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30 tháng 5 2022
a: \(6x^4+25x^3+12x^2-25x+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+18x^2-5x^2-15x+x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(3x-1\right)\left(2x-1\right)=0\)
hay \(x\in\left\{-2;-3;\dfrac{1}{3};\dfrac{1}{2}\right\}\)
b: \(x^5+2x^4+3x^3+3x^2+2x+1=0\)
\(\Leftrightarrow x^5+x^4+x^4+x^3+2x^3+2x^2+x^2+x+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^3+2x^2+x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^4+x^2+x^3+x+x^2+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+x+1\right)\left(x^2+1\right)=0\)
=>x+1=0
hay x=-1
c: \(x^2\left(x^2+2\right)-x^2-2=0\)
\(\Leftrightarrow\left(x^2+2\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
HH
1 tháng 7 2018
a/ \(25x^2-9=0\)
<=> \(\left(5x-3\right)\left(5x+3\right)=0\)
<=> \(\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}}\)
<=> \(\orbr{\begin{cases}5x=3\\5x=-3\end{cases}}\)
<=> \(\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}\)
b/ \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
<=> \(x^2+8x+16-x^2+8x-9=16\)
<=> \(16x+7=16\)
<=> \(16x=9\)
<=> \(x=\frac{9}{16}\)
1 tháng 7 2018
a) \(25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-3=0\\5x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}5x=3\\5x=-3\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{3}{5}\\x=-\frac{3}{5}\end{cases}}}\)
Vậy S = {3/5 ; -3/5}
b) \(\left(x+4\right)^2-\left(x+9\right)\left(x-1\right)=16\)
\(\Leftrightarrow\left(x+4\right)^2-4^2-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+4-4\right)\left(x+4+4\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x\left(x+8\right)-\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow x^2+8x-x^2-8x+9=0\)
\(\Leftrightarrow9=0\left(vl\right)\)
Vậy S = \(\varnothing\)
LT
2
27 tháng 6 2018
a) \(2-25x^2=0\)
\(25x^2=2\)
\(x^2=\frac{2}{25}\)
\(\Rightarrow\orbr{\begin{cases}x=\sqrt{\frac{2}{25}}\\x=-\sqrt{\frac{2}{25}}\end{cases}}\)
Vậy \(x=\sqrt{\frac{2}{25}}\)hoặc \(x=-\sqrt{\frac{2}{25}}\)
b) \(x^2-x+\frac{1}{4}=0\)
\(x^2-2.x.\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\)
\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(x=\frac{1}{2}\)
Vậy \(x=\frac{1}{2}\)
5 tháng 8 2016
a) \(x^4-10x^3+25x^2=0\)
\(\Leftrightarrow x^2\left(x^2-10x+25\right)=0\)
\(\Leftrightarrow x^2\left(x-5\right)^2=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x^2=0\\\left(x-5\right)^2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=0\\x=5\end{array}\right.\)
b) \(x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
HK
3
5 tháng 8 2016
a) x4 - 10x3 + 25x2 = (x2)2 - 2.x2.5x + (5x)2 = (x2 - 5x)2 = 0 => x(x - 5) = 0 => x = 0 hay x - 5 = 0 => x = 0 ; 5
b) x3 + 3x2 + 3x + 1 = x3 + 3.x2.1 + 3.x.12 + 13 = (x + 1)3 = 0 => x + 1 = 0 => x = -1
LC
5 tháng 8 2016
a,x^2(x^2-10x+25)=0
x^2(x-5)^2=0
=> x^2=0 hoac (x-5)^2=0
=>x=0 hoac 5
ND
28 tháng 2 2018
b. sửa đề
\(6x^4+25x^3+12x-25x^2+6=0\)
\(\Leftrightarrow6x^4+12x^3+13x^3+26x^2-14x^2-28x+3x+6=0\)
\(\Leftrightarrow6x^3\left(x+2\right)+13x^2\left(x+2\right)-14x\left(x+2\right)+3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(6x^3+13x^2-14x+3\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+3\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=-2\\x=-3\\x=\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)
Vậy........
K
28 tháng 2 2018
Bài 1 : Giải phương trình
a) (x + 3)4 + (x + 5)4 = 16
Đặt : x + 3 = t
=> x + 5 = x + 3 + 2 = t + 2
Thay x + 3 = t và x + 5 = t + 2 vào phương trình, ta có :
t4 + (t + 2)4 = 16
<=> 2t4 + 8t3 + 24t2 + 32t + 16 = 16
<=> 2(t4 + 4t3 + 12t2 + 16t) = 0
<=> t4 + 4t3 + 12t2 + 16t = 0
<=> (t + 2) . t . (t2 + 2y + 4) = 0
TH1 : t = 0
TH2 : t + 2 = 0 <=> t = -2
TH3 : t2 + 2y + 4 = 0 (vô nghiệm => loại)
Nên t = 0 hoặc t = -2
hay x + 3 = -2 hoặc x + 3 = 0
<=> x = -5 hoặc x = -3
\(S=\left\{-5;-3\right\}\)
b) 6x4 + 25x3 + 12x2 - 25x + 6 = 0
<=> 6x4 + 12x3 + 13x3 + 26x2 - 14x2 - 28x + 3x + 6 = 0
<=> 6x3 (x + 2) + 13x2 (x + 2) - 14x (x + 2) + 3(x + 2) = 0
<=> (x + 2)(6x3 + 13x2 - 14x + 3) = 0
<=> (x + 2)(6x3 + 18x2 - 5x2 - 15x + x + 3) = 0
\(\Leftrightarrow\left(x+2\right)[6x^2\left(x+3\right)-5x\left(x+3\right)+\left(x+3\right)]=0\)
<=> (x + 2)(x + 3) (6x2 - 5x + 1) = 0
<=> (x + 2)(x + 3)(2x - 1)(3x - 1) = 0
TH1 : x + 2 = 0 <=> x = -2
TH2 : x + 3 = 0 <=> x = -3
TH3 : 2x - 1 = 0 <=> 2x = 1 <=> x = \(\dfrac{1}{2}\)
TH4 : 3x - 1 = 0 <=> 3x = 1 <=> 3x = \(\dfrac{1}{3}\)
\(S=\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)
MT
19 tháng 7 2015
a) x3-9x2+27x-27=0
<=>(x-3)3=0
<=>x-3=0
<=>x=3
b) x3-25x=0
<=>x.(x2-25)=0
<=>x.(x-5)(x+5)=0
<=>x=0 hoặc x-5=0 hoặc x+5=0
<=>x=0 hoặc x=5 hoặc x=-5
c)9x2-1=0
<=>(3x-1)(3x+1)=0
<=>3x-1=0 hoặc 3x+1=0
<=>x=1/3 hoặc x=-1/3
19 tháng 7 2015
a, x^3 - 9x^2 + 27x - 27 = 0
=> ( x - 3)^3 = 0
=> x - 3 = 0
=> x = 3
b, x^3 - 25x = 0
=> x(x^2 - 25) = 0
=> x(x-5)(x + 5) = 0
=> x =0 hoặc x - 5 = 0 hoặc x + 5 = 0
=> x= 0 hoặc x =5 hoặc x = -5
c, 9x^2 - 1 = 0
=> (3x)^2 - 1^2 = 0
=> ( 3x- 1)(3x+ 1) = 0
=> 3x - 1 = 0 hoặc 3x + 1 = 0
=> x = 1/3 hoặc x = -1/3
NK
1
10 tháng 10 2020
a) Ta có: \(2-25x^2=0\)
\(\Leftrightarrow25x^2=2\)
\(\Leftrightarrow x^2=\frac{2}{25}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\sqrt{2}}{5}\\x=-\frac{\sqrt{2}}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{\frac{\sqrt{2}}{5};-\frac{\sqrt{2}}{5}\right\}\)
b) Ta có: \(x^2-x+\frac{1}{4}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{2}+\left(\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\)
\(\Leftrightarrow x-\frac{1}{2}=0\)
hay \(x=\frac{1}{2}\)
Vậy: \(x=\frac{1}{2}\)
c) Ta có: \(x\left(x-2\right)+x-2=0\)
\(\Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy: \(x\in\left\{-1;2\right\}\)
d) Ta có: \(5x\left(x-3\right)-x+3=0\)
\(\Leftrightarrow5x\left(x-3\right)-\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\5x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\5x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{1}{5}\end{matrix}\right.\)
Vậy: \(x\in\left\{3;\frac{1}{5}\right\}\)
e) Ta có: \(x^3-\frac{1}{4}x=0\)
\(\Leftrightarrow x\left(x^2-\frac{1}{4}\right)=0\)
\(\Leftrightarrow x\left(x-\frac{1}{2}\right)\left(x+\frac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-\frac{1}{2}=0\\x+\frac{1}{2}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=-\frac{1}{2}\end{matrix}\right.\)
Vậy: \(x\in\left\{0;\frac{1}{2};-\frac{1}{2}\right\}\)
g) Ta có: \(\left(2x-1\right)^2-\left(x+3\right)^2=0\)
\(\Leftrightarrow\left(2x-1-x-3\right)\left(2x-1+x+3\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{4;-\frac{2}{3}\right\}\)
a) \(2-25x^2=0\Leftrightarrow-25x^2=-2\Leftrightarrow x^2=\frac{2}{25}\Leftrightarrow x=\frac{\sqrt{2}}{5}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)