#)Sửa đề : x⁴+2x³+5x²+4x-12=0

#)Giải :

\(x^4+2x^3+5x^2+4x-12=0\)

\(\Leftrightarrow\left(x^4-x^3\right)+\left(3x^3-3x^2\right)+\left(8x^2-8x\right)+\left(12x-12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^3+3x^2+8x+12\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(x^3+2x^2\right)+\left(x^2+2x\right)+\left(6x+12\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x=-2\end{cases}}\)

Giải phương trình:

x⁴+2x³+5x²+4x+4=0

_Sửa đề bài :

2,6g dung dịch 

ez

\(a,\frac{2}{\sqrt{3}-1}-\frac{2}{\sqrt{3}+1}=\frac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\sqrt{3}+2-2\sqrt{3}+2}{3-1}\)

\(=\frac{4}{2}=2\)

\(b,\frac{2+\sqrt{2}}{2-\sqrt{2}}+\frac{2-\sqrt{2}}{2+\sqrt{2}}=\frac{\left(2+\sqrt{2}\right)\left(2+\sqrt{2}\right)+\left(2-\sqrt{2}\right)\left(2-\sqrt{2}\right)}{\left(2-\sqrt{2}\right)\left(2+\sqrt{2}\right)}\)

\(=\frac{4+4\sqrt{2}+2+4-4\sqrt{2}+2}{4-2}\)

\(=\frac{8+4}{2}=\frac{12}{2}=6\)

GIÚP MK NHA CÁC BN

\(\frac{\sqrt{8-\sqrt{15}}}{\sqrt{30}-\sqrt{2}}=\frac{\sqrt{2}\sqrt{8-\sqrt{15}}}{\sqrt{2}\left(\sqrt{15}.\sqrt{2}-\sqrt{2}\right)}=\frac{\sqrt{16-2\sqrt{15}}}{\sqrt{2}.\sqrt{2}\left(\sqrt{15}-1\right)}\)

\(=\frac{\sqrt{15-2\sqrt{15}+1}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{\left(\sqrt{15}-1\right)^2}}{2\left(\sqrt{15}-1\right)}=\frac{\sqrt{15}-1}{2\left(\sqrt{15}-1\right)}=\frac{1}{2}\)

Đặt: \(x^2=t\)

Sao đó giải như pt bậc 2 bình thường 

cops mạng đâu thế :((