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x=99
=>x+1=100
thay x+1=100 và 99=x vào B ta được:
x99-(x+1).x98+(x+1).x97-(x+1).x96+...+(x+1).x-1
=x99-x99-x98+x98+x97-x97-x96+...+x2+x-1
=x-1
=99-1
=98
Vậy B=98
x=99
=>x+1=100
thay x+1=100 và 99=x vào B ta được:
x99-(x+1).x98+(x+1).x97-(x+1).x96+...+(x+1).x-1
=x99-x99-x98+x98+x97-x97-x96+...+x2+x-1
=x+1
=100
Vậy B=100
SỬA
x=99
=>x+1=100
thay x+1=100 và 99=x vào B ta được:
x99-(x+1).x98+(x+1).x97-(x+1).x96+...+(x+1).x-1
=x99-x99-x98+x98+x97-x97-x96+...+x2+x-1
=x-1
=99-1
=98
Vậy B=98
A=\(x^{99}-\left(99+1\right)x^{98}+\left(99+1\right)x^{97}-...-1\)
=\(x^{99}-99x^{98}-x^{98}+99x^{97}+...+99^2+99-1\)mà x =99
nên \(A=99^{99}-99^{99}-99^{98}+99^{98}+99^{97}-99^{97}-...+99-1\)
\(A=99-1=98\)
a) \(\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(=\frac{1}{99}-\left(\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)
đặt \(A=\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\)
\(A=1-\frac{1}{99}\)
\(A=\frac{98}{99}\)
thay A vào, ta được :
\(\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)
b) \(\frac{2}{100.99}-\frac{2}{99.98}-...-\frac{2}{3.2}-\frac{2}{2.1}\)
\(=\frac{2}{100.99}-\left(\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\right)\)
đặt \(A=\frac{2}{99.98}+...+\frac{2}{3.2}+\frac{2}{2.1}\)
\(A=\frac{2}{1.2}+\frac{2}{2.3}+...+\frac{2}{98.99}\)
\(A=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(A=2.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(A=2.\left(1-\frac{1}{99}\right)\)
\(A=2.\frac{98}{99}\)
\(A=\frac{196}{99}\)
Thay A vào, ta được :
\(\frac{2}{100.99}-\frac{196}{99}=\frac{-19598}{9900}\)
Bài 1:
\(M\left(1\right)=a+b+6\)
Mà \(M\left(1\right)=0\)
\(\Rightarrow a+b+6=0\)
\(\Rightarrow a+b=-6\)( * )
\(\Rightarrow2a+2b=-12\) (1)
Ta có: \(M\left(-2\right)=4a-2b+6\)
Mà \(M\left(-2\right)=0\)
\(\Rightarrow4a-2b=-6\)(2)
Lấy (1) cộng (2) ta được:
\(6a=-18\)
\(a=-3\)
Thay a=-3 vào (* ) ta được:
\(b=-3\)
Vậy a=-3 ; b=-3
Bài 2:
a) \(\frac{5}{x}+\frac{y}{4}=\frac{1}{8}\)
\(\Leftrightarrow\frac{1}{8}-\frac{y}{4}=\frac{5}{x}\)
\(\Leftrightarrow\frac{1}{8}-\frac{2y}{8}=\frac{5}{x}\)
\(\Leftrightarrow\frac{1-2y}{8}=\frac{5}{x}\)
\(\Leftrightarrow\left(1-2y\right).x=5.8\)
\(\Leftrightarrow\left(1-2y\right).x=40\)
Vì \(x,y\in Z\Rightarrow1-2y\in Z\)
mà \(40=1.40=40.1=5.8=8.5=\left(-1\right).\left(-40\right)=\left(-40\right).\left(-1\right)=\left(-5\right).\left(-8\right)=\left(-8\right).\left(-5\right)\)
Thử từng TH
Ta có: \(C=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow C=\frac{1}{100}-\left(\frac{1}{99}-\frac{1}{100}\right)-\left(\frac{1}{98}-\frac{1}{99}\right)-...-\left(\frac{1}{2}-\frac{1}{3}\right)-\left(1-\frac{1}{2}\right)\)
\(\Rightarrow C=\frac{1}{100}-\frac{1}{99}+\frac{1}{100}-\frac{1}{98}+\frac{1}{99}-...-\frac{1}{2}+\frac{1}{3}-1+\frac{1}{2}\)
\(\Rightarrow C=\frac{1}{100}+\frac{1}{100}-1\)
\(\Rightarrow C=\frac{2}{100}-\frac{100}{100}\)
\(\Rightarrow C=-\frac{88}{100}=-\frac{22}{25}\)
Vậy \(C=-\frac{22}{25}\)
Chuk bạn hok tốt! 
\(\frac{1}{100.99}-\frac{1}{99.98}-...\frac{1}{2.1}\)
\(=\frac{1}{100.99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...\frac{1}{2.1}\right)\)
\(=\frac{1}{100.99}-\left(\frac{1}{99}-1\right)\)
\(=\frac{1}{9900}-\frac{-98}{99}=\frac{1}{9900}+\frac{98}{99}=\frac{1}{9900}+\frac{9800}{9900}=\frac{9800}{9900}\)
\(\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{2.1}\)
\(\frac{1}{100-99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+..+\frac{1}{2.1}\right)\)
\(\frac{1}{100-99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{98.99}\right)\)
\(\frac{1}{100.99}-\left(\frac{1}{1}-\frac{1}{2}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(\frac{1}{100.99}-\left(\frac{1}{1}-\frac{1}{99}\right)\)
\(\frac{1}{99}-\frac{1}{100}-\frac{98}{99}\)
\(-\frac{97}{99}-\frac{1}{100}\)
\(-\frac{9799}{9900}\)
\(\frac{1}{100\cdot99}-\frac{1}{99\cdot98}-...-\frac{1}{2\cdot1}\)
\(=\frac{1}{100\cdot99}-\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+...+\frac{1}{98\cdot99}\right)\)
\(=\frac{1}{99\cdot100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}\right)\)
\(=\frac{1}{9900}-\frac{98}{99}\)
\(=\frac{-9799}{9900}\)
Bài 2: https://oml.vn/hoi-dap/detail/6465458369.html
Bài 3: https://hoidap247.com/cau-hoi/20162
Bài 1: https://hoidap247.com/cau-hoi/1009171
a, tai x = 5 va y =2
x^2y +5xy^2 = 5^2 . 2 + 5 . 5 . 2^2 = 150
mới học lớp 5 k biết bài lớp 6?