\(\frac{120^3}{40^3}\)
Tính:
a)\(\frac{120^3}{40^3}\)
b)\(\frac{3^2}{\left(0.375\right)^2}\)
a) \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)
b) \(\frac{3^2}{0,375^2}=\left(\frac{3}{0,375}\right)^2=8^2=64\)
HỌC TỐT
Toán lớp 7???
\(a.\)\(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27\)
\(b.\)\(\frac{3^2}{\left(0,375\right)^2}=\left(\frac{3}{0,375}\right)^2=8^2=64\)
~ Rất vui vì giúp đc bn ~
tính
\(\frac{120^3}{40^3}\)
\(\frac{390^4}{130^4}\)
\(\frac{3^2}{\left(0,375\right)^2}\)
(x + 10) x 3 = 120
x + 10 = 120 : 3
x + 10 = 40
x = 40 - 10
x = 30
\(1.\)Tính
\(a)\frac{120^3}{40^3};\frac{390^4}{130^4};\frac{3^2}{0,375^2}\)
\(2.\)Tính giá trị các biểu thức sau :
\(a)\frac{45^{10}.5^{20}}{75^{15}}\) ; \(b)\frac{\left(0,8\right)^5}{\left(0,4\right)^6}\) ; \(c)\frac{2^{15}.9^4}{6^6.8^3}\)
1. sai dấu nhé
2.a, \(\frac{45^{10}.5^{20}}{75^{15}}=\frac{\left(3^2.5\right)^{10}.5^{20}}{\left(5^2.3\right)^{15}}=\frac{3^{20}.5^{30}}{5^{30}.3^{15}}=3^5=243\)
b, \(\frac{\left(0,8\right)^5}{\left(0,4\right)^6}=\frac{\left(\frac{4}{5}\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\cdot2\right)^5}{\left(\frac{2}{5}\right)^6}=\frac{\left(\frac{2}{5}\right)^5\cdot2^5}{\left(\frac{2}{5}\right)^5\cdot\frac{2}{5}}=2^5\div\frac{2}{5}=32\cdot\frac{5}{2}=80\)
c, \(\frac{2^{15}.9^4}{6^6.8^3}=\frac{2^{15}.3^8}{2^6.3^6.2^9}=\frac{2^{15}.3^2}{2^{15}}=3^2=9\)
Tính giá trị của biểu thức \(M=\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{121\sqrt{120}+120\sqrt{121}}\)
Với \(k\in N;k\ne0\) ta có :
\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{\left(k+1\right)}}=\frac{1}{\sqrt{k\left(k+1\right)}\left(\sqrt{k}+\sqrt{k+1}\right)}\)
\(=\frac{\sqrt{k+1}+\sqrt{k}}{\sqrt{k\left(k+1\right)}\left(\sqrt{k+1}-\sqrt{k}\right)\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{k\left(k+1\right)}}\)
\(=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k+1}}\)
Áp dụng ta có :
\(M=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+....+\frac{1}{\sqrt{120}}-\frac{1}{\sqrt{121}}=1-\frac{1}{11}=\frac{10}{11}\)
Tính:
a) \(\frac{120}{40^3}^3\); b) \(\frac{390}{130^4}^4\); c) \(\frac{3}{\left(0,375\right)^2}^2\)
\(\frac{120^3}{40^3}\)= (120:40)3 = 33 = 27
\(\frac{390^4}{130^4}\)=(390 : 130)4=34 = 81
\(\frac{3^2}{\left(0,375\right)^2}\)= (3:0,375)2 = 82 =64
a,\(\frac{120^3}{40^3}=\frac{120}{40}=3\)
b,\(\frac{390^4}{130^4}=\frac{390}{130}=3\)
c,\(\frac{3^2}{\left(0,375\right)^2}=\frac{3}{0,375}=8\)
198 - 3( x + 40 ) = 120
\(198-3\left(x+40\right)=120\\ \Rightarrow3\left(x+40\right)=78\\ \Rightarrow x+40=26\\ \Rightarrow x=-14\)
\(PT\Leftrightarrow3x+120=78\Leftrightarrow3x=-42\Leftrightarrow x=-14\)
Vậy: ...
Tìm x, biết:
a) x - 120: 30 = 40;
b) (x + 120) : 20 = 8;
c) (x + 5). 3 = 300;
d) x.2 + 21 : 3= 27
a) x = 44.
b) x = 40.
c) x = 95.
d) x = 10.
a) x - 120: 30 = 40
x -40 =40
x =40+40
x =80
b) (x + 120) : 20 = 8
(x+ 120) = 8x20
x+120 =160
x = 160-120
x = 40
c) (x + 5). 3 = 300
x+5=300:3
x+5=100
x=100-5
x=95
d) x.2 + 21 : 3= 27
x.2 +7=27
x.2 = 27-7
x.2= 20
x=20:2
x=10
Tính
a,\(\frac{2^{10}\cdot55+2^{10}\cdot26}{2^8\cdot27}\)
b,120:{300:[150-(2.53-23.25)]}
c,\(\left[\left(\frac{40}{130}-\frac{12}{13}\right)\cdot40\%+0,15\right]:\frac{-5}{52}\)
d,\(\frac{0,8:\left(\frac{4}{5}\cdot1,25\right)}{0,64-\frac{1}{25}}+\frac{\left(1,08-\frac{2}{25}\right):\frac{4}{7}}{\left(6\frac{5}{9}-3\frac{1}{4}\right)\cdot2\frac{2}{17}}+\left(1,2+0,5\right):\frac{1}{5}\)