x = 0 

5+x=5-x

<=>x+x=5-5

<=>2x=0

<=>x=0

đặt \(x^2=t\left(t\ge0\right)\)

Khi đó pt trở thành \(2t^2-5t+6=0\)

=> pt vô nghiệm ! 

_Kudo_

Đặt t = x² (t \(\ge\) 0)

Khi đo ta có pt: 2t² - 5t + 6 = 0

<=> 2(t² - \(\frac{5}{2}\)t + 3) = 0

<=> 2(t² - \(\frac{5}{2}\)t +  \(\frac{25}{16}\) + \(\frac{23}{16}\)) = 0

<=> 2(t - \(\frac{5}{4}\))² + \(\frac{23}{8}\) = 0

<=> 2(t - \(\frac{5}{4}\))² = -\(\frac{23}{8}\)(VN)

Vậy pt vô nghiệm

2x⁴-5x²+6=0

Đặt x² = t(t ≥ 0)

Khi đó pt trở thành 2t² − 5t + 6 = 0

=> pt vô nghiệm ! 

Taco:VD:2.3.4.5>24=>x<0

x=-2=>tổng=0

Vayx=-1

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)=24\)

\(\Leftrightarrow\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)=0\)

\(\Leftrightarrow x^2-3x+\frac{1}{2}=0.\)

\(\Leftrightarrow x^2-2.\frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+\frac{1}{2}=0.\)

\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{7}{4}.\)

\(\Rightarrow\orbr{\begin{cases}x-\frac{3}{2}=\frac{\sqrt{7}}{2}\\x-\frac{3}{2}=\frac{-\sqrt{7}}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}+3}{2}\\x=\frac{-\sqrt{7}+3}{2}\end{cases}}}\)

Học tốt

\(x^2-4+x+2=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-2\right)+\left(x+2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)

x²-4+x+2

=(x²-2²)+x+2)

=(x-2)*(x+2)+(x+2)

=(x+2)*(x-2+1)

=(x+2)*(x-1)

x²-2+x=0

x²+x-2=0

x²+2x-x-2=0

x(x+2)-(x+2)=0

(x+2)(x-1)=0

\(\left\{{}\begin{matrix}x+2=0\\x-1=0\end{matrix}\right.\)

\(\left\{{}\begin{matrix}-2\\1\end{matrix}\right.\)

\(x^2-12x-2160=0\)

\(=>x^2-2.x.6+36-2196=0\)

\(=>\left(x-6\right)^2-2196=0\)

\(=>\left(x-6\right)^2=2196\)

\(=>\orbr{\begin{cases}x-6=-2196\\x-6=2196\end{cases}=>\orbr{\begin{cases}x=6-6\sqrt{61}\\x=6+6\sqrt{61}\end{cases}}}\)

Mik thề ko đúng mik sẽ ko bao giờ lên olm nữa

x mũ 2 trừ 12 x bằng 2160

suy ra ta có 2160 chia cho 12 bằng 18

<=> ( 5 - x)² - (2+2x)² = 0

<=> (5 - x - 2 - 2x).(5 - x + 2 + 2x) = 0 

<=> (3- 3x).(7 + x) = 0

<=> 3- 3x = 0 hoặc 7 + x = 0 

+) 3 - 3x = 0 <=> x = 1

+) 7 + x = 0 <=> x = -7

Vậy x = 1 ; x = -7