\(\dfrac{3x+2}{4}=\dfrac{16}{3x+2}\)

`=> (3x+2)^2 =4.16`

`=> (3x+2)^2 = 64`

`=> (3x+2)^2 = +- 8^2`

\(\Rightarrow\left[{}\begin{matrix}3x+2=8\\3x+2=-8\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}3x=6\\3x=-10\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{10}{3}\end{matrix}\right.\)

\(\sqrt{x^4+3x^2-4}+3x=\sqrt{3x^4+16}\)

\(\Leftrightarrow\sqrt{3x^4+16}-\sqrt{x^4+3x^2-4}=3x\)

\(\Leftrightarrow4x^4+3x^2+12-2\sqrt{3x^4+16}.\sqrt{x^4+3x^2-4}=9x^2\)

Đặt \(x^2=a\ge0\)

\(\Leftrightarrow2a^2-3a+6=\sqrt{3a^2+16}.\sqrt{a^2+3a-4}\)

\(\Leftrightarrow\left(2a^2-3a+6\right)^2=\left(3a^2+16\right).\left(a^2+3a-4\right)\)

\(\Leftrightarrow\left(a^2+4\right)\left(a^2-21a^2+25\right)=0\)

Chọn B

1) \(2x^4+3x^3-x^2+3x+2=0\)

\(\Rightarrow2x^4+x^3+2x^3+x^2-2x^2-x+4x+2=0\)

\(\Rightarrow x^3\left(2x+1\right)+x^2\left(2x+1\right)-x\left(2x+1\right)+2\left(2x+1\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x^3+x^2-x+2\right)=0\)

\(\Rightarrow\left(2x+1\right)\left(x^3+2x^2-x^2-2x+x+2\right)=0\)

\(\Rightarrow\left(2x+1\right)\left[x^2\left(x+2\right)-x\left(x+2\right)+\left(x+2\right)\right]=0\)

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

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\\x^2-x+1=0\end{matrix}\right.\)

Ta có:

\(x^2-x+1\)

\(=x^2-2x.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}+1\)

\(=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

Vì \(\left(x-\dfrac{1}{2}\right)^2\ge0\) với mọi x

\(\Rightarrow\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\) với mọi x

\(\Rightarrow x^2-x+1\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}2x+1=0\\x+2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=-2\end{matrix}\right.\)

3) \(\left(x+2\right)^4+\left(x+4\right)^4=16\)

Đặt x + 3 = a, ta được

\(\left(a-1\right)^4+\left(a+1\right)^4=16\)

\(\Rightarrow\left[\left(a-1\right)^2\right]^2+\left[\left(a+1\right)^2\right]^2=16\)

\(\Rightarrow\left(a^2-2a+1\right)^2+\left(a^2+2a+1\right)^2=16\)

\(\Rightarrow a^4+4a^2+1+2a^2-4a^3-4a+a^4+4a^2+1+2a^2+4a^3+4a=16\)

\(\Rightarrow2a^4+2.4a^2+2+2.2a^2=16\)

\(\Rightarrow2a^4+8a^2+4a^2+2=16\)

\(\Rightarrow2a^4+12a^2+2-16=0\)

\(\Rightarrow2a^4+12a^2-14=0\)

\(\Rightarrow2a^4-2a^2+14a^2-14=0\)

\(\Rightarrow2a^2\left(a^2-1\right)+14\left(a^2-1\right)=0\)

\(\Rightarrow\left(a^2-1\right)\left(2a^2+14\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right).2\left(a^2+7\right)=0\)

\(\Rightarrow\left(a-1\right)\left(a+1\right)\left(a^2+7\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\\a^2+7=0\end{matrix}\right.\)

Vì \(a^2\ge0\) với mọi a

\(\Rightarrow a^2+7\ge7\) với mọi a

\(\Rightarrow a^2+7\) vô nghiệm

\(\Rightarrow\left[{}\begin{matrix}a-1=0\\a+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+3-1=0\\x+3+1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x+2=0\\x+4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=-4\end{matrix}\right.\)

(3x + 4) : 4 = 20-16

(3x + 4) : 4 =  4 

3x + 4          =  4 x 4 

3x + 4          = 16

3x                 = 16- 4 

3x                  = 12

x                     = 12 : 3 

x                     = 4

(3x + 4) : 4 = 20-16

(3x + 4) : 4 =  4 

3x + 4          =  4 x 4 

3x + 4          = 16

3x                 = 16- 4 

3x                  = 12

x                     = 12 : 3 

x                     = 4

Xin lỗi bạn nhé, đây là đáp án đúng;
20 - ( 3x + 4 ) / 4 = 16
      ( 3x + 4 ) / 4 = 20 - 16
               3x + 4 = 4 * 4
                     3x = 16 - 4  
                       x = 12 / 3
                       x = 4

Tìm số nguyên x, biết:
1) -16 + 23 + x = - 16

7+x=-16

    x=-16-7

    x=-23
2) 2x – 35 = 15

2x=15+35

2x=50

  x=50:2

  x=25
3) 3x + 17 = 12

3x=12-17

3x=-5

  x=-5/3
4) (2x – 5) + 17 = 6

2x-5=6-17

2x-5=-11

2x=-11+5

2x=-6

  x=-6:2

  x=-3
5) 10 – 2(4 – 3x) = -4

2(4-3x)=10-(-4)

2(4-3x)=14

4-3x=14:2

4-3x=7

3x=4-7

3x=-3

  x=-3:3

  x=-1
6) - 12 + 3(-x + 7) = -18

3(-x+7)=-18-(-12)

3(x+7)=-6

x+7=-6:3

x+7=-2

    x=-2-7

    x=-9

tự đi mà làm

`3x^4+2x^2=16`

Đặt `x^2=t (t >=0)` có:

`3t^2+2t=16`

`<=>3t^2+2t-16=0`

`<=>` \(\left[ \begin{array}{l}t=2(TM)\\t=\dfrac{-8}{3}(L)\end{array} \right.\) 

Với `t=2` : `x^2=2 <=> x= \pm \sqrt2`

Vậy `S={\pm \sqrt2}`.

Ta có: \(3x^4+2x^2=16\)

\(\Leftrightarrow3x^4+2x^2-16=0\)

\(\Leftrightarrow3x^4+6x^2-8x^2-16=0\)

\(\Leftrightarrow3x^2\left(x^2+2\right)-8\left(x^2+2\right)=0\)

\(\Leftrightarrow\left(x^2+2\right)\left(3x^2-8\right)=0\)

mà \(x^2+2>0\forall x\)

nên \(3x^2-8=0\)

\(\Leftrightarrow3x^2=8\)

\(\Leftrightarrow x^2=\dfrac{8}{3}\)

\(\Leftrightarrow x\in\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)

Vậy: \(S=\left\{\dfrac{2\sqrt{6}}{3};-\dfrac{2\sqrt{6}}{3}\right\}\)