\(\frac{2+x}{3}=\frac{4}{5}\)

\(\Leftrightarrow\left(2+x\right).5=3.4\)

\(\Leftrightarrow\left(2+x\right).5=12\)

\(\Leftrightarrow2+x=\frac{12}{5}\)

\(\Leftrightarrow x=\frac{2}{5}\)

x = 2

'-'

\(\frac{5^3.3^5}{5^3.0,5+125.2,5}\)

\(=\frac{5^3.3^5}{5^3.0,5+5^3.2,5}\)

\(=\frac{5^3.3^5}{5^3\left(0,5+2,5\right)}\)

\(=\frac{5^3.3^5}{5^3.3}\)

\(=3^4=81\)

#Học tốt!!!

~NTTH~

\(\frac{13}{38}>\frac{13}{39}=\frac{1}{3}=\frac{29}{87}>\frac{29}{88}\Rightarrow\frac{-13}{38}< \frac{29}{-88}\)

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

Ta có: \(\left|x+\frac{1}{2021}\right|\ge0\) ; \(\left|x+\frac{2}{2021}\right|\ge0\) ; ... ; \(\left|x+\frac{2020}{2021}\right|\ge0\) \(\left(\forall x\right)\)

\(\Rightarrow\left|x+\frac{1}{2021}\right|+\left|x+\frac{2}{2021}\right|+...+\left|x+\frac{2020}{2021}\right|\ge0\left(\forall x\right)\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

Từ đó ta được: \(x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Leftrightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Leftrightarrow x=\frac{\left(2020+1\right)\left[\left(2020-1\right)\div1+1\right]}{2021}\)

\(\Leftrightarrow x=\frac{2021\cdot2020}{2021}=2020\)

Vậy x = 2020

\(\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|=2021x\)

Ta có:\(\left|\frac{x+1}{2021}\right|\ge0;\left|\frac{x+2}{2021}\right|\ge0;....;\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow\left|\frac{x+1}{2021}\right|+\left|\frac{x+2}{2021}\right|+...+\left|\frac{x+2020}{2021}\right|\ge0\forall x\)

\(\Rightarrow2021x\ge0\Rightarrow x\ge0\)

\(\Rightarrow\frac{x+1}{2021}+\frac{x+2}{2021}+...+\frac{x+2020}{2021}=2021x\)

\(\Rightarrow x+\frac{1}{2021}+x+\frac{2}{2021}+...+x+\frac{2020}{2021}=2021x\)

\(\Rightarrow2020x+\frac{1+2+...+2020}{2021}=2021x\)

\(\Rightarrow x=2020\)

a, Ta có (x+2)2≥0(x+2)2≥0

⇒(x+2)2+5≥5⇒(x+2)2+5≥5

⇒30(x+2)2+5≤305=6⇒30(x+2)2+5≤305=6

Hay A≤6A≤6

Dấu = xảy ra ⇔(x+2)2=0⇔x+2=0⇔x=−2⇔(x+2)2=0⇔x+2=0⇔x=−2

b,

Ta có (x−3)2≥0(x−3)2≥0

⇒(x−3)2+4≥4⇒(x−3)2+4≥4

⇒20(x+2)2+5≤204=5⇒20(x+2)2+5≤204=5

Hay A≤5A≤5

Dấu = xảy ra ⇔(x−3)2=0⇔x−3=0⇔x=3⇔(x−3)2=0⇔x−3=0⇔x=3

c,

Ta có (x+1)2≥0(x+1)2≥0

⇒(x+1)2+2≥2⇒(x+1)2+2≥2

⇒10(x+1)2+2≤102=5⇒10(x+1)2+2≤102=5

Hay A≤5A≤5

Dấu = xảy ra ⇔(x+1)2=0⇔x+1=0⇔x=−1⇔(x+1)2=0⇔x+1=0⇔x=−1

A = | 5x + 2 | + 5| x + 1 | 

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

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

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

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

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

Dấu "=" xảy ra khi ab ≥ 0

=> ( 5x + 2 )( -5x - 5 ) ≥ 0

1. \(\hept{\begin{cases}5x+2\ge0\\-5x-5\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}5x\ge-2\\-5x\ge5\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge-\frac{2}{5}\\x\le-1\end{cases}}\)( loại )

2. \(\hept{\begin{cases}5x+2\le0\\-5x-5\le0\end{cases}}\Leftrightarrow\hept{\begin{cases}5x\le-2\\-5x\le5\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le-\frac{2}{5}\\x\ge-1\end{cases}}\Leftrightarrow-1\le x\le-\frac{2}{5}\)

=> MinA = 3 <=> \(-1\le x\le-\frac{2}{5}\)