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F
1
HQ
Hà Quang Minh
Giáo viên
15 tháng 9 2023
\(4x^4-8x^3+4x^3-8x^2+x^2-2x-2x+4\\ =4x^3\left(x-2\right)+4x^2\left(x-2\right)+x\left(x-2\right)-2\left(x-2\right)\\ =\left(x-2\right)\left(4x^3+4x^2+x-2\right)\\ =\left(x-2\right)\left(4x^3-2x^2+6x^2-3x+4x-2\right)\\ =\left(x-2\right)\left[2x^2\left(2x-1\right)+3x\left(2x-1\right)+2\left(2x-1\right)\right]\\ =\left(x-2\right)\left(2x-1\right)\left(2x^2+3x-2\right)\)
T
1
3 tháng 8 2020
1) Ta có: \(\left(x^2-4x+4\right)\left(x^2+4x+4\right)-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)^2\cdot\left(x+2\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left[\left(x-2\right)\left(x+2\right)\right]^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4\right)^2-\left(7x+4\right)^2=0\)
\(\Leftrightarrow\left(x^2-4-7x-4\right)\left(x^2-4+7x+4\right)=0\)
\(\Leftrightarrow\left(x^2-7x-8\right)\left(x^2+7x\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x^2-8x+x-8\right)=0\)
\(\Leftrightarrow x\left(x+7\right)\left[x\left(x-8\right)+\left(x-8\right)\right]=0\)
\(\Leftrightarrow x\left(x+7\right)\left(x-8\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+7=0\\x-8=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-7\\x=8\\x=-1\end{matrix}\right.\)
Vậy: S={0;-7;8;-1}
2) Ta có: \(x^3-8x^2+17x-10=0\)
\(\Leftrightarrow x^3-2x^2-6x^2+12x+5x-10=0\)
\(\Leftrightarrow x^2\left(x-2\right)-6x\left(x-2\right)+5\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-6x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^2-x-5x+5\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-1\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-1=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=5\end{matrix}\right.\)
Vậy: S={2;1;5}
3) Ta có: \(2x^3-5x^2-x+6=0\)
\(\Leftrightarrow2x^3-4x^2-x^2+2x-3x+6=0\)
\(\Leftrightarrow2x^2\left(x-2\right)-x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x^2-3x+2x-3\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left[x\left(2x-3\right)+\left(2x-3\right)\right]=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x-3\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\2x-3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\2x=3\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{3}{2}\\x=-1\end{matrix}\right.\)
Vậy: \(S=\left\{2;\frac{3}{2};-1\right\}\)
4) Ta có: \(4x^4-4x^2-3=0\)
\(\Leftrightarrow4x^4-6x^2+2x^2-3=0\)
\(\Leftrightarrow2x^2\left(2x^2-3\right)+\left(2x^2-3\right)=0\)
\(\Leftrightarrow\left(2x^2-3\right)\left(2x^2+1\right)=0\)
mà \(2x^2+1>0\forall x\in R\)
nên \(2x^2-3=0\)
\(\Leftrightarrow2x^2=3\)
\(\Leftrightarrow x^2=\frac{3}{2}\)
hay \(x=\pm\sqrt{\frac{3}{2}}\)
Vậy: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
9 tháng 2 2017
a) Đặt x^2+2x+2=t
\(\frac{4}{t-1}+\frac{3}{t+1}=\frac{3}{2}\Leftrightarrow\frac{4t+4+3t-3}{t^2-1}=\frac{7t+1}{t^2-1}=\frac{3}{2}\)
\(\Leftrightarrow14t+2=3t^2-3\Leftrightarrow3t^2-14t-5=3t\left(t-5\right)+t-5=0\)\(\Leftrightarrow\left(t-5\right)\left(3t+1\right)=0\Rightarrow\left[\begin{matrix}t=5\\t=-\frac{1}{3}\left(loai\right)\end{matrix}\right.\)
Với t=5 ta có (x+1)^2=4\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
15 tháng 1 2017
a) \(^{x^3}\) - 7x+6=0
\(\Leftrightarrow\) \(^{x^3}\) - x-6x+6=0
\(\Leftrightarrow\) \(\left(x^3-x\right)\) - \(\left(6x-6\right)\) =0
\(\Leftrightarrow\) x\(\left(x^2-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) x\(\left(x+1\right)\)\(\left(x-1\right)\) - 6\(\left(x-1\right)\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left[x-6\left(x+1\right)\right]\) =0
\(\Leftrightarrow\) \(\left(x-1\right)\) \(\left(6-5x\right)\) =0
\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)
Những câu sau dùng phương pháp phân tích đa thức thành nhân tử nhé!
TM
28 tháng 6 2017
a)\(7x\left(y-4\right)^2-\left(4-y\right)^3=7x\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2\left(7x-4+y\right)\)
b)\(\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9\left(8-4x\right)\)
\(=\left(4x-8\right)\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)-9\left(4x-8\right)\)
\(=\left(4x-8\right)\left(x^2-x-10\right)=4\left(x-2\right)\left(x^2-x-10\right)\)
28 tháng 6 2017
a.\(7x.\left(y-4\right)^2-\left(4-y\right)^3\)=\(7x.\left(4-y\right)^2-\left(4-y\right)^3=\left(4-y\right)^2.\left(7x+y-4\right)\)
b.\(\left(4x-8\right).\left(x^2+6\right)-\left(4x-8\right)\left(x+7\right)+9.\left(8-4x\right)\)
=\(\left(4x-8\right)\left(x^2+6-x-7-9\right)=\left(4x-8\right)\left(x^2-x-10\right)\)
NL
3
18 tháng 6 2017
a) \(4x.\left(7x-5\right)-7x\left(4x-2\right)=-12\)
\(\Rightarrow28x^2-20x-28x^2+14x=-12\)
\(\Rightarrow x\left(14-20\right)=-12\)
\(\Rightarrow-6x=12\)
\(\Rightarrow x=-2\)
b) \(3x.\left(2x-4\right)-6x\left(x+5\right)=x-1\)
\(\Rightarrow3x.\left[\left(2x-4\right)-2.\left(x+5\right)\right]=x-1\)
\(\Rightarrow3x.\left(2x-4-2x-10\right)=x-1\)
\(\Rightarrow-42x=x-1\)
\(\Rightarrow-42x-x=-1\)
\(\Rightarrow-43x=-1\)
\(\Rightarrow x=\dfrac{1}{43}\)
DH
18 tháng 6 2017
a, \(4x\left(7x-5\right)-7x\left(4x-2\right)=-12\)
\(\Rightarrow28x^2-20x-28x^2+14=-12\)
\(\Rightarrow-20x=-12-14\)
\(\Rightarrow-20x=-26\Rightarrow x=1,3\)
Vậy \(x=1,3\)
b, \(3x\left(2x-4\right)-6x\left(x+5\right)=x-1\)
\(\Rightarrow6x^2-12x-6x^2-30-x=-1\)
\(\Rightarrow-13x=-1+30\)
\(\Rightarrow-13x=29\Rightarrow x=\dfrac{-29}{13}\)
Vậy \(x=\dfrac{-29}{13}\)
Chúc bạn học tốt!!!
9 tháng 10 2017
4x3 - 4x2 + x
= x.(4x2 - 4x + 1)
= x.[(2x)2 - 2.2.x + 12]
= x.(2x+1)2
x2 - y2 + 4 - 4x
= (x2 - 4x) - (y2 - 4)
= x.(x-4) - (y+4).(y-4)
x2 - 7x + 12
= x2 - 3x - 4x + 12
= x.(x-3) - 4.(x-3)
= (x-3).(x-4)
HN
10 tháng 10 2017
a) 4x3 - 4x2 + x
= x(4x2 - 4x + 1)
= x[(2x)2 - 2.2x.1 + 12)
= x(2x + 1)2
b) x2 - y2 + 4 - 4x
= (x2 - 4x + 4) - y2
= (x2 - 2.x.2 + 22) - y2
= (x - 2)2 - y2
= (x - 2 - y)(x - 2 + y)
c) x2 - 7x + 12
= x2 - 4x - 3x + 12
= (x2 - 4x) - (3x - 12)
= x(x - 4) - 3(x - 4)
= (x - 4)(x - 3).
Bạn cần làm gì với đa thức này?
F(x) = 4x4 - 4x3 - 7x2 - 4x + 4
xét hệ số tự do: 4 ta có: Ư(4) = {-4; -2; -1; 1;2; 4}
F(2) = 4.24 - 4.23 - 7.22 - 4.2 + 4
F(2) = 64 - 32 - 28 - 8 + 4 = 0
theo bezout ta có: F(x) ⋮ (x-2)
F(x) : (x-2) = 4x3 + 4x2 + x - 2
vậy 4x4 - 4x3 - 7x2 - 4x + 4 = (x-2)(4x3 + 4x2 +x-2)