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

2) \(\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)

4) \(\left(\frac{1}{2}-x\right)\left(\frac{1}{4}+\frac{1}{2}x+x^2\right)=\left(\frac{1}{2}\right)^3-x^3=\frac{1}{8}-x^3\)

5) \(\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)=\left(2x\right)^3-\left(3y\right)^3=8x^3-27y^3\)

6) \(\left(2x+5\right)\left(4x^2-10x+25\right)=\left(2x\right)^3+5^3=8x^3+125\)

3) Đề thiếu

a) Vì \(\widehat{B}=\widehat{C}\left(gt\right)\)

Mà BD,CE là tia phân giác của \(\widehat{B}\) và \(\widehat{C}\)

=>\(\widehat{ABD}=\widehat{DBC}=\widehat{ACE}=\widehat{ECB}\)

Xét ΔBCD và ΔCBE có:

         \(\widehat{B}=\widehat{C}\left(gt\right)\)

         BC: cạnh chung

        \(\widehat{DBC}=\widehat{ECB}\)(gt)

=>ΔBCD=ΔCBE(g.c.g)

b)Vì \(\widehat{OBC}=\widehat{OCB}\left(cmt\right)\)

=>ΔOBC cân tại O

=>OB=OC

d) \(100^2-99^2+98^2-97^2+...+2^2-1\)

\(=\left(100+99\right)\left(100-99\right)+\left(98+97\right)\left(98-97\right)+..+\left(2+1\right)\left(2-1\right)\)

\(=100+99+98+97+..+2+1\)

\(=\frac{\left(100+1\right)\cdot100}{2}=5050\)

b) \(\left(x+1\right)^3+\left(x-1\right)^3+x^3-3x\left(x+1\right)\left(x-1\right)\)

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

\(=3x^3+6x-3x^3+3x\)

\(=3x\)

\(\frac{x}{36}=\frac{2}{x+1}\left(ĐK:x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=2\cdot36\)

\(\Leftrightarrow x^2+x-72=0\)

\(\Leftrightarrow x^2-8x+9x-72=0\)

\(\Leftrightarrow x\left(x-8\right)+9\left(x-8\right)=0\)

\(\Leftrightarrow\left(x-8\right)\left(x+9\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-8=0\\x+9=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=8\\x=-9\end{array}\right.\)

\(\left(x-1\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=2\end{array}\right.\)

Vậy x={1;2}

Xe lửa chuyển động như vậy hết:

   24 : 3 x 4 = 32 (phút)

       ĐS: 32 phút

a)Có:\(5x=7y\Rightarrow\)\(\frac{x}{7}=\frac{y}{5}\Rightarrow\)\(\frac{x}{7}=\frac{2y}{10}\)

Áp dụng tính chắt của dãy tỉ số bằng nhau ta có:

\(\frac{x}{7}=\frac{2y}{10}=\frac{x+2y}{7+10}=\frac{51}{17}=3\)

=>\(\frac{x}{7}=x\Rightarrow x=21\)

    \(\frac{2y}{10}=3\Rightarrow2y=30\Rightarrow y=15\)

a)\(x^2+7x+12\)

\(=x^2+x+6x+6\)

\(=x\left(x+1\right)+6\left(x+1\right)\)

\(=\left(x+1\right)\left(x+6\right)\)

a)\(A=\frac{x+1}{x^2-2x+1}:\left(\frac{1}{x^2-x}+\frac{1}{x-1}\right)\left(ĐK:x\ne0;x\ne1\right)\)

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

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

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

\(=\frac{x}{x-1}\)

b)Có:\(x^2+x-2=0\)

\(\Leftrightarrow x^2-x+2x-2=0\)

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

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

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\left(loại\right)\\x=2\end{array}\right.\)

Thay x=2 vào A ta được:

\(A=\frac{2}{2-1}=2\)