10x = 6y => x = 3y/5 
thay vao ta co : 
2(3y/5)^2 - y^2 = -28 
<=> 18y^2/25 - y^2 = -28 
<=> 7y^2 = 700 
<=> y = 10 
=> x = 6 

\(10x=6y\) => \(x=\frac{6y}{10}=\frac{3y}{5}\)

=> \(2x^2-y^2=2\times\left(\frac{3y}{5}\right)^2-y^2=-28\)

<=> \(2\times\frac{9y^2}{25}-y^2=-28\)

<=> \(\frac{18y^2}{25}-y^2=-28\)

<=> \(\frac{-7y^2}{25}=-28\)

<=> \(-7y^2=-700\)

<=> \(y^2=100\)

<=> \(y=10;x=6\) hoặc \(y=-10;x=-6\)

\(10x=6y\Rightarrow\frac{x}{6}=\frac{y}{10}\Rightarrow\frac{x^2}{36}=\frac{y^2}{100}\Rightarrow\frac{2x^2}{72}=\frac{y^2}{100}\)

Theo tính chất dãy tỉ số bằng nhau 

\(\frac{2x^2}{72}=\frac{y^2}{100}=\frac{2x^2-y^2}{72-100}=-\frac{28}{-28}=1\)

\(\Rightarrow2x^2=72\Leftrightarrow x^2=36\Leftrightarrow x=\pm6\)

\(\Rightarrow y^2=100\Rightarrow y=\pm10\)

10x=6y=>x/6=y/10=>2x^2/72=y^2/100

áp dụng tính chất dãy tỉ số bang nhau ta có

\(^{2x^2}_{72}\)=\(^{y^2}_{100}\)=2x^2-y^2/72-100=-28/-28=1

=>x=6,y=10

10x=6y suy ra y/10 = x/6 suy ra y^2/100= x^2/36 suy ra y^2/100=2.x^2/72

2x^2-y^2= -28 nên y^2 -2x^2=28

y^2/100= 2x^2/72 =( y^2 - 2x^2)/(100-72)= 28/28 =1

ý^2/100=1 suy ra y^2=100 suy ra ý=10 hoặc ý=-10

2x^2/72=1 suy ra 2x^2=72 suy ra x^2= 36 suy ra x=6 hoặc x= -6

đặt \(\frac{x}{5}=\frac{y}{3}\text{ }=k\)

\(\Rightarrow\text{ }x=5k\text{ };\text{ }y=3k\)

\(\Rightarrow\left(5k\right)^2-\left(3k\right)^2=4\)

\(\Rightarrow\text{ }25k^2-9k^2=4\)

\(\Rightarrow\text{ }k^2.\left(25-9\right)=4\)

\(\Rightarrow\text{ }k^2.16=4\)

\(\Rightarrow\text{ }k^2=\frac{1}{4}=\left(\frac{1}{2}\right)^2\)

\(\Rightarrow\text{ }\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)

Nếu k = \(\frac{1}{2}\)thì \(x=\frac{5}{2}\text{ };\text{ }y=\frac{3}{2}\)

Nếu k = \(-\frac{1}{2}\)thì \(x=\frac{-5}{2}\text{ };\text{ }y=\frac{-3}{2}\)

10x = 6y

\(\Rightarrow\text{ }\frac{x}{6}=\frac{y}{10}\)

đặt \(\frac{x}{6}=\frac{y}{10}=k\)

\(\Rightarrow\text{ }x=6k\text{ };\text{ }y=10k\)

\(\Rightarrow\text{ }2.\left(6k\right)^2-\left(10k\right)^2=-28\)

\(\Rightarrow\text{ }72k^2-100k^2=-28\)

\(\Rightarrow\text{ }\left(72-100\right).k^2=-28\)

\(\Rightarrow\text{ }\left(-28\right).k^2=\left(-28\right)\text{ }\)

\(\Rightarrow\text{ }k^2=\left(-28\right)\text{ }:\text{ }\left(-28\right)\)

\(\Rightarrow\text{ }k^2=1\)

\(\Rightarrow\text{ }\orbr{\begin{cases}k=1\\k=-1\end{cases}}\)

Nếu k = 1 thì x = 10 ; y = 6

Nếu k = -1 thì x = -10 ; y = -6

Ta có: \(10x=6y\) \(\Leftrightarrow x=\frac{6y}{10}=\frac{3y}{5}\)

\(\Rightarrow2x^2-y^2=2\times\left(\frac{3y}{5}\right)^2-y^2=-28\)

\(\Leftrightarrow2\times\frac{9y^2}{25}-y^2=\frac{18y^2}{25}-y^2=-28\)

\(\Leftrightarrow\frac{-7y^2}{25}=-28\Leftrightarrow-7y^2=-700\Leftrightarrow y^2=100\)

\(\Leftrightarrow x=10\) và \(y=6\) hoặc \(x=-10\) và \(y=-6\)

a.

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

b.

\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)

c.

\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)

\(=\left(x+3\right)^3\)

d.

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

e.

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

\(=\left(x-y\right)\left(x+y-5\right)\)

f.

\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)

g. 10x(x-y)-6y(y-x)

=10x(x-y)+6y(x-y)

=(x-y)(10x+6y)

h.x²-4x-5

=(x-5)(x+1)

i.x⁴-y⁴ = (x²-y²)(x²+y²)

B2.

a.5(x-2)=x-2

⇔5(x-2)-(x-2)=0

⇔4(x-2)=0

⇔x=2

b.3(x-5)=5-x

⇔3(x-5)+(x-5)=0

⇔4(x-5)=0

⇔x=5

c.(x+2)²-(x+2)(x-2)=0

⇔(x+2)[(x+2)-(x-2)]=0

⇔4(x+2)=0

⇔x=-2

\(=>\frac{x}{6}=\frac{y}{10}=\frac{2x^2-y^2}{2\cdot6^2-10^2}=\frac{-28}{-28}=1\)\(1\)

\(=>\hept{\begin{cases}x=1\cdot6=6\\y=1\cdot10=10\end{cases}}\)