a, 3(6x−5)(4x+1)−(8x+3)(9x−2)=2033(6x−5)(4x+1)−(8x+3)(9x−2)=203

⇒3(24x2+6x−20x−5)−(72x2−16x+27x−6)=203⇒3(24x2+6x−20x−5)−(72x2−16x+27x−6)=203

⇒72x2−42x−15−72x2−11x+6=203⇒72x2−42x−15−72x2−11x+6=203

⇒−53x=203−6+15=212⇒−53x=203−6+15=212

nhầm òi 

 3(6x−5)(4x+1)−(8x+3)(9x−2)=2033(6x−5)(4x+1)−(8x+3)(9x−2)=203

⇒3(24x2+6x−20x−5)−(72x2−16x+27x−6)=203⇒3(24x2+6x−20x−5)−(72x2−16x+27x−6)=203

⇒72x2−42x−15−72x2−11x+6=203⇒72x2−42x−15−72x2−11x+6=203

⇒−53x=203−6+15=212⇒−53x=203−6+15=212

⇒x=−4

3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=2033(6x−5)(4x+1)−(8x+3)(9x−2)=203

\Leftrightarrow\left(18x-15\right)\left(4x+1\right)-\left(72x^2+27x-16x-6\right)=203⇔(18x−15)(4x+1)−(72x2+27x−16x−6)=203

\Leftrightarrow72x^2-60x+18x-15-72x^2-27x+16x+6=203⇔72x2−60x+18x−15−72x2−27x+16x+6=203

\Leftrightarrow\left(72x^2-72x^2\right)+\left(18x+16x-60x-27x\right)-\left(15-6\right)=203⇔(72x2−72x2)+(18x+16x−60x−27x)−(15−6)=203

\Leftrightarrow-53x-9=203⇔−53x−9=203

\Leftrightarrow-53x=212⇔−53x=212

\Leftrightarrow x=-4⇔x=−4

Vậy x=-4x=−4

\(2x-1^3+8\)

\(=2x-9\)

\(=\left(\sqrt{2x}\right)^2-3^2\)

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

_________

\(8x^3-12x^2+6x-1\)

\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)

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

_______________

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

\(=8x^3-12x^2+6x-1-1\)

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

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

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

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

________

\(9x^3-12x^2+6x-1\)

\(=x^3+8x^3-12x^2+6x-1\)

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

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

\(=\left(3x-1\right)\left(3x^2-5x+1\right)\)

b: 8x^3-12x^2+6x-1

=(2x)^3-3*(2x)^2*1+3*2x*1^2-1^3

=(2x-1)^3

c: =(8x^3-12x^2+6x-1)-1

=(2x-1)^3-1

=(2x-1-1)[(2x-1)^2+2x-1+1]

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

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

8x³ - 12x² + 6x - 1

= (2x)³ - 3.(2x)².1 + 3.2x.1 - 1³

= (2x - 1)³

--------------------

8x³ - 12x² + 6x - 2

= 8x³ - 12x² + 6x - 1 - 1

= (2x)³ - 3.(2x)².1 + 3.(2x).1 - 1³ - 1³

= (2x - 1)³ - 1³

= (2x - 1 - 1)[(2x - 1)² + (2x - 1).1 + 1]

= (2x - 2)(4x² - 4x + 1 + 2x - 1 + 1)

= 2(x - 1)(4x² - 2x + 1)

--------------------

9x³ - 12x² + 6x - 1

= x³ + 8x³ - 12x² + 6x - 1

= x³ + (2x)³ - 3.(2x)² + 3.2x.1² - 1³

= x³ + (2x - 1)³

= (x + 2x - 1)[x² - x.(2x - 1) + (2x - 1)²]

= (3x - 1)(x² - 2x² + x + 4x² - 4x + 1)

= (3x - 1)(3x² - 3x + 1)

a: =x^4-3x^5+4x^8

b: =2x^3+2x^2+4x

c: =4x^2+8x-5

d: =2x+3x^2+7x^4

a) \(\sqrt[]{x^2-4x+4}=x+3\)

\(\Leftrightarrow\sqrt[]{\left(x-2\right)^2}=x+3\)

\(\Leftrightarrow\left|x-2\right|=x+3\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=x+3\\x-2=-\left(x+3\right)\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}0x=5\left(loại\right)\\x-2=-x-3\end{matrix}\right.\)

\(\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

b) \(2x^2-\sqrt[]{9x^2-6x+1}=5\)

\(\Leftrightarrow2x^2-\sqrt[]{\left(3x-1\right)^2}=5\)

\(\Leftrightarrow2x^2-\left|3x-1\right|=5\)

\(\Leftrightarrow\left|3x-1\right|=2x^2-5\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=2x^2-5\\3x-1=-2x^2+5\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}2x^2-3x-4=0\left(1\right)\\2x^2+3x-6=0\left(2\right)\end{matrix}\right.\)

Giải pt (1)

\(\Delta=9+32=41>0\)

Pt \(\left(1\right)\) \(\Leftrightarrow x=\dfrac{3\pm\sqrt[]{41}}{4}\)

Giải pt (2)

\(\Delta=9+48=57>0\)

Pt \(\left(2\right)\) \(\Leftrightarrow x=\dfrac{-3\pm\sqrt[]{57}}{4}\)

Vậy nghiệm pt là \(\left[{}\begin{matrix}x=\dfrac{3\pm\sqrt[]{41}}{4}\\x=\dfrac{-3\pm\sqrt[]{57}}{4}\end{matrix}\right.\)

a.

ĐKXĐ: \(x\ge-\dfrac{5}{3}\)

\(9x^2-3x-\left(3x+5\right)-\sqrt{3x+5}=0\)

Đặt \(\sqrt{3x+5}=t\ge0\)

\(\Rightarrow9x^2-3x-t^2-t=0\)

\(\Delta=9+36\left(t^2+t\right)=\left(6t+3\right)^2\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3+6t+3}{18}=\dfrac{t+1}{3}\\x=\dfrac{3-6t-3}{18}=-\dfrac{t}{3}\end{matrix}\right.\)

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

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+5}=3x-1\left(x\ge\dfrac{1}{3}\right)\\\sqrt{3x+5}=-3x\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}3x+5=9x^2-6x+1\left(x\ge\dfrac{1}{3}\right)\\3x+5=9x^2\left(x\le0\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

c.

ĐKXĐ: \(x\ge-5\)

\(x^2-3x+2-x-5-\sqrt{x+5}=0\)

Đặt \(\sqrt{x+5}=t\ge0\)

\(\Rightarrow-t^2-t+x^2-3x+2=0\)

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

\(\Rightarrow\left[{}\begin{matrix}t=\dfrac{1+2x-3}{-2}=1-x\\t=\dfrac{1-2x+3}{-2}=x-2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x+5}=1-x\left(x\le1\right)\\\sqrt{x+5}=x-2\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x+5=x^2-2x+1\left(x\le1\right)\\x+5=x^2-4x+4\left(x\ge2\right)\end{matrix}\right.\)

\(\Leftrightarrow...\)

b.

ĐKXĐ: \(x\ge-\dfrac{8}{3}\)

\(\left(3x+2\right)^2-6-\sqrt{3x+8}=0\)

Đặt \(\sqrt{3x+8}=t\ge0\Rightarrow3x+2=t^2-6\)

\(\left(t^2-6\right)^2-6-t=0\)

\(\Leftrightarrow t^4-12t^2-t+30=0\)

\(\Leftrightarrow\left(t^2+t-5\right)\left(t^2-t-6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}t=3\\t=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{3x+8}=3\\\sqrt{3x+8}=\dfrac{\sqrt{21}-1}{2}\end{matrix}\right.\)

\(\Leftrightarrow...\)

a)\(A=x^5-36x^4+37x^3-69x^2+34x+15\)

=\(x^5-35x^4-x^4+35x^3+2x^2-70x^2+x^2-35x+x+15\)

=\(\left(x^4-x^3+x^2+x\right)\left(x-35\right)+x+15\)

=0+35+15=50(do x=35)

b, \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)

\(\Rightarrow3\left(24x^2+6x-20x-5\right)-\left(72x^2-16x+27x-6\right)=203\)

\(\Rightarrow72x^2-42x-15-72x^2-11x+6=203\)

\(\Rightarrow-53x=203-6+15=212\)

\(\Rightarrow x=-4\)

Chúc bạn học tốt!!!

a ) \(4x\left(5x+2\right)-\left(10x-3\right)\left(2x+7\right)=133\)

\(\Leftrightarrow20x^2+8x-\left(20x^2-6x+70x-21\right)=133\)

\(\Leftrightarrow20x^2+8x-20x^2+6x-70x+21=133\)

\(\Leftrightarrow-56x+21=133\)

\(\Leftrightarrow-56x=112\)

\(\Leftrightarrow x=-2\)

Vậy \(x=-2\)

b ) \(3\left(6x-5\right)\left(4x+1\right)-\left(8x+3\right)\left(9x-2\right)=203\)

\(\Leftrightarrow\left(18x-15\right)\left(4x+1\right)-\left(72x^2+27x-16x-6\right)=203\)

\(\Leftrightarrow72x^2-60x+18x-15-72x^2-27x+16x+6=203\)

\(\Leftrightarrow\left(72x^2-72x^2\right)+\left(18x+16x-60x-27x\right)-\left(15-6\right)=203\)

\(\Leftrightarrow-53x-9=203\)

\(\Leftrightarrow-53x=212\)

\(\Leftrightarrow x=-4\)

Vậy \(x=-4\)