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27 tháng 6 2017
a) ... \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)\left(x+2\right)=0\Leftrightarrow\hept{\begin{cases}x=1\\x=2\\x=-2\end{cases}}\)Vậy.....
b) ... \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x^3+10x\right)=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=2\\x^2=-10\Rightarrow x\in\theta\end{cases}}\)(\(\theta\)là rỗng) Vậy.........
c) ... \(\Leftrightarrow2x-3=x+5\Leftrightarrow x=8\)Vậy.......
d) ... \(\Leftrightarrow x\left(x^2-16\right)=0\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\Leftrightarrow\hept{\begin{cases}x=0\\x=4\\x=-4\end{cases}}\)Vậy......
23 tháng 6 2017
\(\left(x+1\right)^2=x+1\)
\(\left(x+1\right)^2-\left(x+1\right)=0\)
\(\left(x+1\right)\left(x+1-1\right)=0\)
\(\left(x+1\right)x=0\)
\(\orbr{\begin{cases}x+1=0\\x=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=-1\\x=0\end{cases}}\)vậy.....
\(x\left(x-5\right)^2-4x+20=0\)
\(x\left(x-5\right)^2-4\left(x-5\right)=0\)
\(\left(x-5\right)\left[x\left(x-5\right)-4\right]=0\)
\(\left(x-5\right)\left(x^2-5x-4\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2-5x-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-0,7015621187\end{cases}}}\)vậy.........
\(x\left(x+6\right)-7x-42=0\)
\(x\left(x+6\right)-7\left(x+6\right)=0\)
\(\left(x+6\right)\left(x-7\right)=0\)
\(\orbr{\begin{cases}x+6=0\\x-7=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-6\\x=7\end{cases}}}\) vậy....
\(x^3-5x^2+x-5=0\)
\(x^2\left(x-5\right)+\left(x-5\right)=0\)
\(\left(x-5\right)\left(x^2+1\right)=0\)
\(\orbr{\begin{cases}x-5=0\\x^2+1=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x^2=-1\Rightarrow x\in\Phi\end{cases}}}\)vậy........
\(x^4-2x^3+10x^2-20x=0\)
\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\left(x-2\right)\left(x^3+10x\right)=0\)
\(\orbr{\begin{cases}x-2=0\\x^3+10x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=2\\x=0\end{cases}}}\)vậy..............
nhớ chọn mk nha
27 tháng 10 2020
Bài 3:
a) Ta có: \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4;-4\right\}\)
b) Ta có: \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)
c) Ta có: \(\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\)
d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
AN
12 tháng 11 2018
(*)\(\dfrac{15x\left(x+5\right)}{20x^2\left(x+5\right)}=\dfrac{3}{4x}\)
(*)\(\dfrac{x^3-4x^2}{y\left(x-4\right)}=\dfrac{x^2\left(x-4\right)}{y\left(x-4\right)}=\dfrac{x^2}{y}\)
(*)\(\dfrac{5\left(a-2c\right)^2}{2a^2-4ac}=\dfrac{5\left(a-2c\right)^2}{2a\left(a-2c\right)}=\dfrac{5\left(a-2c\right)}{2a}\)
4 tháng 3 2018
a)\(x\left(x+1\right)\left(x+2\right)\left(x+3\right)=24\)
\(\Leftrightarrow x\left(x+3\right)\left(x+1\right)\left(x+2\right)-24=0\)
\(\Leftrightarrow\left(x^2+3x\right)\left(x^2+3x+2\right)-24=0\)
Đặt \(x^2+3x+1=t\)
\(\Leftrightarrow\left(t-1\right)\left(t+1\right)-24=0\)
\(\Leftrightarrow t^2-25=0\)
\(\Leftrightarrow\left(t-5\right)\left(t+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)
TH1:t=5\(\Rightarrow x^2+3x+1=5\)
\(\Leftrightarrow x^2+3x-4=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-4\end{matrix}\right.\)
TH2:t=-5\(\Rightarrow x^2+3x+1=-5\)
\(\Leftrightarrow x^2+3x+6=0\)(vô nghiệm)
Vậy ...
4 tháng 3 2018
b)\(\Leftrightarrow2\left(x^4-10x^2+9\right)=0\)
\(\Leftrightarrow x^4-9x^2-x^2+9=0\)
\(\Leftrightarrow x^2\left(x^2-9\right)-\left(x^2-9\right)=0\)
\(\Leftrightarrow\left(x^2-9\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-3\\x=1\\x=-1\end{matrix}\right.\)
3 tháng 6 2022
Câu 2:
a: Để f(x) chia hết cho g(x) thì \(2x^3+3x^2-x+4⋮2x+1\)
\(\Leftrightarrow2x^3+x^2+2x^2+x-2x-1+5⋮2x+1\)
\(\Leftrightarrow2x+1\in\left\{1;-1;5;-5\right\}\)
hay \(x\in\left\{0;-1;2;-3\right\}\)
b: Để f(x) chia hết cho g(x) thì \(3x^3-x^2+6x⋮3x-1\)
\(\Leftrightarrow3x^3-x^2+6x-2+2⋮3x-1\)
\(\Leftrightarrow3x-1\in\left\{1;-1;2;-2\right\}\)
hay \(x\in\left\{\dfrac{2}{3};0;1;-\dfrac{1}{3}\right\}\)
Y
1 tháng 8 2018
a) \(5x\left(3x-7\right)-15x\left(x-1\right)=3\)
\(\Rightarrow15x^2-35x-15x^2+15x=3\)
\(\Rightarrow-20x=3\)
\(\Rightarrow x=-\dfrac{3}{20}\)
b) \(\left(4x+2\right)\left(6x-3\right)-\left(8x+5\right)\left(3x-4\right)=2\)
\(\Rightarrow24x^2+12x-12x-6-24x^2-15x+24x+20=2\)
\(\Rightarrow9x+14=2\)
\(\Rightarrow9x=-12\)
\(\Rightarrow x=-\dfrac{4}{3}\)
c) \(7x^2-21x=0\)
\(\Rightarrow7x\left(x-3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}7x=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)
d) \(9x^2-6x+1=0\)
\(\Rightarrow\left(3x\right)^2-2.3x+1=0\)
\(\Rightarrow\left(3x-1\right)^2=0\)
\(\Rightarrow3x-1=0\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\dfrac{1}{3}\)
e) \(16x^2-49=0\)
\(\Rightarrow\left(4x\right)^2-7^2=0\)
\(\Rightarrow\left(4x-7\right)\left(4x+7\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}4x-7=0\\4x+7=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}4x=7\\4x=-7\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{7}{4}\\x=-\dfrac{7}{4}\end{matrix}\right.\)
f) \(5x^3-20x=0\)
\(\Rightarrow5x\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=0\\x^2-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x^2=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
HH
25 tháng 6 2018
\(x^3+9x=0\)
<=> \(x\left(x^2+9\right)=0\)
<=> \(\orbr{\begin{cases}x=0\\x^2+9=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=0\\x\in\varnothing\end{cases}}\)
<=> \(x=0\)
\(9x^2-4-2\left(3x-2\right)^2=0\)
<=> \(\left(9x^2-4\right)-2\left(3x-2\right)^2=0\)
<=> \(\left[\left(3x\right)^2-2^2\right]-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left(3x+2\right)-2\left(3x-2\right)^2=0\)
<=> \(\left(3x-2\right)\left[\left(3x+2\right)-2\left(3x-2\right)\right]=0\)
<=> \(\left(3x-2\right)\left(3x+2-6x+4\right)=0\)
<=> \(\left(3x-2\right)\left(-3x+6\right)=0\)
<=> \(\left(3x-2\right)3\left(-x+2\right)=0\)
<=> \(3\left(3x-2\right)\left(2-x\right)=0\)
<=> \(\orbr{\begin{cases}3x-2=0\\2-x=0\end{cases}}\)
<=> \(\orbr{\begin{cases}3x=2\\x=2\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{2}{3}\\x=2\end{cases}}\)
\(\left(x^3-x^2\right)-4x+8x-4=0\)
<=> \(\left(x^3-x^2\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2+4\right)=0\)
<=> \(\orbr{\begin{cases}x-1=0\\x^2+4=0\end{cases}}\)
<=> \(x=1\)
\(\left(25x^2-10x\right):\left(-5x\right)-3\left(x-2\right)=4\)
<=> \(5x\left(5x-2\right)\left(-\frac{1}{5x}\right)-3\left(x-2\right)=4\)
<=> \(-\left(5x-2\right)-3\left(x-2\right)=4\)
<=> \(\left(5x-2\right)+3\left(x-2\right)=-4\)
<=> \(5x-2+3x-6=-4\)
<=> \(8x-8=-4\)
<=> \(8\left(x-1\right)=-4\)
<=> \(x-1=-\frac{1}{2}\)
<=> \(x=-\frac{3}{2}\)
với x=0 pt vô nghiệm
pt tương đương
(x+4)(x+6)(x+8)(x+12)=15x2
<=>[(x+4)(x+12)][(x+6)(x+8)]=15x2
<=>(x2+16x+48)(x2+14x+48)=15x2
chia 2 vế cho x2 ta được:
\(\left(x+16+\frac{48}{x}\right)\left(x+14+\frac{48}{x}\right)=15\)
Đặt t=x+48/x pt trở thành:
(t+16)(t+14)=15
<=>t2+30t+209=0
<=>t=-11 hoặc t=-19
với t=-11 không có giá trị x
với t=-19 =>x=-3 hoặc x=-16