Giải PT
\(\left(x^2-1\right)\left(x^4+4x+3\right)=192\)
giải pt:\(192-\)\(\left(x^2-1\right)\left(x^2+4x+3\right)\)\(=0\)
\(192-\left(x^2-1\right)\left(x^2+4x+3\right)=0\)
\(\Leftrightarrow192-\left(x-1\right)\left(x+1\right)\left(x+1\right)\left(x+3\right)=0\)
\(\Leftrightarrow192-\left[\left(x-1\right)\left(x+3\right)\right]\left[\left(x+1\right)\left(x+1\right)\right]=0\)
\(\Leftrightarrow192-\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
Đặt \(x^2+2x-3=a\)
\(pt\Leftrightarrow192-a\left(a+4\right)=0\)
\(\Leftrightarrow192-a^2-4a=0\)
\(\Leftrightarrow-a^2-16a+12a+192=0\)
\(\Leftrightarrow-a\left(a+16\right)+12\left(a+16\right)=0\)
\(\Leftrightarrow\left(a+16\right)\left(-a+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=-16\\a=12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x-3=-16\\x^2+2x-3=12\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+13=0\\x^2+2x-15=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x^2+2x+1+12=0\\x^2+5x-3x-15=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}\left(x+1\right)^2=-12\\x\left(x+5\right)-3\left(x+5\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\\left(x+5\right)\left(x-3\right)=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=3\end{cases}}\)
Vậy.....
giải pt :
a, \(\left(2x-6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :a,\(\left(2x+6\right)\sqrt{x+4}-\left(x-5\right)\sqrt{2x+3}=3\left(x-1\right)\)
b, \(\left(4x+1\right)\sqrt{x+2}-\left(4x-1\right)\sqrt{x-2}=21\)
c, \(\left(4x+2\right)\sqrt{x+1}-\left(4x-2\right)\sqrt{x-1}=9\)
d, \(\left(2x-4\right)\sqrt{3x-2}+\sqrt{x+3}=5x-7+\sqrt{3x^2+7x-6}\)
giải pt :
a,\(\left(6x-5\right)\sqrt{x+1}-\left(6x+2\right)\sqrt{x-1}+4\sqrt{x^2-1}=4x-3\)
b, \(\left(9x-2\right)\sqrt{3x-1}+\left(10-9x\right)\sqrt{3-3x}-4\sqrt{-9x^2+12x-3}=4\)
c, \(\left(13-4x\right)\sqrt{2x-3}+\left(4x-3\right)\sqrt{5-2x}=2+8\sqrt{-4x^2+16x-15}\)
Giải pt:
\(\left(3x^2+4x-4\right)\sqrt{x-1}=x\left(x^2-3x+3\right)\)
Giải phương trình
\(\left(x^2-1\right)\left(x^4+4x+3\right)=192\)
(x² - 1)(x² + 4x + 3) = 192
<=> (x - 1)(x + 1)(x + 1)(x + 3) = 192
<=> (x - 1)(x + 3)(x + 1)² = 192
<=> (x² + 2x - 3)(x² + 2x + 1) = 192
Đặt t = x² + 2x + 1 => x² + 2x - 3 = t - 4
ta có pt: (t - 4)t = 192
<=> t² - 4t - 192 = 0
<=> t = - 12 hoặc t = 16
*t = x² + 2x + 1 = -12: vn
*t = x² + 2x + 1 = 16
<=> (x+1)² = 16
<=> x = -5 hoặc x = 3
Mãi mãi có một tương lai tươi sáng
Giải PT:
1)\(\left(x^2+4x+2\right)\cdot\left(1-\frac{1}{x}\right)+\frac{36x^2}{\left(x-2\right)^2}=0\)
2)\(\left(x^2-x+1\right)^3-6\left(x+1\right)^3=\left(x^3+1\right)\left(6x^2-17x-5\right)\)
3)\(\left(x^3+4x-4\right)^3+4x^3+15x-20=0\)
giải pt
1.\(\left(x^2+3x-4\right)^3+\left(3x^3+7x+4\right)^3=\left(4x^2+10x\right)^3\)
2.\(x\left(x+2\right)+a^2=2a\left(x+1\right)\)
Áp dụng BĐT\(a^3+b^3+c^3=3abc\) ta có (cái này bạn phải cm mới được áp dụng\(\left(x^2+3x-4\right)^3+\left(3x^3+7x+4\right)^3-\left(4x^2+10x\right)^3=-3\left(x^2+3x-4\right)\left(3x^3+7x+4\right)\left(4x^2+10x\right)=0\)
Sau đó bạn chia 3 trường hợp ra rồi giải pt tìm x
k mk nha