Bài 3:

1, Áp dụng t/c dtsbn:

\(\dfrac{x}{6}=\dfrac{y}{4}=\dfrac{z}{3}=\dfrac{z-x}{3-6}=\dfrac{-21}{-3}=7\\ \Rightarrow\left\{{}\begin{matrix}x=42\\y=28\\z=21\end{matrix}\right.\)

2, Áp dụng t/c dtsbn:

\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{z}{7}=\dfrac{2x+3y-z}{6+15-7}=\dfrac{-14}{14}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-3\\y=-5\\z=-7\end{matrix}\right.\)

Bài 4: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{x}{\dfrac{1}{2}}=\dfrac{y}{\dfrac{1}{3}}=\dfrac{z}{\dfrac{1}{4}}=\dfrac{x+y+z}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}}=\dfrac{130}{\dfrac{13}{12}}=120\)

Do đó: x=60; y=40; z=30

giúp mk bài 4 với ạ

100 Miss white has a small oval face

Task 2

1 Mr and Mrs Brown are Peter's parents

3 How many classroom does Julia's school have?

mấy câu còn lại mk làm r

Bài 1 :

\(a,6^3.6^7.6^5=6^{3+7+5}=6^{15}\)

\(b,17^9:17^5:17^2=17^{9-5-2}=17^2\)

\(c,=\left(3^3\right)^3.3^3=3^9.3^3=3^{9+3}=3^{12}\)

\(d,=\left(2^4\right)^3.\left(2^6\right)^5=2^{12}.2^{30}=2^{12+30}=2^{42}\)

Bài 2 :

\(a,11^{60}:11^{58}=11^{60-58}=11^2=121\)

\(b,8^{10}:8^5:8^4=8^{10-5-4}=8^1=8\)

\(c,=\left(5^2\right)^9:\left(5^3\right)^5=5^{18}:5^{15}=5^{18-15}=5^3=125\)

\(d,=\left(2^4\right)^5:\left(2^2\right)^6:\left(2^3\right)^2=2^{20}:2^{12}:2^6=2^{20-12-6}=2^2=4\)

\(e,=10^5.\left(10^2\right)^5.\left(10^3\right)^2=10^5.10^{10}.10^6=10^{5+10+6}=10^{21}\)

Bài 3:

a)\(58.75+58.50-58.25\)

=\(58.\left(75+50-25\right)\)

=\(58.100\)

=\(5800\)

b)\(27.39+27.63-2.27\)

=\(27.\left(39+63-2\right)\)

=\(27.100\)

=\(2700\)

c)\(156.25+5.156+156.14+36.156\)

=\(156.\left(25+5+14+36\right)\)

=\(156.80\)
=\(12480\)

d)\(12.35+35.182-35.94\)

=\(35.\left(12+182-94\right)\)

=\(35.100\)

=\(3500\)

e)\(48.19+48.115+67.104\)

=\(48.\left(19+115\right)+67.104\)

=\(48.134+67.104\)

=\(48.67+48.67+67.104\)

=\(67.\left(48+48+104\right)\)

=\(67.200\)

=\(13400\)

f)\(128.72+128.67+128.72+11.72\)

=\(128.\left(72+67\right)+72.\left(128+11\right)\)

=\(128.139+72.139\)

=\(139.\left(72+128\right)\)

=\(139.200\)

=\(27800\)

\(a.=x\)

\(b.=y^3\)

\(c.=3xy\)

\(d.=-\frac{5}{2}a\)

\(e.=3yz\)

\(f.=-3xy\)

làm cả bước giải ra đc ko ạ?

Ví dụ 1 cái nhé: 



\(x^2yz:xyz=x^2:x=x^{\left(2-1\right)}=x^1=x\)

Còn \(yz:yz=1\)

\(\Rightarrow x^2yz:xyz=x\)

\(2,\\ a,=2x^2+4x-3x-6-2x^2-4x-2=-3x-8\\ b,=\left[x-2+2\left(x+1\right)\right]^2=\left(x-2+2x+2\right)^2=9x^2\)