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

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

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

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

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

a, Đặt \(2^x=t,t>0\)

Pt trở thành: \(t^2-10t+16=0\Leftrightarrow\left(t-2\right)\left(t-8\right)=0\Leftrightarrow\orbr{\begin{cases}t=2\\t=8\end{cases}\left(tm\right)}\)

Nếu t=2 => x=1

nếu t=8=> x=3

Vậy x=...

b, Đặt: \(2x^2-3x-1=t\)

pt trở thành: \(t^2-3\left(t-4\right)-16=0\Leftrightarrow t^2-3t-4=0\Leftrightarrow\left(t+1\right)\left(t-4\right)=0\Leftrightarrow\orbr{\begin{cases}t=-1\\t=4\end{cases}}\)

* Nếu t=-1 <=> \(2x^2-3x-1=-1\Leftrightarrow x\left(2x-3\right)=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{3}{2}\end{cases}}\)

* Nếu t=4 <=> \(2x^2-3x-1=4\Leftrightarrow2x^2-3x-5=0\Leftrightarrow\left(x+1\right)\left(2x-5\right)=0\Leftrightarrow\orbr{\begin{cases}x=-1\\x=\frac{5}{2}\end{cases}}\)

Vậy x=...

Đặt x+1/x=t thì x^2+1/x^2=t^2-2

Phương trình viết lại:3(t^2-2)-16t(+-)26=0

Sau đó quy về giải phương trình bậc hai là được

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

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

\(\Leftrightarrow4x+6=0\)

\(\Leftrightarrow4x=-6\Leftrightarrow x=-\dfrac{3}{2}\)

Vậy ...

b) sửa đề: \(9x^2-16-x\left(3x+4\right)=0\)

\(\Leftrightarrow\left(3x\right)^2-4^2-\left(3x+4\right)=0\)

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

\(\Leftrightarrow\left(3x+4\right)\left(3x-4-1\right)=0\)

\(\Leftrightarrow\left(3x+4\right)\left(3x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+4=0\\3x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{4}{3}\\x=\dfrac{5}{3}\end{matrix}\right.\)

Vậy ...

\(\left\{{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left(x-16\right)\left(3x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-16=0\\3x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=16\\x=\dfrac{1}{3}\end{matrix}\right.\)

=>x-16=0 hoặc 3x-1=0

    x     = 16         x   = ⅓

\(a,\dfrac{x-3}{x}=\dfrac{x-3}{x+3}\)\(\left(đk:x\ne0,-3\right)\)

\(\Leftrightarrow\dfrac{x-3}{x}-\dfrac{x-3}{x+3}=0\)

\(\Leftrightarrow\dfrac{\left(x-3\right)\left(x+3\right)-x\left(x-3\right)}{x\left(x+3\right)}=0\)

\(\Leftrightarrow x^2-9-x^2+3x=0\)

\(\Leftrightarrow3x-9=0\)

\(\Leftrightarrow3x=9\)

\(\Leftrightarrow x=3\left(n\right)\)

Vậy \(S=\left\{3\right\}\)

\(b,\dfrac{4x-3}{4}>\dfrac{3x-5}{3}-\dfrac{2x-7}{12}\)

\(\Leftrightarrow\dfrac{4x-3}{4}-\dfrac{3x-5}{3}+\dfrac{2x-7}{12}>0\)

\(\Leftrightarrow\dfrac{3\left(4x-3\right)-4\left(3x-5\right)+2x-7}{12}>0\)

\(\Leftrightarrow12x-9-12x+20+2x-7>0\)

\(\Leftrightarrow2x+4>0\)

\(\Leftrightarrow2x>-4\)

\(\Leftrightarrow x>-2\)

20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

1) -16 + 23 + x = - 16

ó x = -23
2) 2x – 35 = 15

ó 2x = 50

ó x = 25
3) 3x + 17 = 12

ó 3x = -5

ó x = -5/3
4) │x - 1│= 0

=> x – 1 = 0 ó x = 1
5) -13 .│x│ = -26

ó |x| = 2

ó x = 2 hoặc x = -2

1     -16 + 23 + x = -16

       7 + x = -16     

               x = -16 - 7

                  x = -23

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

 Vì -5 ko chia hết cho 3 nên=> ko có x

4)/x-1/=0

Vì chỉ có /0/ là=0 nên x-1=0

 x-1=0

    x=0+1

    x=1

5)-13./x/= -26

        /x/= -26:(-13)

        /x/=2

   Vì /2/ và /-2/ đều = 2 nên =>x thuộc{2;-2}