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a,113-3+20
=1328+20
=1348
b,300+7781+1425
=9506
c,7+36-40
=43-40=3
a,
115 : 112 - 33 : ( 11 + 23 ) + 22 x 30 x 5
= 11( 5 - 2 ) - 27 : ( 1 + 8 ) + 4 x 1 x 5
= 113 - 27 : 9 + 4 x 5
= 1331 - 3 + 20
= 1328 + 20
= 1348
b,
12 x 25 + 31 x 251 + 57x 25
= 12 x 25 + 57 x 25 + 31 x 251
= ( 12 + 57 ) x 25 + 31 x 251
= 69 x 25 + 31 x 251
= 1725 + 7781
= 9506
c,
15 - 23 + 4 x 32 - 5 x 8
= 15 - 8 + 4 x 9 - 40
= 15 - 8 + 36 - 40
= 7 + 36 - 40
= 43 - 40
= 3
Thực hiện phép tính
1. [ 36 . 4 - 4 . ( 82 - 7 . 11 )2] : 4 - 20190
= [ 36 . 4 - 4 . 25 ] : 4 - 1
= [ 4.(36 - 25)] : 4 - 1
= 44 : 4 - 1
= 11 - 1
= 10
2. 500 - { 5 . [ 409 - ( 23. 3 - 21 )3] - 1724 }
a/ =60+4-6^2
=64-36
=28
b/ = (9+8.5):7
=(9+40):7
= 49:7
= 7
c/ = 11^2 -243:9 - 60
= 121-27-60
= 34
Ta có : A = 1 + 2 + 3 + ... + 2008
\(A=\frac{\left(2008+1\right)\left[\left(2008-1\right)\div1+1\right]}{2}\)
\(A=\frac{2009.2008}{2}\)
\(A=2017036\)
Ta có: B = 1 + 2 + 3 + ... + 1010
\(B=\frac{\left(1010+1\right)\left[\left(1010-1\right):1+1\right]}{2}\)
\(B=\frac{1011.1010}{2}\)
\(B=510555\)
\(A=1+2+3+4+5+...+2008\)
\(A=\left(2008+1\right)\left(\left(2008-1\right):1+1\right):2=2009.2008:2\)
\(=2009.1004=2017036\)
\(B=1+2+3+4+...+1010\)
\(B=\left(1010+1\right)\left(\left(1010-1\right):1+1\right):2=1011.\left(1010:2\right)\)
\(=1011.505=510555\)
\(C=2+5+8+11+...+302\)
\(C=\left(302+2\right)\left(\left(302-2\right):3+1\right):2=304.101:2\)
\(=15352\)
\(D=3+3^2+3^3+3^4+...+3^{2019}\)
\(3D=3^2+3^3+3^4+...+3^{2020}\)
\(3D-D=\left(3^2+3^3+3^4+...+3^{2020}\right)-\left(3+3^2+3^3+3^4+...+3^{2019}\right)\)
\(2D=3^{2020}-3\)
\(\Rightarrow D=\frac{3^{2020}-3}{2}\)
\(E=4^{10}+4^{11}+4^{12}+...+4^{100}\)
\(4E=4^{11}+4^{12}+4^{13}+...+4^{101}\)
\(4E-E=\left(4^{11}+4^{12}+4^{13}+...+4^{101}\right)-\left(4^{10}+4^{11}+4^{12}+...+4^{100}\right)\)
\(3E=4^{101}-4^{10}\)
\(E=\frac{4^{101}-4^{10}}{3}\)
a, \(3^4\div3^2-\left[120-\left(2^6.2+5^2.2\right)\right]\)
\(=3^2-\left\{120-\text{[}2.\left(2^6+5^2\right)\text{]}\right\}\)
\(=3^2-\left(120-2\cdot89\right)\)
\(=9--58=9+58=67\)
1. \(a,3^4:3^2-\left[120-(2^6\cdot2+5^2\cdot2)\right]\)
\(=3^2-\left[120-\left\{(2^6+5^2)\cdot2\right\}\right]\)
\(=3^2-\left[120-\left\{(64+25)\cdot2\right\}\right]\)
\(=9-\left[120-89\cdot2\right]\)
\(=9-\left[120-178\right]=9-(-58)=67\)
b, Tương tự như bài a
2.a,\(4^x\cdot5+4^2\cdot2=2^3\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+16\cdot2=8\cdot7+56\)
\(\Leftrightarrow4^x\cdot5+32=56+56\)
\(\Leftrightarrow4^x\cdot5+32=112\)
\(\Leftrightarrow4^x\cdot5=80\)
\(\Leftrightarrow4^x=16\Leftrightarrow4^x=4^2\Leftrightarrow x=2\)
\(b,24:(2x-1)^3-2=1\)
\(\Leftrightarrow24:(2x-1)^3=3\)
\(\Leftrightarrow(2x-1)^3=8\)
\(\Leftrightarrow(2x-1)^3=2^3\)
\(\Leftrightarrow2x-1=2\)
Làm nốt là xong thôi
YẾN ƠI TUI BẠN BÀ NÈ
NHƯNG TUI HỔNG PHẢI LỚP 6 THÔNG CẢM NHA
a) 42 . 57 + 43 . 42 - 600
= 2394 + 1806 - 600
= 4200 - 600
= 3600.
b) 22 . 52 - 64 : 23
= 4 . 25 - 64 : 8
= 100 - 8
= 92
c) 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - 33 ]
= 120 - [ 100 - 27 ]
= 120 - 73
= 47.
1.Thực hiện các phép tính sau
a)42.57+43.42-600
b)22.52-64: 23
c) 120-[100-(5-2)3]
d)(102+112+122):(132+142)
a) 42 . 57 + 43 . 42 - 600
= 2394 + 1806 - 600
= 4200 - 600
= 3600.
b) 22 . 52 - 64 : 23
= 4 . 25 - 64 : 8
= 100 - 8
= 92
c) 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - ( 5 - 2 )3 ]
= 120 - [ 100 - 33 ]
= 120 - [ 100 - 27 ]
= 120 - 73
= 47.
a: \(\left(-2\right)^3-45:\left(-3\right)^2+\left(-2019\right)^0\cdot1^{2019}\)
\(=-8-45:9+1\)
\(=-8-5+1\)
=-13+1
=-12
b: \(11^{25}:11^{13}-3^5:\left(1^{10}+2^3\right)-60\)
\(=11^{25-13}-3^5:3^2-60\)
\(=11^{12}-27-60\)
\(=11^{12}-87\)
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