Tìm x ,biết
a.(x-2)(x-3)=0
b.x^2-3x=0
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NP
3
31 tháng 10 2017
\(b,x^3-x^2-x+1=0\)
\(\Rightarrow\left(x^3-x^2\right)-\left(x-1\right)=0\)
\(\Rightarrow x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Rightarrow\left(x-1\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x-1\right)^2\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(c,2x^2-5x-7=0\)
\(\Rightarrow2x^2+2x-7x-7=0\)
\(\Rightarrow\left(2x^2+2x\right)-\left(7x+7\right)=0\)
\(\Rightarrow2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\Rightarrow\left(x+1\right)\left(2x-7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x+1=0\\2x-7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{7}{2}\end{matrix}\right.\)
31 tháng 10 2017
a)\(3\left(x-1\right)^2-3x\left(x-5\right)-2=0\)
\(3\left(x^2-2x+1\right)-\left(3x^2-15x\right)-2=0\)
\(3x^2+6x+3-3x^2+15x-2=0\)
\(9x+1=0\)
=>\(9x=1\)=>\(x=\dfrac{-1}{9}\)
Vậy...
b)\(x^3-x^2-x+1=0\)
\(\left(x^3-x^2\right)-\left(x-1\right)=0\)
\(x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\left(x-1\right).\left(x^2-1\right)=0\)
\(\left(x-1\right)\left(x-1\right)\left(x+1\right)=0\)
\(\left(x-1\right)^2\left(x+1\right)=0\)
=>\(\left[{}\begin{matrix}\left(x-1\right)^2=0\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy...
c)\(2x^2-5x-7=0\)
\(2x^2+2x-7x-7=0\)
\(\left(2x^2+2x\right)-\left(7x+7\right)=0\)
\(2x\left(x+1\right)-7\left(x+1\right)=0\)
\(\left(2x-7\right)\left(x+1\right)=0\)
=)\(\left[{}\begin{matrix}2x-7=0\\x+1=0\end{matrix}\right.=>\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=-1\end{matrix}\right.\)
Vậy...
CN
1
21 tháng 10 2017
b.\(x^3+6x^2+11x+6=0\)
\(\Leftrightarrow x^3+x^2+5x^2+5x+6x+6=0\)
\(\Leftrightarrow x^2\left(x+1\right)+5x\left(x+1\right)+6\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+5x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+2x+3x+6\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)=0\)
\(\Leftrightarrow\)\(x+1=0\)hoặc \(x+2=0\)hoặc \(x+3=0\)
\(\Leftrightarrow\)...... tự viết nha bn
CN
1
KT
21 tháng 10 2017
a) (6x5 - 3x2):3x - (4x2 + 8x):4x = 5
\(\Rightarrow\)2x4 - x - x - 2 = 5
\(\Rightarrow\)2(x4 - x -1) = 5
\(\Rightarrow\)x4 - 2x2 + 1 + 2x2 - 2 = 2.5
\(\Rightarrow\)(x2 - 1)2 + 2(x2 - 1) + 1 - \(\frac{7}{2}\) = 0
\(\Rightarrow\)x4 = \(\frac{7}{2}\)
\(\Rightarrow\)x = \(\pm\)\(\sqrt[4]{\frac{7}{2}}\)
b) x3 + 6x2 + 11x +6 = 0
\(\Rightarrow\)x3 + 6x2 + 12x + 8 - x - 2 = 0
\(\Rightarrow\)(x + 2)3 - (x + 2) = 0
\(\Rightarrow\)(x + 2)(x-1)(x+3)=0
\(\Rightarrow\)x + 2 = 0 \(\Rightarrow\)x = -2
x - 1 =0 \(\Rightarrow\)x = 1
x + 3 = 0 \(\Rightarrow\)x = -3
Vay.....
30 tháng 10 2017
Bài 1: Tìm x , biết :
\(a,x^2-3x=0\)
\(\Rightarrow x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
\(b,x^3-x=0\)
\(\Rightarrow x\left(x^2-1\right)=0\)
\(\Rightarrow x\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-1\end{matrix}\right.\)
30 tháng 10 2017
Bài 2: Phân tích đã thức thành nhân tử
\(a,3x-6y+xy-2y\)
\(=\left(3x-6y\right)+\left(xy-2y\right)\)
\(=3\left(x-2\right)+y\left(x-2\right)\)
\(=\left(x-2\right)\left(3+y\right)\)
\(b,x^2-2x-y^2+1\)
\(=\left(x^2-2x+1\right)-y^2\)
\(=\left(x-1\right)^2-y^2\)
\(=\left(x-1-y\right)\left(x-1+y\right)\)
\(c,x^2-4x+3\)
\(=x^2-3x-x+3\)
\(=\left(x^2-3x\right)-\left(x-3\right)\)
\(=x\left(x-3\right)-\left(x-3\right)\)
\(=\left(x-3\right)\left(x-1\right)\)
ND
3
29 tháng 7 2018
\(x^2+6x-x-6=0 \)
\(\Leftrightarrow x\left(x+6\right)-\left(x+6\right)=0\)
\(\Leftrightarrow \left(x+6\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+6=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-6\\x=1\end{cases}}\)
D
29 tháng 7 2018
\(a,x^2+6x-x-6=0\)
\(\left(x^2-x\right)+\left(6x-6\right)=0\)
\(x\left(x-1\right)+6\left(x-1\right)=0\)
\(\left(x+6\right)\left(x-1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+6=0\\x-1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=1\end{cases}}}\)
Vậy x = - 6 hoặc x = 1
\(b,x^2+2x-3x-6=0\)
\(x\left(x+2\right)-3\left(x+2\right)=0\)
\(\left(x-3\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-3=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=3\\x=-2\end{cases}}}\)
Vậy x = 3 hoặc x = -2
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
\(a,\left(x-2\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-2=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
\(b,x^2-3x=0\\ \Rightarrow x\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)