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F
1
HQ
Hà Quang Minh
Giáo viên
15 tháng 9 2023
\(4x^4-8x^3+4x^3-8x^2+x^2-2x-2x+4\\ =4x^3\left(x-2\right)+4x^2\left(x-2\right)+x\left(x-2\right)-2\left(x-2\right)\\ =\left(x-2\right)\left(4x^3+4x^2+x-2\right)\\ =\left(x-2\right)\left(4x^3-2x^2+6x^2-3x+4x-2\right)\\ =\left(x-2\right)\left[2x^2\left(2x-1\right)+3x\left(2x-1\right)+2\left(2x-1\right)\right]\\ =\left(x-2\right)\left(2x-1\right)\left(2x^2+3x-2\right)\)
TT
0
KT
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: \(S=\left\{\sqrt{\frac{3}{2}};-\sqrt{\frac{3}{2}}\right\}\)
2 tháng 1 2023
a: \(\dfrac{x+10}{4x-8}\cdot\dfrac{4-2x}{x+2}\)
\(=\dfrac{x+10}{4\left(x-2\right)}\cdot\dfrac{-2\left(x-2\right)}{x+2}=\dfrac{-\left(x+10\right)}{2\left(x+2\right)}\)
b: \(\dfrac{1-4x^2}{x^2+4x}:\dfrac{2-4x}{3x}\)
\(=\dfrac{\left(2x-1\right)\left(2x+1\right)}{x\left(x+4\right)}\cdot\dfrac{3x}{2\left(x-2\right)}\)
\(=\dfrac{3\left(2x-1\right)\left(2x+1\right)}{2\left(x-2\right)\left(x+4\right)}\)
c: \(=\dfrac{4y^2}{7x^4}\cdot\dfrac{35x^2}{-8y}=\dfrac{5}{x^2}\cdot\dfrac{-1}{2}\cdot y=\dfrac{-5y}{2x^2}\)
d: \(=\dfrac{\left(x-2\right)\left(x+2\right)}{3\left(x+4\right)}\cdot\dfrac{x+4}{2\left(x-2\right)}=\dfrac{x+2}{6}\)
LT
1
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\(\Rightarrow\left[\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=1\\x=-3\end{matrix}\right.\)
LA
1
ND
Nguyễn Đức Trí
VIP
11 tháng 8 2023
a) \(4x^2+16x+3=0\)
\(\Delta'=84-12=72\Rightarrow\sqrt[]{\Delta'}=6\sqrt[]{2}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+6\sqrt[]{2}}{4}\\x=\dfrac{-8-6\sqrt[]{2}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2\left(4-3\sqrt[]{2}\right)}{4}\\x=\dfrac{-2\left(4+3\sqrt[]{2}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-\left(4-3\sqrt[]{2}\right)}{2}\\x=\dfrac{-\left(4+3\sqrt[]{2}\right)}{2}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3\sqrt[]{2}-4}{2}\\x=\dfrac{-3\sqrt[]{2}-4}{2}\end{matrix}\right.\)
b) \(7x^2+16x+2=1+3x^2\)
\(4x^2+16x+1=0\)
\(\Delta'=84-4=80\Rightarrow\sqrt[]{\Delta'}=4\sqrt[]{5}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-8+4\sqrt[]{5}}{4}\\x=\dfrac{-8-4\sqrt[]{5}}{4}\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-4\left(2-\sqrt[]{5}\right)}{4}\\x=\dfrac{-4\left(2+\sqrt[]{5}\right)}{4}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\left(2-\sqrt[]{5}\right)\\x=-\left(2+\sqrt[]{5}\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-2+\sqrt[]{5}\\x=-2-\sqrt[]{5}\end{matrix}\right.\)
c) \(4x^2+20x+4=0\)
\(\Leftrightarrow4\left(x^2+5x+1\right)=0\)
\(\Leftrightarrow x^2+5x+1=0\)
\(\Delta=25-4=21\Rightarrow\sqrt[]{\Delta}=\sqrt[]{21}\)
Phương trình có 2 nghiệm
\(\left[{}\begin{matrix}x=\dfrac{-5+\sqrt[]{21}}{2}\\x=\dfrac{-5-\sqrt[]{21}}{2}\end{matrix}\right.\)
25 tháng 8 2021
a: Ta có: \(x^2-4x\left(3x-4\right)+7x-5\)
\(=x^2-12x^2+16x+7x-5\)
\(=-11x^2+23x-5\)
b: Ta có: \(7x\left(x^2-5\right)-3x^2y\left(xy-6y^2\right)\)
\(=7x^3-35x-3x^3y^2+18x^2y^3\)
c: Ta có: \(\left(5x+4\right)\left(2x-7\right)\)
\(=10x^2-35x+8x-28\)
\(=10x^2-27x-28\)
2 tháng 1 2023
a.(x+10) /(4*x)-8* 4 -(2*x)/x+2
-(127*x-10)/(4*x)
(5/2-127*x/4)/x
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
\(\Leftrightarrow\) \(\left[\begin{matrix}x-1=0\\6-5x=0\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\5x=-6\end{matrix}\right.\)
\(\Leftrightarrow\) \(\left[\begin{matrix}x=1\\x=-\frac{6}{5}\end{matrix}\right.\)
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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)