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Câu 1:
\(A=\frac{2^5.7+2^5}{2^5.3-2^5}\)= \(\frac{2^5.8}{2^5.2}\)= 4
Vậy A = 4
Câu 2:
\(B=2^3.5^3-3.\left\{400-\left[673-2^3.\left(7^8:7^6+7^0\right)\right]\right\}\)
\(B=8.125-3.\left\{400-\left[673-8.\left(7^2+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.\left(49+1\right)\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-8.50\right]\right\}\)
\(B=1000-3.\left\{400-\left[673-400\right]\right\}\)
\(B=1000-3.\left\{400-273\right\}\)
\(B=1000-3.127\)
\(B=1000-381\)
\(B=619\)
Vậy B = 619
sao đề kì quá sao mik ko tính ra hay là mik tính sai nhỉ! mik tính được ra là 77 nhưng thuer lại thì thấy kì
\(2^3\cdot5^3-3\left\{400-\left[673-2^3(7^8:7^6+7^0)\right]\right\}\)
\(=8\cdot125-3\left\{400-\left[673-8(7^{8-6}+7^0)\right]\right\}\)
\(=8\cdot125-3\left\{400-\left[673-8(7^2+7^0)\right]\right\}\)
\(=8\cdot125-3\left\{400-\left[673-8(49+1)\right]\right\}\)
\(=8\cdot125-3\left\{400-\left[673-8\cdot50\right]\right\}\)
\(=8\cdot125-3\left\{400-\left[673-400\right]\right\}\)
\(=8\cdot125-3\left\{400-273\right\}\)
\(=8\cdot125-3\cdot127\)
\(=619\)
\(2^3.5^3-3\left\{400-\left[673-2^3\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=8.125-3\left\{400-\left[673-8\left(49+1\right)\right]\right\}\)
\(=1000-3\left\{400-\left[673-8.50\right]\right\}\)
\(=1000-3\left\{400-\left[673-400\right]\right\}\)
\(=1000-3\left\{400-273\right\}\)
\(=1000-3.\left(127\right)\)
\(=1000-381\)
\(=619\)
ta có
\(S_2=\left(1-3\right)+\left(5-7\right)+..+\left(1997-1999\right)+2001\)
ha y \(S_2=-2-2-2..+2001=-2.500+2001=1001\)
\(S_3=\left(1-2-3+4\right)+\left(5-6-7+8\right)+..+\left(1997-1998-1999+2002\right)\)
hay \(S_3=0+0+..+0=0\)
\(S_2=\left(1-3\right)+\left(5-7\right)+...+\left(1997-1999\right)+2001\)
\(=\left(-2\right)+\left(-2\right)+....+\left(-2\right)+2001=\left(-2\right).500+2001=-1000+2001=1001\)
\(S_3=\left(0+1-2-3\right)+\left(4+5-6-7\right)+...+\left(1996+1997-1998-1999\right)+2000\)
\(=-4+\left(-4\right)+...+\left(-4\right)+2000=\left(-4\right).500+2000=0\)
23 . 25 - 3 { 400 - [ 673 - 23 ( 78 : 76 +70)]}
= 28 - 3 { 400 - [ 673 - 8 ( 72 +1 )]}
= 256 - 3 { 400 - [ 673 - 8 . 50 ]}
= 256 - 3 { 400 - [ 673 - 400 ]}
= 256 - 3 { 400 - 273 }
= 256 - 3 . 127
= 256 - 381
= - 125
C=1+2+3-4+5+6+7-8+...+1999-2000+2001
C=1+2+(-1)+5+6+(-1)+...+1997+1998+(-1)+2001
C=1+1+5+5+...+1997+1997+2001
C=2001+(1.2+5.2+...+1997.2)
C=2001+2(1+5+...+1997)
Tong 1+5+...+1997={(1997+1)[(1997-1):4+1]}:2=499500
C=2001+2.499500
C=2001+999000
C=1001001
he he
\(b,2^3x5^3-3x\left\{400-\left[673-2^3x\left(7^8:7^6+7^0\right)\right]\right\}\)
\(=1000-3x\left\{400-\left[673-8x\left(7^2+1\right)\right]\right\}\)
\(=1000-3x\left\{400-\left[673-8x50\right]\right\}\)
\(=1000-3x\left\{400-\left[673-400\right]\right\}\)
\(=1000-3x\left\{400-273\right\}\)
\(=1000-3x127\)
\(=1000-381\)
\(=619\)
\(a,1968:16+5136:16-704:16\)
\(=\left(1968+5136-704\right):16\)
\(=6400:16\)
\(=400\)
1. Ta có: A = 2^1+ 2^2 +2^3+2^4+....2^10
A= ( 2^1 + 2^2) + ( 2^3+2^4) +....( 2^9+ 2^10)
A= 3.( 2^1+2^3+2^5+...+2^1005)
Do 3 \(⋮\)3 => A\(⋮\)3
Ta có: A =.....
A= Ghép 3 số lại
A= 7. (2^1+ 2^4+...+2^670)
Do 7 \(⋮\)7 => A \(⋮\)7
2;3;4 đều ghép 2 hoặc 3 số như tke và phần trog ngoặc cx y hệt như tke, ko thay đổi
Duyệt nhanh....