a,  1, Vì |x - 2019| ≥ 0 ; (y - 1)²⁰²⁰ ≥ 0 => |x - 2019| + (y - 1)²⁰²⁰ ≥ 0 => |x - 2019| + (y - 1)²⁰²⁰​ + (-2) ≥ (-2) => A ≥ -2

Dấu " = " xảy ra <=> \(\hept{\begin{cases}x-2019=0\\y-1=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2019\\y=1\end{cases}}\)

Vậy GTNN A = -2 khi x = 2019 và y = 1

2, Ta có: |x - 3| = |3 - x|

Vì |x - 3| + |x + 4| ≥ |x - 3 + x + 4| = |1| = 1

=> C ≥ 1 - 5 => C ≥ -4

Dấu " = " xảy ra <=> (3 - x)(x + 4) ≥ 0

+) Th1: \(\hept{\begin{cases}3-x\ge0\\x+4\ge0\end{cases}\Rightarrow}\hept{\begin{cases}x\le3\\x\ge-4\end{cases}\Rightarrow}-4\le x\le3\)

+) Th2: \(\hept{\begin{cases}3-x\le0\\x+4\le0\end{cases}\Rightarrow}\hept{\begin{cases}x\ge3\\x\le-4\end{cases}}\)(Vô lý)

Vậy GTNN của C = -4 khi -4 ≤ x ≤ 3

b,

1, Vì |x² - 25| ≥ 0 => 4|x² - 25| ≥ 0 => 3² - 4|x² - 25| ≤ 3² = 9

Dấu " = " xảy ra <=> x² - 25 = 0 <=> x² = 25 <=> x = 5 hoặc x = -5

Vậy GTLN B = 9 khi x = 5 hoặc x = -5

2, Đk: x ≠ 5

 \(D=\frac{x-4}{x-5}=\frac{\left(x-5\right)+1}{x-5}=1+\frac{1}{x-5}\)

Để D mang giá trị lớn nhất <=> \(\frac{1}{x-5}\)mang giá trị lớn nhất <=> x - 5 mang giá trị nhỏ nhất <=> x - 5 = 1 <=> x = 6

=> \(D=1+1=2\)

Vậy GTLN của D = 2 khi x = 6

Gấp gáp chi em cuộc sống vẫn rực rỡ sắc màu

Chim vẫn reo ca và môi hôn đang đứng đợi

Hoa vẫn nở và xuân thì đương tới

Hãy trải lòng xao xuyến với tình yêu.

2346 : (25 + x) = 23

=> 25 + x = 2346 : 23 = 102

=> x = 102 - 25  =  77

Vậy x = 77

2(x-3) + 5(x+4) = 49

=> 2x - 6 + 5x + 20 = 49

=> 7x + (20 - 6) = 49

=> 7x + 14 = 49

=> 7x = 49 - 14 = 35

=> x = 35 : 7 = 5

Vậy x = 5

(2x-10)³ . 25¹⁰⁰⁷ = \(5^2.5^{2015}\)

\(\left(2x-10\right)^3.5^{2014}=5^{2014}.5^3\)

\(\left(2x-10\right)^3=5^3\)

2x - 10 = 5

=> 2x = 5 + 10 = 15

=> x = 15  2 = 7,5

Vậy x = 7,5

1. a. 2346:(25+x)=23

=> 25+x=2346:23

=> 25+x=102

=> x=102-25

=> x=77

b. 2(x-3)+5(x+4)=49

=> 2x-6+5x+20=49

=> 7x+14=49

=> 7x=49-14

=> 7x=35

=> x=35:7

=> x=5

c. (2x-10)³.25¹⁰⁰⁷=5².5²⁰¹⁵

=> (2x-10)³.(5²)¹⁰⁰⁷=5²⁰¹⁷

=> (2x-10)³.5²⁰¹⁴=5²⁰¹⁷

=> (2x-10)³=5²⁰¹⁷:5²⁰¹⁴

=> (2x-10)³=5³

=> 2x-10=5

=> 2x=5+10

=> 2x=15

=> x=15:2

=> x=7,5

=997866

a)

\(3:\left(\dfrac{9}{4}\right)=\dfrac{3}{4}:\left(6.x\right)\\ \Rightarrow3.6.x=\dfrac{3}{4}.\dfrac{9}{4}\\ x=\dfrac{3}{4}.\dfrac{9}{4}.\dfrac{1}{3}.\dfrac{1}{6}\\ x=\dfrac{3}{4.4.2}\\ x=\dfrac{3}{32}\)

b)

\(4,5:0,3=\left(5.0,09\right):\left(0,01.x\right)\\ 0,01.x.4,5=5.0,09.0,3\\ x=5.\dfrac{9}{100}.\dfrac{3}{10}.100.\dfrac{10}{45}\\ x=3\)

d)

\(\left(\dfrac{1}{9}.x\right)=\dfrac{7}{4}:\dfrac{2}{25}\\ \left(\dfrac{1}{9}.x\right)=\dfrac{7}{4}.\dfrac{25}{2}\\ x:\dfrac{7}{4}=\dfrac{25}{2}:\dfrac{1}{9}\\ x=\dfrac{25}{2}.9.\dfrac{7}{4}\\ x=\dfrac{1575}{8}\\ x=196\dfrac{7}{8}\)

e)

\(\dfrac{-2}{x}=\dfrac{-x}{\dfrac{8}{25}}\\ -x.x=-2.\dfrac{8}{25}\\ -x^2=-\dfrac{16}{25}=-\dfrac{4^2}{5^2}\\ -x^2=-\left(\dfrac{4}{5}\right)^2\\ \Rightarrow x=\dfrac{4}{5}\)

Chúc bạn học tốt

Tìm x thuoc z:

1) \(26-\left|x+9\right|=-13\)

\(\Leftrightarrow\left|x+9\right|=26-\left(-13\right)\)

\(\Leftrightarrow\left|x+9\right|=39\)

\(\Leftrightarrow\left[{}\begin{matrix}x+9=39\\x+9=-39\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=39-9=30\\x=-39-9=-48\end{matrix}\right.\)

Vậy: \(x\in\left\{30;-48\right\}\)

2) \(\left|x+7\right|-13=25\)

\(\Leftrightarrow\left|x+7\right|=25+13=38\)

\(\Leftrightarrow x+7\in\left\{38;-38\right\}\)

\(\Leftrightarrow x\in\left\{31;-45\right\}\)

Vậy:.................

tim x biet

\(1)123-3.\left(x+4\right)=23\)

\(\Leftrightarrow3\left(x+4\right)=123-23\)

\(\Leftrightarrow3\left(x+4\right)=100\)

\(\Leftrightarrow x+4=\frac{100}{3}\)

\(\Leftrightarrow x=\frac{100}{3}-4=\frac{100-12}{3}=\frac{88}{3}\)

Vậy:................

2) Tương tự

(-25)-(5.x-3)=4

5.x-3 = -25 - 4

5.x-3=-29

5.x=-26

x=-26/5

x=-5,2

5.x-6=20-(12-6)

5.x-6=20-6

5.x-6=14

5.x=20

x=20:5

x=4

vậy x=4

1/
-25-(5x-3)=4
-25-4=5x-3
-25-4+3=5x
5x= -26
x= -26/5
2/
5x-6=20-(12-6)
5x = 20-12+6+6
5x = 20
x = 20/5 = 4
3/
-2x -40 = (5-x) -(-15+60
-2x -40 = 5-x +15-60
-2x +x = 5+15-60+40
-x = 0
x=0

a) <=> 15-5x-20=-12-3

<=> -5x=-12-3-15+20=-10

=>x=-10:(-5)=2

b)<=>7-x-25-7=-25

<=> -x=-25-7+25+7=0 =>x=0

c) /x+2/=0 => x+2=0 =>x=-2

d) sai đề

e)<=> /x-5/ = 7

<=> \(\orbr{\begin{cases}x-5=7\\x-5=-7\end{cases}}\)

<=> \(\orbr{\begin{cases}x=12\\x=-2\end{cases}}\)

g) <=> -x-20-8+2x=-15

<=> x=-15+20+8=13

d là | x + 3 | = 7 - ( - 2 )

         \(x^2-5x-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=5\\x=4\end{cases}}\)

Vậy....

\(2x\left(x+6\right)=7x+42\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7x-42=0\)

\(\Leftrightarrow\)\(2x\left(x+6\right)-7\left(x+6\right)=0\)

\(\Leftrightarrow\)\(\left(x+6\right)\left(2x-7\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+6=0\\2x-7=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-6\\x=\frac{7}{2}\end{cases}}\)

Vậy......

\(x^3-5x^2+x-5=0\)

\(\Leftrightarrow\)\(x^2\left(x-5\right)+\left(x-5\right)=0\)

\(\Leftrightarrow\)\(\left(x-5\right)\left(x^2+1\right)=0\)

\(\Leftrightarrow\)\(x-5=0\)

\(\Leftrightarrow\)\(x=5\)

\(x^4-2x^3+10x^2-20x=0\)

\(\Leftrightarrow\)\(x^3\left(x-2\right)+10x\left(x-2\right)=0\)

\(\Leftrightarrow\)\(x\left(x-2\right)\left(x^2+10\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

Vậy...