Gọi số học sinh 4 khối là a, b, c, d. Ta có:

\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}\) và b- d=70

Áp dụng tính chất của dãy tỉ số bằng nhau, Ta có:

\(\dfrac{a}{9}=\dfrac{b}{8}=\dfrac{c}{7}=\dfrac{d}{6}=\dfrac{b-d}{8-6}=\dfrac{70}{2}=35\)

Khi đó:\(\dfrac{a}{9}=35\Rightarrow a=35.9=315\)

\(\dfrac{b}{8}=35\Rightarrow b=35.8=280\)

\(\dfrac{c}{7}=35\Rightarrow c=35.7=245\)

\(\dfrac{d}{6}=35\Rightarrow d=35.6=210\)

Vậy số học sinh 4 khối 6, 7, 8, 9 lần lượt là:315; 280; 245; 210(em)

a, \(10^9+10^8+10^7⋮222\)

Ta có:\(10^9+10^8+10^7=10^7.\left(10^2+10+1\right)\)

\(=10^7.111=5^7.2^7.111=5^7.2^6.2.111=5^7.2^6.222\)

Vì 222\(⋮222\Rightarrow5^7.2^6.222⋮222\)

Vậy \(10^9+10^8+10^7⋮222\)

b) 81⁷ - 27⁹ - 9¹³ 45

\(\)Ta có: \(81^7-27^9-9^{13}=\left(3^4\right)^7-\left(3^3\right)^9-\left(3^2\right)^{13}\)

\(=3^{28}-3^{27}-3^{26}=3^{26}.\left(3^2-3-1\right)\)

\(=3^{26}.5=3^{24}.3^2.5=3^{24}.45\)

Vì \(45⋮45\Rightarrow3^{24}.45⋮45\)

Vậy \(81^7-27^9-9^{13}⋮45\)

CHÚC BẠN HỌC TỐT!!

Gọi là ab , ab8 đc

ab+ab8=514

Nên b+8=14

Nên b=6

a+1+b=11 nên a=4(Số 1 là số nhớ)

Vậy số bé là 46 , lớn là 468

a, \(\left(\dfrac{1}{3}x\right):\dfrac{2}{3}=\dfrac{13}{4}:\dfrac{2}{5}\)

\(\left(\dfrac{1}{3}x\right).\dfrac{3}{2}=\dfrac{13}{4}.\dfrac{5}{2}\)

\(\dfrac{1}{2}x=\dfrac{65}{8}\Rightarrow x=\dfrac{65}{8}:\dfrac{1}{2}\Rightarrow x=\dfrac{65}{4}\)

b, \(4,5:0,3=2,25:\left(0,1x\right)\)

\(\Rightarrow2,5:\left(0,1x\right)=15\)

\(\Rightarrow0,1x=\dfrac{3}{20}\Rightarrow x=\dfrac{3}{2}\)

c,\(8:\left(\dfrac{1}{4}x\right)=2:0,02\)

\(8:\left(\dfrac{1}{4}x\right)=100\) \(\Rightarrow\dfrac{1}{4}x=\dfrac{8}{100}\Rightarrow x=\dfrac{8}{100}:\dfrac{1}{4}=0,32\)

CHÚC BẠN HỌC TỐT!!!!

Tởm vãi ***** Chi ak

Tìm x trong các tỉ lệ thức sau:

a)

b)

c) 0,01: 2,5 = (0,75x): 0,75

d)

Giải

b)

c)

d)

Hê hê, hơi dài

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)

a, Ta có: \(\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk+b\right)^2}{\left(dk+d\right)^2}\)

\(\Rightarrow\dfrac{b^2.k}{d^2.k}=\dfrac{\left[b.\left(k+1\right)\right]^2}{\left[d.\left(k+1\right)\right]^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{\left(a+b\right)^2}{\left(c+d\right)^2}\)

b, Ta có:\(\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{bk.b}{dk.d}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}\)

\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.k^2+b^2}{d^2.k^2+d^2}\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2.\left(k^2+1\right)}{d^2.\left(k^2+1\right)}\)

\(\Rightarrow\dfrac{b^2}{d^2}=\dfrac{b^2}{d^2}\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2+b^2}{c^2+d^2}\)

CHÚC BẠN HỌC TỐT!!

Ta có:

\(A=\left(2^1+2^2+2^3+....+2^{2016}+2^{2017}\right)-2^{2018}\)

\(\Rightarrow2A-A=\left[\left(2^2+2^3+2^4+...+2^{2017}+2^{2018}-2^{2019}\right)\right]-\left[\left(2+2^2+2^3+...+2^{2016}+2^{2017}\right)-2^{2018}\right]\)

\(A=2+2.2^{2018}+2^{2019}=2+2.2^{2019}=2+2^{2020}\)