a) 6 . x - 15 = 15

6 . x = 15 + 15

6 . x = 30

x = 30 : 6

x = 5

b) 2 . x + 3 = 9

2 . x = 9 - 3

2 . x = 6

x = 6 : 2

x = 3

c) 17 - 6 : x = 15

6 : x = 17 - 15

6 : x = 2

x = 6 : 2

x = 3

d) x : 3 + 12 = 14

x : 3 = 14 - 12

x : 3 = 2

x = 2 . 3

x = 6

a) 6 . x -15 = 15 

⇒ 6 .x = 15 + (-15)

⇒ 6 .x = 0

⇔ x = 0

Vậy x = 0.

b) 2.x + 3 = 9 

⇒2.x = 9 -3 

⇒  2.x = 6

⇔ x = 3

Vậy x = 3.

Áp dụng bất đẳng thức AM - GM:

\(\sqrt{\left(x^2-15\right)\left(x-3\right)}\le\dfrac{x^2-15+x-3}{2}=\dfrac{x^2+x-18}{2};\sqrt{x^2-15}\le\dfrac{x^2-15+1}{2}=\dfrac{x^2-14}{2};\sqrt{x-3}\le\dfrac{x-3+1}{2}=\dfrac{x-2}{2}\).

Do đó \(F\ge x^2+x-\dfrac{x^2+x-18}{2}-\dfrac{x^2-14}{2}-\dfrac{x-2}{2}-38=-21\).

Đẳng thức xảy ra khi x = 4.

Vậy...

a: =>x/-3=3

hay x=-9

b: =>x/9=-1/9

hay x=-1

c: =>x+1/5=-1/3

hay x=-8/15

d: =>-7/x=-7/9

hay x=9

a, \(\dfrac{x}{-3}=3\Leftrightarrow x=-9\)

b, \(\dfrac{x}{9}=-\dfrac{1}{9}\Rightarrow x=-1\)

c, \(\dfrac{x+3}{15}=-\dfrac{6}{15}\Rightarrow x=-9\)

d, \(\dfrac{42}{-54}=-\dfrac{42}{6x}\Rightarrow6x=54\Leftrightarrow x=9\)

Bài 2:

a: \(\Leftrightarrow x-1\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

=>\(x\in\left\{2;0;3;-1;4;-2;7;-5\right\}\)

b: \(\Leftrightarrow x+3\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{-2;-4;0;-6;2;-8;12;-18\right\}\)

c: \(\Leftrightarrow x+3\in\left\{1;-1;2;-2;3;-3;4;-4;6;-6;12;-12\right\}\)

=>\(x\in\left\{-2;-4;-1;-5;0;-6;1;-7;3;-9;9;-15\right\}\)

d: =>x+1+15 chia hết cho x+1

=>\(x+1\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

=>\(x\in\left\{0;-2;2;-4;4;-6;14;-16\right\}\)

d, x=-110

\(a,3\left(x-5\right)-4\left(x-3\right)=-12\)

\(\Leftrightarrow3x-15-4x+12=-12\)

\(\Leftrightarrow-x=-9\)

\(\Leftrightarrow x=9\)

Vậy x=9

\(c,20-\left(x-15\right)=x-15\)

\(\Leftrightarrow20-x+15=x-15\)

\(\Leftrightarrow2x=50\)

\(\Leftrightarrow x=25\)

Vậy x=25

a

\(x^2\left(2x+15\right)+4\left(2x+15\right)=0\\ \Leftrightarrow\left(2x+15\right)\left(x^2+4\right)=0\\ \Leftrightarrow2x+15=0\left(x^2+4>0\forall x\right)\\ \Leftrightarrow2x=-15\\ \Leftrightarrow x=-\dfrac{15}{2}\)

b

\(5x\left(x-2\right)-3\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(5x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\5x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0+2=2\\x=\dfrac{0+3}{5}=\dfrac{3}{5}\end{matrix}\right.\)

c

\(2\left(x+3\right)-x^2-3x=0\\ \Leftrightarrow2\left(x+3\right)-\left(x^2+3x\right)=0\\ \Leftrightarrow2\left(x+3\right)-x\left(x+3\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x+3=0\\2-x=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0-3=-3\\x=2-0=2\end{matrix}\right.\)

a: =>(2x+15)(x^2+4)=0

=>2x+15=0

=>2x=-15

=>x=-15/2

b; =>(x-2)(5x-3)=0

=>x=2 hoặc x=3/5

c: =>(x+3)(2-x)=0

=>x=2 hoặc x=-3

\(1.x-\dfrac{2}{3}\times\left(x+9\right)=1\)

\(x-\dfrac{2}{3}\times x-6=1\)

\(x\times\left(1-\dfrac{2}{3}\right)=7\)

\(x\times\dfrac{1}{3}=7\)

\(x=21\)

\(2.x-\dfrac{11}{15}=\dfrac{3+x}{5}\)

\(\dfrac{15x}{15}-\dfrac{11}{15}=\dfrac{9+3x}{15}\)

\(15x-11=9+3x\)

\(12x=20\)

\(x=\dfrac{5}{3}\)

(X-15)⁵=(x-15)³

(x-15)⁵-(x-15)³=0

(X-15)^5-3=0( bạn có thể bỏ bước này)

(X-15)²=0

=>x-15=0

=>x=15