Mình nghĩ là 1

mik nghĩ vậy nhưng chưa bít trình bày í

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a) 1007-0/2=503.5

b)

\(S=1-2+3-4+...+199-200+201\)

\(=\left(1-2\right)+\left(3-4\right)+...+\left(199-200\right)+201\)

\(=1+1+...+1+201\)

\(=\dfrac{200}{2}+201\)

\(=301\)

A = 1^2 + 3^2 + ... + 97^2 + 99^2

= 1.1 + 3.3 + ... + 97.97 + 99.99

> 1.2 + 2.3 + ... + 97.98 + 98.99

= 1.99 = 99

Suy ra A > 1

Xét \(A=\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}a>0\)

Ta có: \(A^2=1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}=\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}\)

\(\frac{a^4+2a^2\left(a+1\right)+\left(a+1\right)^2}{a^2\left(a+1\right)^2}=\frac{\left(a^2+a+1\right)^2}{a^2\left(a+1\right)^2}\)

Vì a>0, D>0  nên \(A=\frac{a^2+a+1}{a\left(a+1\right)}=1+\frac{1}{a}-\frac{1}{a+1}\)

Áp dụng ta có: \(D=\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{99^2}+\frac{1}{100^2}}\)

\(=\left(1+\frac{1}{1}-\frac{1}{2}\right)+\left(1+\frac{1}{2}-\frac{1}{3}\right)+...+\left(1+\frac{1}{99}-\frac{1}{100}\right)=100-\frac{1}{100}=99,99\)

1 ) 3yx - 6xy² 

= 3xy ( 1 - 2y )

2 ) 5ab² - 20a³b²

= 5ab² ( 1 - 4a² )

= 5ab² ( 1 - 2a ) ( 1 + 2a )

3 ) 3x - 3b - y ( b - x )

= 3 ( x - b ) + y ( x - b )

= ( x - b ) ( 3 + y ) 

1)3xy-6xy²=3xy(1-2y)

2)5ab²-20a³b²=5ab²(1-4a²)=5ab²[1²-(2a)²]=5ab²(1+2a)(1-2a)

3)3x-3b-y(b-x)=3x-3b-by+xy=(3x+xy)-(3b+by)=3x(1+y)-3b(1+y)=3(1+y)(x-b)

\(3xy-6\left(xy\right)^2=3xy.\left(1-2xy\right)\)

\(5\left(ab\right)^2-20a^3b^2=5a^2b^2\left(1-4a\right)\)

\(3x-3b-y\left(b-x\right)=3.\left(x-b\right)+y.\left(x-b\right)=\left(x-b\right)\left(3+y\right)\)

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