Gửi em

\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=0\\ \Leftrightarrow\left[\left(x+2\right)\left(x+5\right)\right]\cdot\left[\left(x+3\right)\left(x+4\right)\right]=24\\ \Leftrightarrow\left(x^2+7x+10\right)\left(x^2+7x+12\right)=24\)

đặt \(t=x^2+7x+11\) khi đó ta có

\(\left(t-1\right)\left(t+1\right)=24\\ \Leftrightarrow t^2-1-24=0\\ \Leftrightarrow\left(t-5\right)\left(t+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=5\\t=-5\end{matrix}\right.\)

Trở về ẩn x ta có

Với t=5

\(x^2+7x+11=5\Leftrightarrow x^2+7x+6\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-6\end{matrix}\right.\)

Với t=-5

\(x^2+7x+11=-5\\\Leftrightarrow x^2+7x+16=0\\ \Leftrightarrow\left(x+3,5\right)^2+3,75=0\)

Voi \(\left(x+3,5\right)^2\ge0\Rightarrow\varnothing\)

Vậy ...................

\(\Leftrightarrow\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2=5\\x=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{5}\\x=\sqrt{5}\\x=-3\end{matrix}\right.\)

\(=>\left[{}\begin{matrix}x^2-5=0\\x+3=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=\pm\sqrt{5}\\x=-3\end{matrix}\right.\)

vậy.....

⇔[x2−5=0x+3=0⇔[x2−5=0x+3=0

⇔[x2=5x=−3⇔[x2=5x=−3

⇔⎡⎢⎣x=−√5x=√5x=−3

(x² + x)² + 4(x² + x) = 12

⇔ (x² + x + 2)² = 16

⇔ (x² + x + 2)² - 16 = 0

⇔ (x² + x + 2 - 4)(x² + x + 2 + 4) = 0

⇔ (x² + x - 2)(x² + x + 6) = 0

⇔ [(x² + 2x) - (x + 2)](x² + x + 6) = 0

⇔ [x(x + 2) - (x + 2)](x² + x + 6) = 0

⇔ (x + 2)(x - 1)(x² + x + 6) = 0

Vì x² + x + 6 = (x² + 2.\(\frac{1}{2}\)x + \(\frac{1}{4}\)) + \(\frac{21}{4}\) = (x + \(\frac{1}{2}\))² + \(\frac{21}{4}\) ≥ \(\frac{21}{4}\) > 0

Nên suy ra \(\left[{}\begin{matrix}x-1=0\\x+2=0\end{matrix}\right.\) ⇔ \(\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Vậy nghiệm của pt là x = 1; x = -2

\(\left(x^2+x\right)^2+4\left(x^2+x\right)=12\)

\(\Leftrightarrow\left(x^2+x\right)^2+4\left(x^2+x\right)+4=16\)

\(\Leftrightarrow\left(x^2+x+2\right)^2=16\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=4\\x^2+x+2=-4\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+x+2=0\\x^2+x+6=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x+\frac{1}{2}\right)^2+\frac{7}{4}>0\\\left(x+\frac{1}{2}\right)^2+\frac{23}{4}>0\end{matrix}\right.\)

Vậy PTVN

a) \(\frac{2}{x+1}-\frac{1}{x-2}=\frac{3x-11}{\left(x+1\right).\left(x-2\right)}\) Đk : x \(\ne-1\) ; x \(\ne2\)

\(\Leftrightarrow\frac{2.\left(x-2\right)}{\left(x+1\right).\left(x-2\right)}-\frac{1.\left(x+1\right)}{\left(x+1\right).\left(x-2\right)}=3x-11\)

\(\Leftrightarrow2x-4-x-1=3x-11\)

\(\Leftrightarrow2x-3x-x=-11+4+1\)

\(\Leftrightarrow-2x=-6\)

\(\Leftrightarrow x=3\)

Vậy S = \(\left\{3\right\}\)

a: (x-3)(x-2)<0

=>x-2>0 và x-3<0

=>2<x<3

b: \(\left(x+3\right)\left(x+4\right)\left(x^2+2\right)\ge0\)

=>(x+3)(x+4)>=0

=>x+3>=0 hoặc x+4<=0

=>x>=-3 hoặc x<=-4

c: \(\dfrac{x-1}{x-2}\ge0\)

=>x-2>0 hoặc x-1<=0

=>x>2 hoặc x<=1

d: \(\dfrac{x+3}{2-x}>=0\)

=>\(\dfrac{x+3}{x-2}< =0\)

=>x+3>=0 và x-2<0

=>-3<=x<2