\(x^2+x=12\)

\(\Rightarrow x^2+x-12=0\)

\(\Rightarrow x\left(x-3\right)+4\left(x-3\right)=0\Rightarrow\left(x-3\right)\left(x+4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\)

a) Ta có B (12) = {0;12;24;36;48;60;72,..}. Mà 20 ≤ x ≤ 50 ;

=> xϵ {24;36;48}.

b) x ϵ {10;20}.

c) x ϵ {0;7;14;21;28;35;42;49}.

d) x ϵ {1;2;3;4;6;12

a)

\(\sqrt{9x^2}=12\\ < =>\left(\sqrt{9x^2}\right)^2=12^2\\ < =>9x^2=144\\ < =>x^2=16\\ < =>\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)

b)

\(\sqrt{25x^2}=\left|-50\right|\\ < =>\sqrt{25x^2}=50\left(vì-50< 0\right)\)

\(< =>\left(\sqrt{25x^2}\right)^2=50^2\\ =>25x^2=2500\\ < =>x^2=100\\ < =>\left[{}\begin{matrix}x=10\\x=-10\end{matrix}\right.\)

a, x = 4

b, x = 10 

\(a)\sqrt{9x^2}=12\\ \Leftrightarrow\sqrt{\left(3x\right)^2}=12\\ \Leftrightarrow\left|3x\right|=12\\ \Leftrightarrow\left|x\right|=\dfrac{12}{3}\\ \Leftrightarrow x=4\)

Vậy x=4

\(b)\sqrt{25x^2}=\left|-50\right|\\ \Leftrightarrow\sqrt{\left(5x\right)^2}=\left|-50\right|\\ \Leftrightarrow\left|5x\right|=50\\ \Leftrightarrow\left|x\right|=\dfrac{50}{5}\\ \Leftrightarrow x=10\)

Vậy x=10

( 2x + 1 / 2 ) * ( 3 / 4 - x ) = 0

=> 2x + 1 / 2 = 0 hoặc 3 / 4 - x = 0

               2x  = - 1 / 2             x = 3 / 4 

                 x  = - 1 / 4                

Vậy x = -1 / 4 hoặc x = 3 / 4

2 / 10 * 12 + 2 / 12 * 14 + 2 / 14 * 16 + ... + 2 / 48 * 50

= 1 / 10 - 1 / 12 + 1 / 12 - 1 / 14 + 1 / 14 + 1 / 16 + .... + 1 / 48 - 1/ 50

= 1 / 10 - 1 / 50 

= 2 / 25

Tìm x : \(\left(2x+\frac{1}{2}\right)\left(\frac{3}{4}-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+\frac{1}{2}=0\\\frac{3}{4}-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=\frac{3}{4}\end{cases}}}\)

Tính nhanh : \(\frac{2}{10.12}+\frac{2}{12.14}+...+\frac{2}{48.50}\)

\(=2\left(\frac{1}{10.12}+\frac{1}{12.14}+...+\frac{1}{48.50}\right)\)

\(=2.\frac{1}{2}\left(\frac{1}{10}-\frac{1}{12}+\frac{1}{12}-\frac{1}{14}+...+\frac{1}{48}-\frac{1}{50}\right)\)

\(=\frac{1}{10}-\frac{1}{50}\)

\(=\frac{2}{25}\)

\(x-\dfrac{4}{12}=-\dfrac{5}{6}\)

⇔\(x-\dfrac{1}{3}=-\dfrac{5}{6}\)

⇔\(x=-\dfrac{1}{2}\)

\(\Leftrightarrow x=-\dfrac{1}{2}\)

=>(x-4).6=12.(-5)

=>6x-24=-60

=>6x=-36

=x=-6

2x - \(\frac{1}{2}-\frac{1}{6}-\frac{1}{12}-....-\frac{1}{49.50}\)= 7-\(\frac{1}{50}\)+x

2x - x - \(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\right)\)= 7 - \(\frac{1}{50}\)

x - \(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\right)\)= \(\frac{349}{50}\)

x - \(\left(1-\frac{1}{50}\right)\)=\(\frac{349}{50}\)

x - \(\frac{49}{50}\)=\(\frac{349}{50}\)

x = \(\frac{349}{50}+\frac{49}{50}\)

x = \(\frac{199}{25}\)

a) Có: 12 ⋮ x; 30 ⋮ x; 18 ⋮ x

⇒ x ∈ ƯC(12; 30; 18)

Lại có:

12 = 2² . 3

30 = 2 . 3 . 5 

18 = 2 . 3²

⇒ ƯCLN(12; 30; 18) = 2 . 3 = 6

⇒ ƯC(12; 30; 18) = Ư(6) = {1; 2; 3; 6; -1; -2; -3; -6}

⇒ x ∈ {1; 2; 3; 6; -1; -2; -3; -6}

Mà: x ≤ 3 nên x ∈ {1; 2; 3; -1; -2; -3; -6}

b) -34 - (x - 2)² = -50

⇒ (x - 2)² = -34 - (-50)

⇒ (x - 2)² = 16

⇒ (x - 2)² = 4² hoặc (x - 2)² = (-4)²

⇒ x - 2 = 4 hoặc x - 2 = -4

⇒ x = 6 hoặc x = -2

a) 12 ⋮ x ; 30 ⋮ x ; 18 ⋮ x và x ≤ 3

Gọi số cần tìm là x, theo đề bài, ta có:

12 ⋮ x ; 30 ⋮ x ; 18 ⋮ x ⇒ x ϵ ƯC(12,30,18)

⇒ Ta có:

12 = 2².3

30 = 2.3.5

18 = 2.3²

⇒ ƯC(12,30,18) = 2.3 = 6

⇒ Ư(6) = {1;2;3;6}

Mà x ≤ 3 ⇒ x = 2 hoặc x = 3

⇒ x = 2 hoặc x = 3

b) -34 - (x - 2)² = -50

(x - 2)² = -34 - (-50)

(x - 2)² = -34 + 50

(x - 2)² = 16

(x - 2)² = 4² (cùng số mũ)

⇒ x - 2 = 4

x = 4 + 2

x = 6

(2x + 1).(3y - 2) = 12

Lập bảng:

Vậy...

\(1+2+3+...+x=50\Leftrightarrow\frac{\left(x+1\right).x}{2}=50\Leftrightarrow\left(x+1\right).x=100??\)

1,tự lập bảng rồi tìm

2,\(1+2+3+..+x=50\Leftrightarrow\frac{x\left(x+1\right)}{2}=50\Leftrightarrow x\left(x+1\right)=100\)

sai đề?

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