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15^3+5.15^2-5^3 / 18^3+6.18^2-6^3
=3375+5.225-125 /5832+6.324-216
=3375+1125-125 /5832+1944-216
=4500-125 /7826-216
=4375 /7610
=875 /1522
mk ko bt viết phần nên viết phần bằng dấu gạch / nhé.mk cx ko bt đúng hay sai đâu nhé
\(15^3+5.15^2-5^3.\frac{1}{8}^3+6.18^2-6^3\)
\(=15^2.\left(15+5\right)-\left(5.\frac{1}{8}\right)^3+6.\left(18^2-6^2\right)\)
\(=225.20-\left(\frac{5}{8}\right)^3+6.\left(324-36\right)\)
\(=4500-\frac{125}{512}+6.288\)
\(=\left(4500+1728\right)-\frac{125}{512}\)
\(=6228-\frac{125}{512}\)
\(=...........\)
sorry mk chỉ giải đk câu a thui nhé
(2.5)^2= =100 ; (6.5)^3 =27000
k cho mk vs dù sao thì mk cũng giúp đk bn mà ^ ^
\(\dfrac{15^3+5.15^2-5^3}{18^3+6.18^2-6^3}\)
= \(\dfrac{\left(5.3\right)^3+5.\left(5.3\right)^2-5^3}{\left(6.3\right)^3+6.\left(6.3\right)^2-6^3}\)
= \(\dfrac{5^3.3^3+5^3.3^2-5^3}{6^3.3^3+6^3.3^2-6^3}\)
= \(\dfrac{5^3.\left(3^3+3^2-1\right)}{6^3.\left(3^3+3^2-1\right)}\)
=\(\dfrac{5^3}{6^3}=\dfrac{125}{216}\)
Tính nhanh \(A=\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}\)
\(A=\frac{25^3.5^5}{6.5^{10}}=\frac{\left(5^2\right)^3.5^5}{6.5^{10}}=\frac{5^6.5^5}{6.5^{10}}=\frac{5^{11}}{6.5^{10}}=\frac{5}{6}\)
\(B=\frac{2^5.6^3}{8^2.9^2}=\frac{2^5.2^3.3^3}{\left(2^3\right)^2.\left(3^2\right)^2}=\frac{2^8.3^3}{2^6.3^4}=\frac{4}{3}\)
Mình mới lớp 6 nên chỉ làm được câu thôi.
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}=\frac{3.\left\{\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right\}}{11.\left\{\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right\}}\)
\(=\frac{3}{11}\)
cho mình nha các bạn.
Ta có :
\(A=\frac{1}{3}+\frac{2}{3^2}+......+\frac{100}{3^{100}}\) \(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+.....+\frac{100}{3^{99}}\)
\(\Rightarrow3A-A=1+\frac{1}{3}+\frac{1}{3^2}+.....+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)= 2A
Đặt \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\) \(\Rightarrow3B=3+1+\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{98}}\)
\(\Rightarrow3B-B=3-\frac{1}{3^{99}}=2B\) \(\Rightarrow B=\frac{3}{2}-\frac{1}{3^{99}.2}\)
\(\Rightarrow2A=\frac{3}{2}-\frac{1}{3^{99}.2}-\frac{100}{3^{100}}\)\(\Rightarrow A=\frac{3}{4}-\frac{1}{3^{99}.4}-\frac{100}{3^{100}}< \frac{3}{4}\Rightarrow\left(đpcm\right)\)
Ta có :
\(C=1+3+3^2+....+3^{100}\) \(\Rightarrow C-1=3+3^2+....+3^{100}\)
\(\Rightarrow3\left(C-1\right)=3^2+3^3+.....+3^{101}\)\(\Rightarrow3C-3-\left(C-1\right)=3^{101}-3\)
\(\Rightarrow2C-2=3^{101}-3\Rightarrow2C=3^{101}-1\)\(\Rightarrow C=\frac{3^{101}-1}{2}\)
Ta có :
\(D=2^{100}-2^{99}+2^{98}-.....-2\) \(\Rightarrow2D=2^{101}-2^{100}+2^{99}-.....-2^2\)
\(\Rightarrow2D+D=2^{101}-2=3D\) \(\Rightarrow D=\frac{2^{101}-2}{3}\)
Gọi giá trị trên là : A
\(A=3^{100}-3^{99}+3^{98}+....+3^2-3+1\)
\(\Rightarrow3A=3^{101}-3^{100}+3^{99}-3^{98}+......+3^3-3^2+3\)
\(\Rightarrow3A+A=3^{101}+1\)
\(\Rightarrow4A=3^{101}+1\Rightarrow A=\frac{3^{101}+1}{4}\)
33×53+5×32×52-53/ 63+33+ 6×62×32-63=53×(33+32- 1)/63×(33+32-1)=53/63=125/216