a: Ta có: \(10-\left\{\left[\left(x:3+17\right):10+3\cdot2^4\right]:10\right\}=5\)

\(\Leftrightarrow\left[\left(x:3+17\right):10+48\right]:10=5\)

\(\Leftrightarrow\left(x:3+17\right):10+48=50\)

\(\Leftrightarrow\left(x:3+17\right):10=2\)

\(\Leftrightarrow x:3+17=20\)

\(\Leftrightarrow x:3=3\)

hay x=9

b: Ta có: \(2448:\left[119-\left(x-6\right)\right]=24\)

\(\Leftrightarrow119-\left(x-6\right)=102\)

\(\Leftrightarrow x-6=17\)

hay x=23

c: Ta có: \(165-\left(35:x+3\right)\cdot19=13\)

\(\Leftrightarrow\left(35:x+3\right)\cdot19=152\)

\(\Leftrightarrow\left(35:x+3\right)=8\)

\(\Leftrightarrow35:x=5\)

hay x=7

Bài 3 là hỗn số hả em?

Bài 1: 

a) Ta có: \(\dfrac{2}{5}\cdot x+\dfrac{1}{3}=\dfrac{1}{5}\)

\(\Leftrightarrow\dfrac{2}{5}\cdot x=\dfrac{1}{5}-\dfrac{1}{3}=\dfrac{-2}{15}\)

\(\Leftrightarrow x=\dfrac{-2}{15}:\dfrac{2}{5}=\dfrac{-2}{15}\cdot\dfrac{5}{2}\)

hay \(x=-\dfrac{1}{3}\)

Vậy: \(x=-\dfrac{1}{3}\)

b) Ta có: \(\dfrac{1}{5}+\dfrac{5}{3}:x=\dfrac{1}{2}\)

\(\Leftrightarrow\dfrac{5}{3}:x=\dfrac{1}{2}-\dfrac{1}{5}=\dfrac{3}{10}\)

\(\Leftrightarrow x=\dfrac{5}{3}:\dfrac{3}{10}=\dfrac{5}{3}\cdot\dfrac{10}{3}\)

hay \(x=\dfrac{50}{9}\)

Vậy: \(x=\dfrac{50}{9}\)

c) Ta có: \(\dfrac{4}{9}-\dfrac{5}{3}\cdot x=-2\)

\(\Leftrightarrow\dfrac{5}{3}x=\dfrac{4}{9}+2=\dfrac{22}{9}\)

\(\Leftrightarrow x=\dfrac{22}{9}:\dfrac{5}{3}=\dfrac{22}{9}\cdot\dfrac{3}{5}\)

hay \(x=\dfrac{22}{15}\)

Vậy: \(x=\dfrac{22}{15}\)

d) Ta có: \(\dfrac{5}{7}:x-3=\dfrac{-2}{7}\)

\(\Leftrightarrow\dfrac{5}{7}:x=\dfrac{-2}{7}+3=\dfrac{19}{21}\)

\(\Leftrightarrow x=\dfrac{5}{7}:\dfrac{19}{21}=\dfrac{5}{7}\cdot\dfrac{21}{19}\)

hay \(x=\dfrac{15}{19}\)

Vậy:\(x=\dfrac{15}{19}\)

a)   x + 118 = 218

x = 218 – 118

x = 100

b)   18 – x = -17

X = 18 – (-17)

X = 35

c)    (-18) – x = -17

X = (-18) – (-17)

X = -1

d)124 + (118 – x) = 217

118 – x = 93

X = 118 – 93

X = 25

e) (3x – 6).3 = 3⁴

(3x – 6).3 = 81

3x – 6 = 27

3x = 33

ð X = 11

g) 23.(42 – x) = 46

42 – x = 2

ð X = 40

h) 84 – (3x + 8) = 34

3x + 8 = 84 – 34

3x + 8 = 50

3x = 50 – 8

3x = 42

X = 42 : 3

ð X = 14

i)               3.(2x + 1) – 19 = 14

3.(2x + 1) = 14 + 19

3.(2x + 1) = 33

2x + 1 = 33 : 3

2x + 1 = 11

2x = 11 – 1

2x = 10

X = 10 : 2

ð X = 5

k) 6x – 302 = 2³.5

6x – 302 = 40

6x = 40 + 302

6x = 342

X = 342 : 6

ð X = 57

m) (x – 1)² = 81

(x – 1)² = 9²

ð X – 1 = 9

ð X = 9 + 1

ð X = 10

n) 2^x+1 – 5 = 27

2^x+1 = 27 + 5

2^x+1 = 32

2^x+1 = 2⁵

ð X = 4

a: -29-(15-x)=34

=>-29-15+x=34

=>x-44=34

=>x=34+44

=>x=78

b: 15-(x-17)=-12

=>x-17=15+12=27

=>x=27+17

=>x=44

c: -(x-6+85)=(x+51)-54

=>-(x+79)=x+51-54

=>x-3=-x-79

=>2x=-79+3=-76

=>x=-38

\(\text{ a) 12+(4−x)=−5}\\ 4-x=-5-12\\ 4-x=-17\\ x=4-\left(-17\right)\\ x=21\)

viết tắt ai mà hiểu được :V 

d: \(\frac{3}{x}+\frac{y}{3}=\frac56\)

=>\(\frac{3}{x}=\frac56-\frac{y}{3}=\frac{5-2y}{6}\)

=>x(5-2y)=18

=>x(2y-5)=-18

mà 2y-5 lẻ và x>0

nên (x;2y-5)∈{(18;1);(2;9);(6;3)}

=>(x;2y)∈{(18;6);(2;14);(6;8)}

=>(x;y)∈{(18;3);(2;7);(6;4)}

c: \(\frac{x}{9}-\frac{3}{y}=\frac{1}{18}\)

=>\(\frac{x}{9}-\frac{1}{18}=\frac{3}{y}\)

=>\(\frac{2x-1}{18}=\frac{3}{y}\)

=>y(2x-1)=54

mà 2x-1 lẻ

nên (2x-1;y)∈{(1;54);(3;18);(9;6);(27;2)}

=>(2x;y)∈{(2;54);(4;18);(10;6);(28;2)}

=>(x;y)∈{(1;54);(2;18);(5;6);(14;2)}

b: \(\frac{x}{6}-\frac{2}{y}=\frac{1}{30}\)

=>\(\frac{x}{6}-\frac{1}{30}=\frac{2}{y}\)

=>\(\frac{5x-1}{30}=\frac{2}{y}\)

=>y(5x-1)=60

mà 5x-1 chia 5 dư 4

nên (5x-1;y)∈{(4;15)}

=>(5x;y)∈{(5;15)}

=>(x;y)∈{(1;15)}

a: \(\frac{x}{2}-\frac{4}{y}=\frac15\)

=>\(\frac{x}{2}-\frac15=\frac{4}{y}\)

=>\(\frac{5x-2}{10}=\frac{4}{y}\)

=>y(5x-2)=40

mà 5x-2 chia 5 dư 3

nên (5x-2;y)∈{(8;5)}

=>(5x;y)∈{(10;5)}

=>(x;y)∈{(2;5)}

\(\frac{a}{b}\cdot4+\frac{1}{6}=\frac{17}{6}\)

\(\frac{a}{b}\cdot4=\frac{17}{6}-\frac{1}{6}\)

\(\frac{a}{b}\cdot4=\frac{8}{3}\)

\(\frac{a}{b}=\frac{8}{3}:4\)

\(\frac{a}{b}=\frac{2}{3}\)

   a/b*4+1/6=17/6

=a/b*4       =17/6-1/6 

=a/b                =16/6:4  

=a/b                =2/3

a: Ta có: \(-x^2+4x+5\)

\(=-\left(x^2-4x-5\right)\)

\(=-\left(x^2-4x+4-9\right)\)

\(=-\left(x-2\right)^2+9\le9\forall x\)

Dấu '=' xảy ra khi x=2

b: Ta có: \(-x^2-7x+4\)

\(=-\left(x^2+7x-4\right)\)

\(=-\left(x^2+2\cdot x\cdot\dfrac{7}{2}+\dfrac{49}{4}-\dfrac{65}{4}\right)\)

\(=-\left(x+\dfrac{7}{2}\right)^2+\dfrac{65}{4}\le\dfrac{65}{4}\forall x\)

Dấu '=' xảy ra khi \(x=-\dfrac{7}{2}\)

Cảm ơn bn nhiều