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(2x - 1)2 + (x + 3)2 - 5 (x + 7) (x - 7) = 0
<=> 4x2 - 4x + 1 + x2 + 6x + 9 - 5 (x2 - 49) = 0
<=> 4x2 - 4x + 1 + x2 + 6x + 9 - 5x2 + 245 = 0
<=> 2x + 255 = 0
<=> 2x = -255
<=> x = -255/2
x^2 -2x = 24
=> x^2 - 2x - 24=0
=>x^2 -8x+6x - 24 = 0
=> ( x^2- 8x)+( 6x-24) = 0
=> x(x-8) + 6(x-8) = 0
=> (x+6)(x-8)=0
=>\(\orbr{\begin{cases}x=-6\\x=8\end{cases}}\)
a/ (x-3)2 - 4 = 0
=> (x-3-2)(x-3+2)=0
=> (x-5)(x-1)=0
=> x = 5 hoặc x=1
a) x^2+3x=0
<=> x(x+3)=0
<=> x+3=0
---> X=-3
b)x.(x-7).(x+7)=0
<=>x.(x^2-7^2)=0
<=> X^2-7^2=0
==>x= 7 và x=-7
c) x^3-9x=0
<=> x(x^2-3^2)=0
<=> x^2-3^2=0
~~> x = 3 và x=-3
d) x^2-5x-6=0
<=> x^2-5x-5-1=0
<=> (x^2-1)-(5x-5) =0
<=> x(x-1) - 5(x-1)=0
<=> (x-1)(x-5)=0
~~> x-1 = 0 ~> x=1
~~> x-5=0 ~~> x=5
Vậy x=1 và x=5
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(=4x^2+1-4x+\left(x^2+9+6x\right)-5\left(x^2-7^2\right)=0\)
\(=4x^2+1-4x+x^2+9+6x-5x^2+245=0\)
\(=\left(4x^2+x^2-5x^2\right)-\left(4x-6x\right)+\left(1+9+245\right)=0\)
\(=2x+255=0\)
\(\Rightarrow2x=-255\)
\(x=-127,5\)
(2x-1)^2+(x+3)^2-5(x+7)(x-7)=0
=>4x2-4x+1+x2+6x+9+245-5x2=0
=>(4x2+x2-5x2)+(6x-4x)+(1+9+245)=0
=>2x+255=0
=>2x=-255 <=>x=-255/2
(x-1)2+(x+3)2-5(x+7)(x-7)=0
\(\Leftrightarrow x^2-2x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow x^2-2x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow-3x^2+4x+255=0\)
\(\Leftrightarrow-3\left(x^2-\frac{4}{3}x\right)+255=0\)
\(\Leftrightarrow-3\left(x^2-2.x.\frac{2}{3}+\frac{4}{9}\right)+3.\frac{4}{9}+255=0\)
\(\Leftrightarrow-3\left(x-\frac{2}{3}\right)^2+\frac{769}{3}\)
\(\Leftrightarrow-3\left(x-\frac{2}{3}\right)^2=-\frac{769}{3}\)
\(\Leftrightarrow\left(x-\frac{2}{3}\right)^2=\frac{769}{9}\)
\(\Leftrightarrow\orbr{\begin{cases}x-\frac{2}{3}=\sqrt{\frac{769}{9}}\\x-\frac{2}{3}=-\sqrt{\frac{769}{9}}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\sqrt{\frac{769}{9}}+\frac{2}{3}=\frac{\sqrt{769}+2}{3}\\x=-\sqrt{\frac{769}{9}}+\frac{2}{3}=\frac{2-\sqrt{769}}{3}\end{cases}}\)
Vậy \(\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{769}+2}{3}\\x=\frac{2-\sqrt{769}}{3}\end{cases}}\)
a)
\(\left(x+2\right)^2-9=0\)
\(\Rightarrow\left(x+2\right)^2=9=3^2\)
\(\Rightarrow x+2=\pm3\)
\(\Rightarrow x=-5;1\)
b)
\(25x^2-10x+1=0\)
\(\left(5x\right)^2-2\cdot5x+1^2=0\)
\(\Rightarrow\left(5x+1\right)^2=0\)
\(\Rightarrow5x+1=0\)
\(\Rightarrow5x=-1;x=\dfrac{-1}{5}\)
c)
\(x^2+14x+49=0\)
\(\Rightarrow x^2+2\cdot7x+7^2=0\)
\(\Rightarrow\left(x+7\right)^2=0;x+7=0\)
\(\Rightarrow x=-7\)
d)
\(\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(4x^2-4x+1+x^2+6x+9-5x^2+5\cdot49=0\)
\(\Rightarrow5x^2-5x^2-4x+6x+10+245=0\)
\(\Rightarrow2x+255=0\)
\(\Rightarrow2x=-255\)
\(\Rightarrow x=\dfrac{-255}{2}\)
1) \(\Rightarrow x^2+4x+4-x^2+1=9\)
\(\Rightarrow4x=4\Rightarrow x=1\)
2) \(\Rightarrow x\left(2x+7\right)+2\left(2x+7\right)=0\)
\(\Rightarrow\left(2x+7\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=-2\end{matrix}\right.\)
3) \(\Rightarrow x^3+3x^2+3x+1-x^3-3x^2=2\)
\(\Rightarrow3x=1\Rightarrow x=\dfrac{1}{3}\)
(2x_1)2+(x+3)2_5(x+7)(x_7)=0
=>4x2-4x+1+x2+6x+9-5x2+245=0
=>(4x2+x2-5x2)+(-4x+6x)+(9+245)=0
=>2x+255=0
=>2x=-255
=>x=-255/2