chịu

\(-6x=5y\Rightarrow\frac{x}{5}=\frac{-y}{6}\)

Áp dụng t/c dãy tỉ số = nha ta có :

 \(\frac{x}{5}=\frac{-y}{6}=\frac{x-y}{5+6}=-\frac{22}{11}=-2\)

\(\Rightarrow\hept{\begin{cases}\frac{x}{5}=-2\Rightarrow x=-10\\\frac{-y}{6}=-2\Rightarrow y=12\end{cases}}\)

-6x = 5y => x = -5/6y

Ta có: x - y = -22

=> -5/6y - y = -22

=> -11/6y = -22

=> y = -22 : (-11/6)

=> y = -22 × (-6/11)

=> y = 12

=> x = -5/6 × 12 = -10

a) \(2x^2+5y^2+8x-10y+13=0\)

\(\Leftrightarrow\left(2x^2+8x+8\right)+\left(5y^2-10y+5\right)=0\)

\(\Leftrightarrow2\left(x+2\right)^2+5\left(y-1\right)^2=0\)

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

Vậy x=-2;y=1

b) \(3x^2+5y^2-6x+20y+23=0\)

\(\Leftrightarrow\left(3x^2-6x+3\right)+\left(5y^2+20y+20\right)=0\)

\(\Leftrightarrow3\left(x-1\right)^2+5\left(y+2\right)^2=0\)

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

Vậy x=1;y=-2

Bài 1:

\(\left(2x-1\right)^2-4x^2\left(2x-3\right)=5\)

\(\Rightarrow4x^2-4x+1-8x^3+12x^2-5=0\)

\(\Rightarrow-8x^3+16x^2-4x-4=0\)

\(\Rightarrow-4\left(2x^3-4x^2+x+1\right)=0\)

\(\Rightarrow2x^3-4x^2+x+1=0\)

\(\Rightarrow2x^3-2x^2-x-2x^2+2x+1=0\)

\(\Rightarrow x\left(2x^2-2x-1\right)-\left(2x^2-2x-1\right)=0\)

\(\Rightarrow\left(x-1\right)\left(2x^2-2x-1\right)=0\)

\(\Rightarrow\left[\begin{matrix}x-1=0\\2x^2-2x-1=0\end{matrix}\right.\)\(\Rightarrow\left[\begin{matrix}x=1\\\Delta_{2x^2-2x-1}=\left(-2\right)^2-\left(-4\left(2.1\right)\right)=12\end{matrix}\right.\)

\(\Rightarrow\left[\begin{matrix}x=1\\x_{1,2}=\frac{2\pm\sqrt{12}}{4}\end{matrix}\right.\)

Bài 2:

a)\(x^2+6x-y^2+9\)

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

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

\(=\left(x+3-y\right)\left(x+3+y\right)\)

b)\(2x^2-3xy-2y^2+5x+5y-3\)

\(=2x^2+xy-x-4xy-2y^2+2y+6x+3y-3\)

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

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

1) ta có: \(x:3=y.15\Rightarrow x\cdot\frac{1}{3}=y.15\Rightarrow\frac{x}{15}=\frac{y}{\frac{1}{3}}\)

ADTCDTSBN

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2) bn ghi thiếu đề r

3) ta có: \(3x=7y\Rightarrow\frac{x}{7}=\frac{y}{3}=k\Rightarrow\hept{\begin{cases}x=7k\\y=3k\end{cases}}\)

mà xy = 189 => 7k.3k = 189

                          21 k² = 189

                                 k² = 9 = 3² = (-3)² => k = 3 hoặc k  = - 3

TH1: k = 3

x = 7.3 => x  = 21

y = 3.3 => y = 9

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4) ta có: \(4x=5y\Rightarrow\frac{x}{5}=\frac{y}{4}\Rightarrow\frac{x^2}{25}=\frac{y^2}{16}\)

ADTCDTSBN

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a ) \(\frac{3x+1}{5y+2}=\frac{6x+3}{10y+6}\)

\(\Leftrightarrow\left(3x+1\right).\left(10y+6\right)=\left(5y+2\right).\left(6x+3\right)\)

\(\Leftrightarrow30xy+18x+10y+6=30xy+15y+12x+6\)

\(\Leftrightarrow6x-5y=0\)

kHÔNG CÓ X,Y THÕA MÃN

cÂU B TƯƠNG TỰ

\(x^2-4xy+5y^2=169\)

\(x^2-4xy+4y^2+y^2-169=0\)

\(\left(x^2-4xy+4y^2\right)+\left(y^2-13^2\right)=0\)

\(\left(x-2y\right)^2+\left(y-13\right)\left(y+13\right)=0\)

b/    \(\Leftrightarrow x^2-4xy+4y^2+y^2=13^2\)

        \(\Leftrightarrow\left(x-2y\right)^2=\left(13^2-y^2\right)\)

        \(\Rightarrow y^2\le13^2\)và    \(13^2-y^2\)là số chính phương .  Do đó :

      \(y^2=0\)hay  \(y=0\)

     Thay vào ta có các nghiệm sau   \(\left(13,0\right);\left(-13;0\right)\)

\(b,x^2-4xy+5y^2=169.\)

\(\Rightarrow x^2-2.x.2y+\left(2y\right)^2+y^2-169=0\)

\(\Rightarrow\left(x-2\right)^2+\left(y^2-13^2\right)=0\)

\(\Rightarrow\left(x-2\right)^2+\left(y-13\right).\left(y+13\right)=0\)

\(\Rightarrow\hept{\begin{cases}\left(x-2\right)^2=0\\y-13=0\\y+13=0\end{cases}}\Rightarrow\hept{\begin{cases}x-2=0\\y=13\\y=-13\end{cases}}\Rightarrow\hept{\begin{cases}x=2\\y=13\\y=-13\end{cases}}\)