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TT
1
1 tháng 10 2016
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=5\end{array}\right.\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=7\end{array}\right.\)
d) \(x^2-9x+8=0\)
\(\Leftrightarrow x^2-x-8x+8=0\)
\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=8\end{array}\right.\)
g) \(3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{2}{3}\end{array}\right.\)
MR
1
KT
22 tháng 7 2018
\(x^2-5x-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
Vậy....
\(2x\left(x+6\right)=7x+42\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)
Vậy......
\(x^3-5x^2+x-5=0\)
\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow\)\(x-5=0\)
\(\Leftrightarrow\)\(x=5\)
\(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)
Vậy...
TT
0
19 tháng 10 2014
2.(x+5) - x2 - 5x = 0
2(x+5) - x(x+5) = 0
(x+5)(2-x) = 0
=> x+5=0 hoặc 2-x=0
=> x=-5 hoặc x=2
DT
25 tháng 6 2016
a)\(3x\left(x-1\right)+x-1=0\Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\Leftrightarrow\hept{\begin{cases}x-1=0\\3x-1=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\x=\frac{1}{3}\end{cases}}}\)
\(S=\left\{1;\frac{1}{3}\right\}\)
b)\(2\left(x+3\right)-x^2-3x=0\)
\(\Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\)
\(\Leftrightarrow\left(2-x\right)\left(x+3\right)=0\Leftrightarrow\hept{\begin{cases}2-x=0\\x+3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(S=\left\{2;-3\right\}\)
TQ
3
29 tháng 11 2018
\(8x^3+12x^2+6x+1=0.\)
\(\Leftrightarrow8x^2\left(x+\frac{1}{2}\right)+8x\left(x+\frac{1}{2}\right)+2\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left(x+\frac{1}{2}\right)\left(8x^2+8x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\2\left(4x^2+4x+1\right)=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-\frac{1}{2}\\2\left(2x+1\right)^2=0\Leftrightarrow x=-\frac{1}{2}\end{cases}}\)
Vậy pt có 1 No là...
29 tháng 11 2018
\(2\left(x+5\right)-x^2-5x=0.\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
GV
2
4 tháng 8 2017
a)\(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow x^2+2xy+y^2+y^2-2y+1=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Leftrightarrow\hept{\begin{cases}y-1=0\\x+y=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=1\\x=-y=-1\end{cases}}\)
Vậy x=-1 y=1
TA
4 tháng 8 2017
a) \(x^2+2y^2+2xy-2y+1=0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(y^2-2y+1\right)=0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(y-1\right)^2=0\)
\(\Rightarrow\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-1\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x+y=0\\y-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=1\end{cases}\Rightarrow}x=-1;y=1}\)
b) \(5x^2+3y^2+z^2-4x+6xy+4z+6=0\)
\(\Leftrightarrow\left(2x^2-4x+2\right)+\left(3x^2+6xy+3y^2\right)+\left(z^2+4z+4\right)=0\)
\(\Leftrightarrow2.\left(x-1\right)^2+3.\left(x+y\right)^2+\left(z+2\right)^2=0\)
\(\Rightarrow\) \(\left(x-1\right)^2=0\Rightarrow x-1=0\Rightarrow x=1\)
\(\left(x+y\right)^2=0\Rightarrow x+y=0\Rightarrow y=-x=-1\)
\(\left(z+2\right)^2=0\Rightarrow z+2=0\Rightarrow z=-2\)
\(2\left(x+5\right)-x^2-5x=0\\ 2\left(x+5\right)-\left(x^2+5x\right)=0\\ 2\left(x+5\right)-\left[x\left(x+5\right)\right]=0\\ \left(2-x\right)\left(x+5\right)=0\\ \Rightarrow\left\{{}\begin{matrix}2-x=0\\x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
2(x + 5) - x^2 - 5x = 0
=> 2(x+5)-(2x+5x) = 0
=> 2(x+5) - x(x+5) = 0
=> (x+5).(2-x) = 0
=> x+5=0 => x = -5
2-x=0 => x = 2
Vậy x=-5 và x=2