Bài 1:

\(a,=15x^4-12x^3+9x^2\\ b,=-15x^3y^2+25x^2y^2-5xy^3\\ c,=5x^3-15x^2-4x^2+12x=5x^3-19x^2+12x\\ d,=3x^3-9x^2y+xy^2-3y^3+5x^2y-15xy^2=3x^3-3y^3-4x^2y-14xy^2\)

Bài 2:

\(a,=x^2+4x-21-x^2-4x+5=-16\\ b,=x^2+16x+64-2x^2-12x+32+x^2-4x+4=100\\ c,=x^4-16x^2-x^4+1=1-16x^2\\ d,=x^3+1-x^3+1=2\)

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b1

a) \(\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\)

\(=\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\)

\(=\dfrac{1}{5}-\dfrac{1}{10}\)

\(=\dfrac{2}{10}-\dfrac{1}{10}\)

\(=\dfrac{1}{10}\)

b) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{99.100}\)

\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=\dfrac{1}{1}-\dfrac{1}{100}\)

\(=\dfrac{99}{100}\)

c) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+\dfrac{2}{7.9}+\dfrac{2}{9.11}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\)

\(=\dfrac{1}{3}-\dfrac{1}{11}\)

\(=\dfrac{8}{33}\)

d) \(\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)

\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=\dfrac{1}{3}-\dfrac{1}{101}\)

\(=\dfrac{98}{303}\)

1: Sửa đề: 3x-5

\(=\dfrac{-x^2\left(3x-5\right)-3\left(3x-5\right)}{3x-5}=-x^2-3\)

2: \(=\dfrac{5x^4-5x^3+14x^3-14x^2+12x^2-12x+8x-8}{x-1}\)

=5x^2+14x^2+12x+8

3: \(=\dfrac{5x^3+10x^2+4x^2+8x+4x+8}{x+2}=5x^2+4x+4\)

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

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

\(a,\frac{2}{5}:\frac{6}{5}+\frac{3}{4}.\frac{12}{5}\)

=\(\frac{2}{5}.\frac{5}{6}+\frac{9}{5}\)

=\(\frac{1}{3}+\frac{9}{5}\)

=\(\frac{32}{5}\)

b)  \(\frac{17}{5}.\frac{3}{5}.\frac{17}{5}.\frac{2}{5}\)

=\(\frac{17}{5}.\left(\frac{3}{5}.\frac{2}{5}\right)\)

=\(\frac{17}{5}.\frac{6}{25}\)

=\(\frac{102}{125}\)

c)\(\frac{1}{4}+\frac{3}{4}.\frac{2}{3}-\frac{1}{2}\)

=\(\frac{1}{4}+\frac{1}{2}-\frac{1}{2}\)

=\(\frac{1}{4}\)

d)\(\frac{6}{7}+\frac{5}{8}:5\)

=\(\frac{6}{7}+\frac{5}{8}.\frac{1}{5}\)

=\(\frac{6}{7}+\frac{1}{8}\)

=\(\frac{48}{56}+\frac{7}{56}\)

=\(\frac{55}{56}\)

tk nhé bn!

\(1,\)\(\left(2x+3\right)^2=4x^2+12x+9\)

\(2,\)\(\left(3x+2y\right)^2=9x^2+12xy+4x^2\)

\(3,\)\(\left(3a-1\right)^2=9x^2-6x+1\)

\(4,\)\(\left(a-2\right)^2=a^2-4a+4\)

\(5,\)\(\left(1-5a\right)^2=1-10a+25a^2\)

\(6,\)\(\left(x-4\right)^3=x^3-12a^2+48a-64.\)

\(7,\)\(\left(x^2-2y\right)^2=x^4-4x^2y-4y^2\)

\(8,\)\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)

\(9,\)\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^4-49\)

\(10,\)\(\left(x-1\right)\left(x^2+x+1\right)=x^3-1\)

\(11,\)\(\left(x^3-2\right)\left(x^6+2x^3+4\right)=x^9-8\)

\(12,\)\(\left(3x+2\right)\left(9x^2-6x+4\right)=27x^3+8\)

\(13,\)\(\left(x^2+3\right)\left(x^4-3x^2+9\right)=x^6+27\)

1, ( 2x + 3 )² = 4x² + 12x + 9

2, ( 3x + 2y )² = 9x² +12xy + 4y²

3 ( 3a - 1 )² = 9a² - 6x + 1

4, ( a - 2 )² = a² - 4a + 4

5, ( 1 - 5a )² = 1 - 10a + 25a²

6,  ( x- 4 )³ = x³ - 12x² + 48x - 64

7, ( x² - 2y )² = x⁴ - 4x²y + 4y²

8, ( 5X² - 2 ).( 5X² + 2 ) = 25X² - 4

9, ( 2a² - 7 ).( 2a² + 7 ) = 4a⁴ - 49

10, ( x - 1 ).( x² + x + 1 ) = x³ - 1

các bạn ơi đừng làm tắt được kg nhờ các bạn đó

2 = 1 + 1     6 = 2 + 4     8 = 5 + 3     10 = 8 + 2

3 = 1 + 2     6 = 3 + 3     8 = 4 + 4     10 = 7 + 3

4 = 3 + 1     7 = 1 + 6     9 = 8 + 1     10 = 6 + 4

4 = 2 + 2     7 = 5 + 2     9 = 6+ 3     10 = 5 + 5

5 = 4 + 1     7 = 4 + 3     9 = 7 + 2     10 = 10 + 0

5 = 3 + 2     8 = 7 + 1     9 = 5 + 4     10 = 0 + 10

6 = 5 + 1     8 = 6 + 2     10 = 9 + 1     1 = 1 + 0

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