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\(x^4-3x^3+4x^2-3x-1=0\)
\(\Leftrightarrow x^4+x^3+2x^3+2x^2+2x^2+2x+x+1=0\)
\(\Leftrightarrow x^3\left(x+1\right)+2x^2\left(x+1\right)+2x\left(x+1\right)+\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x^3+2x^2+2x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow(x^3+x^2+x^2+x+x+1)\left(x+1\right)=0\)
\(\Leftrightarrow[x^2\left(x+1\right)+x\left(x+1\right)+\left(x+1\right)]\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\left(x^2+x+1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}(x+1)^2=0\\x^2+x+1=0\end{cases}}\Rightarrow\hept{\begin{cases}x+1=0\\\varnothing\end{cases}}\Rightarrow x=-1\)
bạn đã học giải pt bậc 2 chưa có công thức bài nào cũng giải đc
a) x^2+3x=0
<=> x(x+3)=0
<=> x=0 hoặc x+3=0
<=> x=0 hoặc x=-3
S={0;-3}
b) x^2-x-42=0
<=> x^2-7x+6x-42=0
<=> x(x-7)+6(x-7)=0
<=> (x-7)(x+6)=0
<=> x-7=0 hoac x+6=0
<=> x=7,x=-6
c) ,d) tương tự
e) 2x^3+3x^2-x-1=0
<=> 2x^3+x^2+2x^2+x-2x-1=0
<=> x^2(2x+1)+x(2x+1)-(2x+1)=0
<=> (2x+1)(x^2+x-1)=0
<=>2x+1=0 hoặc x^2+x-1=0
<=> x=-1/2 ,x=-1+căn5/2,x=-1-căn5/2
a) (5x - 1)(2x + 1) = (5x -1)(x + 3)
<=> (5x - 1)(2x + 1) - (5x -1)(x + 3) = 0
<=> (5x - 1)(2x + 1 - x - 3) = 0
<=> (5x - 1)(x - 2) = 0
<=> \(\orbr{\begin{cases}5x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0,2\\x=2\end{cases}}\)
Vậy x = 0,2 ; x = 2 là nghiệm phương trình
b) x3 - 5x2 - 3x + 15 = 0
<=> x2(x - 5) - 3(x - 5) = 0
<=> (x2 - 3)(x - 5) = 0
<=> \(\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)\left(x-5\right)=0\)
<=> \(x-\sqrt{3}=0\text{ hoặc }x+\sqrt{3}=0\text{ hoặc }x-5=0\)
<=> \(x=\sqrt{3}\text{hoặc }x=-\sqrt{3}\text{hoặc }x=5\)
Vậy \(x\in\left\{\sqrt{3};\sqrt{-3};5\right\}\)là giá trị cần tìm
c) (x - 3)2 - (5 - 2x)2 = 0
<=> (x - 3 + 5 - 2x)(x - 3 - 5 + 2x) = 0
<=> (-x + 2)(3x - 8) = 0
<=> \(\orbr{\begin{cases}-x+2=0\\3x-8=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{8}{3}\end{cases}}\)
Vậy tập nghiệm phương trình \(S=\left\{2;\frac{8}{3}\right\}\)
d) x3 + 4x2 + 4x = 0
<=> x(x2 + 4x + 4) = 0
<=> x(x + 2)2 = 0
<=> \(\orbr{\begin{cases}x=0\\\left(x+2\right)^2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}\)
Vậy tập nghiệm phương trình S = \(\left\{0;-2\right\}\)
a) \(x^3+3x^3+4x+4\)=0
=>\(x^3\)(x+1) + 4 ( x+1) = 0
=>(x+1)(\(^{x^3}\)+4) = 0
=>\(\hept{\begin{cases}x+1=0\\x^3+4=0\end{cases}}\)
=> \(\hept{\begin{cases}x=-1\\x^3=-4\end{cases}}\)
\(A.x^2-16x=0\)
\(x^2-\left(4x\right)^2=0\)
\(\left(x-4x\right)\left(x+4x\right)=0\)
\(\left(-3x\right)\left(5x\right)=0\)
\(\Rightarrow\) \(-3x=0\) hoặc \(5x=0\)
\(x=\dfrac{0}{-3}\) hoặc \(x=\dfrac{0}{5}\)
Vậy \(x=0\) hoặc \(x=0.\)
B. 4x2 - 4x + 1 = 0
(2x)2 - (2x)2 + 12 = 0
(2x - 2x + 1 ) (2x + 2x +1) = 0
1 (4x + 1) =0
=> 1 (4x + 1) =0
4x + 1 = 0
4x = 0-1
Vậy x = \(\dfrac{-1}{4}.\)
C. (3x-1)2 - (2x+3)2 = 0
(3x -1 -2x +3) (3x -1 +2x +3) = 0
(x + 2)(5x + 2) = 0
=> x + 2 =0 hoặc 5x + 2 =0
x = 0 - 2 hoặc 5x = 0 - 2
Vậy x = -2 hoặc x = \(\dfrac{-2}{5}.\)
Còn về câu d thì mình hơi phân vân, tại mình dốt toán lắm
a/ \(x^2-16x=0\)
\(\Leftrightarrow x\left(x-16\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-16=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=16\end{matrix}\right.\)
Vậy...
b/ \(4x^2-4x+1=0\)
\(\Leftrightarrow\left(2x-1\right)^2=0\)
\(\Leftrightarrow2x-1=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy..
c/ \(\left(3x-1\right)^2-\left(2x+3\right)^2=0\)
\(\Leftrightarrow\left(3x-1-2x-3\right)\left(3x-1+2x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\5x+2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\dfrac{2}{5}\end{matrix}\right.\)
Vậy...
d/ \(2013x^2-2014x+1=0\)
\(\Leftrightarrow2013x^2-x-2013x+1=0\)
\(\Leftrightarrow x\left(2013x-1\right)-\left(2013x-1\right)=0\)
\(\Leftrightarrow\left(2013x-1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2013x-1=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2013}\\x=1\end{matrix}\right.\)
Vậy..
Bài 1:
a) Với x=1 thì:
(2.1+3m)(3.1-2m-1)=0
\(\Leftrightarrow\)(2+3m)(3-2m-1)=0
\(\Leftrightarrow\)(2+3m)(2-2m)=0
\(\Leftrightarrow\)2(2+3m)(1-m)=0
\(\Rightarrow\left\{\begin{matrix}2+3m=0\\1-m=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}3m=-2\\m=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}m=-\frac{2}{3}\\m=1\end{matrix}\right.\)
Bài 1:
b)Với \(m=-\frac{2}{3}\), ta có:
\(\left[2x+3\cdot\left(-\frac{2}{3}\right)\right]\left[3x-2\cdot\left(-\frac{2}{3}\right)-1\right]=0\)
\(\Leftrightarrow\left(2x-2\right)\left(3x+\frac{4}{3}-1\right)=0\)
\(\Leftrightarrow2\left(x-1\right)\left(3x+\frac{1}{3}\right)=0\)
\(\Leftrightarrow\left\{\begin{matrix}x-1=0\\3x+\frac{1}{3}=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=1\\3x=-\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left\{\begin{matrix}x=1\\x=-\frac{1}{9}\end{matrix}\right.\)
Với m=1, ta có:
\(\left(2x+3\cdot1\right)\left(3x-2\cdot1-1\right)=0\)
\(\Leftrightarrow\left(2x+3\right)\left(3x-3\right)=0\)
\(\Leftrightarrow3\left(2x+3\right)\left(x-1\right)=0\)
\(\Rightarrow\left\{\begin{matrix}2x+3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}2x=-3\\x=1\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-\frac{3}{2}\\x=1\end{matrix}\right.\)
a) \(x^2-2x-7=0\)
\(\Rightarrow x^2-2.x.1+1^2-8=0\)
\(\Rightarrow\left(x-1\right)^2-8=0\)
\(\Rightarrow\left(x-1\right)^2-\sqrt{8}^2=0\)
\(\Rightarrow\left(x-1+\sqrt{8}\right)\left(x-1-\sqrt{8}\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1+2\sqrt{2}\\x=-1-2\sqrt{2}\end{matrix}\right.\)
b) \(4x^2+4x-3=0\)
\(\Rightarrow\left(x-\dfrac{1}{2}\right)\left(x+\dfrac{3}{2}\right)=0\) (tách hạng tử )
Rồi dùng máy tính nhấn. Mấy câu kia tương tự nhé!
3x2 - 2x - 1 = 0
=>3x2+x-3x-1=0
=>x(3x+1)-1(3x+1)=0
=>(x-1)(3x+1)=0
=> \(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\)=> \(\left[{}\begin{matrix}x=1\\3x=-1\Rightarrow x=-\dfrac{1}{3}\end{matrix}\right.\)
vậy x=1 hoặc x=-\(\dfrac{1}{3}\)