\(\Rightarrow\left(x-4\right)\left(2x+x-4\right)=0\\ \Rightarrow\left(x-4\right)\left(3x-4\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{4}{3}\end{matrix}\right.\)

      \(8\left(x-\frac{1}{2}\right)\left(x^2+\frac{1}{2}x+\frac{1}{4}\right)-4x\left(1-x+2x^2\right)+2=0\)

\(\Leftrightarrow8\left[x^3-\left(\frac{1}{2}\right)^3\right]-4x+4x^2-8x^3+2=0\)

\(\Leftrightarrow8x^3-1-4x+4x^2-8x^3+2=0\)

\(\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\)

\(\Leftrightarrow x=\frac{1}{2}\)

8(x-1/2)(x^2+1/2x+1/4)  -  4x(1-x+2x^2)+2=0

=>   8𝑥^3 − 1  −  8𝑥^3 + 4𝑥2 − 4𝑥 + 2  =  0

=>   4𝑥2 − 4𝑥 + 1 = 0

=>  ( 2x - 1 )^2  = 0

=>  2x - 1 = 0

=>  2x  =  1

=>  x  = 1/2

<=> 8( x³ - 1/8 ) - 4x + 4x² - 8x³ + 2 = 0

<=> 8x³ - 1 - 4x + 4x² - 8x³ + 2 = 0

<=> 4x² - 4x + 1 = 0

<=> ( 2x - 1 )² = 0 <=> x = 1/2

\(1.\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}.\Leftrightarrow\dfrac{x-1-3x}{3}=\dfrac{x-2}{2}.\Leftrightarrow\dfrac{-2x-1}{3}-\dfrac{x-2}{2}=0.\)

\(\Leftrightarrow\dfrac{-4x-2-3x+6}{6}=0.\Rightarrow-7x+4=0.\Leftrightarrow x=\dfrac{4}{7}.\)

\(2.\left(x-2\right)\left(2x-1\right)=x^2-2x.\Leftrightarrow\left(x-2\right)\left(2x-1\right)-x\left(x-2\right)=0.\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1-x\right)=0.\Leftrightarrow\left(x-2\right)\left(x-1\right)=0.\)

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

\(3.3x^2-4x+1=0.\Leftrightarrow\left(x-1\right)\left(x-\dfrac{1}{3}\right)=0.\Leftrightarrow\left[{}\begin{matrix}x=1.\\x=\dfrac{1}{3}.\end{matrix}\right.\)

\(4.\left|2x-4\right|=0.\Leftrightarrow2x-4=0.\Leftrightarrow x=2.\)

\(5.\left|3x+2\right|=4.\Leftrightarrow\left[{}\begin{matrix}3x+2=4.\\3x+2=-4.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}.\\x=-2.\end{matrix}\right.\)

\(1,\dfrac{x-1}{3}-x=\dfrac{2x-4}{4}\\ \Leftrightarrow\dfrac{x-1}{3}-x=\dfrac{x-2}{2}\\ \Leftrightarrow\dfrac{2\left(x-1\right)-6x}{6}=\dfrac{3\left(x-2\right)}{6}\\ \Leftrightarrow2\left(x-1\right)-6x=3\left(x-2\right)\\ \Leftrightarrow2x-2-6x=3x-6\\ \Leftrightarrow-4x-2=3x-6\)

\(\Leftrightarrow3x-6+4x+2=0\\ \Leftrightarrow7x-4=0\\ \Leftrightarrow x=\dfrac{4}{7}\)

\(2,\left(x-2\right)\left(2x-1\right)=x^2-2x\\ \Leftrightarrow2x^2-4x-x+2=x^2-2x\\ \Leftrightarrow x^2-3x+2=0\\ \Leftrightarrow\left(x^2-2x\right)-\left(x-2\right)=0\\ \Leftrightarrow x\left(x-2\right)-\left(x-2\right)=0\\ \Leftrightarrow\left(x-1\right)\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

\(3,3x^2-4x+1=0\\ \Leftrightarrow\left(3x^2-3x\right)-\left(x-1\right)=0\\ \Leftrightarrow3x\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)\left(3x-1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{3}\end{matrix}\right.\)

\(4,\left|2x-4\right|=0\\ \Leftrightarrow2x-4=0\\ \Leftrightarrow2x=4\\ \Leftrightarrow x=2\)

\(5,\left|3x+2\right|=4\\ \Leftrightarrow\left[{}\begin{matrix}3x+2=4\\3x+2=-4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=2\\3x=-6\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-2\end{matrix}\right.\)

\(6,\left|2x-5\right|=\left|-x+2\right|\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=-x+2\\2x-5=x-2\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}3x=7\\x=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=3\end{matrix}\right.\)

a: P(x)=2x^5-2x^5+4x^4-3x^4+5=x^4+5

Q(x)=-5x^4+2x^4-x^3+3x^2-10x+2

=-3x^4-x^3+3x^2-10x+2

b: P(x)+Q(x)

=x^4+5-3x^4-x^3+3x^2-10x+2

=-2x^4-x^3+3x^2-10x+7

Q(x)-P(x)

=-3x^4-x^3+3x^2-10x+2-x^4-5

=-4x^4-x^3+3x^2-10x-3

P(x)-Q(x)=-(Q(x)-P(x))

=4x^4+x^3-3x^2+10x+3

a)x^m+4+x^m+3-x-1

=(x^m+4-x)+(x^m+3-1)

=x(x^m+3-1)+(x^m+3-1)

=(x+1)(x^m+3-1)

Với x=-2 ta có:... bn tự thay

b)x⁶-x⁴+2x³+2x²=x⁶-2x⁵+2x⁴+2x⁵-4x⁴+4x³+x⁴-2x³+2x²

=x⁴(x²-2x+2)+2x³(x²-2x+2)+x²(x²-2x+2)

=(x⁴+2x³+x²)(x²-2x+2)

=[x²(x²+2x+1)](x²-2x+2)

=x²(x+1)²(x²-2x+2)

Với x=-2 bn tự thay nhé h mk bận

\(\left(-x-2\right)^2+\left(x-2\right)\left(x^2+2x+4\right)-x^2\left(x-6\right)\)

\(=-x^3-6x^2-12x-8+x^3-8-x^3+6x^2\)

\(=-x^3-12x-16\)

a) Ta có: \(\sqrt{2x-1}\)

Biểu thức này có nghĩa là: \(2x-1\ge0\Leftrightarrow x\ge\dfrac{1}{2}\)

b) Ta có: \(\sqrt{4-x}\)

Biểu thức này có nghĩa là: \(4-x\ge0\Leftrightarrow x\le4\)

c) Ta có: \(\sqrt{\dfrac{3x}{2}}\)

Biểu thức này có nghĩa là: \(\dfrac{3x}{2}\ge0\Leftrightarrow x\ge0\)

d) Ta có: \(\sqrt{2x^2}\)

Biểu thức có nghĩa là: \(2x^2\ge0\Leftrightarrow x^2\ge0\) với mọi x

mk chỉnh lại đề nhé:  

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

\(\Leftrightarrow\) \(\left(\frac{x+1}{x-2}\right)^2+\frac{x+1}{x-4}-12\left(\frac{x-2}{x-4}\right)^2=0\)

Đặt:    \(\frac{x+1}{x-2}=a;\)   \(\frac{x-2}{x-4}=b\)

\(\Rightarrow\)\(a.b=\frac{x+1}{x-2}.\frac{x-2}{x-4}=\frac{x+1}{x-4}\)

Khi đó phương trình trở thành:

   \(a^2-ab-12b^2=0\)

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

đến đây bn thay trở lại rồi tìm nghiệm

+)  \(a=3b\) thì  phương trình vô nghiệm

+)  \(a=-4b\)thì phương trình có tập nghiệm     \(S=\left\{3;\frac{4}{5}\right\}\)

P/S: bn tham khảo nhé

bạn chắt mình giải đúng chứ

gio con noc ha ?!

<=> 2x^2 +x-4x-2-5x-15=2x^2-6x+4+8x-2-2x

      2x^2-8x-17-2x^2-2=0

     -8x-19=0

x=-19/8

(x-2).(2x+1)-5(x+3)=2x(x-3)+4(1+2x)-2(1+x)

3x-4x+x-2-5x-15=3x-6x+4+8x-2-2x

-5x-17=3x+2

-19=8x

-19/8=x

Vậy x=-19/8