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=>x^2>=5/2
=>\(\left(x-\dfrac{\sqrt{10}}{4}\right)\left(x+\dfrac{\sqrt{10}}{4}\right)>=0\)
=>\(\left[{}\begin{matrix}x>=\dfrac{\sqrt{10}}{4}\\x< =-\dfrac{\sqrt{10}}{4}\end{matrix}\right.\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25+12=0\\ \Leftrightarrow4x+38=0\\ \Leftrightarrow x=-\dfrac{19}{2}\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\\ \Leftrightarrow4x=-38\Leftrightarrow x=-\dfrac{19}{2}\)
\(\Leftrightarrow3x^2+2x+x^2+2x+1-4x^2+25+12=0\\ \Leftrightarrow4x+38=0\\ \Leftrightarrow x=-\dfrac{19}{2}\)
\(\Rightarrow3x^2+2x+x^2+2x+1-4x^2+25=-12\)
\(\Rightarrow4x=-38\Rightarrow x=-\dfrac{19}{2}\)
Bài 1:
$2x(x+3)+(2x+3)(5-x)=2$
$\Leftrightarrow 2x^2+6x+(10x-2x^2+15-3x)=2$
$\Leftrightarrow 2x^2+6x+7x-2x^2+15=2$
$\Leftrightarrow 13x+15=2$
$\Leftrightarrow 13x=2-15=-13$
$\Leftrightarrow x=-13:13=-1$
Bài 2:
$x-y=4\Rightarrow x=y+4$. Thay vào $xy=5$ thì:
$(y+4)y=5$
$\Leftrightarrow y^2+4y-5=0$
$\Leftrightarrow (y-1)(y+5)=0$
$\Leftrightarrow y=1$ hoặc $y=-5$
Nếu $y=1$ thì $x=y+4=5$. Khi đó $x^3+y^3=5^3+1^3=126$
Nếu $y=-5$ thì $x=y+4=-1$. Khi đó: $x^3+y^3=(-1)^3+(-5)^3=-126$
Ta có \(x\left(4x+2\right)-\left(2x-5\right)\left(2x+5\right)=1\)
\(\Leftrightarrow4x^2+2x-\left(4x^2-25\right)=1\)
\(\Leftrightarrow4x^2+2x-4x^2+25=1\)
\(\Leftrightarrow2x=-24\)
\(\Leftrightarrow x=-12\)
Vậy x=-12
a: \(\left(x+5\right)^2-\left(x-5\right)^2-2x+1=0\)
=>\(x^2+10x+25-\left(x^2-10x+25\right)-2x+1=0\)
=>\(x^2+8x+26-x^2+10x-25=0\)
=>18x+1=0
=>\(x=-\dfrac{1}{18}\)
b: \(\left(2x-7\right)^2-\left(x+3\right)^2=3x^2+6\)
=>\(4x^2-28x+49-\left(x^2+6x+9\right)-3x^2-6=0\)
=>\(x^2-28x+43-x^2-6x-9=0\)
=>34-34x=0
=>34x=34
=>x=1
c: \(\left(3x+2\right)^2-9\left(x-5\right)\left(x+5\right)=225-5x\)
=>\(9x^2+12x+4-9\left(x^2-25\right)-225+5x=0\)
=>\(9x^2+17x+4-225-9x^2+225=0\)
=>17x+4=0
=>x=-4/17
a: Ta có: \(2x\left(x-1\right)-2x^2=-6\)
\(\Leftrightarrow2x^2-2x-2x^2=-6\)
\(\Leftrightarrow x=3\)
b: Ta có: \(2x\left(x-3\right)+5\left(x-3\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(2x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{5}{2}\end{matrix}\right.\)