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DB
2
LT
6 tháng 7 2017
(x-2)3+6.(x+1)2-x3+12=0
<=> x3-6x2+12x-8 +6(x2+2x+1) - x3+12=0
<=>24x+10=0
<=> 24x=-10
<=> x= -5/12
vậy x = - 5/12
14 tháng 10 2018
1) \(2\left(x+2\right)-\left(3x+1\right)\left(x+2\right)=0\)
\(\left(x+2\right)\left(2-3x-1\right)=0\)
\(\left(x+2\right)\left(1-3x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+2=0\\1-3x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-2\\x=\frac{1}{3}\end{cases}}}\)
2) \(3x\left(x-3\right)-\left(2x-6\right)=0\)
\(3x\left(x-3\right)-2\left(x-3\right)=0\)
\(\left(x-3\right)\left(3x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\3x-2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=\frac{2}{3}\end{cases}}}\)
3) \(\left(2x-1\right)^2=\left(3x-5\right)^2\)
\(\left(2x-1\right)^2-\left(3x-5\right)^2=0\)
\(\left(2x-1-3x+5\right)\left(2x-1+3x-5\right)=0\)
\(\left(4-x\right)\left(5x-6\right)=0\)
\(\Rightarrow\orbr{\begin{cases}4-x=0\\5x-6=0\end{cases}\Rightarrow\orbr{\begin{cases}x=4\\x=\frac{6}{5}\end{cases}}}\)
4) \(\left(4x+3\right)\left(x-1\right)=x^2-1\)
\(\left(4x+3\right)\left(x-1\right)=\left(x+1\right)\left(x-1\right)\)
\(\left(4x+3\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x+3-x-1\right)=0\)
\(\left(x-1\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\3x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=\frac{-2}{3}\end{cases}}}\)
5) \(6-4x-\left(2x-3\right)\left(x-3\right)=0\)
\(-2\left(2x-3\right)-\left(2x-3\right)\left(x-3\right)=0\)
\(\left(2x-3\right)\left(-2-x+3\right)=0\)
\(\left(2x-3\right)\left(1-x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}2x-3=0\\1-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{3}{2}\\x=1\end{cases}}}\)
6) \(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+1=0\\2x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=\frac{7}{2}\end{cases}}}\)
7) \(x^2-x-12=0\)
\(x^2+3x-4x-12=0\)
\(x\left(x+3\right)-4\left(x+3\right)\)
\(\left(x+3\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+3=0\\x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=4\end{cases}}}\)
8) \(3x^2+14x-5=0\)
\(3x^2+15x-x-5=0\)
\(3x\left(x+5\right)-\left(x+5\right)=0\)
\(\left(x+5\right)\left(3x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+5=0\\3x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-5\\x=\frac{1}{3}\end{cases}}}\)
T
1 tháng 11 2015
a) \(\Leftrightarrow36+3-11x=0\)
\(\Leftrightarrow-11x=-39\)
\(\Leftrightarrow x=\frac{39}{11}\)
b) \(x^2-2x\frac{1}{4}+\frac{1}{16}-\frac{81}{16}=0\)
\(\left(x-\frac{1}{4}\right)^2=\frac{81}{16}\)
\(x-\frac{1}{4}=\frac{9}{4}\)
\(x=\frac{10}{4}=\frac{5}{2}\)
c) \(x^2\left(x-3\right)-4\left(x-3\right)=0\)
\(\left(x^2-4\right)\left(x-3\right)=0\)
\(\left(x-2\right)\left(x+2\right)\left(x-3\right)=0\)
x = 2 hoặc x = - 2 hoặc x = 3
MV
1 tháng 11 2015
a) \(\frac{8}{2}\)
b) \(\frac{5}{2}\)
c) x=2 hoạc x=-2 hoặc x=3
14 tháng 8 2016
a) (x-2)3 - 6(x+1)2 - x3 + 12 = 0
<=> x3-6x2+12x-8-6(x2+2x+1)-x3+12=0
<=> x3-6x2+12x-8-6x2-12x-6-x3+12=0
<=> -12x2+4=0
<=> \(x=\frac{1}{\sqrt{3}},x=-\frac{1}{\sqrt{3}}\)
vậy pt có 2 nghiệm....
b) x3 - 6x2 + 12x - 8 = 0
<=> (x3-2x2)-(4x2-8x)+(4x+8)=0
<=> (x-2)(x2-4x+4)=(x-2)3=0
=> x=2 là nghiệm
c) 8x3 - 12x2 + 6x - 1 = 0
<=> (2x-1)3=0
<=> x=1/2
14 tháng 8 2016
a) \(\left(x-2\right)^3-6\left(x+1\right)^2-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-6\left(x^2+2x+1\right)-x^3+12=0\)
\(\Leftrightarrow x^3-6x^2+12x-8-6x^2-12x-6-x^3+12=0\)
\(\Leftrightarrow-12x^2-2=0\)
\(\Leftrightarrow-2\left(6x^2+1\right)=0\)
\(\Leftrightarrow6x^2+1=0\) (vô nghiệm)
Vậy không có giá trị nào của x thỏa mãn pt
b) \(x^3-6x^2+12x-8=0\)
\(\Leftrightarrow\left(x-2\right)^3=0\)
\(\Leftrightarrow x-2=0\)
\(\Leftrightarrow x=2\)
Vậy x=2
c) \(8x^3-12x^2+6x-1=0\)
\(\Leftrightarrow\left(2x-1\right)^3=0\)
\(\Leftrightarrow2x-1=0\Leftrightarrow x=\frac{1}{2}\)
Vậy \(=\frac{1}{2}\)
11 tháng 9 2017
a) \(x^3\)\(-\)\(\frac{1}{4}x\)\(=\)\(0\)
\(x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2-\frac{1}{4}=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x^2=0,5^2\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=+-0,5\end{cases}}\)
Vậy .............................
b) \(\left(2x-1\right)^2\)\(-\)\(\left(x+3\right)^2\)\(=\)\(0\)
\(\left(2x-1+x+3\right)\left(2x-1-x-3\right)=0\)
\(\left(3x+2\right)\left(x-4\right)=0\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}3x+2=0\\x-4=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}3x=-2\\x=4\end{cases}}\)\(\orbr{\begin{cases}x=\frac{-2}{3}\\x=4\end{cases}}\)
Vậy ................................
c) \(x^2\)\(\left(x-3\right)\)\(+\)\(12\)\(-\)\(4x\)\(=\)\(0\)
\(x^2\)\(\left(x-3\right)\)\(-\)\(4\)\(\left(x-3\right)\)\(=\)\(0\)
\(\left(x^2-4\right)\left(x-3\right)\)
\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x-3=0\end{cases}-4=0}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x^2\\x=3\end{cases}=2^2}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=+-2\\x=3\end{cases}}\)
11 tháng 9 2017
a)\(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\hept{\begin{cases}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}}\)
A
14 tháng 10 2018
1/\(\left(x^2-6x+15\right):\left(x-3\right)\)
Đặt cột dọc ta được x-3 dư 6
2/a/\(p=\left(x+1\right)^3+\left(x+1\right)\left(6-x^2\right)-12\)
\(=x^3+3x^2+3x+1+6x-x^3+6-x^2-12\)
\(=2x^2+9x-11\)
b/thay x = -1/2 ta đc \(2.-\left(\frac{1}{2}\right)^2+9.-\frac{1}{2}-11\)
\(=\frac{1}{2}+\left(-\frac{9}{2}\right)-11\)
\(=\left(-15\right)\)
YN
12 tháng 12 2021
Answer:
\(3x^2-4x=0\)
\(\Rightarrow x\left(3x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{4}{3}\end{cases}}\)
\(\left(x^2-5x\right)+x-5=0\)
\(\Rightarrow x\left(x-5\right)+\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
\(x^2-5x+6=0\)
\(\Rightarrow x^2-2x-3x+6=0\)
\(\Rightarrow\left(x^2-2x\right)-\left(3x-6\right)=0\)
\(\Rightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=2\\x=3\end{cases}}\)
\(5x\left(x-3\right)-x+3=0\)
\(\Rightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}5x-1=0\\x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=3\end{cases}}\)
\(x^2-2x+5=0\)
\(\Rightarrow\left(x^2-2x+1\right)+4=0\)
\(\Rightarrow\left(x-1\right)^2=-4\) (Vô lý)
Vậy không có giá trị \(x\) thoả mãn
\(x^2+x-6=0\)
\(\Rightarrow x^2+3x-2x-6=0\)
\(\Rightarrow x.\left(x+3\right)-2\left(x+3\right)=0\)
\(\Rightarrow\left(x-2\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-2=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=-3\end{cases}}}\)
\(x^3-6x^2+12x-8+6\left(x^2+2x+1\right)-x^3+12=0\)
\(4-6x^2+12x+6x^2+12x+6=0\)
\(x=-\frac{10}{24}\)