x^2-25x^4=0
Giúp mình với nhanh nhanh nhé, cảm ơn a) ( x^2 + x )^2 + 2( x^2 + x ) - 8 = 0 b) ( x^2 - 4x +3 ) ( x^2 +6x + 8 ) + 24 = 0 c) 6x^4 + 25x^3 + 12x^2 - 25x + 6 = 0 d) ( x - 2 )^4 + ( x- 3 )^4 = 0
a: \(\left(x^2+x\right)^2+2\left(x^2+x\right)-8=0\)
\(\Leftrightarrow\left(x^2+x+4\right)\left(x^2+x-2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)
hay \(x\in\left\{-2;1\right\}\)
b: \(\Leftrightarrow\left(x-1\right)\left(x-3\right)\left(x+2\right)\left(x+4\right)+24=0\)
\(\Leftrightarrow\left(x^2+x-2\right)\left(x^2+x-12\right)+24=0\)
\(\Leftrightarrow\left(x^2+x\right)^2-14\left(x^2+x\right)+48=0\)
\(\Leftrightarrow\left(x^2+x-6\right)\left(x^2+x-8\right)=0\)
hay \(x\in\left\{-3;2;\dfrac{-1+\sqrt{33}}{2};\dfrac{-1-\sqrt{33}}{2}\right\}\)
Tìm x biết:
a, 6x4 + 25x3 + 12x2 - 25x +6 = 0
b, x5 + 2x4 + 3x3 + 3x2 + 2x +1 = 0
c, x2 (x2 + 2) - x2 - 2 = 0
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
a) 9-64x^2=0
=> 64x^2 = 8
=> \(x^2=\frac{8}{64}=\frac{1}{8}\)
=> \(x=\frac{1}{\sqrt{8}}\)
b ) 25x^2 - 3 = 0
=> 25x^2 = 3
=> \(x^2=\frac{3}{25}\)
=> \(x=\frac{\sqrt{3}}{5}\)
C) 7 - 16x^2 =0
=> 16x^2 = 7
=> \(x^2=\frac{7}{16}\)
=> \(x=\frac{\sqrt{7}}{4}\)
d) 4x^2 - (x-4)^2 = 0
=> 4x^2 - x^2 + 8x - 16 =0
=> 3x^2 + 8x -16 = 0
=> ( 3x^2 + 12x ) - ( 4x +16 ) = 0
=> 3x( x + 4 ) - 4( x + 4 ) = 0
=>( x + 4 )( 3x - 4 ) = 0
=> \(\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}\)
e) ( 3x + 4 )^2 - ( 2x - 5 )^2 = 0
=> ( 3x + 4 + 2x - 5 )( 3x + 4 - 2x + 5 ) = 0
=> ( 5x -1 ) ( x + 9 ) = 0
=> \(\orbr{\begin{cases}5x-1=0\\x+9=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-9\end{cases}}\)
Trả lời:
a, \(9-64x^2=0\)
\(\Leftrightarrow\left(3-8x\right)\left(3+8x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-8x=0\\3+8x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{8}\\x=-\frac{3}{8}\end{cases}}}\)
Vậy x = 3/8; x = - 3/8 là nghiệm của pt.
b, \(25x^2-3=0\)
\(\Leftrightarrow\left(5x-\sqrt{3}\right)\left(5x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-\sqrt{3}=0\\5x+\sqrt{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{5}\\x=-\frac{\sqrt{3}}{5}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{3}}{5}\)
c, \(7-16x^2=0\)
\(\Leftrightarrow\left(\sqrt{7}-4x\right)\left(\sqrt{7}+4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{7}-4x=0\\\sqrt{7}+4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{4}\\x=-\frac{\sqrt{7}}{4}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{7}}{4}\)
d, \(4x^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(2x-x+4\right)\left(2x+x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}}\)
Vậy x = - 4; x = 4/3 là nghiệm của pt.
e, \(\left(3x+4\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(3x+4-2x+5\right)\left(3x+4+2x-5\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = - 9; x = 1/5 là nghiệm của pt.
tim x : 4/9-25x^2=0
\(\dfrac{4}{9}-25x^2=0\)
\(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
\(\left(\dfrac{2}{3}+5x\right)\left(\dfrac{2}{3}-5x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{2}{3}+5x=0\\\dfrac{2}{3}-5x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{2}{3}\\-5x=-\dfrac{2}{3}\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{2}{15}\\x=\dfrac{2}{15}\end{matrix}\right.\)
\(\dfrac{4}{9-25x^2}=0\\ \Leftrightarrow\dfrac{4}{\left(3-5x\right)\left(3+5x\right)}=0\\ \Leftrightarrow\left(3-5x\right)\left(3+5x\right)=0\\ \)
\(\Leftrightarrow\left[{}\begin{matrix}3-5x=0\\3+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
Vậy....
tìm x
x^2(x+1)+2x(x+1)=0
4/9-25x^2=0
\(x^2\left(x+1\right)+2x\left(x+1\right)=0\Leftrightarrow\left(x^2+2x\right)\left(x+1\right)=0\Leftrightarrow x\left(x+2\right)\left(x+1\right)=0\left\{{}\begin{matrix}x=0\\x+2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=-2\\x=-1\end{matrix}\right.\)
a, \(x^2.\left(x+1\right)+2x\left(x+1\right)=0\)
=> ( x + 1 ) ( \(x^2\) + 2x ) = 0
=> ( x + 1 ) x (x + 2 ) = 0
=>\(\left[{}\begin{matrix}x=0\\x+1=0\\x+2=0\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
b, \(\dfrac{4}{9}-25x^2=0\)
=> \(\left(\dfrac{2}{3}\right)^2-\left(5x\right)^2=0\)
=> \(\left(\dfrac{2}{3}-5x\right)\left(\dfrac{2}{3}+5x\right)=0\)
=>\(\left[{}\begin{matrix}\dfrac{2}{3}-5x=0\\\dfrac{2}{3}+5x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{15}\\x=\dfrac{-2}{15}\end{matrix}\right.\)
b, \(\dfrac{4}{9}-25x^2=0\Leftrightarrow\left(5x\right)^2=-\dfrac{4}{9}\Leftrightarrow\left(5x\right)^2=-\left(\dfrac{2}{3}\right)^2\Leftrightarrow5x=-\dfrac{2}{3}\Leftrightarrow x=-\dfrac{2}{15}\)
\(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(x+\sqrt{5-4x}=0\)
\(\sqrt{1-2x^2}=x-1\)
a: ta có: \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\)
\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)
\(\Leftrightarrow\sqrt{x-1}=1\)
hay x=2
c: Ta có: \(\sqrt{1-2x^2}=x-1\)
\(\Leftrightarrow1-2x^2=x^2-2x+1\)
\(\Leftrightarrow-3x^2+2x=0\)
\(\Leftrightarrow-x\left(3x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(loại\right)\\x=\dfrac{2}{3}\left(loại\right)\end{matrix}\right.\)
Tìm x:
a) 2-25x2=0
b) x2-x+1/4=0
a) \(2-25x^2=0\Leftrightarrow-25x^2=-2\Leftrightarrow x^2=\frac{2}{25}\Leftrightarrow x=\frac{\sqrt{2}}{5}\)
b) \(x^2-x+\frac{1}{4}=0\Leftrightarrow x^2-2\cdot x\cdot\frac{1}{2}+\frac{1}{4}=0\Leftrightarrow\left(x-\frac{1}{2}\right)^2=0\Leftrightarrow x-\frac{1}{2}=0\Leftrightarrow x=\frac{1}{2}\)
Bài 1: Giải phương trình
a) (x+3)4 + (x+5)4 = 16
b) 6x4 + 25x3 + 12x - 25x+ 6= 0
c) 9x4 - 15x3 + 28x2 -20x+16 = 0
d) x4 + 7x2 - 12x+5 =0
e) x5= x4 + x3 + x2 + x+2
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........
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 = \(\dfrac{1}{3}\)
\(S=\left\{-2;-3;\dfrac{1}{2};\dfrac{1}{3}\right\}\)
\(\text{a) }\left(x+3\right)^4+\left(x+5\right)^4=16\\ \Leftrightarrow\left(x^2+6x+9\right)^2+\left(x^2+10x+25\right)^2=16\\ \Leftrightarrow x^4+36x^2+81+12x^3+18x^2+108x+x^4+100x^2+625+20x^3+50x^2+500x=16\\ \Leftrightarrow2x^4+32x^3+204x^2+608x+690=0\\ \Leftrightarrow x^4+16x^3+102x^2+304x+345=0\\ \Leftrightarrow x^4+5x^3+11x^3+55x^2+47x^2+235x+373x+69x+345=0\\ \Leftrightarrow\left(x^4+5x^3\right)+\left(11x^3+55x^2\right)+\left(47x^2+235x\right)+\left(69x+345\right)=0\\ \Leftrightarrow x^3\left(x+5\right)+11x^2\left(x+5\right)+47x\left(x+5\right)+69\left(x+5\right)=0\\ \Leftrightarrow\left(x^3+11x^2+47x+69\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x^3+3x^2+8x^2+24x+23x+69\right)\left(x+5\right)=0\\ \Leftrightarrow\left[\left(x^3+3x^2\right)+\left(8x^2+24x\right)+\left(23x+69\right)\right]\left(x+5\right)=0\\ \Leftrightarrow\left[x^2\left(x+3\right)+8x\left(x+3\right)+23\left(x+3\right)\right]\left(x+5\right)=0\\ \Leftrightarrow\left(x^2+8x+23\right)\left(x+3\right)\left(x+5\right)=0\)\(\Leftrightarrow\left(x^2+8x+16+7\right)\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left[\left(x+4\right)^2+7\right]\left(x+3\right)\left(x+5\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x+5\right)=0\left(\text{Vì }\left(x+4\right)^2+7\ne0\right)\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-5\end{matrix}\right.\)
Vậy tập nghiệm phương trình là \(S=\left\{-3;-5\right\}\)
tìm x
5x^2 - 25x - 4 = 0
\(5x^2-25x-4=0\)
\(\Leftrightarrow x^2-5x-\frac{4}{5}=0\)
\(\Leftrightarrow x^2-5x+\frac{25}{4}=\frac{141}{20}\)
\(\Leftrightarrow\left(x-\frac{5}{2}\right)^2=\frac{141}{20}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{5}{2}=\frac{\sqrt{705}}{10}\\x-\frac{5}{2}=-\frac{\sqrt{705}}{10}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{25+\sqrt{705}}{10}\\x=\frac{25-\sqrt{705}}{10}\end{cases}}\)