Eat/ healthy/ diet /and do exercise/ regularly /help you/ stay/ healthy 

=>Eat healthy diet and do exercise regularly help you stay healthy

We/ need/ calories/ or/ energy /do /things everyday 

=>We need calories or energy to do things everyday

We /should/ balance /calories /we/ get/ from /food/ with /calories /we /physical /activity 

Eat /healthy/ balanced/ diet /be/ important /part /maintain/ good /healthy

=>Eating healthy and balanced diet is an important part in maintaining a good heathy

Eat/ less /sweet/ food /and leaf /more/ fruit/ vegetables

=>Eat less sweet food and eat more fruits and vegetables.

Drink /lots /water/ be/ good /our/ health 

=>Drink lots of water is good for our health

 Eat heathily on the diet and do exercise regularly help you stay your health.

 We need calories or energy to do things everyday.

We should balance calories we get from food with calories we use for physical activity.

 Eat healthily and balanced diet are important parts maintain good health.

  Eat less sweet food and eat more fruits and vegetables.

 Drink lots of water is good for our health.

a) \(1+2+3+...+n=\frac{n\left(n+1\right)}{2}\)

b) \(1^2+2^2+...+n^2\)

\(=1\left(2-1\right)+2\left(3-1\right)+...+n\left[\left(n+1\right)-1\right]\)

\(=1.2+2.3+...+n\left(n+1\right)-\left(1+2+...+n\right)\)

\(=\frac{1.2.3+2.3.\left(4-1\right)+...+n.\left(n+1\right).\left[\left(n+2\right)-\left(n-1\right)\right]}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{1.2.3-1.2.3+2.3.4-...-\left(n-1\right)n\left(n+1\right)+n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=\frac{n\left(n+1\right)\left(n+2\right)}{3}-\frac{n\left(n+1\right)}{2}\)

\(=n\left(n+1\right)\left(\frac{n+2}{3}-\frac{1}{2}\right)\)

\(=\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)

câu c đâu bạn

| x - 3 | + | x + 5 | = 8

Ta có : 

| x - 3 | + | x + 5 |

= | -( x - 3 ) | + | x + 5 |

= | 3 - x | + | x + 5 |

Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta có :

| 3 - x | + | x + 5 | ≥ | 3 - x + x + 5 | = | 8 | = 8

Dấu "=" xảy ra ( tức | x - 3 | + | x + 5 | = 8 ) khi ab ≥ 0

=> ( 3 - x )( x + 5 ) ≥ 0

=> -5 ≤ x ≤ 3

Vậy | x - 3 | + | x + 5 | = 8 <=> -5 ≤ x ≤ 3

\(\left|x-3\right|+\left|x+5\right|=3\)

\(\left|x-3\right|-\left|5+x\right|=3\)

AD BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\)ta đc:

\(\left|x-3\right|-\left|5+x\right|\ge\left|x-3-5+x\right|=3\)

Dấu"=" xảy ra khi \(ab\ge0\)

Tự lm tiếp

| x - 2 | + | x - 5 | = 3 

Xét | x - 2 | + | x - 5 | 

= | x - 2 | + | -( x - 5 ) |

= | x - 2 | + | 5 - x |

Áp dụng bất đẳng thức | a | + | b | ≥ | a + b | ta được :

| x - 2 | + | 5 - x | ≥ | x - 2 + 5 - x | = | 3 | = 3

Dấu "=" xảy ra khi ab ≥ 0

=> ( x - 2 )( 5 - x ) ≥ 0

=> 2 ≤ x ≤ 5

Vậy | x - 2 | + | x - 5 | = 3 <=> 2 ≤ x ≤ 5

Ta có: \(\left|x-2\right|+\left|x-5\right|=3\)

Xét \(x< 2\) : \(2-x+5-x=3\)

\(\Leftrightarrow2x=4\)

\(\Rightarrow x=2\left(ktm\right)\) => loại

Xét \(2\le x< 5\) : \(x-2+5-x=3\)

\(\Leftrightarrow3=3\left(tm\right)\)

Vậy \(2\le x< 5\)

Xét \(x\ge5\) : \(x-2+x-5=3\)

\(\Leftrightarrow2x=10\)

\(\Rightarrow x=5\left(tm\right)\)

Vậy \(2\le x\le5\)

| -x + 2 | - | x + 7 | = 0

<=> | -( -x + 2 ) | - | x + 7 | = 0

<=> | x - 2 | - | x + 7 | = 0 (1)

Xét 3 trường hợp :

1. x < -7

(1) <=> -( x - 2 ) - [ -( x + 7 ) ] = 0

      <=> -x + 2 - ( -x - 7 ) = 0

      <=> -x + 2 + x + 7 = 0

      <=> 9 = 0 ( vô lí )

2. -7 ≤ x < 2

(1) <=> -( x - 2 ) - ( x + 7 ) = 0

     <=> -x + 2 - x - 7 = 0

     <=> -2x - 5 = 0

     <=> -2x = 5

     <=> x = -5/2 ( thỏa mãn )

3. x ≥ 2

(1) <=> ( x - 2 ) - ( x + 7 ) = 0

     <=> x - 2 - x - 7 = 0

     <=> -9 = 0 ( vô lí )

Vậy x = -5/2

b) | 2x - 1 | + | 2 + y | = 0

Ta có : | 2x - 1 | ≥ 0 ∀ x ; | 2 + y | ≥ 0 ∀ y

=> | 2x - 1 | + | 2 + y | ≥ 0 ∀ x, y

Dấu "=" xảy ra <=> x = 1/2 ; y = -2

1) Ta có: \(\left|9y-1\right|+\left(2x+3\right)^2=0\)

Mà \(\hept{\begin{cases}\left|9y-1\right|\ge0\\\left(2x+3\right)^2\ge0\end{cases}}\left(\forall x,y\right)\)

=> \(\left|9y-1\right|+\left(2x+3\right)^2\ge0\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|9y-1\right|=0\\\left(2x+3\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}9y-1=0\\2x+3=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)

Vậy \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{1}{9}\end{cases}}\)

2)

a) Ta có: \(\left[\left(-\frac{1}{3}\right)^7\right]^4=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)

và \(\left[\left(-\frac{1}{2}\right)^{14}\right]^2=\left(\frac{1}{2}\right)^{28}=\frac{1}{2^{28}}\)

Vì \(\frac{1}{3^{28}}< \frac{1}{2^{28}}\Rightarrow\left[\left(-\frac{1}{3}\right)^7\right]^4< \left[\left(-\frac{1}{2}\right)^{14}\right]^2\)

b) Ta có: \(\left(-\frac{2}{3}\right)^{12}=\left[\left(-\frac{2}{3}\right)^2\right]^6=\left(\frac{4}{9}\right)^6\)

Ta thấy \(0< \frac{4}{9}< 1\)\(\Rightarrow\left(\frac{4}{9}\right)^6>\left(\frac{4}{9}\right)^7\)

\(\Rightarrow\left(-\frac{2}{3}\right)^{12}>\left(\frac{4}{9}\right)^7\)

\(7^{24}=\left(7^3\right)^8=343^8\)

\(3^{30}=\left(3^5\right)^6=243^6\)

\(\Rightarrow7^{24}>3^{30}\)

Ta có : 7²⁴=7^4.6=(7⁴)⁶=2401⁶

            3³⁰=3^5.6=(3⁵)⁶=243⁶

Mà 2401⁶>243⁶

=> 7²⁴>3³⁰

Nối A với C theo đề bài có AB//CD

=> \(\widehat{BAC}+\widehat{ACD}=180^o\) (2 góc trong cùng phí bù nhau)

Xét tam giác AEC có

\(\widehat{AEC}+\widehat{EAC}+\widehat{ECA}=180^o\) (Tổng các góc trong của 1 tg bằng 180)

\(\Rightarrow\widehat{BAC}+\widehat{ACD}+\widehat{AEC}+\widehat{EAC}+\widehat{ECA}=180^o\)

\(\Rightarrow\left(\widehat{BAC}+\widehat{EAC}\right)+\widehat{AEC}+\left(\widehat{ACD}+\widehat{ECA}\right)=180^o\)

\(\Rightarrow\widehat{A}+\widehat{E}+\widehat{C}=180^o\left(dpcm\right)\)