Bỏ ngoặc ra, tìm nghiệm theo pt bậc 3.

= x^2(X-1) - 4(x^2-2x+1)

=x^2(x-1)-4(x-1)^2

=(x-1)(x^2-4x+4)

=(x-1)(x-2)^2

Ta có: \(\left(\frac23x-\frac12\right)\cdot\left(\frac{x}{4}-\frac18\right)=0\)

=>\(\left[\begin{array}{l}\frac23x-\frac12=0\\ \frac{x}{4}-\frac18=0\end{array}\right.\Rightarrow\left[\begin{array}{l}\frac23x=\frac12\\ \frac{x}{4}=\frac18\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac12:\frac23=\frac12\cdot\frac32=\frac34\\ x=\frac48=\frac12\end{array}\right.\)

a: \(\dfrac{x}{6}=\dfrac{8}{3}\)

=>\(x=6\cdot\dfrac{8}{3}=\dfrac{6}{3}\cdot8=8\cdot2=16\)

b: \(\dfrac{5}{x}=\dfrac{4}{9}\)

=>\(x=\dfrac{5\cdot9}{4}=\dfrac{45}{4}\)

c: \(\dfrac{x+3}{-4}=\dfrac{5}{20}\)

=>\(x+3=\dfrac{-4\cdot5}{20}=-1\)

=>x=-1-3=-4

d: \(\dfrac{7}{3+4x}=\dfrac{-2}{9}\)

=>\(4x+3=\dfrac{9\cdot7}{-2}=-\dfrac{63}{2}\)

=>\(4x=-\dfrac{63}{2}-3=-\dfrac{69}{2}\)

=>\(x=-\dfrac{69}{8}\)

f: ĐKXĐ: x<>1

\(\dfrac{3}{x-1}=\dfrac{x-1}{27}\)

=>\(\left(x-1\right)^2=3\cdot27=81\)

=>\(\left[{}\begin{matrix}x-1=9\\x-1=-9\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=10\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)

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tớ chịu.

hi hi.

a) Ta có: \(8x\left(2x-3\right)-4x\left(4x+3\right)=72\)

\(\Leftrightarrow16x^2-24x-16x^2-12x=72\)

\(\Leftrightarrow-36x=72\)

hay x=-2

b) Ta có: \(\left(x+2\right)\left(x+4\right)-x\left(x+2\right)=104\)

\(\Leftrightarrow x^2+6x+8-x^2-2x=104\)

\(\Leftrightarrow4x=96\)

hay x=24

c) Ta có: \(\left(x-1\right)\left(x+4\right)-x\left(x-1\right)=308\)

\(\Leftrightarrow x^2+3x-4-x^2+x=308\)

\(\Leftrightarrow4x=312\)

hay x=78

d) Ta có: \(15x\left(2x-3\right)-\left(5x+2\right)\left(6x-5\right)=-22\)

\(\Leftrightarrow30x^2-45x-30x^2+25x-12x+10=-22\)

\(\Leftrightarrow-32x=-32\)

hay x=1

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

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

Vậy a = 2; b = 1; c = 1.

Làm rõ hơn đi bạn