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\(\left(x+1\right)^2=4\left(x^2-2x+1\right)^2\\\Leftrightarrow\left(x+1\right)^2=4\left(x-1\right)^2\\\Leftrightarrow \left(x+1\right)^2-4\left(x-1\right)^2=0\\\Leftrightarrow \left(x+1\right)^2-\left(2x-2\right)^2=0\\\Leftrightarrow \left[\left(x+1\right)+\left(2x-2\right)\right]\left[\left(x+1\right)-\left(2x-2\right)\right] =0\\ \Leftrightarrow\left(x+1+2x-2\right)\left(x+1-2x+2\right)=0\\\Leftrightarrow \left(3x-1\right)\left(3-x\right)=0\\\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\3-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{1}{3}\\x=3\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{1}{3};3\right\}\)
\(\left(2x+7\right)^2=9\left(x+2\right)^2\\ \Leftrightarrow\left(2x+7\right)^2-9\left(x+2\right)^2=0\\ \Leftrightarrow\left(2x+7\right)^2-\left(3x+6\right)^2=0\\ \Leftrightarrow\left[\left(2x+7\right)+\left(3x+6\right)\right]\left[\left(2x+7\right)-\left(3x+6\right)\right]=0\\ \Leftrightarrow\left(2x+7+3x+6\right)\left(2x+7-3x-6\right)=0\\ \Leftrightarrow\left(5x+13\right)\left(1-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}5x+13=0\\1-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-13}{5}\\x=1\end{matrix}\right.\)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-13}{5};1\right\}\)
\(4\left(2x+7\right)^2=9\left(x+3\right)^2\\\Leftrightarrow 4\left(2x+7\right)^2-9\left(x+3\right)=0\\ \Leftrightarrow\left(4x+14\right)^2-\left(3x+9\right)^2=0\\\Leftrightarrow \left[\left(4x+14\right)+\left(3x+9\right)\right]\left[\left(4x+14\right)-\left(3x+9\right)\right]=0\\\Leftrightarrow \left(4x+14+3x+9\right)\left(4x+14-3x-9\right)=0\\\Leftrightarrow \left(7x+23\right)\left(x+5\right)=0\\\Leftrightarrow\left[{}\begin{matrix}7x+23=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{-23}{7}\\x=-5\end{matrix}\right. \)
Vậy phương trình có tập nghiệm \(S=\left\{\frac{-23}{7};-5\right\}\)
1,x^2-(x+1)(x-1)=0
x^2-x^2+1+0
1=0(vô lý)
2,5x^3+3x^2+3x+1=4x^2
x^3+3x^2+3x+1=0
(x+1)=0
x=-1
3,x^3+x^2=0
x^2(x+1)=0
x=0 or x=-1
4,2x^3-12x^2+18x=0
x^3-6x^2+9x=0
x(x^2-6x+9)=0
x(x-3)^2=0
x=0 or x=3
5,5x^2-4(x^2-2x+1)+20=0
5x^2-4x^2+8x-4+20=0
x^2+8x+16=0
(x+4)^2=0
x=-4
6,5x(x-3)+7x-21=0
5x(x-3)+7(x-3)=0
(5x+7)(x-3)=0
5x-7=0 or x-3=0
x=7/5 or x=3
7,2x^3-50x=0
2x(x^2-25)=0
2x(x-5)(x+5)=0
x=0 or x=5 or x=-5
8,(4x-1)^2-9(x+3)^2=0
(4x-1)^2-3^2*(x+3)^2=0
(4x-1)^2-(3x+9)^2=0
(4x-1-3x-9)(4x-1+3x+9)=0
(x-10)(7x+8)=0
x=10 or x=-8/7
9,3(x-2)^2-x+2=0
3*(x-2)*(x-2)-(x-2)=0
(3x-6)(x-2)-(x-2)=0
(x-2)(3x-6-1)=0
(x-2)(3x-7)=0
x=2 or x=7/3
10,9x^2+6x-8=0
9x^2+12x-6x-8=0
3x(3x-2)+4(3x-2)=0
(3x+4)(3x-2)=0
3x+4=0 or 3x-2=0
x=-4/3 or x=2/3
a, (a, (x + 2)2 - 9 = 0
⇒ (x + 2)2 = 0 + 9 = 9
⇒ (x + 2)2 = \(\left(\pm3\right)^2\)
⇒ x + 2 = \(\pm3\)
\(\Rightarrow\left\{{}\begin{matrix}x+2=3\\x+2=-3\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=3-2\\x=-3-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=1\\x=-5\end{matrix}\right.\)
Vậy x ∈ {1; -5}
b, \(\left(x+2\right)^2-x^2+4=0\)
⇒ x2 + 4x + 4 - x2 + 4 =0
⇒ 4x + 8 = 0
⇒ 4 (x + 2) = 0
⇒ x + 2 = 0
⇒ x = 0 - 2
⇒ x = -2
Vậy x = -2
c, (x - 3)2 = (2 - 3x)2
⇒ (x - 3)2 - (2 - 3x)2 = 0
⇒ x2 - 6x + 9 - 4 + 12x - 9x2 = 0
⇒ 6x - 8x2 + 5 = 0
⇒2 \(\left(3x-4x^2+\dfrac{5}{2}\right)\)= 0
⇒ 3x - 4x2 + \(\dfrac{5}{2}\) = 0
⇒ - (4x2- 3x + \(\dfrac{9}{16}+\dfrac{31}{16}\)) = 0
⇒ - (4x2 - 3x + \(\dfrac{9}{16}\)) - \(\dfrac{31}{16}\) = 0
⇒ - (2x - \(\dfrac{3}{4}\))2 = \(\dfrac{31}{16}\) (vô lí)
Vậy x ∈ ∅
ý bạn là như thế này đúng không ạ:
a/ \(x^2-6x+5=0\)
\(x^2-5x-x+5=0\)
\(x\left(x-5\right)-\left(x-5\right)=0\)
\(\left(x-5\right)\left(x-1\right)=0\)
\(\orbr{\begin{cases}x-5=0\rightarrow x=5\\x-1=0\rightarrow x=1\end{cases}}\)
b/\(2x^2+7x+9=0\)
?!
c/ \(4x^2-7x+3=0\)
\(4x^2-4x-3x+3=0\)
\(4x\left(x-1\right)-3\left(x-1\right)=0\)
\(\left(x-1\right)\left(4x-3\right)=0\)
\(\orbr{\begin{cases}x-1=0\Rightarrow x=1\\4x-3=0\Rightarrow x=\frac{3}{4}\end{cases}}\)
d/ \(2\left(x+5\right)=2x+10\)
-,- mik ko rõ đề ạ, sai thì ibox ạ.Cảm ơn
Lần sau đăng thì chia thành nhiều câu hỏi nhé
\(16^2-9.\left(x+1\right)^2=0\)
\(16^2-\text{ }\left[3.\left(x+1\right)\right]^2=0\)
\(\left[16-3.\left(x+1\right)\right].\left[16+3\left(x+1\right)\right]=0\)
\(\left[16-3x-3\right]\left[16+3x+3\right]=0\)
\(\left[13-3x\right].\left[19+3x\right]=0\)
\(\Rightarrow\orbr{\begin{cases}13-3x=0\\19+3x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}3x=13\\3x=-19\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{13}{3}\\x=-\frac{19}{3}\end{cases}}}\)
KL:..............................
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) \(5x^3-20x=0\)
\(\Rightarrow5x\left(x^2-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x=0\\x^2-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x^2=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=5\\x=2\\x=-2\end{matrix}\right.\)
A, => (x-2)(x+2)(x-3)(x-3)=0
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\x+2=0\\x-3=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=-2\\x=3\\x=-3\end{matrix}\right.\)
Vậy tập no của pt ..
B, \(\Leftrightarrow x^2+2x+5x+10=0\)
\(\Leftrightarrow x\left(x+2\right)+5\left(x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x+2=0\\x+5=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=-2\\x=-5\end{matrix}\right.\)
Vậy ...............
C, \(\Leftrightarrow3x^2-6x-6x+12=0\)
\(\Leftrightarrow3x\left(x-2\right)-6\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x-6\right)=0\)
\(\Leftrightarrow\left[\begin{matrix}x-2=0\\3x-6=0\end{matrix}\right.\Leftrightarrow\left[\begin{matrix}x=2\\x=2\end{matrix}\right.\)
Vậy tập no của pt là S{2}