 6x + 30
= ( x3 - 5x2 ) - ( x2 - 5x ) - ( 6x - 30 )
= x2( x - 5 ) - x( x - 5 ) - 6( x - 5 )
= ( x - 5 )( x2 - x - 6 )
= ( x - 5 )( x2 - 3x + 2x - 6 )
= ( x - 5 )[ ( x2 - 3x ) + ( 2x - 6 ) ]
= ( x - 5 )[ x( x - 3 ) + 2( x - 3 ) ]
= ( x - 5 )( x - 3 )( x + 2 )
\(x^3-6x^2-x+30\)
\(=x^3-5x^2-x^2+5x-6x+30\)
\(=\left(x^3-5x^2\right)-\left(x^2-5x\right)-\left(6x-30\right)\)
\(=x^2\left(x-5\right)-x\left(x-5\right)-6\left(x-5\right)\)
\(=\left(x-5\right)\left(x^2-x-6\right)\)
\(=\left(x-5\right)\left(x^2-3x+2x-6\right)\)
\(=\left(x-5\right)\left[\left(x^2-3x\right)+\left(2x-6\right)\right]\)
\(=\left(x-5\right)\left[x\left(x-3\right)+2\left(x-3\right)\right]\)
\(=\left(x-5\right)\left(x-3\right)\left(x+2\right)\)