K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
TT
1
16 tháng 1 2020
\( a,\dfrac{2}{{x - 3}} = \dfrac{1}{{x + 2}}\left( {x \ne 3;x \ne - 2} \right)\\ \Leftrightarrow 2\left( {x + 2} \right) = x - 3\\ \Leftrightarrow 2x + 4 = x - 3\\ \Leftrightarrow x = - 7\left( {TM} \right)\\ b,\dfrac{5}{{3x - 2}} - \dfrac{1}{{x - 4}} = 0\left( {x \ne \frac{2}{3}; \ne 4} \right)\\ \Leftrightarrow 5\left( {x - 4} \right) - \left( {3x - 2} \right) = 0\\ \Leftrightarrow 5x - 20 - 3x + 2 = 0\\ \Leftrightarrow 2x = 18\\ \Leftrightarrow x = 9\left( {TM} \right)\\ c,\dfrac{3}{{x + 4}} = \dfrac{2}{{2x + 1}}\left( {x \ne - 4;x \ne - \frac{1}{2}} \right)\\ \Leftrightarrow 3\left( {2x + 1} \right) = 2\left( {x + 4} \right)\\ \Leftrightarrow 6x + 3 = 2x + 8\\ \Leftrightarrow 4x = 5\\ \Leftrightarrow x = \dfrac{5}{4}\left( {TM} \right)\\ d,\dfrac{7}{{3x - 4}} - \dfrac{3}{{3x - 3}} = 0\left( {x \ne \frac{4}{3};x \ne 1} \right)\\ \Leftrightarrow 7\left( {3x - 3} \right) - 3\left( {3x - 4} \right) = 0\\ \Leftrightarrow 21x - 21 - 9x + 12 = 0\\ \Leftrightarrow 12x = 9\\ \Leftrightarrow x = \dfrac{3}{4}\left( {TM} \right) \)
20 tháng 2 2017
a)
\(\left(x^2-1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left[\left(x+2\right)^2-1\right]=\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)\)
\(\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=\left(x^2+2x-3\right)\left(x^2+2x+1\right)\)
dặt x^2+2x-1=t(*)
(a) \(\Leftrightarrow\left(t-2\right)\left(t+2\right)=192\) \(\Leftrightarrow t^2-4=192\Rightarrow t^2=196\Rightarrow\left\{\begin{matrix}t=-14\\t=14\end{matrix}\right.\)
Thay t vào (*) => x (tự làm)
20 tháng 2 2017
a) (x-1)(x+1)(x+1)(x+3)=192. \(\Leftrightarrow\) (x+1)2(x-1)(x+3)=192 \(\Leftrightarrow\) (x2+2x+1) (x2+2x-3)=192 Đặt x2+2x+1=t thì x2+2x-3=t-4 ta có t(t-4)=192 \(\Leftrightarrow\) t2-4t-192=0 \(\Leftrightarrow\) t=-12 hoặc t=16 Với t=-12 thì (x+1)2=-12 ( vô lí ) Với t=16 thì (x+1)2=16 \(\Leftrightarrow\) x=-5 hoặc x=3 b) x\(^5\)+x4-2x4-2x3+5x3+5x2-2x2-2x+x+1=0 \(\Leftrightarrow\) x4(x+1)-2x3(x+1)+5x2(x+1)-2x(x+1)+(x+1)=0 \(\Leftrightarrow\) (x+1)(x4-2x3+5x2-2x+1)=0 \(\Leftrightarrow\) x=-1 ( CM x4-2x3+5x2-2x+1 vô nghiệm ) c) x4-x3-2x3+2x2+2x2-2x-x+1=0 \(\Leftrightarrow\) x3(x-1)-2x2(x-1)+2x(x-1)-(x-1)=0 \(\Leftrightarrow\) (x-1)(x3-2x2+2x-1)=0 \(\Leftrightarrow\) (x-1)(x-1)(x2-x+1)=0 \(\Leftrightarrow\) x-1=0 ( vì x2-x+1=(x-\(\frac{1}{2}\))2+\(\frac{3}{4}\)>0 với mọi x) \(\Leftrightarrow\) x=1
PT
6 tháng 2 2019
ta có : x^5+2x^4+3x^3+3x^2+2x+1=0
\(\Leftrightarrow\)x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0
\(\Leftrightarrow\)(x^5+x^4)+(x^4+x^3)+(2x^3+2x^2)+(x^2+x)+(x+1)=0
\(\Leftrightarrow\)x^4(x+1)+x^3(x+1)+2x^2(x+1)+x(x+1)+(x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+2x^2+x+1)=0
\(\Leftrightarrow\)(x+1)(x^4+x^3+x^2+x^2+x+1)=0
\(\Leftrightarrow\)(x+1)[x^2(x^2+x+1)+(x^2+x+1)]=0
\(\Leftrightarrow\)(x+1)(x^2+x+1)(x^2+1)=0
VÌ x^2+x+1=(x+\(\dfrac{1}{2}\))^2+\(\dfrac{3}{4}\)\(\ne0\) và x^2+1\(\ne0\)
\(\Rightarrow\)x+1=0
\(\Rightarrow\)x=-1
CÒN CÂU B TỰ LÀM (02042006)
14 tháng 2 2023
b: x^4+3x^3-2x^2+x-3=0
=>x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0
=>(x-1)(x^3+4x^2+2x+3)=0
=>x-1=0
=>x=1
SP
2
27 tháng 8 2018
a) \(x^3+3x^2+3x+2=0\)
<=> \(x^3+x^2+x+2x^2+2x+2=0\)
<=> \(x\left(x^2+x+1\right)+2\left(x^2+x+1\right)=0\)
<=> \(\left(x+2\right)\left(x^2+x+1\right)=0\)
tự làm
b) \(x^4-2x^3+2x-1=0\)
<=> \(\left(x^4-3x^3+3x^2-x\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(x\left(x^3-3x^2+3x-1\right)+\left(x^3-3x^2+3x-1\right)=0\)
<=> \(\left(x^3-3x^2+3x-1\right)\left(x+1\right)=0\)
<=> \(\left(x-1\right)^3\left(x+1\right)=0\)
tự làm
27 tháng 8 2018
c) \(x^4-3x^3-6x^2+8x=0\)
<=> \(x\left(x^3-3x^2-6x+8\right)=0\)
<=> \(x\left[\left(x^3+x^2-2x\right)-\left(4x^2+4x-8\right)\right]=0\)
<=>\(x\left[x\left(x^2+x-2\right)-4\left(x^2+x-2\right)\right]=0\)
<=> \(x\left(x-4\right)\left(x^2+x-2\right)=0\)
<=> \(x\left(x-4\right)\left(x-1\right)\left(x+2\right)=0\)
tự làm
3 tháng 2 2019
a) \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow x^4\left(x+1\right)+x^3\left(x+1\right)+2x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+x^2+x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
Dễ thấy \(x^2+x+1>0\forall x;x^2+1>0\forall x\)
\(\Rightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy....
3 tháng 2 2019
b) \(x^4+3x^3-2x^2+x-3=0\)
\(\Leftrightarrow x^4-x^3+4x^3-4x^2+2x^2-2x+3x-3=0\)
\(\Leftrightarrow x^3\left(x-1\right)+4x^2\left(x-1\right)+2x\left(x-1\right)+3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^3+4x^2+2x+3\right)=0\)
...
\(\Leftrightarrow x=1\)
p/s: có bác nào giải đc pt \(x^3+4x^2+2x+3=0\)thì giúp nhé :))
SP
5
KB
27 tháng 8 2018
a ) \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow x^3+3x^2+3x+1+1=0\)
\(\Leftrightarrow\left(x+1\right)^3+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=-1\)
\(\Leftrightarrow x+1=-1\)
\(\Leftrightarrow x=-2\)
Vậy \(x=-2\)
b ) \(x^4-2x^3+2x-1=0\)
\(\Leftrightarrow x^4-1-2x\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+1-2x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^3\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(x-1\right)^3=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
27 tháng 8 2018
a, \(x^3+3x^2+3x+2=0\)
\(\Leftrightarrow\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x^2+x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=0\end{matrix}\right.\)
b, \(x^4-2x^3+2x-1=0\)
\(\Leftrightarrow\left(x^4-x^3\right)-\left(x^3-x^2\right)-\left(x^2-x\right)+\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left[\left(x^3-x^2\right)\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)^3=0\)
\(\Leftrightarrow x=1\)
HY
8 tháng 2 2020
\(2.\left(x+3\right)\left(x+5\right)+\left(x+3\right)\left(3x-4\right)=0\\ \Leftrightarrow x^2+5x+3x+15+3x^2-4x+9x-12=0\\ \Leftrightarrow x^2+3x^2+5x+3x-4x+9x+15-12=0\\\Leftrightarrow 4x^2+13x+3=0\\\Leftrightarrow 4\left(x^2+\frac{13}{4}x+\frac{3}{4}\right)=0\\\Leftrightarrow x^2+\frac{13}{4}x+\frac{3}{4}=0\\ \Leftrightarrow x^2+\frac{1}{4}x+3x+\frac{3}{4}=0\\\Leftrightarrow x\left(x+\frac{1}{4}\right)+3\left(x+\frac{1}{4}\right)=0\\\Leftrightarrow \left(x+3\right)\left(x+\frac{1}{4}\right)=0\\\Leftrightarrow \left[{}\begin{matrix}x+3=0\\x+\frac{1}{4}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-3\\x=-\frac{1}{4}\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là: \(S=\left\{-3;-\frac{1}{4}\right\}\)
HY
8 tháng 2 2020
\(3.\left(x+6\right)\left(3x-1\right)+x+6=0\\ \Leftrightarrow3x^2-x+18x-6+x+6=0\\ \Leftrightarrow3x^2+18x=0\\ \Leftrightarrow3x\left(x+6\right)=0\\\Leftrightarrow \left[{}\begin{matrix}3x=0\\x+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-6\end{matrix}\right.\)
Vậy tập nghiệm của phương trình trên là \(S=\left\{0;-6\right\}\)
\(\left(x-4\right)\left(-3x+3\right)+3x\left(x+3\right)=0\)
\(\Rightarrow-3x^2+3x+12x-12+3x^2+9x=0\)
\(\Rightarrow24x-12=0\Leftrightarrow24x=12\Leftrightarrow x=\frac{1}{2}\)
Vậy ...
\(\left(x-4\right)\left(-3x+3\right)+3x\left(x+3\right)=0\)
\(-3x^2+3x+12x-12+3x^2+9x=0\)
\(24x-12=0\)
\(24x=12\Leftrightarrow x=\frac{1}{2}\)