a) \(4x^4+y^4=\left(4x^4+4x^2y^2+y^4\right)-4x^2y^2=\left(2x^2+y^2\right)^2-\left(2xy\right)^2=\left(2x^2+2xy+y^2\right)\left(2x^2-2xy+y^2\right)\)

b) \(x^4y^4+4=\left(x^4y^4+4x^2y^2+4\right)-4x^2y^2=\left(x^2y^2+2\right)^2-\left(2xy\right)^2=\left(x^2y^2+2xy+2\right)\left(x^2y^2-2xy+2\right)\)

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

d) \(x^8+x^7+1=\left(x^8+x^7+x^6\right)-\left(x^6-1\right)=x^6\left(x^2+x+1\right)-\left(x^3+1\right)\left(x-1\right)\left(x^2+x+1\right)\)

\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)

\(\dfrac{11}{333}\) giữ nguyên

\(\dfrac{17}{3}=\dfrac{17\times111}{3\times111}=\dfrac{1887}{333}\)

\(\dfrac{3}{8}-\dfrac{3}{21}=\dfrac{63}{168}-\dfrac{24}{168}=\dfrac{39}{168}=\dfrac{13}{56}\\ \dfrac{7}{2}-\dfrac{7}{10}=\dfrac{35}{10}-\dfrac{7}{10}=\dfrac{28}{10}=\dfrac{14}{5}\)

\(\dfrac{11}{333}\) giữ nguyên

\(\dfrac{17}{37}=\dfrac{17\times9}{37\times9}=\dfrac{629}{333}\)

a) \(\sqrt{x}=15\Rightarrow\left\{{}\begin{matrix}x\ge0\\x=15^2\end{matrix}\right.\Rightarrow x=225\)

b) \(2\sqrt{x}=14\Rightarrow\sqrt{x}=7\Rightarrow\left\{{}\begin{matrix}x\ge0\\x=7^2\end{matrix}\right.\Rightarrow x=49\)

c) \(\sqrt{x}< \sqrt{2}\Rightarrow\left\{{}\begin{matrix}x\ge0\\x< 2\end{matrix}\right.\Rightarrow0\le x< 2\)

d) \(\sqrt{2x}< 4\Rightarrow\left\{{}\begin{matrix}x\ge0\\2x< 16\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x\ge0\\x< 8\end{matrix}\right.\Rightarrow0\le x< 8\)

Xét \(\left(m+1\right)x^2-2\left(m-1\right)x+m-2=0\)

Với \(m+1=0\Rightarrow m=-1\) ta được phương trình bậc nhất, không thể có 2 nghiệm.

\(\Rightarrow m\ne-1\)

Khi đó, \(\Delta'=\left(m-1\right)^2-\left(m+1\right)\left(m-2\right)=m^2-2m+1-m^2+m+2=-m+3\)

a) Để PT có 2 nghiệm phân biệt thì \(\Delta'>0\Rightarrow-m+3>0\Rightarrow m< 3\)

Vậy \(m< 3,m\ne-1\) thì PT có 2 nghiệm phân biệt

b) Khi đó, áp dụng hệ thức Vi-et ta có: \(\left\{{}\begin{matrix}x_1+x_2=\dfrac{2\left(m-1\right)}{m+1}\\x_1x_2=\dfrac{m-2}{m+1}\end{matrix}\right.\)

Khi 1 nghiệm \(x_1=2\) ta có: 

\(4\left(m+1\right)-4\left(m-1\right)+m-2=0\Rightarrow m+6=0\Rightarrow m=-6\) 

\(\Rightarrow x_1+x_2=\dfrac{2\left(-6-1\right)}{-6+1}=\dfrac{14}{5}\Rightarrow x_2=\dfrac{4}{5}\)

c) Để \(x_1^2+x_2^2=2\Rightarrow\left(x_1+x_2\right)^2-2x_1x_2=2\)

\(\Rightarrow\dfrac{4\left(m-1\right)^2}{\left(m+1\right)^2}-\dfrac{4\left(m-1\right)}{m+1}=2\)

\(\Rightarrow4\left(m-1\right)^2-4\left(m-1\right)\left(m+1\right)-2\left(m+1\right)^2=0\)

\(\Rightarrow4m^2-8m+4-4m^2+4-2m^2-4m-1=0\\ \Rightarrow-2m^2-12m+7=0\Rightarrow\left[{}\begin{matrix}m=\dfrac{-6+5\sqrt{2}}{2}\left(TM\right)\\m=\dfrac{-6-5\sqrt{2}}{2}\left(TM\right)\end{matrix}\right.\)

a) \(x^3-6x^2+9x=x\left(x^2-6x+9\right)=x\left(x-3\right)^2\)

b) \(a^2+b^2-2ab+a-b=\left(a-b\right)^2+\left(a-b\right)=\left(a-b\right)\left(a-b+1\right)\)

c) Sửa đề \(4a^2-4b^2-4a+1=\left(4a^2-4a+1\right)-4b^2=\left(2a-1\right)^2-\left(2b\right)^2=\left(2a+2b-1\right)\left(2a-2b-1\right)\)