Bài 2: 

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

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

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

4) \(9-25x^2=3^2-\left(5x\right)^2=\left(3-5x\right)\left(3+5x\right)\)

5) \(4x^2-25=\left(2x\right)^2-5^2=\left(2x+5\right)\left(2x-5\right)\)

6) \(9x^2-36=\left(3x\right)^2-6^2=\left(3x-6\right)\left(3x+6\right)\)

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

8) \(x^2-\left(2y\right)^2=\left(x-2y\right)\left(x+2y\right)\)

9) \(\left(2x\right)^2-y^2=\left(2x-y\right)\left(2x+y\right)\)

10) \(\left(3x\right)^2-9y^4=\left(3x\right)^2-\left(3y^2\right)^2=\left(3x-3y^2\right)\left(3x+3y^2\right)\)

Bài 2: 

21) \(\left(\dfrac{x}{3}-\dfrac{y}{4}\right)\left(\dfrac{x}{3}+\dfrac{y}{4}\right)=\left(\dfrac{x}{3}\right)^2-\left(\dfrac{y}{4}\right)^2=\dfrac{x^2}{9}-\dfrac{y^2}{16}\)

22) \(\left(\dfrac{x}{y}-\dfrac{2}{3}\right)\left(\dfrac{x}{y}+\dfrac{2}{3}\right)=\left(\dfrac{x}{y}\right)^2-\left(\dfrac{2}{3}\right)^2=\dfrac{x^2}{y^2}-\dfrac{4}{9}\)

23) \(\left(\dfrac{x}{2}+\dfrac{y}{3}\right)\left(\dfrac{x}{2}-\dfrac{y}{3}\right)=\left(\dfrac{x}{2}\right)^2-\left(\dfrac{y}{3}\right)^2=\dfrac{x^2}{4}-\dfrac{y^2}{9}\)

24) \(\left(2x-\dfrac{2}{3}\right)\left(\dfrac{2}{3}+2x\right)=\left(2x-\dfrac{2}{3}\right)\left(2x+\dfrac{2}{3}\right)=\left(2x\right)^2-\left(\dfrac{2}{3}\right)^2=4x^2-\dfrac{4}{9}\)

25) \(\left(2x+\dfrac{3}{5}\right)\left(\dfrac{3}{5}-2x\right)=\left(\dfrac{3}{5}+2x\right)\left(\dfrac{3}{5}-2x\right)=\left(\dfrac{3}{5}\right)^2-\left(2x\right)^2=\dfrac{9}{25}-4x^2\)

26) \(\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{4}{3}+\dfrac{1}{2}x\right)=\left(\dfrac{1}{2}x-\dfrac{4}{3}\right)\left(\dfrac{1}{2}x+\dfrac{4}{3}\right)=\left(\dfrac{1}{2}x\right)^2-\left(\dfrac{4}{3}\right)^2=\dfrac{1}{4}x^2-\dfrac{16}{9}\)

27) \(\left(\dfrac{2}{3}x^2-\dfrac{y}{2}\right)\left(\dfrac{2}{3}x^2+\dfrac{y}{2}\right)=\left(\dfrac{2}{3}x^2\right)^2-\left(\dfrac{y}{2}\right)^2=\dfrac{4}{9}x^4-\dfrac{y^2}{4}\)

28) \(\left(3x-y^2\right)\left(3x+y^2\right)=\left(3x\right)^2-\left(y^2\right)^2=9x^2-y^4\)

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

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

Bài `3`

\(1,\left(2x+1\right)^2+\left(2x-1\right)^2\\ \left[\left(2x\right)^2+4x+1^2\right]+\left[\left(2x\right)^2-4x+1^2\right]\\ =4x^2+4x+1+4x^2-4x+1\\ =8x^2+2\)

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

\(3,\left(x+2y\right)^2-\left(x-2y\right)^2\\ =\left[\left(x+2y\right)+\left(x-2y\right)\right]\left[\left(x+2y\right)-\left(x-2y\right)\right]\\ =\left(x+2y+x-2y\right)\left(x+2y-x+2y\right)\\ =2x.4y=8xy\)

\(4,\left(3x+y\right)^2+\left(x-y\right)^2\\ =\left[\left(3x\right)^2+6xy+y^2\right]+\left[\left(x^2-2xy+y^2\right)\right]\\ =6x^2+6xy+y^2+x^2-2xy+y^2\\ =7x^2+4xy+2y^2\)

\(5,-\left(x+5\right)^2-\left(x-3\right)^2\\ =-\left(x^2+10x+5^2\right)-\left(x^2-6x+3^2\right)\\ =-x^2-10x-25-x^2+6x-9\\ =-2x^2+16x-34\)

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

`@ Kidd`

\(E=2.3+3.4+4.5+3.6+2.7+4.15=2\left(3+7\right)+3\left(4+6\right)+4\left(5+15\right)=2.10+3.10+4.20=20+30+80=130\)

\(F=3\left(12+13+14+15\right)+3\left(8+7+6+5\right)=3\left(12+8+13+7+14+6+15+5\right)=3\left(20+20+20+20\right)=3.80=240\)

f: Ta có: \(E=3\cdot\left(12+13+14+15\right)+3\left(8+7+6+5\right)\)

\(=3\left(12+13+14+15+8+7+6+5\right)\)

\(=3\cdot80=240\)

a,

A=1−3−5−7−9−...−97−99a)A=1−3−5−7−9−...−97−99 

=1−(3+5+7+...+99)=1−(3+5+7+...+99)

=1−(99+3).[(99−3):2+1]2=1−(99+3).[(99−3):2+1]2
=1−2499=−2498=1−2499=−2498

b)B=1+3−5−7+9+...+97−99b)B=1+3−5−7+9+...+97−99
=(−8)+(−8)+(−8)+...+(−8)+97−99=(−8)+(−8)+(−8)+...+(−8)+97−99
=(−8).12+(−2)=−98=(−8).12+(−2)=−98

c)C=1−3−5+7+9−11−13+15+...+97−99c)C=1−3−5+7+9−11−13+15+...+97−99
=0+0+0+0+0+...+0−99=0+0+0+0+0+...+0−99
=−99

a: Ta có: \(\dfrac{4}{\sqrt{7}-\sqrt{3}}+\dfrac{6}{3+\sqrt{3}}+\dfrac{\sqrt{7}-7}{\sqrt{7}-1}\)

\(=\sqrt{7}+\sqrt{3}+3-\sqrt{3}-\sqrt{7}\)

=3

Đề bảo gì thế bạn

\(\left(3+\frac{1}{3}\right):4,8=0,5:7\)

\(\left(3+\frac{1}{3}\right):4,8=\frac{1}{14}\)

\(3+\frac{1}{3}=\frac{1}{14}.4,8=\frac{12}{35}\)

=> Đề bài bị lỗi hở bạn ???

Đề yêu cầu gì bạn ?

\(a,\frac{62}{7}:x=\frac{29}{9}:\frac{3}{56}\)

\(\frac{62}{7}:x=\frac{1624}{27}\)

\(x=\frac{62}{7}:\frac{1624}{27}=\frac{837}{5684}\)

\(b,\frac{1}{5}:x=\frac{1}{5}-\frac{1}{7}\)

\(\frac{1}{5}:x=\frac{2}{35}\)

\(x=\frac{1}{5}:\frac{2}{35}=\frac{7}{2}\)

\(c,\frac{2}{3}.x-\frac{4}{7}=\frac{1}{7}\)

\(\frac{2}{3}.x=\frac{1}{7}+\frac{4}{7}=\frac{5}{7}\)

\(x=\frac{5}{7}:\frac{2}{3}=\frac{15}{14}\)

\(d,\frac{2}{7}-\frac{8}{9}.x=\frac{2}{3}\)

\(\frac{8}{9}.x=\frac{2}{7}-\frac{2}{3}=-\frac{8}{21}\)

\(x=-\frac{8}{21}:\frac{8}{9}=-\frac{3}{7}\)

\(e,\frac{4}{7}+\frac{5}{9}:x=\frac{1}{5}\)

\(\frac{5}{9}:x=\frac{1}{5}-\frac{4}{7}=-\frac{13}{35}\)

\(x=\frac{5}{9}:-\frac{13}{35}=\frac{175}{117}\)

\(i,\frac{2}{5}-\frac{2}{5}.x=\frac{2}{5}\)

\(\frac{2}{5}.\left(1-x\right)=\frac{2}{5}\)

\(1-x=\frac{2}{5}:\frac{2}{5}=1\)

\(x=1-1=0\)

\(g,\frac{2}{3}+\frac{1}{3}:x=-1\)

\(\frac{1}{3}:x=-1-\frac{2}{3}=-\frac{5}{3}\)

\(x=\frac{1}{3}:-\frac{5}{3}=-\frac{1}{5}\)

học tốt nha

\(\left(x-2\right)^3+\left(5-2x\right)^3=0\)

\(\Leftrightarrow\left(x-2+5-2x\right)\left[\left(x-2\right)^2-2\left(x-2\right)\left(5-2x\right)+\left(5-2x\right)^2\right]=0\)

\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4-2\left(5x-2x^2-10+4x\right)+25-20x+4x^2\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(x^2-4x+4-10x+4x^2+20-8x+25-20x+4x^2\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(9x^2-42x+49\right)=0\)

\(\Leftrightarrow\left(3-x\right)\left(x-\frac{7}{3}\right)^2=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(3-x\right)=0\\\left(x-\frac{7}{3}\right)^2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\frac{7}{3}\end{matrix}\right.\)

Vậy...

a, \(3,5.\dfrac{-7}{5}=\dfrac{7}{2}.\dfrac{-7}{5}=\dfrac{-49}{10}\)

b, \(\dfrac{-5}{23}:\left(-2\right)=\dfrac{-5}{23}.\dfrac{-1}{2}=\dfrac{5}{46}\)