a/ \(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x+4\right)^2+2\left(x^2+5x+4\right)-24=0\)
\(\Leftrightarrow\left(x^2+5x\right)\left(x^2+5x+10\right)=0\)
b/ ĐKXĐ; ...
\(\Leftrightarrow\frac{x^2}{x^2+4x+4}+12x+5=3x^2+6x+2\)
\(\Leftrightarrow\frac{x^2+\left(12x+5\right)\left(x^2+4x+4\right)}{x^2+4x+4}=3x^2+6x+2\)
\(\Leftrightarrow\frac{\left(4x+10\right)\left(3x^2+6x+2\right)}{x^2+4x+4}=3x^2+6x+2\)
\(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\\frac{4x+10}{x^2+4x+4}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}3x^2+6x+2=0\\x^2=6\end{matrix}\right.\)
a) Ta có :
(x2 + 3x + 2)(x2 + 7x + 12) = 24
⇔ ( x + 1 ) ( x + 2 ) (x + 3 ) ( x + 4 ) = 24
⇔ ( x + 1 ) ( x + 4 ) ( x + 2 ) ( x + 3 ) - 24 = 0
⇔ ( x2 + 5x + 4 ) ( x2 + 5x + 6 ) - 24 = 0
Đặt t = x2 + 5x + 4, ta có :
t ( t + 2 ) - 24 = 0
⇔ t2 + 2t +1 - 25 = 0
⇔ ( t + 1 )2 - 52 = 0
⇔ ( t - 4 ) ( t + 6 ) = 0
\(\Leftrightarrow\left[{}\begin{matrix}t-4=0\\t+6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}t=4\\t=-6\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+5x+4=4\\x^2+5x+4=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x^2+5x=0\\x^2+5x+10=0\end{matrix}\right.\)
Sau đó tìm x bạn tự làm nha
Ý b) là - 3 à !?
\(a)(x^2 + 3x + 2)(x^2 + 7x + 12) = 24\)
\(\Leftrightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)=24\)
\(\Leftrightarrow\left(x+1\right)\left(x+4\right)\left(x+3\right)\left(x+2\right)=24\)
\(\Leftrightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)=24\)
Đặt: \(x^2+5x+5=y\) ta được: \(\left(y-1\right)\left(y+1\right)=24\)
\(\Leftrightarrow y^2=25\Leftrightarrow\left[{}\begin{matrix}y=-5\\y=5\end{matrix}\right.\)
Với: \(y=-5\Rightarrow x^2+5x+5=-5\Leftrightarrow x^2+5x+10=0\)
Có: \(\Delta=-15< 0\) vô nghiệm.
Với: \(y=5\Rightarrow x^2+5x+5=5\Leftrightarrow\left(x+5\right)x=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)
Vậy pt có tập \(n_0S=\left\{0;-5\right\}\)
\(b,\frac{x^2}{\left(x+2\right)^2}=3x^2-6x-3\left(Đkxđ:x\ne-2\right)\)
\(\Leftrightarrow x^2=\left(x+2\right)^2\left(3x^2-6x-3\right)\)
\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2-6x-3\right)-x^2=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+1\right)+\left(x^2+4x+4\right)\left(-12x-5\right)-x^2=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+2\right)-\left(12x^3+48x^2+48x+5x^2+20x+20+x^2\right)=0\)
\(\Leftrightarrow\left(x^2+4x+4\right)\left(3x^2+6x+2\right)-2\left(2x+5\right)\left(3x^2+6x+2\right)=0\)
\(\Leftrightarrow\left(3x^2+6x+2\right)\left(x^2-6\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=\pm\sqrt{6}\\x=-1\pm\frac{1}{\sqrt{3}}\end{matrix}\right.\)
Vậy ..............
