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a) \(125\cdot\left(-24\right)+24\cdot225\)
\(=\left(225-125\right)\cdot24\)
\(=100\cdot24\)
\(=2400\)
b) \(26\cdot\left(-125\right)-125\cdot\left(-36\right)\)
\(=\left(36-26\right)\cdot125\)
\(=10\cdot125\)
\(=1250\)
Bài 1:
\(a.\left(-356+57\right)-\left(27-356\right)=-356+57-27+356=\left(-356+356\right)+\left(57-27\right)=30\) \(b.125.\left(-24+24.225\right)=125.\left(-24+5400\right)=125.\left(-24\right)+125.5400=-3000+675000=672000\)
\(c.26.\left(-125\right)-125.\left(-36\right)=-125.\left(26-36\right)=-125.\left(-10\right)=1250\)
Bài 2:
\(a.\left(2x-4\right)^2=0\)
\(\Rightarrow2x-4=0\)
\(\Rightarrow2x=4\)
\(\Rightarrow x=2\)
\(b.\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\)
Để (x+5) chia hết cho (x+3) thì 2 phải chia hết cho (x+3)
\(\Rightarrow x+3\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)
\(x+3=1\Rightarrow x=-2\)
\(x+3=-1\Rightarrow x=-4\)
\(x+3=2\Rightarrow x=-1\)
\(x+3=-2\Rightarrow x=-5\)
Vậy \(x\in\left\{-2;-4;-1;-5\right\}\)
Bài 2:
a)\(\left(2x-4\right)^2=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\Leftrightarrow x=2\)
b)\(\frac{x+5}{x+3}=\frac{x+3+2}{x+3}=\frac{x+3}{x+3}+\frac{2}{x+3}=1+\frac{2}{x+3}\in Z\)
Suy ra \(2⋮x+3\Rightarrow x+3\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5\right\}\)
a) \(125^3:25^4=\left(5^3\right)^3:\left(5^2\right)^4=5^9:5^8=5\)
b) \(\left(10^3+10^4+125^2\right):5^3\)
\(=\left[\left(5.2\right)^3+\left(5.2\right)^4+\left(5^3\right)^2\right]:5^3\)
\(=\left(5^3.2^3+5^4.2^4+5^6\right):5^3\)
\(=\frac{5^3.2^3}{5^3}+\frac{5^4.2^4}{5^3}+\frac{5^6}{5^3}\)
\(=2^3+5.2^4+5^3=213\)
c) \(24^4:3^4-32^{12}:16^{12}\)
\(=\left(3.8\right)^4:3^4-\left(16.2\right)^{12}:16^{12}\)
\(=3^4.8^4:3^4-2^{12}.16^{12}:16^{12}\)
\(=8^4-2^{12}=\left(2^3\right)^4-2^{12}=2^{12}-2^{12}=0\)
a) \(3^{7} \cdot 27^{5} \cdot 81^{3}\)
- Đổi về cơ số 3: \(27 = 3^{3} , \textrm{ }\textrm{ } 81 = 3^{4}\).
\(3^{7} \cdot \left(\right. 3^{3} \left.\right)^{5} \cdot \left(\right. 3^{4} \left.\right)^{3} = 3^{7} \cdot 3^{15} \cdot 3^{12} = 3^{7 + 15 + 12} = 3^{34} .\)
b) \(100^{6} \cdot 1000^{5} \cdot 10 \textrm{ } 000^{3}\)
- Đổi về cơ số 10: \(100 = 10^{2} , \textrm{ }\textrm{ } 1000 = 10^{3} , \textrm{ }\textrm{ } 10 \textrm{ } 000 = 10^{4}\).
\(\left(\right. 10^{2} \left.\right)^{6} \cdot \left(\right. 10^{3} \left.\right)^{5} \cdot \left(\right. 10^{4} \left.\right)^{3} = 10^{12} \cdot 10^{15} \cdot 10^{12} = 10^{39} .\)
c) \(\frac{36^{5}}{18^{5}}\)
- Do cùng số mũ:
\(\left(\left(\right. \frac{36}{18} \left.\right)\right)^{5} = 2^{5} = 32.\)
d) \(24 \cdot 5^{2} + 5^{2} \cdot 5^{3}\)
\(= 24 \cdot 25 + 25 \cdot 125 = 600 + 3125 = 3725.\)
e) \(\frac{125^{4}}{5^{8}}\)
- Đổi về cơ số 5: \(125 = 5^{3}\).
\(\left(\right. 5^{3} \left.\right)^{4} : 5^{8} = 5^{12} : 5^{8} = 5^{12 - 8} = 5^{4} = 625.\)
TÓM lại là a) \(3^{34}\)
b) \(10^{39}\)
c) \(32\)
d) \(3725\)
e) \(625\)
bạn muốn chép đáp án hay sem cách làm///??
a: \(3^7\cdot27^5\cdot81^3\)
\(=3^7\cdot\left(3^3\right)^5\cdot\left(3^4\right)^3\)
\(=3^7\cdot3^{15}\cdot3^{12}=3^{7+15+12}=3^{34}\)
b: \(100^6\cdot1000^5\cdot10000^3\)
\(=\left(10^2\right)^6\cdot\left(10^3\right)^5\cdot\left(10^4\right)^3\)
\(=10^{12}\cdot10^{15}\cdot10^{12}=10^{15+12+12}=10^{39}\)
c: \(36^5:18^5=\left(\frac{36}{18}\right)^5=2^5=32\)
d: \(24\cdot5^2+5^2\cdot5^3\)
\(=5^2\left(24+5^3\right)\)
\(=25\cdot\left(24+125\right)=25\cdot149=3725\)
e: \(125^4:5^8=\left(5^3\right)^4:5^8=5^{12}:5^8=5^{12-8}=5^4\)
\(S_2=10+12+14+...+210=\frac{\left(210-10\right)}{2}\cdot\left(210+10\right):2=\frac{200\cdot220}{4}=100\cdot110=11000.\)
\(S_3=21+23+25+...+101=\frac{101-21}{2}\cdot\left(101+21\right):2=\frac{80\cdot122}{4}=40\cdot61=2440\)
v.v
Các bài sau bn cứ tính theo công thức: tổng dãy có quy luật=(số đầu - số cuối) : 2 x (số đầu + số cuối) :2
học tốt ^_^
125(−24)+24×225
= \(- 24 \times 125 + 24 \times 225\)
= \(24 \times \left(\right. 225 - 125 \left.\right)\)
= \(24\times100=2400\)
2400 nha bạn