Áp dụng tính chất a² - b² = a² - ab + ab - b² = a(a - b) + b(a - b) = (a + b)(a - b)

B =\(\left(200^{-2}-1\right)\left(199^{-2}-1\right)...\left(101^{-2}-1\right)=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)

\(=\frac{1-200^2}{200^2}.\frac{1-199^2}{199^2}...\frac{1-101^2}{101^2}=\frac{1^2-200^2}{200^2}.\frac{1^2-199^2}{199^2}....\frac{1^2-101^2}{101^2}\)

\(=\frac{\left(1-200\right)\left(1+200\right)}{200^2}.\frac{\left(1-199\right)\left(1+199\right)}{199^2}...\frac{\left(1-101\right)\left(1+101\right)}{101^2}\)

\(=-\left(\frac{199.201}{200^2}.\frac{198.200}{199^2}...\frac{100.102}{101^2}\right)=-\frac{199.201.198.200..100.102}{200.200.199.199...101.101}\)

\(=-\frac{\left(199.198...100\right)\left(201.200...102\right)}{\left(200.199...101\right).\left(200.199...101\right)}=-\frac{100.201}{200.101}=-\frac{201}{202}\)

                                          Bài giải

\(B=\left(200^{-2}-1\right)\left(199^{-2}-1\right)\left(198^{-2}-1\right)...\left(101^{-2}-1\right)\)

\(B=\left(\frac{1}{200^2}-1\right)\left(\frac{1}{199^2}-1\right)\left(\frac{1}{198^2}-1\right)...\left(\frac{1}{101^2}-1\right)\)

\(B=\left[\left(\frac{1}{200}\right)^2-1^2\right]\left[\left(\frac{1}{199}\right)^2-1^2\right]\left[\left(\frac{1}{198}\right)^2-1^2\right]...\left[\left(\frac{1}{101}\right)^2-1^2\right]\)

\(B=\left(\frac{1}{200}+1\right)\left(\frac{1}{200}-1\right)\left(\frac{1}{199}+1\right) \left(\frac{1}{199}-1\right)..\left(\frac{1}{101}-1\right)\left(\frac{1}{101}+1\right)\)

\(B=\frac{201}{200}\cdot\frac{-199}{200}\cdot\frac{200}{199}\cdot\frac{-198}{199}\cdot...\cdot\frac{-100}{101}\cdot\frac{102}{101}\)

\(B=\frac{201\cdot\left(-199\right)\cdot200\cdot\left(-198\right)\cdot...\cdot\left(-100\right)\cdot102}{200\cdot200\cdot199\cdot199\cdot...\cdot101\cdot101}=\frac{100\cdot201}{200\cdot101}=\frac{201}{202}\)

Bài làm:

Ta có: \(A=\frac{1}{1.3.5}+\frac{1}{3.5.7}+...+\frac{1}{47.49.51}\)

\(A=\frac{1}{4}\left(\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{4}\left(\frac{1}{3}-\frac{1}{49.51}\right)\)

\(A=\frac{1}{12}-\frac{1}{4.49.51}< \frac{1}{12}\)

Vậy \(A< \frac{1}{12}\)

Từ đề bài suy ra\(4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{47.49.51}\)

\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{47.49}-\frac{1}{49.51}=\frac{1}{3}-\frac{1}{49.51}< \frac{1}{3}\)

\(\Rightarrow A< \frac{1}{12}\left(đpcm\right)\)

\(A=\left(0,3.5-0,5:\frac{1}{3}\right)\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(A=\left(0,3.5-0,5.3\right)\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(A=\left(1,5-1,5\right)\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(A=0.\left(\frac{1}{2006^2}+\frac{1}{2008^2}\right)\)

\(A=0\)

VẬY    \(A=0\)

(0,3.5-0,5:1/3).(1/2006^2+1/2008^2)

(1,5-1,5).(1/200^2+1/2008^2)

0.(1/2006^2+1/2008^2)

0

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

\(\Rightarrow3x+\frac{1}{4}=\frac{16}{3}\)

\(\Rightarrow3x=\frac{16}{3}-\frac{1}{4}\)

\(\Rightarrow3x=\frac{61}{12}\)

\(\Rightarrow x=\frac{61}{12}:3\)

\(\Rightarrow x=\frac{61}{36}\)

Vậy \(x=\frac{61}{36}.\)

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

\(3x+\frac{1}{4}=\frac{16}{3}\)

\(3x=\frac{16}{3}-\frac{1}{4}\)

\(3x=\frac{61}{12}\)

\(x=\frac{61}{12}:3\)

\(x=\frac{61}{36}\)

Học tốt

\(-3\frac{1}{2}-2x=0,8\)

\(\Rightarrow-\frac{7}{2}-2x=\frac{4}{5}\)

\(\Rightarrow-2x=\frac{4}{5}+\frac{7}{2}\)

\(\Rightarrow-2x=\frac{43}{10}\)

\(\Rightarrow x=\frac{43}{10}:\left(-2\right)\)

\(\Rightarrow x=-\frac{43}{20}\)

Vậy  \(x=-\frac{43}{20}.\)

-7/2-2x=4/5

2x=-7/2-4/5

2x=-43/10

x=-43/10:2

x=-43/20

vậy x=-43/20

\(\frac{2}{3}+2\left(x-1\right)=\frac{4}{5}\)

=> \(2\left(x-1\right)=\frac{4}{5}-\frac{2}{3}\)

=> \(2\left(x-1\right)=\frac{2}{15}\)

=> \(x-1=\frac{1}{15}\)

=> \(x=\frac{16}{15}\)

2/3+2(x+1)=4/5

2(x+1)=4/5-2/3

2(x+1)=2/15

(x+1)=2/15:2

(x+1)=1/15

x=1/15-1

x=-14/15

vậy x=-14/15

Giả sử \(\frac{a}{b}=\frac{c}{d}\)Suy ra  điều ta cần chứng minh là \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)

Theo tính chất của dãy tỉ số bằng nhau :

\(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\\\frac{a}{b}=\frac{3c}{3d}=\frac{a+3c}{b+3d}\end{cases}}< =>\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)

Vậy ta có điều phải chứng minh

Ta có : \(\frac{a+3c}{b+3d}=\frac{a+c}{b+d}\)

=> (a + 3c)(b+ d) = (b + 3d)(a + c)

=> ab +ad + 3bc + 3cd = ab + bc + 3ad + 3cd

=> ad + 3bc  = bc + 3ad

=> 3bc - bc = 3ad - ad

=> 2bc = 2ad

=> bc = ad

=> \(\frac{a}{b}=\frac{c}{d}\) (đpcm)

Ta có :\(\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\)

=> (2a + b)(c - 2d) = (a - 2b)(2c + d)

=> 2ac - 4ad + bc - 2bd = 2ac + ad - 4bc  - 2bd

=> -4ad + bc = ad - 4bc

=> -4ad - ad = -4bc - bc

=> -5ad = - 5bc

=> ad = bc

=> \(\frac{a}{b}=\frac{c}{d}\)(đpcm)

Theo bài ra ta có : 

\(\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\Leftrightarrow\left(2a+b\right)\left(c-2d\right)=\left(2c+d\right)\left(a-2b\right)\)

\(\Leftrightarrow2ac-4ad+bc-2db=2ca-4bc+da-2bd\)

\(\Leftrightarrow-5ad+5bc=0\Leftrightarrow-5ab=-5bc\)

\(\Leftrightarrow ad=bc\Leftrightarrow\frac{a}{b}=\frac{c}{d}\left(đpcm\right)\)

a) A = \(\frac{5}{1.4}+\frac{29}{4.7}+\frac{71}{7.10}+....+\frac{10301}{100.103}\) (có 34 số hạng)

A = \(\frac{4+1}{1.4}+\frac{4.7+1}{4.7}+\frac{7.10+1}{7.10}+....+\frac{100.103+1}{103.100}\)

A = \(1+\frac{1}{1.4}+1+\frac{1}{4.7}+1+\frac{1}{7.10}+....+1+\frac{1}{100.103}\)

A = \(1.34+\frac{1}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{100.103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{100}-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}.\left(1-\frac{1}{103}\right)\)

A = \(34+\frac{1}{3}\cdot\frac{102}{103}\)

A = \(34+\frac{34}{103}=\frac{3536}{103}\)

bạn làm hộ mik câu B với