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b) 3^2 . [(5^2 - 3 ) : 11 ] - 2^4 + 2.10^3
= 9 . [(25 - 3 ) : 11 ] - 16 + 2.1000
= 9 . [22 : 11 ] - 16 + 2000
= 9 . 2 - 16 + 2000
= 18 - 16 + 2000
= 2 + 2000
= 2002
(72005 + 72004) : 72004
= 72005 : 72004 + 72004 : 72004
= 72005 - 2004 + 1
= 71 + 1
= 7 + 1
= 8
a) ( 3^5 . 3^7 ) : 3^10 + 5.2^4 - 7^3 : 7
= 3^10 : 3^10 + 80 - 7^2
= 1 + 80 - 49
= 32
1)\(79-5\left(11-x\right)=34\)
\(\Rightarrow79-55+5x=34\)
\(\Rightarrow24+5x=34\)
\(\Rightarrow5x=-10\)
\(\Rightarrow x=-2\)
Vậy \(x=-2\)
2)\(32+2\left(7-x\right)=40\)
\(\Rightarrow32+14-2x=40\)
\(\Rightarrow46-2x=40\)
\(\Rightarrow2x=6\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
3)\(\left(166-2x\right).8^9=2.8^{11}\)
\(\Rightarrow\left(83-x\right).2.8^9=2.8^{11}\)
\(\Rightarrow83-x=8^3\)
\(\Rightarrow83-x=512\)
\(\Rightarrow x=-429\)
Vậy \(x=-429\)
4)\(5^2.x-2^3.x=51\)
\(\Rightarrow x\left(5^2-2^3\right)=51\)
\(\Rightarrow x\left(25-8\right)=51\)
\(\Rightarrow17x=51\)
\(\Rightarrow x=3\)
Vậy \(x=3\)
5)\(3^x+4.3^x=5.3^7\)
\(\Rightarrow3^x\left(1+4\right)=5.3^7\)
\(\Rightarrow5.3^x=5.3^7\)
\(\Rightarrow3^x=3^7\)
\(\Rightarrow x=7\)
Vậy \(x=7\)
6)\(7.2^x-2^x=6.32\)
\(\Rightarrow2^x\left(7-1\right)=6.2^5\)
\(\Rightarrow6.2^x=6.2^5\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
7)\(15^{3-x}=225\)
\(\Rightarrow15^{3-x}=15^2\)
\(\Rightarrow3-x=2\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
8)\(4.5^x-3=97\)
\(\Rightarrow4.5^x=100\)
\(\Rightarrow5^x=25\)
\(\Rightarrow5^x=5^2\)
\(\Rightarrow x=2\)
Vậy \(x=2\)
9)\(171-3.2^x=123\)
\(\Rightarrow3.2^x=48\)
\(\Rightarrow2^x=16\)
\(\Rightarrow2^x=2^4\)
\(\Rightarrow x=4\)
Vậy \(x=4\)
10)\(180-4.x^5=32\)
\(\Rightarrow4.x^5=148\)
\(\Rightarrow x^5=37\)//Đề có lỗi không ???
a) 5^x=5^78:5^14(lấy 78-14)
5^x=5^64
=> x=64
b) 7^x.7^2=7^21
7^x=7^21:7^2
7^x=7^19
=> x=19
B1:
Ta có: \(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}-2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\frac{2^{12}.3^{10}-2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.\left(1-5\right)}{2^{11}.3^{11}.\left(2.3-1\right)}\)
\(=\frac{2.\left(-4\right)}{3.5}=-\frac{8}{15}\)
B2:
Ta có: \(1+3+5+...+x=1600\)
\(\Leftrightarrow\frac{\left(x+1\right)\cdot\left(\frac{x-1}{2}+1\right)}{2}=1600\)
\(\Leftrightarrow\left(x+1\right)\cdot\frac{x+1}{2}=3200\)
\(\Leftrightarrow\left(x+1\right)^2=6400\)
Xét theo dãy tăng tiến ta thấy được giá trị của x càng tăng
=> x dương => x + 1 dương
\(\Rightarrow x+1=80\)
\(\Rightarrow x=79\)
\(3,1+5^2+5^4+...+5^{26}\)
\(=\left(1+5^2\right)+\left(5^4+5^6\right)+...+\left(5^{24}+5^{26}\right)\)
\(=\left(1+5^2\right)+5^4\left(1+5^2\right)+...+5^{24}\left(1+5^2\right)\)
\(=26+5^4.26+...+5^{24}.26\)
\(=26\left(5^4+...+5^{24}\right)\)
Vì \(26⋮26\)
\(\Rightarrow26\left(5^4+...+5^{24}\right)⋮26\)
\(\Rightarrow1+5^2+5^4+...+5^{26}⋮26\)
\(4,1+2^2+2^4+...+2^{100}\)
\(=\left(1+2^2+2^4\right)+...+\left(2^{98}+2^{99}+2^{100}\right)\)
\(=\left(1+2^2+2^4\right)+....+2^{98}\left(1+2^2+2^4\right)\)
\(=21+2^6.21...+2^{98}.21\)
\(=21\left(2^6+...+2^{98}\right)\)
Có : \(21\left(2^6+...+2^{98}\right)⋮21\)
\(\Rightarrow1+2^2+2^4+...+2^{100}⋮21\)