x^2+ 5x+6=0 tim x
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15 tháng 1 2016
b/ x2 + x + 6 = 0
=> x2 + 2.1/2 .x + (1/4) - (1/4) + 6 = 0
=> (x + 1/2)2 + 23/4 = 0
mà (x + 1/2)2 + 23/4 > 0 => vô nghiệm
LS
7 tháng 1 2016
a,x2-5x+4=0
x^2-x-4x+4=0
(x^2-x)-(4x-4)=0
x(x-1)-4(x-1)=0
(x-1)(x-4)=0
x-1=0. x-4=0
x=1 x=4
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...
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
\(2\left(x+5\right)-x^2-5x=0.\)
\(\Leftrightarrow2x+10-x^2-5x=0\)
\(\Leftrightarrow x^2+3x-10=0\)
\(\Leftrightarrow x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-4\end{cases}}}\)
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) \(3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{2}{3}\end{array}\right.\)
IA
18 tháng 3 2020
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)
Vậy ....
21 tháng 7 2016
a,x2+5x-6=x2+6x-x-6=x(x+6)-(x+6)=(x-1)(x+6)=0suy ra x=1;-6
b,x2-x-6=x2-3x+2x-6=x(x-3)+2(x-3)=(x+2)(x-3)=0suy ra x=-2;3
b2 ko có nghiệm
TT
0
B
1
23 tháng 9 2016
e) \(2\left(x+5\right)-x^2-5x=0\)
\(=>2\left(x+5\right)-x\left(x+5\right)=0\)
\(=>\left(x+5\right)\left(2-x\right)=0\)
\(=>\hept{\begin{cases}x+5=0\\2-x=0\end{cases}}\)
\(=>\hept{\begin{cases}x=-5\\x=2\end{cases}}\)
f) \(x^2-2x-3=0\)
\(=>x^2-3x+x-3=0\)
\(=>x\left(x-3\right)+\left(x-3\right)=0\)
\(=>\left(x+1\right)\left(x-3\right)=0\)
\(=>\hept{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
\(=>\hept{\begin{cases}x=-1\\x=3\end{cases}}\)
g) \(2x^2+5x-3=0\)
\(=>2x^2-6x+x-3=0\)
\(=>2x\left(x-3\right)+\left(x-3\right)=0\)
\(=>\left(2x+1\right)\left(x-3\right)=0\)
\(=>\hept{\begin{cases}2x+1=0\\x-3=0\end{cases}}\)
\(=>\hept{\begin{cases}x=\frac{-1}{2}\\x=3\end{cases}}\)
h) \(x^2+x-6=0\)
\(=>x^2-2x+3x-6=0\)
\(=>x\left(x-2\right)+3\left(x-2\right)=0\)
\(=>\left(x+3\right)\left(x-2\right)=0\)
\(=>\hept{\begin{cases}x+3=0\\x-2=0\end{cases}}\)
\(=>\hept{\begin{cases}x=-3\\x=2\end{cases}}\)
\(x^2+5x+6=0< =>x^2+2x+3x+6=0< =>x\left(x+2\right)+3\left(x+2\right)=0< =>\left(x+3\right)\left(x+2\right)=0< =>\left[{}\begin{matrix}x+2=0\\x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\\x=-3\end{matrix}\right.\)
cố quên một người: chưa hiểu chỗ nào vậy