Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(x^4-3x^3-6x^2+3x+1\)
\(=x^4-2x^2+1-3x^3+3x-4x^2\)
\(=\left(x^2-1\right)^2-3x\left(x^2-1\right)-4x^2\)
đặt \(a=x^2-1\) khi đó biểu thức trở thành
\(a^2-3ax-4x^2\)
\(=a^2+ax-4ax-4x^2\)
\(=\left(a+x\right)\left(a-4x\right)\)
\(=\left(x^2+x-1\right)\left(x^2-4x+1\right)\)
\(x^2-6x+9=-y^2-10y-20.\)
\(\left(x-3\right)^2=-y^2-10y-20\)
\(\left(x-3\right)^2=-y^2-10y-20\)
\(\left(x-3\right)^2=-\left(y^2+2.5y+25\right)+5\)
\(\left(x-3\right)^2=-\left(y+5\right)^2+5\)
\(\hept{\begin{cases}x=3\\y+5=\sqrt{5}\Leftrightarrow y=\sqrt{5}-5\end{cases}}\)
b)
\(\left(4x^2-4x+1\right)=-y^2-x^2-2xy\)
\(\left(2x-1\right)^2=-\left(y+x\right)^2\)
\(x=\frac{1}{2}\Leftrightarrow y=-\frac{1}{2}\)
1)\(ĐKXĐ:x\ne2;x\ne-2\)
đầu bài..
.\(\Rightarrow\left(1-6x\right)\left(x+2\right)+\left(9x+4\right)\left(x-2\right)=x\left(3x-2\right)+1\)
\(\Leftrightarrow-6x^2-11x+2+9x^2-14x-8=3x^2-2x+1\)
\(\Leftrightarrow-23x=7\Leftrightarrow x=-\frac{7}{23}\)(nhận)
Vậy...........
2).......\(ĐKXĐ:x\ne2;x\ne7\)
\(\Rightarrow\left(x+1\right)\left(x-7\right)=x-2\)
\(\Leftrightarrow x^2-6x-7=x-2\)
\(\Leftrightarrow x^2-7x-5=0\)..............
Vậy........
3)ĐKXĐ:\(x\ne1\)
.........\(\Rightarrow3\left(7x-3\right)=2\left(x-1\right)\)
\(\Leftrightarrow21x-9=2x-2\)
\(\Leftrightarrow19x=7\Leftrightarrow x=\frac{7}{19}\)(nhận)
4)ĐKXĐ:\(x\ne-1\)
.........\(\Rightarrow2\left(3-7x\right)=1+x\)
\(\Leftrightarrow6-14x=1+x\)
\(\Leftrightarrow15x=5\Leftrightarrow x=\frac{1}{3}\)(nhận)
Vậy...................
\(1,\left(x-2\right)\left(x+2\right)\left(x^2+4\right)-\left(x^2-3\right)\left(x^2+3\right)\)
\(=\left(x^2-4\right)\left(x^2+4\right)-\left(x^2-9\right)\)
\(=x^2-16-x^2+9\)
\(=-7\)
\(2,\left(6x+1\right)^2+\left(6x-1\right)^2-2\left(1+6x\right)\left(6x-1\right)\)
\(=\left(6x+1-6x+1\right)^2\)
\(=2^2=4\)
à quên thêm +12
nhầm =12