`x/2+x+x/3+x+x+x/4=5 3/4`

`=>3x+x/2+x/3+x/4=23/4`

`=>49/12x=23/4`

`=>x=69/49`

Vậy `x=69/49`

giúp mik với xin các bạn đó mik đnag cần gấp

Soạn đề bằng thanh công cụ cho rõ nha bạn

\(\left(x-2\right)\left(x^2+2x+4\right)-x\left(x-3\right)\left(x+3\right)=10\\ \Leftrightarrow x^3-8-x\left(x^2-9\right)=10\\ \Leftrightarrow x^3-8-x^3-9x=10\\ \Leftrightarrow-9x=18\\ \Leftrightarrow x=-2\)

x=2

(x−2)(x2+2x+4)−x(x−3)(x+3)=10⇔x3−8−x(x2−9)=10⇔x3−8−x3−9x=10⇔−9x=18⇔x=−2

70 - 11.[(5².3 - 11) : 12 - 2] = 70 - 11.[(25.3 - 11) : 12 - 2] = 70 - 11.[(75 - 11) : 12 - 2] = 70 - 11.[64 : 12 - 2] = 70 - 11.[64/12 - 12/6] = 70 - 11.[32/6 - 12/6] = 70 - 11.[20/6] = 70 - 11.[10/3] = 210/3 - 110/3 = 100/3. HT

Ta có :\(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=\left(-\frac{3}{4}+\frac{5}{22}+\frac{3}{26}\right)\)

=> \(\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right).\left(2x-2\right)=-\frac{1}{2}\left(\frac{3}{2}-\frac{5}{11}-\frac{3}{13}\right)\)

=> \(2x-2=-\frac{1}{2}\)

=> \(2x=\frac{3}{2}\)

=> \(x=\frac{3}{4}\)

\(B=\left|x-4\right|\left(2-\left|x-4\right|\right)\ge0\forall x\)

Dấu '=' xảy ra khi x=4

x=80

x + 20 \(⋮\) 10 

\(\Rightarrow\) CSTC của x là 0

Mà x \(⋮\) 8

\(\Rightarrow\) x \(\in\) {0; 80; 160; 240; 320; 400; ...}

Mà x < 300

\(\Rightarrow\) x \(\in\) {0; 80; 160; 240}

Mà x + 1 \(⋮\) 9

\(\Rightarrow\) x + 1 = 81

\(\Rightarrow\) x = 80

\(4^{x+3}+4^{x+2}+4^{x+1}+4^x=5440\)

\(\Rightarrow4^x.4^3+4^x.4^2+4^x.4+4^x=5440\)

\(\Rightarrow4^x\left(4^3+4^2+4+1\right)=5440\)

\(\Rightarrow4^x.\left(64+16+4+1\right)=5440\)

\(\Rightarrow4^x.85=5440\)

\(\Rightarrow4^x=5440:85\)

\(\Rightarrow4^x=64=4^3\)

\(\Rightarrow x=3\)

\(x+4⋮2x+1\)

=>\(2x+8⋮2x+1\)

=>\(2x+1+7⋮2x+1\)

=>\(7⋮2x+1\)

=>\(2x+1\in\left\{1;-1;7;-7\right\}\)

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

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

Ta có:

(x + 4) ⋮ (2x + 1)

⇒ 2(x + 4) ⋮ (2x + 1)

⇒ (2x + 8) ⋮ (2x + 1)

⇒ (2x + 1 + 7) ⋮ (2x +1)

⇒ 7 ⋮ (2x + 1)

⇒ 2x + 1 ∈ Ư(7) = {-7; -1; 1; 7}

⇒ 2x ∈ {-8; -2; 0; 6}

⇒ x ∈ {-4; -1; 0; 3}

Bài 1:

ta có: A = 11^9+11^8+..+11+1

=> 11A = 11^10+11^9+...+11^2+11

=> 11A-A = 11^10-1

10A = 11^10 -1

mà (11^10)-1 = (...1) - 1 = (...0) chia hết cho 10

=> A = (11^10-1):10 sẽ chia hết

=> A chia hết cho 5

Bài 2:

ta