Tính :
B = \(\dfrac{3737.43-4343.37}{2+4+6+....+100}\)
\(\dfrac{3737.43-4343.37}{2+4+6+...+100}\)
`(3737.43-4343.37)/(2+4+...+100)`
`= (101.37.43-101.43.37)/(2+4+...+100)`
`=0/(2+4+...+100) = 0`
\(\dfrac{3737.43-4343.37}{2+4+6+...+100}\)
Gọi \(A=2+4+6+...+100\)
Số số hạng: \(\left(100-2\right):2+1=50\)
Tổng dạy: \(\left(2+100\right).50:2=2550\)
Gọi \(B=3737.43-4343.37\)
\(B=101.37.43-101.43.37=0\)
\(\Rightarrow\dfrac{3737.43-4343.37}{2+4+6+...+100}=\dfrac{0}{2550}=0\)
\(A =\)\(\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
\(B=\) \(\dfrac{3737.43-4343.37}{2+4+6+...+100}\)
Làm cách lớp 6 thôi ah
\(A=\dfrac{101\cdot\dfrac{102}{2}}{\left(101-100\right)+99-98+...+3-2+1}\)
\(=\dfrac{101\cdot51}{1+1+...+1}=\dfrac{101\cdot51}{51}=101\)
\(B=\dfrac{37\cdot43\left(101-101\right)}{2+4+...+100}=0\)
a, \(A=\dfrac{101+100+99+98+...+3+2+1}{101-100+99-98+...+3-2+1}\)
Ta có: \(T=101+100+99+98+...+3+2+1\) \(=\dfrac{\left(101+1\right).101}{2}\)
\(=\dfrac{102.101}{2}\Leftrightarrow51.101\)
\(M=101-100+99-98+...+3-2+1\)
Ta có: \(101:2=50\) (dư \(1\))
\(\Rightarrow M=\left(101-100\right)+\left(99-98\right)+...+\left(3-2\right)+1\)
Có \(50\) dấu ngoặc tròn "\(\left(\right)\)"
\(\Rightarrow M=1+1+...+1+1=51.1=51\)
\(M\) có \(51\) số \(1\)
\(\Rightarrow A=\dfrac{T}{M}=\dfrac{51.101}{51}=101\)
Vậy \(A=101\)
b, \(B=\dfrac{3737.43-4343.37}{2+4+6+...100}\)
Ta có: \(T=3737.43-4343.37\)
\(T=37.101.43-43.101.37\)
\(T=0\)
\(\Rightarrow\) \(B=\dfrac{T}{2+4+6+...+100}=\dfrac{0}{2+4+6+...+100}\) \(=0\)
Vậy \(B=0\)
tính nhanh ;\(\frac{3737.43-4343.37}{2+4+6+......+100}\)
ta có
3737 . 43 - 4343 . 37 = ( 3700 + 37) . 43 - (4300+43) . 37
= 3700 . 43 + 37 .43 - 4300 . 37 + 43 . 37
= 43 . 37 . ( 3700 . 43 - 4300 . 37)
= 1591 . (159100 - 159100)
= 1591 . 0
= 0
vậy \(\frac{3737.43-4343.37}{2+4+6+....+100}\)= 0
đap số 0
1 nha bạn
Chúc các bạn học giỏi
Tết vui vẻ nha
tính
a. A= 101+100+99+98+....+3+2+1/101-100+99-98+...+3-2+1
b,B= 3737.43-4343.37/2+4+6+...+100
tính D=\(\frac{3737.43-4343.37}{2+4+6+...+100}\)
giúp mk nhé
\(D=\frac{3737.43-4343.37}{2+4+6+...+100}\)
\(\Rightarrow D=\frac{37.101.43-4343.37}{2+4+6+...+100}\)
\(\Rightarrow D=\frac{37.4343-4343.37}{2+4+6+...+100}\)
\(\Rightarrow D=\frac{4343.\left(37-37\right)}{2+4+6+...+100}\)
\(\Rightarrow D=\frac{4343.0}{2+4+6+...+100}\)
\(\Rightarrow D=\frac{0}{2+4+6+...+100}\)
\(\Rightarrow D=0\)
Vậy D = 0
\(D=\frac{3737.43-4343.37}{2+4+6+...+100}\)
Biến đổi tử số ta có :
\(D_1=3737.43-4343.37\)
\(\Rightarrow D_1=37.101.43-4343.37\)
\(\Rightarrow D_1=\left(43.101\right).37-4343.37\)
\(\Rightarrow D_1=4343.37-4343.37\)
\(\Rightarrow D_1=4343\left(37-37\right)=4343.0=0\)
Thay vào , ta có :
\(\frac{0}{2+4+6+...+100}=0\div\left(2+4+6+...+100\right)=0\)
Vậy : D = 0
\(b=\frac{3737.43-4343.37}{2+4+6+...+100}\)= ?
\(b=\frac{3737\cdot43-4343\cdot37}{2+4+6+...+100}=\frac{37\cdot101\cdot43-43\cdot101\cdot37}{2+4+6+...+100}=\frac{0}{2+4+6+...+100}=0\)
\(B=\frac{3737.43-4343.37}{2+4+6+...+100}\)
\(B=\frac{3737\cdot43-4343\cdot37}{2+4+6+...+100}=\frac{37\cdot101\cdot43-43\cdot101\cdot37}{2+4+6+...+100}=\frac{0}{2+4+6+...+100}=0\)
Tính \(D=\frac{3737.43-4343.37}{2+4+6+...+100}\)
\(D=\frac{37.101.43-43.101.37}{2+4+6+..+100}=\frac{0}{2+4+...+100}=0\)
\(\frac{3737.43-4343.37}{2+4+6+...+100}\)?
B=\(\frac{3737.43-4343.37}{2+4+6+...+100}\)=\(\frac{101.37.43-101.43.37}{2+4+6+...+100}\)=\(\frac{101\left(37.43-43.37\right)}{2+4+6+...+100}\)=\(\frac{0}{2+4+6+...+100}\)=0