1) Chỉ tìm được Max thôi nhé

a) \(C=\frac{4}{5}+\frac{20}{\left|3x+5\right|+\left|4y+5\right|+8}\le\frac{4}{5}+\frac{20}{8}=\frac{33}{10}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left|3x+5\right|=0\\\left|4y+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-\frac{5}{3}\\y=-\frac{5}{4}\end{cases}}\)

b) \(E=\frac{2}{3}+\frac{21}{\left(x+3y\right)^2+5\left|x+5\right|+14}\le\frac{2}{3}+\frac{21}{14}=\frac{13}{6}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+3y\right)^2=0\\5\left|x+5\right|=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=\frac{5}{3}\end{cases}}\)

2) Thì chỉ tìm được GTNN thôi nhé

a) \(A=5+\frac{-8}{4\left|5x+7\right|+24}\ge5-\frac{8}{24}=\frac{14}{3}\left(\forall x\right)\)

Dấu "=" xảy ra khi: \(4\left|5x+7\right|=0\Rightarrow x=-\frac{7}{5}\)

Vậy Min(A) = 14/3 khi x = -7/5

b) \(B=\frac{6}{5}-\frac{14}{5\left|6y-8\right|+35}\ge\frac{6}{5}-\frac{14}{35}=\frac{4}{5}\left(\forall y\right)\)

Dấu "=" xảy ra khi: \(5\left|6y-8\right|=0\Rightarrow x=\frac{4}{3}\)

Vậy Min(B)  = 4/5 khi x = 4/3

\(\frac{1}{3}x+\frac{2}{5}=\frac{6}{5}+\frac{1}{2}x\) 

\(\frac{1}{3}x-\frac{1}{2}x=\frac{6}{5}-\frac{2}{5}\) 

\(-\frac{1}{6}x=\frac{4}{5}\) 

\(x=\frac{4}{5}:\left(-\frac{1}{6}\right)\) 

\(x=-\frac{24}{5}\)

Ta có: \(\frac{1}{3}x+\frac{2}{5}=\frac{6}{5}+\frac{1}{2}x\)

\(\Leftrightarrow\frac{1}{2}x-\frac{1}{3}x=\frac{2}{5}-\frac{6}{5}\)

\(\Leftrightarrow\frac{1}{6}x=-\frac{4}{5}\)

\(\Rightarrow x=-\frac{24}{5}\)

Ta có: \(\frac{4}{5}+\left|x-\frac{1}{3}\right|=7\)

\(\Leftrightarrow\left|x-\frac{1}{3}\right|=\frac{31}{5}\)

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{31}{5}\\x-\frac{1}{3}=-\frac{31}{5}\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{98}{15}\\x=-\frac{88}{15}\end{cases}}\)

I = | x + 1 | + | x + 4 | + | x + 3 |

= | x + 3 | + ( | x + 1 | + | x + 4 | )

Ta có :

+) | x + 3 | ≥ 0 ∀ x (1)

+) | x + 1 | + | x + 4 |

= | x + 1 | + | -( x + 4 ) |

= | x + 1 | + | -x - 4 | ≥ | x + 1 - x - 4 | = | -3 | = 3 (2)

Cộng (1) với (2) theo vế

=> | x + 3 | + ( | x + 1 | + | x + 4 | ) ≥ 3 ∀ x

Dấu "=" xảy ra <=> \(\hept{\begin{cases}\left|x+3\right|=0\\\left(x+1\right)\left(-x-4\right)\ge0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-3\\-4\le x\le-1\end{cases}}\Leftrightarrow x=-3\)

=> Min_I = 3 <=> x = -3

\(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)

=> \(\frac{6x}{11}=\frac{9y}{2}=\frac{18z}{5}\)

=> \(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}\)

Mà -x + z = -196

=> -(x - z) = -196

=> x - z = 196

Áp dụng t/c dãy tỉ số bằng nhau ta có :

\(\frac{x}{\frac{11}{6}}=\frac{y}{\frac{2}{9}}=\frac{z}{\frac{5}{18}}=\frac{x-z}{\frac{11}{6}-\frac{5}{18}}=\frac{196}{\frac{14}{9}}=126\)

=> \(\hept{\begin{cases}\frac{x}{\frac{11}{6}}=126\\\frac{y}{\frac{2}{9}}=126\\\frac{z}{\frac{5}{18}}=126\end{cases}}\Rightarrow\hept{\begin{cases}x=231\\y=28\\z=35\end{cases}}\)

Vậy x = 231,y = 28,z = 35

Ta có: \(\frac{6}{11}x=\frac{9}{2}y=\frac{18}{5}z\)

\(\Leftrightarrow\frac{x}{33}=\frac{y}{4}=\frac{z}{5}\)

Áp dụng t/c dãy tỉ số bằng nhau ta được:

\(\frac{x}{33}=\frac{y}{4}=\frac{z}{5}=\frac{-x+z}{-33+5}=\frac{-196}{-28}=7\)

\(\Rightarrow\hept{\begin{cases}x=231\\y=28\\z=35\end{cases}}\)

Ta có: \(A=2+2^2+...+2^{120}\)

\(\Rightarrow2A=2^2+2^3+...+2^{121}\)

\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{121}\right)-\left(2+2^2+...+2^{120}\right)\)

\(\Leftrightarrow A=2^{121}-2\)

Xét \(2^{121}=2^{120}\cdot2=\left(2^4\right)^{30}\cdot2=\overline{.....6}\cdot2=\overline{.....2}\)

\(\Rightarrow A=\overline{.....2}-2=\overline{.....0}\)

Vậy A có cstc là 0

\(A=2+2^2+2^3+........+2^{120}\)

\(\Rightarrow2A=2^2+2^3+2^4+......+2^{121}\)

\(\Rightarrow2A=A=2^{121}-2\)

mà \(A=2^n-2\)

\(\Rightarrow2^n-2=2^{121}-2\)

\(\Leftrightarrow2^n=2^{121}\)\(\Leftrightarrow n=121\)

Vậy \(n=121\)

Ta có: \(A=2+2^2+..+2^{120}\)

\(\Rightarrow2A=2^2+2^3+...+2^{121}\)

\(\Rightarrow2A-A=\left(2^2+2^3+...+2^{121}\right)-\left(2+2^2+...+2^{120}\right)\)

\(\Leftrightarrow A=2^{121}-2\)

Mà \(A=2^n-2=2^{121}-2\)

\(\Rightarrow n=121\)

A = 2 + 2² + ... + 2¹²⁰

Chứng minh chia hết cho 3

A = ( 2 + 2² ) + ( 2³ + 2⁴ ) + ... + ( 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 ) + 2³( 1 + 2 ) + ... + 2¹¹⁹( 1 + 2 )

= 2.3 + 2³.3 + ... + 2¹¹⁹.3

= 3( 2 + 2³ + ... + 2¹¹⁹ ) chia hết cho 3 ( đpcm )

Chứng minh chia hết cho 7

A = ( 2 + 2² + 2³ ) + ( 2⁴ + 2⁵ + 2⁶ ) + ... + ( 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 + 2² ) + 2⁴( 1 + 2 + 2² ) + ... + 2¹¹⁸( 1 + 2 + 2² )

= 2.7 + 2⁴.7 + ... + 2¹¹⁸.7

= 7( 2 + 2⁴ + ... + 2¹¹⁸ ) chia hết cho 7 ( đpcm )

Chứng minh chia hết cho 15

A = ( 2 + 2² + 2³ + 2⁴ ) + ( 2⁵ + 2⁶ + 2⁷ + 2⁸ ) + ... + ( 2¹¹⁷ + 2¹¹⁸ + 2¹¹⁹ + 2¹²⁰ )

= 2( 1 + 2 + 2² + 2³ ) + 2⁵( 1 + 2 + 2² + 2³ ) + ... + 2¹¹⁷( 1 + 2 + 2² + 2³ )

= 2.15 + 2⁵.15 + ... + 2¹¹⁷.15

= 15( 2 + 2⁵ + ... + 2¹¹⁷ ) chia hết cho 15 ( đpcm )

1) Ta có: \(A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{119}+2^{120}\right)\)

\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{119}\left(1+2\right)\)

\(A=3\left(2+2^3+...+2^{119}\right)\) chia hết cho 3

2) Ta có: \(A=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{118}+2^{119}+2^{120}\right)\)

\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{118}\left(1+2+2^2\right)\)

\(A=7\left(2+2^4+...+2^{118}\right)\) chia hết cho 7

3) Ta có: \(A=\left(2+2^2+2^3+2^4\right)+...+\left(2^{117}+2^{118}+2^{119}+2^{120}\right)\)

\(A=2\left(1+2+2^2+2^3\right)+...+2^{117}\left(1+2+2^2+2^3\right)\)

\(A=15\left(2+2^5+...+2^{117}\right)\) chia hết cho 15

A = 2 + 2² + ... + 2¹²⁰

2A = 2( 2 + 2² + ... + 2¹²⁰ )

     = 2² + 2³ + ... + 2¹²¹

A = 2A - A

   = 2² + 2³ + ... + 2¹²¹ - ( 2 + 2² + ... + 2¹²⁰ )

   = 2² + 2³ + ... + 2¹²¹ - 2 - 2² - ... - 2¹²⁰ 

   = 2¹²¹ - 2