`(4x-1)^3=27`

`=>(4x-1)^3=3^3`

`=>4x-1=3`

`=>4x=4=>x=1`

Vậy `x=1`

     \(\left(4x-1\right)^3=27\)

⇔ \(\left(4x-1\right)^3=3^3\)

⇒      \(4x-1=3\)

⇔             \(x=1\)

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

4x-1=3 hoặc 4x-1=-3

x=1 hoặc x=-1/2

Bài 1:

\(3a.\left(2a^2-ab\right)=6a^3-3a^2b\)

\(\left(4-7b^2\right).\left(2a+5b\right)=8a+20b-14ab^2-35b^3\)

Bài 2:

\(2x^2-6x+xy-3y=2x.\left(x-3\right)+y.\left(x-3\right)=\left(x-3\right).\left(2x+y\right)\)

Bài 3: Tại x = 3/2, y =1/3 thì Q = 67/9

Bài 4:

 \(\left(\frac{1}{x+1}+\frac{2x}{1-x^2}\right).\left(\frac{1}{x-1}\right)\) \(\frac{1}{\left(x+1\right).\left(x-1\right)}+\frac{2x}{\left(1-x^2\right).\left(x-1\right)}=\frac{x-1}{\left(x+1\right).\left(x-1\right)^2}+\frac{-2x}{\left(x-1\right)^2.\left(x+1\right)}\)  

= \(\frac{x-1-2x}{\left(x+1\right).\left(x-1\right)^2}=\frac{-\left(x+1\right)}{\left(x+1\right).\left(x-1\right)^2}=\frac{-1}{\left(x-1\right)^2}\)

\(\left(4x-1\right)^3=-\dfrac{1}{27}\)

\(\Rightarrow4x-1=-\dfrac{1}{3}\)

\(\Rightarrow4x=\dfrac{2}{3}\Rightarrow x=\dfrac{1}{6}\)

\(\left(4x-1\right)^3=\left(-\dfrac{1}{27}\right)\Leftrightarrow4x-1=-\dfrac{1}{3}\Leftrightarrow4x=\dfrac{2}{3}\Leftrightarrow x=\dfrac{1}{6}\)

c \(\frac{\left(2x-4\right)\left(x-3\right)}{\left(x-2\right)\left(3x^2-27\right)}=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x^2-9\right)}\)

\(=\frac{2\left(x-2\right)\left(x-3\right)}{3\left(x-2\right)\left(x-3\right)\left(x+3\right)}=\frac{2}{3\left(x+3\right)}\)

d, \(\frac{x^2+5x+6}{x^2+4x+4}=\frac{\left(x+2\right)\left(x+3\right)}{\left(x+2\right)^2}=\frac{x+3}{x+2}\)

Tương tự với a ; b 

0

6456 x 0=0

0

a) 2(x-1)² - 4(x+3)² + 2x(x-5)

= 2(x² -2x +1)- 4(x² + 6x +9) + 2x² -10x

= 2x² - 4x + 2 -4x² - 24x - 36 + 2x² - 10x

= (2x² + 2x² - 4x²) - (4x + 24x+10x) +(2-36)

= -38x-34

b) 2(2x+5)²  -3(4x+1)(1-4x)

= 2(4x² + 20x + 25) + 3(4x+1)(4x-1)

= 8x² +40x + 50 + 3(16x² -1)

= 8x² + 40x + 50 + 48x² - 3

=56x² +40x + 47

a, \(2\left(x-1\right)^2-4\left(x+3\right)^2+2x\left(x-5\right)\)

\(=2\left(x^2-2x+1\right)-4\left(x^2+6x+9\right)+2x\left(x-5\right)\)

\(=2x^2-4x+2-4x^2-24x-36+2x^2-10=-28x-44\)

b, \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)

\(=2\left(4x^2+20x+25\right)-3\left(1-16x^2\right)\)

\(=8x^2+40x+50-3+48x^2=56x^2+40x+47\)

mình sửa dòng cuối phần a nhé 

\(=2x^2-4x+2-4x^2-24x-36+2x^2-10x=-38x-34\)

\(4\left(x+3\right)\left(3x-2\right)-3\left(x-1\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow\left(4x+12\right)\left(3x-2\right)-\left(3x-3\right)\left(4x-1\right)=-27\)

\(\Leftrightarrow12x^2+36x-8x-24-12x^2+12x+3x-3=-27\)

\(\Leftrightarrow43x=0\)

\(\Leftrightarrow x=0\)

Vậy...

\(a,9\left(2x+1\right)=4\left(x-5\right)^2\)

\(4x^2-40x+100=18x+9\)

\(4x^2-58x+91=0\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{29+3\sqrt{53}}{4}\\x=\frac{29-3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}}\)

\(c,x^3+3x^2-6x-8=0\)

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

\(Th1:x+4=0\Leftrightarrow x=-4\)

\(Th2:x-2=0\Leftrightarrow x=2\)

\(Th3:x+1=0\Leftrightarrow x=-1\)

\(a,9.\left(2x+1\right)=4.\left(x-5\right)^2\)

\(< =>4x^2-40x+100=18x+9\)

\(< =>4x^2+58x+91=0\)

\(< =>\orbr{\begin{cases}x=\frac{29-3\sqrt{53}}{4}\\x=\frac{29+3\sqrt{53}}{4}\end{cases}}\)

\(b,x^3-4x^2-12x+27=0\)

\(< =>\left(x+3\right)\left(x^2-7x+9\right)=0\)

\(< =>\orbr{\begin{cases}x+3=0\\x^2-7x+9=0\end{cases}}\)

\(< =>\orbr{\begin{cases}x=-3\\x=\frac{7\pm\sqrt{13}}{2}\end{cases}}\)