\(\Leftrightarrow\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-2\right)\left(x-5\right)}+\frac{9x}{x^2-7x+10}=10\)

\(\Leftrightarrow\frac{3x^2-15x-x^2+2x+9x}{\left(x-2\right)\left(x-5\right)}=10\)

\(\Leftrightarrow2x^2-4x=10x^2-70x+100\)

\(\Leftrightarrow8x^2-66+100=0\)

\(\Leftrightarrow4x^2-33x+50=0\)

\(\Leftrightarrow4x\left(x-2\right)-25\left(x-2\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(4x-25\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=\frac{25}{4}\end{matrix}\right.\)

b) [(x-7)(x-2)][(x-4)(x-5)]=72

<=> (x²-9x+14)(x²-9x+20)=72

Đặt x²-9x+17=a

=> (a+3)(a-3)=72

<=> a²-9=72

<=> a²=81

=> a=\(\left\{9;-9\right\}\)

TH1: a=9

=> x²-9x+17=9

<=> x²-9x+8=0

<=> (x-1)(x-8)=0

=> x=\(\left\{1;8\right\}\)

TH2: a=-9

=> x²-9x+17=-9

<=> x²-9x+26=0

<=> x²-9x+20,25+5,75=0

<=> (x-4,5)²+5,75=0

=> x\(\in\varnothing\)

Vậy x=\(\left\{1;8\right\}\)

a.x²-7x+10

⇔x²-2x-5x+10

⇔x(x-2)-5(x-2)

⇔(x-2)(x-5)

b.\(\left(12x^6y^4+9x^5y^3-15x^2y^3\right):3x^2y^3\)

=\(4x^4y+3x^3-5\)

=

x²-7x+10

= x²-5x-2x+10

=x(x-5)-2(x-5)

=(x-5)(x-2)

(12x⁶y⁴+9x⁵y³-15x²y³): 3x²y³

=4x⁴y+3x³-3

\(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{3x^2-15x-x^2+2x+9x}{x^2-7x+10}=10\)

\(\Rightarrow\frac{2x^2-4x}{x^2-7x+10}=10\)

\(\Rightarrow2x^2-4x=10x^2-70x+100\)

\(\Rightarrow8x^2-66x+100=0\)

Ta có \(\Delta=66^2-4.8.100=1156,\sqrt{\Delta}=34\)

\(\Rightarrow\orbr{\begin{cases}x=\frac{66+34}{16}=\frac{25}{4}\\x=\frac{66-34}{16}=2\end{cases}}\)

a) \(\frac{3x}{x-2}-\frac{x}{x-5}+\frac{9x}{x^2-7x+10}=10\)

<=> \(\frac{3x\left(x-5\right)}{\left(x-2\right)\left(x-5\right)}-\frac{x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}+\frac{9x}{\left(x-2\right)\left(x-5\right)}=10\)

<=> \(\frac{3x^2-15x-x^2+2x+9x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x^2-4x}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(\frac{2x\left(x-2\right)}{\left(x-5\right)\left(x-2\right)}=10\)

<=> \(2x=10\left(x-5\right)\)

<=> 2x - 10x = -50

<=> -8x = -50

<=>x = 6,25

Vậy S = {6,25}

b) (x - 7)(x - 2)(x - 4)(x - 5) = 72

<=> (x² - 9x + 14)(x² - 9x + 20) = 72

Đặt x² - 9x + 14 = t <=> t(t + 6) = 72

<=> t² + 6t - 72 = 0

<=> t² + 12t - 6t - 72 = 0

<=> (t + 12)(t - 6) = 0

<=> \(\orbr{\begin{cases}t+12=0\\t-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x^2-9x+14+12=0\\x^2-9x+14-6=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-9x+20,25\right)+5,75=0\\x^2-9x+8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x-4,5\right)^2+5,75=0\left(vn\right)\\x^2-x-8x+8=0\end{cases}}\)

<=> (x - 1)(x - 8) = 0

<=> \(\orbr{\begin{cases}x-1=0\\x-8=0\end{cases}}\)

<=> \(\orbr{\begin{cases}x=1\\x=8\end{cases}}\)

Vậy S = {1; 8}

9x² + y² + 2z² - 18x + 4z - 6y + 20 = 0

<=> 9x² - 18x + 9 + y² - 6y + 9 + 2x² + 4z + 2 = 0

<=> 9(x² - 2x + 1) + (y - 3)² + 2(z² + 2z + 1) = 0

<=> 9(x - 1)² + (y - 3)² + 2(z + 1)² = 0

<=> \(\left\{\begin{matrix}x-1=0\\y-3=0\\z+1=0\end{matrix}\right.\)

<=> \(\left\{\begin{matrix}x=1\\y=3\\z=-1\end{matrix}\right.\)

    \(2x^2-4xy+8y^2+7x+6y-15.\)

=  \(x^2+x^2-4xy+4y^2+4y^2+7x+6y-15\)   

=  \(\left(x^2-4xy+4y^2\right)+\left[x^2+7x+\left(\frac{7}{2}\right)^2\right]+\left[4y^2+6y+\left(\frac{3}{2}\right)^2\right]-\left(\frac{7}{2}\right)^2-\left(\frac{3}{2}\right)^2-15\)

=  \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\)

Vì \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2\ge0\forall x;y\)

=> \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\ge0-\frac{59}{2}\forall x;y\)

=> \(\left(x-2y\right)^2+\left(x+\frac{7}{2}\right)^2+\left(2y+\frac{3}{2}\right)^2-\frac{59}{2}\ge-\frac{59}{2}\)

Vậy GTNN của bt là  \(\frac{-59}{2}\Leftrightarrow\hept{\begin{cases}x-2y=0\\x+\frac{7}{2}=0\\2y+\frac{3}{2}=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=2y\Rightarrow\orbr{\begin{cases}x=-\frac{7}{4}\\y=-\frac{3}{2}\end{cases}}\\x=-\frac{7}{2}\\y=-\frac{3}{4}\end{cases}}\)

8) \(y^2-y-30=y^2+5y-6y-30=y\left(y+5\right)-6\left(y+5\right)=\left(y-6\right)\left(y+5\right)\)

9) \(y^2-8y+15=y^2-3y-5y+15=y\left(y-3\right)-5\left(y-3\right)=\left(y-5\right)\left(y-3\right)\)

10) \(y^2+y-6=y^2-2y+3y-6=y\left(y-2\right)+3\left(y-2\right)=\left(y+3\right)\left(y-2\right)\)

11) \(y^2-y-12=y^2+3y-4y-12=y\left(y+3\right)-4\left(y+3\right)=\left(y-4\right)\left(y+3\right)\)

12) \(x^2-5x+6=x^2-2x-3x+6=x\left(x-2\right)-3\left(x-2\right)=\left(x-3\right)\left(x-2\right)\)

13) \(u^2+u-42=u^2+7u-6u-42=u\left(u+7\right)-6\left(u+7\right)=\left(u-6\right)\left(u+7\right)\)

14) \(2x^2+x-6=2x^2+4x-3x-6=2x\left(x+2\right)-3\left(x+2\right)=\left(2x-3\right)\left(x+2\right)\)

15) \(7x^2+50x+7=7x^2+49x+x+7=7x\left(x+7\right)+\left(x+7\right)=\left(7x+1\right)\left(x+7\right)\)

16) \(12x^2+7x-12=12x^2+16x-9x-12=4x\left(3x+4\right)-3\left(3x+4\right)=\left(4x-3\right)\left(3x+4\right)\)

17) \(15x^2+7x-2=15x^2-3x+10x-2=3x\left(5x-1\right)+2\left(5x-1\right)=\left(3x+2\right)\left(5x-1\right)\)

18) \(2x^2-y^2+xy=2x^2+2xy-xy-y^2=2x\left(x+y\right)-y\left(x+y\right)=\left(2x-y\right)\left(x+y\right)\)

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