| x + 1/3 | - 4 = -1

<=> | x + 1/3 | = 3

<=> x + 1/3 = 3 hoặc x + 1/3 = -3

<=> x = 8/3 hoặc x = -10/3

\(\frac{5}{x-2}=\frac{15}{6}\)

<=> 5 . 6 = ( x - 2 ).15

<=> 30 = 15x - 30

<=> 30 + 30 = 15x

<=> 60 = 15x

<=> x = 4 

a) \(\left|x+\frac{1}{3}\right|-4=-1\)

=> \(\left|x+\frac{1}{3}\right|=-1+4=3\)

=> \(\orbr{\begin{cases}x+\frac{1}{3}=3\\x+\frac{1}{3}=-3\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{8}{3}\\x=-\frac{10}{3}\end{cases}}\)

b) \(\frac{5}{x-2}=\frac{15}{6}\)

=> \(\frac{15}{3\left(x-2\right)}=\frac{15}{6}\)

=> 3x - 6 = 6

=> 3x = 12

=> x = 4

a) \(x+\frac{1}{5}-\frac{2}{7}=4\frac{1}{2}\)

=> \(x+\frac{1}{5}-\frac{2}{7}=\frac{9}{2}\)

=> \(x=\frac{9}{2}+\frac{2}{7}-\frac{1}{5}=\frac{321}{70}\)

b) \(1\frac{3}{4}+x-\frac{7}{8}=5\)

=> \(\frac{7}{4}+x-\frac{7}{8}=5\)

=> \(\frac{7}{4}+x=\frac{47}{8}\)

=> \(x=\frac{33}{8}\)

c) \(\frac{1}{2}-\frac{1}{3}\cdot x=\frac{3}{4}\)

=> \(\frac{1}{3}\cdot x=\frac{1}{2}-\frac{3}{4}=-\frac{1}{4}\)

=> \(x=\left(-\frac{1}{4}\right):\frac{1}{3}=\left(-\frac{1}{4}\right)\cdot3=-\frac{3}{4}\)

Một cách khác mà hôm nay ngủ dạy lại nghĩ ra :))

Áp dụng liên tiếp BĐT Svacxo cho 3 các số dương ta được :

\(\left(a+b\right)^4+\left(b+c\right)^4+\left(c+a\right)^4\)

\(=\frac{\left(a+b\right)^4}{1}+\frac{\left(b+c\right)^4}{1}+\frac{\left(c+a\right)^4}{1}\ge\frac{\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\right]^2}{1+1+1}\)

\(=\frac{\left[\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\right]^2}{3}=\frac{\left[\frac{\left(a+b\right)^2}{1}+\frac{\left(b+c\right)^2}{1}+\frac{\left(c+a\right)^2}{1}\right]^2}{3}\)

\(\ge\frac{\left[\frac{\left(a+b+b+c+c+a\right)^2}{3}\right]^2}{3}=\frac{\left(\frac{2^2}{3}\right)^2}{3}=\frac{16}{27}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Ta đi chứng minh BĐT : \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\) 

BĐT trên tương đương : \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( Đúng )

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

+) Ta xét : \(x^4+y^4+z^4=\left(x^2\right)^2+\left(y^2\right)^2+\left(z^2\right)^2\)\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{3}\) (*)

Lại có : \(x^2+y^2+z^2\ge\frac{\left(x+y+z\right)^2}{3}\)

Nên từ (*) suy ra \(x^4+y^4+z^4\ge\frac{\left(\frac{\left(x+y+z\right)^2}{3}\right)^2}{3}=\frac{\left(x+y+z\right)^4}{27}\)

Áp dụng vào bài toán với \(\hept{\begin{cases}x=a+b\\y=b+c\\z=c+a\end{cases}}\) ta có :

\(\left(a+b\right)^4+\left(b+c\right)^4+\left(c+a\right)^4\ge\frac{\left(a+b+b+c+c+a\right)^4}{27}=\frac{2^4}{27}=\frac{16}{27}\)

Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)

Vậy BĐT được chứng minh !

.

BCNN(54,126)

* Phân tích 2 thừa số nguyên tố :

54 = 2.3³

126 = 2.3².7

=> BCNN(54,126) = 2.3³.7 = 378

\(54=2\cdot3^3\)    

\(126=2\cdot3^2\cdot7\)     

\(BCNN\left(54;126\right)=2\cdot3^3\cdot7=378\)

- What your name ? 

- My name is Diep . 

- I'm Diep .

c1 - What is your name?

- My name's Châu.

c2 -How do I call you?

- Just call me Châu

11.9.2=11.18

45.3.5=15.45

6.3.11=11.18

9.5.15=15.45

vậy(1) 11.8=11.9.2=6.3.11

(2)15.45=45.3.5=9.5.15

i was born in Viet Nam

i am Vietnamese

- I come from Vietnam

- I am Vietnamese

\(\text{a)Số thập phân gấp 10 lần x:3057,65 }\)

\(\text{Số thập phân bằng }\frac{1}{10}x:30,5765\)

\(\text{b)}x.100+x.0,01=30576,5+3,05765=30579,55765\)

Bài 3 :

a) \(\left(\frac{1}{25}-0,6\right)^2:\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\cdot2\frac{2}{17}\right]\)

\(=\left(\frac{1}{25}-\frac{3}{5}\right)^2\cdot\frac{125}{49}+\left[\left(\frac{13}{4}-\frac{59}{9}\right)\cdot\frac{36}{17}\right]\)

\(=\left(-\frac{14}{25}\right)^2\cdot\frac{125}{49}+\left[\left(-\frac{119}{36}\right)\cdot\frac{36}{17}\right]\)

\(=-\frac{196}{625}\cdot\frac{125}{49}+\left(-7\right)=-\frac{4}{5}+\left(-7\right)=-\frac{39}{5}\)

Trả lời :

\(\left(\frac{1}{25}-0,6\right)^2\div\frac{49}{125}+\left[\left(3\frac{1}{4}-6\frac{5}{9}\right)\times2\frac{2}{17}\right]\)

\(=\left(\frac{1}{25}-\frac{3}{5}\right)^2\div\frac{49}{125}+\left[\frac{-119}{36}\times\frac{36}{17}\right]\)

\(=\left(\frac{-14}{25}\right)^2\div\frac{49}{125}-7\)

\(=\frac{4}{5}-7\)

\(=\frac{-31}{5}\)