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\(75-\left(3\cdot5^2-4\cdot2^3\right)\)
\(=75-\left(75-24\right)\)
\(=24\)
\(72:\left[32+\left(100-2^2\cdot17\right):8\right]\)
\(=72:\left[32+32:8\right]\)
\(=72:36\)
\(=2\)
\(75-\left(3.5^2-4.2^3\right)\)
\(=75-\left(3.25-4.8\right)\)
\(=75-75+32\)
\(=32\)
\(72\div\left[32+\left(100-2^2.17\right)\div8\right]\)
\(=72\div\left[32+\left(100-4.17\right)\div8\right]\)
\(=72\div\left[32+\left(100-68\right)\div8\right]\)
\(=72\div\left[32+32\div8\right]=72\div\left[32+4\right]=72\div36=2\)
Bài 2:
a)105-(x+7)=27:25
=>105-(x+7)=22
=>x+7=105-4
=>x+7=101
=>x=94
b. (2x -8) . 2 = 24
=>4x-16=16
=>4x=32
=>x=8
bài 2:
a) \(105-\left(x+7\right)=2^7:2^5\)
\(105-\left(x+7\right)=2^2\)
\(x+7=105-4\)
\(x+7=101\)
\(x=101-7\)
\(x=94\)
b) \(\left(2x-8\right).2=2^4\)
\(\left(2x-8\right).2=16\)
\(2x-8=16:2\)
\(2x-8=8\)
\(2x=8+8\)
\(2x=16\)
\(x=8\)
bài 1:
a) \(24:\left\{390:\left[500-\left(160+30.7\right)\right]\right\}\)
\(=24:\left\{390:\left[500-\left(160+210\right)\right]\right\}\)
\(=24:\left\{390:\left[500-370\right]\right\}\)
\(=24:\left\{390:130\right\}\)
\(=24:3=8\)
b) \(120-\left[98-\left(16-9\right)^2\right]\)
\(=120-\left[98-\left(256-81\right)\right]\)
\(=120-\left[98-175\right]\)
\(=120-\left(-77\right)\)
\(=120+77=197\)
bài 3 từ từ suy nghĩ hãng like nha
Giả sử: (3n+2;5n+3)=d
->(3n+2)chc d =>5(3n+2)chc d=>(15n+10)chc d
->(5n+3)chc d =>3(5n+3)chc d=>(15n+9)chc d
=>1 chc d
=>d=1
Vậy hai số đó nguyên tố cùng nhau
Bài 1 :
\(a)\) \(A=78.31+78.24+17.78+22.72\)
\(A=78\left(31+24+17\right)+22.72\)
\(A=78.72+22.72\)
\(A=72\left(78+22\right)\)
\(A=72.100\)
\(A=7200\)
\(b)\) \(B=3^4.109-3^6+\left(1+2+3+...+2018\right)\left(199199.198-198198-199\right)\)
\(B=3^4\left(109-3^2\right)+\left(1+2+3+...2018\right)\left(199.1000.198-198.1000.199\right)\)
\(B=81.100+\left(1+2+3+...+2018\right).0\)
\(B=8100\)
\(c)\) \(C=1.2+2.3+3.4+...+99.100\)
\(3C=1.2.3+2.3.3+3.4.3+...+99.100.3\)
\(3C=1.2.\left(3-0\right)+2.3.\left(4-1\right)+3.4.\left(5-2\right)+...+99.100.\left(101-98\right)\)
\(3C=\left(1.2.3+2.3.4+3.4.5+...+99.100.101\right)-\left(0.1.2+1.2.3.+2.3.4+...+98.99.100\right)\)
\(3C=99.100.101-0.1.2\)
\(3C=999900\)
\(C=\frac{999900}{3}\)
\(C=333300\)
Chúc bạn học tốt ~