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

\(\Leftrightarrow\left(x^2+1\right)^2=16x^2\Leftrightarrow x^4+2x^2+1=16x^2\Rightarrow x^4+1=14x^2\)(2)

\(\Leftrightarrow\left(x^2+1\right)^3=64x^3\Leftrightarrow x^6+3x^4+3x^2+1=64x^3\)

\(\Leftrightarrow x^6+3x^2\left(x^2+1\right)+1=64x^3\Leftrightarrow x^6+12x^3+1=64x^3\)

\(\Rightarrow x^6+1=52x^3\)(3)

Thay (1);(2);(3) vào T ta dược :

\(T=\left(\frac{x^2+1}{x}\right)^2+\left(\frac{x^4+1}{x^2}\right)^2+\left(\frac{x^6+1}{x^3}\right)^2\)

\(=\left(\frac{4x}{x}\right)^2+\left(\frac{14x^2}{x^2}\right)^2+\left(\frac{52x^3}{x^3}\right)^2=4^2+14^2+52^2=2916\)

a)

\(A=3\left(x-y\right)^2-2\left(x+y\right)^2-\left(x-y\right)\left(x+y\right)\)\(2A=\left[\left(x-y\right)-\left(x+y\right)\right]^2+5\left(x-y\right)^2-5\left(x+y\right)^2\)

\(2A=4y^2+5\left[\left(x-y\right)-\left(x+y\right)\right]\left[\left(x-y\right)+\left(x+y\right)\right]\)\(2A=4y^2+5\left[-2y\right]\left[2x\right]=4y^2-20xy=4y\left(y-5x\right)\\ \)\(A=2y\left(y-5x\right)\)

A. \(\left(x+2\right)\left(x+3\right)-\left(x-2\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left(x^2+3x+2x+6\right)-\left(x^2+5x-2x-10\right)=0\)
\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)
\(\Leftrightarrow x^2+3x+2x-x^2-5x+2x=-6-10\)
\(\Leftrightarrow2x=-16\)
\(\Leftrightarrow x=-8\) .Vậy \(S=\left\{-8\right\}\)

B. \(\left(2x+3\right)\left(x-4\right)+\left(x-5\right)\left(x-2\right)=\left(3x+5\right)\left(x-4\right)\)
\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x+5x-20\)
\(\Leftrightarrow2x^2-8x+3x+x^2-2x-5x-3x^2+12x-5x=12-10-20\)
\(\Leftrightarrow-5x=-18\)
\(\Leftrightarrow x=\dfrac{18}{5}\) . Vậy \(S=\left\{\dfrac{18}{5}\right\}\)

C. \(\left(8-4x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4\left(x^2+x-2x-2\right)=0\)
\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)
\(\Leftrightarrow8x-4x^2-8x+4x^2+4x-8x=-16+8\)
\(\Leftrightarrow-4x=-8\)
\(\Leftrightarrow x=2\) . Vậy \(S=\left\{2\right\}\)

D. \(\left(2x-3\right)\left(8x+2\right)=\left(4x+1\right)\left(4x-1\right)-3\)
\(\Leftrightarrow16x^2+4x-24x-6=16x^2+1^2-3\)
\(\Leftrightarrow16x^2+4x-24x-16x^2=6+1-3\)
\(\Leftrightarrow-20x=4\)
\(\Leftrightarrow x=-\dfrac{1}{5}\) . Vậy \(S=\left\{-\dfrac{1}{5}\right\}\)

a)(x+2)(x+3)-(x-2)(x+5)=0

\(\Leftrightarrow x^2+3x+2x+6-x^2-5x+2x+10=0\)

<=>2x=-16

<=>x=-8

b)(2x+3)(x-4)+(x-5)(x-2)=(3x-5)(x-4)

\(\Leftrightarrow2x^2-8x+3x-12+x^2-2x-5x+10=3x^2-12x-5x+20\)

\(\Leftrightarrow3x^2-12x-2=3x^2-17x+20\)

\(\Leftrightarrow5x=22\Leftrightarrow x=\dfrac{22}{5}\)

c)(8-4x)(x+2)+4(x-2)(x+1)=0

\(\Leftrightarrow8x+16-4x^2-8x+4x^2+4x-8x-8=0\)

\(\Leftrightarrow-4x=-8\Leftrightarrow x=2\)

d)(2x-3)(8x+2)=(4x+1)(4x-1)-3

\(\Leftrightarrow16x^2+4x-24x-6=16x^2-4x+4x-1-3\)

\(\Leftrightarrow-20x=-2\Leftrightarrow x=\dfrac{-1}{10}\)

1)⇔x²⁺1x-3x+3=0

⇔x(x+1)-3(x+1)=0

⇔(x+1)(x-3)=0

⇔x+1=0 hoặc x-3=0

⇔x=-1 hoặc x=3

4)⇔x(1+5x)=0

⇔x=0 hoặc 1+5x=0

⇔x=0 hoặc 5x=-1

⇔x=0 hoặc x=-0.2

a.\(x^3-x=0 \)

\(x(x^2-1)=0\)

x=0 hay x²-1=0

x=0 hay x²=1

x=0 hay x=1

Vậy x=0 hay x=1

b.\(x^3+1=0\)

\(x(x^2+1)=0\)

\(x=0 hay x^2+1=0\)

\(x=0 hay x^2=-1\)(vô lí vì x²≥0)

Vậy x=0

c.\(x^2-4x=0\)

\(x(x-4)=0\)

x=0 hay x-4=0

x=0 hay x=4

Vậy x=0 hay x=4

d.\(x(x-1)-2(1-x)=0\)

\(x(x-1)+2(x-1)=0 \)

\((x-1)(x+2)=0\)

x-1=0 hay x+2=0

x=1 hay x=-2

Vậy x=1 hay x=-2

e.\(2x(x-2)-(2-x)^2=0\)

\(2x(x-2)+(x-2)^2=0\)

\((x-2)(2x+x-2)=0\)

\((x-2)(3x-2)=0\)

x-2=0 hay 3x-2=0

x=2 hay 3x=2

x=2 hay x=2/3

Vậy x=2 hay x=2/3

f.\(4x(x+1)=8(x+1)\)

\(4x(x+1)-8(x+1)=0\)

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

_{4(x+1)=0 hay x-2=0}

_{x+1=0 hay x=2}

_{x=-1 hay x=2}

_{Vậy x=-1 hay x=2}

_{g.\(5x(x-2)-x+2=0\)}

\(5x(x-2)-(x-2)=0\)

\((x-2)(5x-1)=0\)

x-2=0 hay 5x-1=0

x=2 hay 5x=1

x=2 hay x=1/5

Vậy x=2 hay x=1/5

h.\((x+1)=(x+1)^2\)

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

\((x+1)(1-x-1)=0\)

\((x+1)(-x)=0\)

x+1= 0 hay -x=0

x=-1 hay x=0

Vậy x=-1 hay x=0

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