giả pt 2x+x+12=0
gấp
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\sqrt{\frac{1-x}{x}}=\frac{2x+x^2}{1+x^2}\)
\(\Leftrightarrow\sqrt{\frac{1-x}{x}}-1=\frac{2x+x^2}{1+x^2}-1\)
\(\Leftrightarrow\frac{-\left(2x-1\right)}{\sqrt{\frac{1-x}{x}}+1}-\frac{2x-1}{1+x^2}=0\)
\(\Leftrightarrow\left(2x-1\right)\left(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}\right)=0\)
Dễ thấy: \(\frac{-1}{\sqrt{\frac{1-x}{x}}+1}-\frac{1}{1+x^2}< 0\)
\(\Rightarrow2x-1=0\Rightarrow x=\frac{1}{2}\)
\(\left(x^2+2x\right)^2-6x^2+12x+9=0\Leftrightarrow x^4+4x^3+4x^2-6x^2+12x+9=0\\ \Leftrightarrow x^4+4x^3-2x^2+12x+9=0\Leftrightarrow x^2+4x-2+\frac{12}{x}+\frac{9}{x^2}=0\\ \Leftrightarrow\left(x^2+\frac{9}{x^2}\right)+4\left(x+\frac{3}{x}\right)-2=0\)
Đặt \(k=x+\frac{3}{x}\Rightarrow x^2+\frac{9}{x^2}=k^2-6\)
Ta đc \(k^2-6+4k-2=0\Leftrightarrow k^2+4k-8=0\)
\(\left(x^2+2x\right)^2\)\(-6x^2\)\(+12x+9\)=0
⇔\(\left(x^2\right)^2\)\(+2.2x.x^2\)+\(2x^2\)-6x2+12x+9=0
⇔ x4+ 4x3+2x2-6x2+12x+9=0
⇔ x2+4x3-4x2 +12x=-9
⇔x2+ 4x(x-x+3)=-9
⇔x2+12x=-9
⇔x(x+12)=-9
⇔ {x=-9 hoặc x+12=-9}
⇔ {x=-9 hoặc x=-21}
S={-9;-21}
Ta có \(\frac{12}{x^2+2x+4}-\frac{5}{x^2+2x+5}=2\)
<=>\(12\left(x^2+2x+5\right)-5\left(x^2+2x+4\right)=2\left(x^2+2x+5\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow12x^2+24x+60-5x^2-10x-20=2x^4+8x^3+26x^2+36x+40\)
\(\Leftrightarrow7x^2+14x+40=2x^4+8x^3+26x^2+36x+40\)
\(\Leftrightarrow2x^4+8x^3+19x^2+22x=0\)
\(\Leftrightarrow x\left(2x^3+8x^2+19x+22\right)=0\)
\(\Leftrightarrow x\left(2x^3+4x^2+4x^2+8x+11x+22\right)=0\)
\(\Leftrightarrow x\left[2x^2\left(x+2\right)+4x\left(x+2\right)+11\left(x+2\right)\right]=0\)
\(\Leftrightarrow x\left(x+2\right)\left(2x^2+4x+11\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x+2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-2\end{cases}}}\)
Vậy PT có nghiệm duy nhất S ={0 ; -2 } vì( \(2x^2+4x+11\ne0\))
a/ \(2x-3=5x+2\)
\(\Leftrightarrow5x-2x=-3-2\)
\(\Leftrightarrow3x=-5\Leftrightarrow x=-\dfrac{5}{3}\)
Vậy..
b. \(2x\left(x-1\right)=2x+2\)
\(\Leftrightarrow2x^2-4x-2=0\)
\(\Leftrightarrow x^2-2x-1=0\)
\(\Leftrightarrow\left(x-1+\sqrt{2}\right)\left(x-1-\sqrt{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1-\sqrt{2}\\x=1+\sqrt{2}\end{matrix}\right.\)
Vậy...
c/ ĐKXĐ : \(x\ne\pm2\)
\(\dfrac{x+2}{x-2}-\dfrac{x^2}{x^2-4}=\dfrac{6}{\left(x+2\right)}\)
\(\Leftrightarrow\dfrac{\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}=\dfrac{6\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)
\(\Leftrightarrow x^2+4x+4-x^2=6x-12\)
\(\Leftrightarrow2x-16=0\)
\(\Leftrightarrow x=8\)
Vậy..
y-x=12 <=> y = x+12
2x+y=126 <=> 2x + x + 12=126
<=> 3x+12=126
<=> x = 38
<=> y = 50
\(\left\{{}\begin{matrix}2x+y=126\\y-x=12\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x=114\\y-x=12\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=38\\y=50\end{matrix}\right.\)
2x + x + 12 = 0
3x + 12 = 0
3x = 0 - 12
3x = -12
x = -12 : 3
x = -4
\(2x+x+12=0\)
\(\Leftrightarrow3x=-12\)
\(\Leftrightarrow x=-4\)