Supose all of the skewers contain \(6\)meatballs. 

Then the number of mearballs used is: 

\(6\times32=192\)(meatballs) 

In order to have \(150\)meatballs total, the number of skewers have only \(4\)meatballs is: 

\(\left(192-150\right)\div\left(6-4\right)=21\)(skewers) 

tư vấn à ? tui khuyên bồ nên mở  : " Trung tâm tư vấn tình yêu quả sung " 

mk ko co nhung ban noi xem (gui tin nhan cung dc)

Bài 1:

a)

\(\dfrac{x-1}{9}=\dfrac{8}{3}\\ \Leftrightarrow\dfrac{x-1}{9}=\dfrac{24}{9}\\ \Leftrightarrow x-1=24\\ x=24+1\\ x=25\)

b)

\(\left(\dfrac{3x}{7}+1\right):\left(-4\right)=\dfrac{-1}{8}\\ \dfrac{3x}{7}+1=\dfrac{-1}{8}\cdot\left(-4\right)\\ \dfrac{3x}{7}+1=\dfrac{1}{2}\\ \dfrac{3x}{7}=\dfrac{1}{2}-1\\ \dfrac{3x}{7}=\dfrac{-1}{2}\\ 3x=\dfrac{-1}{2}\cdot7\\ 3x=\dfrac{-7}{2}\\ x=\dfrac{-7}{2}:3\\ x=\dfrac{-7}{6}\)

c)

\(x+\dfrac{7}{12}=\dfrac{17}{18}-\dfrac{1}{9}\\ x+\dfrac{7}{12}=\dfrac{5}{6}\\ x=\dfrac{5}{6}-\dfrac{7}{12}\\ x=\dfrac{1}{4}\)

d)

\(0,5x-\dfrac{2}{3}x=\dfrac{7}{12}\\ \dfrac{1}{2}x-\dfrac{2}{3}x=\dfrac{7}{12}\\ x\cdot\left(\dfrac{1}{2}-\dfrac{2}{3}\right)=\dfrac{7}{12}\\ \dfrac{-1}{6}x=\dfrac{7}{12}\\ x=\dfrac{7}{12}:\dfrac{-1}{6}\\ x=\dfrac{-7}{2}\)

e)

\(\dfrac{29}{30}-\left(\dfrac{13}{23}+x\right)=\dfrac{7}{46}\\ \dfrac{29}{30}-\dfrac{13}{23}-x=\dfrac{7}{46}\\ \dfrac{277}{690}-x=\dfrac{7}{46}\\ x=\dfrac{277}{690}-\dfrac{7}{46}\\ x=\dfrac{86}{345}\)

f)

\(\left(x+\dfrac{1}{4}-\dfrac{1}{3}\right):\left(2+\dfrac{1}{6}-\dfrac{1}{4}\right)=\dfrac{7}{46}\\ \left(x-\dfrac{1}{12}\right):\dfrac{23}{12}=\dfrac{7}{46}\\ x-\dfrac{1}{12}=\dfrac{7}{46}\cdot\dfrac{23}{12}\\ x-\dfrac{1}{12}=\dfrac{7}{24}\\ x=\dfrac{7}{24}+\dfrac{1}{12}\\ x=\dfrac{3}{8}\)

g)

\(\dfrac{13}{15}-\left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{13}{15}-\dfrac{7}{10}\\ \left(\dfrac{13}{21}+x\right)\cdot\dfrac{7}{12}=\dfrac{1}{6}\\ \dfrac{13}{21}+x=\dfrac{1}{6}:\dfrac{7}{12}\\ \dfrac{13}{21}+x=\dfrac{2}{7}\\ x=\dfrac{2}{7}-\dfrac{13}{21}\\ x=\dfrac{-1}{3}\)

h)

\(2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|-\dfrac{3}{2}=\dfrac{1}{4}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{1}{4}+\dfrac{3}{2}\\ 2\cdot\left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{4}:2\\ \left|\dfrac{1}{2}x-\dfrac{1}{3}\right|=\dfrac{7}{8}\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\\dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\end{matrix}\right.\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{7}{8}\\ \dfrac{1}{2}x=\dfrac{7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{29}{24}\\ x=\dfrac{29}{24}:\dfrac{1}{2}\\ x=\dfrac{29}{12}\\ \dfrac{1}{2}x-\dfrac{1}{3}=\dfrac{-7}{8}\\ \dfrac{1}{2}x=\dfrac{-7}{8}+\dfrac{1}{3}\\ \dfrac{1}{2}x=\dfrac{-13}{24}\\ x=\dfrac{-13}{24}:\dfrac{1}{2}\\ x=\dfrac{-13}{12}\)

i)

\(3\cdot\left(3x-\dfrac{1}{2}\right)^3+\dfrac{1}{9}=0\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=0-\dfrac{1}{9}\\ 3\cdot\left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{9}:3\\ \left(3x-\dfrac{1}{2}\right)^3=\dfrac{-1}{27}\\ \left(3x-\dfrac{1}{2}\right)^3=\left(\dfrac{-1}{3}\right)^3\\ \Leftrightarrow3x-\dfrac{1}{2}=\dfrac{-1}{3}\\ 3x=\dfrac{-1}{3}+\dfrac{1}{2}\\ 3x=\dfrac{1}{6}\\ x=\dfrac{1}{6}:3\\ x=\dfrac{1}{18}\)

a) 8²⁰ và 7²⁰

\(\Rightarrow\) 8²⁰ > 7²⁰

b) 4²⁰ và 16¹⁰

4²⁰

16¹⁰ = (4²)¹⁰ = 4²⁰

\(\Rightarrow\) 4²⁰ = 16¹⁰

c) 27⁷ và 81⁵

27⁷ = (3³)⁷ = 3²¹

81⁵ = (3⁴)⁵ = 3²⁰

\(\Rightarrow\) 27⁷ > 81⁵

Đ:a,b,c,d,f,g,h,i,l,m,n,o,p,q

S:e,j,k,r

tks bạn

a) 8²⁰ và 7²⁰

vì 8>7 nên 8²⁰>7²⁰

b) 4²⁰ và 16²⁰

vì 4<16 nên 4²⁰<16²⁰

c) 27⁷= (3³)⁷= 3²¹

81⁵=( 3⁴)⁵=3²⁰

vì 21>20 nên 3²¹>3²⁰ hay 27⁷> 81⁵

e) 5²¹= 5²⁰ . 5

vì 5²⁰ . 5>5²⁰ . 4 nên 5²¹> 4 . 5²⁰

Bài 1 :

a,8²⁰ > 7²⁰

b, 4²⁰ = 16¹⁰

c, 27⁷ > 81⁵

d , 5⁵⁴ > 3⁸¹

e, 5²¹ > 4 . 5²⁰

f, 2²⁰ > 7.2¹⁷

Câu a:

18 - |x - 1| = 2^2

18 - |x - 1| = 4

|x - 1| = 18 - 4

|x - 1| = 14

x - 1 = -14 hoặc x - 1 = 14

Th1: x - 1 = -14

x = - 14 + 1

x = -13

Th2: x - 1 = 14

x = 14 + 1

x = 15

Vậy x ∈ {-13; 15}

Câu b:

x - {55 - [49 + (-28 - x)]} = 13 - {47 + [25 - (32 - x)]}

x - {55 - [49 - 28 - x]} = 13 - {47 + [25 - 32 + x]}

x - {55 - 49 + 28 + x} = 13 - {47 + 25 - 32 + x}

x - {6 + 28 + x} = 13 - {72 - 32 + x}

x - {34 + x} = 13 - {40 + x}

x - 34 - x = 13 - 40 - x

(x - x) + x = 13 - 40 + 34

0 + x = - 27 + 34

x = 7

Vậy x = 7