Question

Giải phương trình bằng cách đặt ẩn phụ
1) (x²+x)²+4(x²+x)-12=0
2) (x²+2x+3)²-9(x+1)²-28=0
3) (x-2)(x+2)(x²-10)=72
4) x(x+1)(x²+x+1)=42

Answer 1

1) Đặt: `x^2+x=t` 

Pt trở thành: 

`t^2+4t-12=0`

\(\Leftrightarrow t^2-2t+6t-12=0\\ \Leftrightarrow t\left(t-2\right)+6\left(t-2\right)=0\\ \Leftrightarrow\left(t-2\right)\left(t+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=2\\t=6\end{matrix}\right.\) 

Với `t=2` \(\Leftrightarrow x^2+x=2\Leftrightarrow x^2+x-2=0\Leftrightarrow x^2-x+2x-2=0\)

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

Với `t=6` \(\Leftrightarrow x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow x^2-2x+3x-6=0\)

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

Vậy: ... 

2) \(\left(x^2+2x+3\right)^2-9\left(x+1\right)^2-28=0\)

\(\Leftrightarrow\left[\left(x+1\right)^2+2\right]^2-9\left(x+1\right)^2-28=0\\ \Leftrightarrow\left(x+1\right)^4+4\left(x+1\right)^2+4-9\left(x+1\right)^2-28=0\\ \Leftrightarrow\left(x+1\right)^4-5\left(x+1\right)^2-24=0\\ \Leftrightarrow\left(x+1\right)^4-8\left(x+1\right)^2+3\left(x+1\right)^2-24=0\\ \Leftrightarrow\left(x+1\right)^2\left[\left(x+1\right)^2-8\right]+3\left[\left(x+1\right)^2-8\right]=0\\ \Leftrightarrow\left[\left(x+1\right)^2+3\right]\left[\left(x+1\right)^2-8\right]=0\\ \Leftrightarrow\left[{}\begin{matrix}\left(x+1\right)^2+3\ge3>0\left(L\right)\\\left(x+1\right)^2-8=0\end{matrix}\right.\\ \Leftrightarrow\left(x+1\right)^2 =8\\ \Leftrightarrow\left[{}\begin{matrix}x+1=2\sqrt{2}\\x+1=-2\sqrt{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=-2\sqrt{2}-1\end{matrix}\right.\)

Vậy: ... 

Answer 2

a, Đặt x^2 + x = t 

\(t^2+4t-12=0\Leftrightarrow t=2;t=-6\)

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

b, \(\left(x^2+2x+3\right)^2-9\left(x^2+2x+1\right)-28=0\)

Đặt x^2 + 2x + 1 = t ( t >= 0 ) 

\(\left(t+2\right)^2-9t-28=0\Leftrightarrow t^2-5t-24=0\Leftrightarrow t=8;t=-3\left(l\right)\)

\(\Rightarrow\left(x+1\right)^2=8\Leftrightarrow\left[{}\begin{matrix}x=2\sqrt{2}-1\\x=-2\sqrt{2}-1\end{matrix}\right.\)

c, \(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\)

Đặt x^2 - 4 = t 

\(t\left(t-6\right)=72\Leftrightarrow t^2-6t-72=0\Leftrightarrow t=12;t=-6\)

\(\left[{}\begin{matrix}x^2-4=12\\x^2-4=-6\left(l\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

d, \(x\left(x+1\right)\left(x^2+x+1\right)=42\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt x^2 + x = t 

\(t\left(t+1\right)=42\Leftrightarrow t^2+t-42=0\Leftrightarrow t=6;t=-5\)

\(\left[{}\begin{matrix}x^2+x=6\\x^2+x=-5\left(l\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

 

Answer 3

3) 

\(\left(x-2\right)\left(x+2\right)\left(x^2-10\right)=72\\ \Leftrightarrow\left(x^2-4\right)\left(x^2-10\right)=72\\ \Leftrightarrow\left[\left(x^2-7\right)+3\right]\left[\left(x^2-7\right)-3\right]=72\\ \Leftrightarrow\left(x^2-7\right)^2-9=720\\ \Leftrightarrow\left(x^2-7\right)^2=81\\ \Leftrightarrow\left[{}\begin{matrix}x^2-7=-9\\x^2-7=9\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x^2=-2\left(L\right)\\x^2=16\end{matrix}\right.\\ \Leftrightarrow x^2=16\\ \Leftrightarrow x=\pm4\)

Vậy: ... 

4) \(x\left(x+1\right)\left(x^2+x+1\right)=42\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x+1\right)=42\)

Đặt: `x^2+x=t` 

Pt trở thành: 

\(t\left(t+1\right)=42\)

\(\Leftrightarrow t^2+t-42=0\\\Leftrightarrow t^2-6t+7t-42=0\\ \Leftrightarrow t\left(t-6\right)+7\left(t-6\right)=0\\ \Leftrightarrow\left(t-6\right)\left(t+7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}t=6\\t=-7\end{matrix}\right.\)

Với \(t=6\Leftrightarrow x^2+x=6\Leftrightarrow x^2+x-6=0\Leftrightarrow\left(x-2\right)\left(x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-3\end{matrix}\right.\)

Với \(t=-7\Leftrightarrow x^2+x=-7\Leftrightarrow x^2+x+7=0\Leftrightarrow\left(x+\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)

Vậy: ...