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

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

2) \(=\left(x+1\right)^3-\left(2y\right)^3=\left(x+1-2y\right)\left[\left(x+1\right)^2+\left(x+1\right).2y+4y^2\right]\)

\(=\left(x-2y+1\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)Đến đây bạn phân tích tiếp nha

CHÚC BẠN HỌC TỐT 

T I C K ủng hộ nha

1)P\(=9\left(x+3\right)^2-4\left(x-2\right)^2\)\(=\left(3x+9\right)^2-\left(2x-4\right)^2\)

      \(=\left(3x+9+2x-4\right)\left(3x+9-2x+4\right)\)(hằng đẳng thức số 3)

       \(=\left(5x+5\right)\left(x+13\right)\)

        \(=5\left(x+1\right)\left(x+13\right)\)

2)P\(=25\left(2x-y\right)^2-16\left(x+2y\right)^2\)\(=\left(10x-5y\right)^2-\left(4x+8y\right)^2\)

      \(=\left(14x+3y\right)\left(6x-13y\right)\)(tương tự câu 1)

Đặt \(A=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-xy-2y^2}\)

      \(B=\frac{4x^4+4x^2y+y^2-4}{x^2+y+xy+x}\)

    \(C=\frac{x+1}{2x^2+y+2}\)

Ta có: 

A = \(\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{x^2-y^2-xy-y^2}=\frac{x-y}{2y-x}-\frac{x^2+y^2+y-2}{\left(x-2y\right)\left(x+y\right)}=\frac{\left(x-y\right)\left(x+y\right)+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)

=>A=\(\frac{x^2-y^2+x^2+y^2+y-2}{\left(2y-x\right)\left(x+y\right)}=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}\)

B=\(\frac{\left(2x^2\right)^2+2.2x^2.y+y^2-4}{x^2+xy+x+y}=\frac{\left(2x^2+y\right)^2-4}{x\left(x+y\right)+\left(x+y\right)}=\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+1\right)\left(x+y\right)}\)

=>\(P=\left(A:B\right):C\)

       \(=\left[\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}:\frac{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}{\left(x+y\right)\left(x+1\right)}\right]:\frac{x+1}{2x^2+y+2}\)

       \(=\frac{2x^2+y-2}{\left(2y-x\right)\left(x+y\right)}.\frac{\left(x+y\right)\left(x+1\right)}{\left(2x^2+y+2\right)\left(2x^2+y-2\right)}.\frac{2x^2+y+2}{x+1}\)

        \(=\frac{1}{2y-x}\)

=>\(P=\frac{1}{2y-x}\)

Thế x=-1,76 và y=3/25 vào P

=>\(P=\frac{1}{2.\frac{3}{25}-1,76}=\frac{1}{2}\)

P= 125x^3-8y^3

  =5^3x^3-2^3y^3

  =(5x)^3-(2y)^3

  =(5x-2y)(25x^2+10xy+4y^2)

P=4x(x-2y)+8y(2y-x)

  =4x(x-2y)-8y(x-2y)

  =(4x-8y)(x-2y)

  =4(x-2y)(x-2y)

  =4(x-2y)^2

(2x+1)^2-(x-1)^2=(2x+1-x+1)(2x+1+x-1)

                       =(x+2)3x

K NHA!

đề bài của bài 1 là rút gọn

ai giải giùm đi

\(A=\left(x^2+2x+1\right)+4=\left(x+1\right)^2+4Ma:\hept{\begin{cases}\left(x+1\right)^2\ge0\\4>0\end{cases}\Rightarrow\left(x+1\right)^2+4>0}\)với mọi x

\(B=\left(x^2-2.3x+9\right)+1=\left(x-3\right)^2+1\\ Ma:\hept{\begin{cases}\left(x-3\right)^2\ge0\\1>0\end{cases}\Rightarrow\left(x-3\right)^2+1}\)(dấu phái sau là do lỗi nha )

\(C=\left(x^2-2.\frac{3}{2}x+\frac{9}{4}\right)+\frac{11}{4}=\left(x-\frac{3}{2}\right)^2+\frac{11}{4}\)

\(Ma:\hept{\begin{cases}\left(x-\frac{3}{2}\right)^2\ge0\\\frac{11}{4}>0\end{cases}\Rightarrow\left(x-\frac{3}{2}\right)^2+\frac{11}{4}>0}\)

\(D=2\left(x^2-\frac{5}{2}x+7\right)=2\left(x^2-2.\frac{5}{4}x+\frac{25}{16}+\frac{87}{16}\right)=2\left(x-\frac{5}{4}\right)^2+\frac{87}{8}\)

\(Ma:\hept{\begin{cases}\left(x-\frac{5}{4}\right)^2\ge0\\\frac{87}{4}>0\end{cases}\Rightarrow2\left(x-\frac{5}{4}\right)^2+\frac{87}{8}>0}\)

Học tốt nha bạn 

1 T I C K nha