\(c,\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ d,\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\\ e,\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

c: \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}12x+4y=40\\12x-9y=27\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}13y=13\\3x+y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=1\\x=3\end{matrix}\right.\)

d: \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-x=-4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=\dfrac{22-4x}{3}=\dfrac{22-4\cdot4}{3}=2\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

e. \(\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\4x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

\(a,\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-y=5\\2x+y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\\ b,\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\23y=46\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=2\end{matrix}\right.\)

\(c,\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\ d,\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

\(e,\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a. \(\left\{{}\begin{matrix}3x-y=5\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}6x-2y=10\\4x+2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}10x=20\\6x-2y=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

b. \(\left\{{}\begin{matrix}5x+2y=9\\x+5y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x+2y=9\\5x+25y=55\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}23y=46\\5x+2y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=1\end{matrix}\right.\)

c. \(\left\{{}\begin{matrix}3x+y=10\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x+3y=30\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}13x=39\\4x-3y=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)

d. \(\left\{{}\begin{matrix}4x+3y=22\\5x+3y=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\4x+3y=22\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=2\end{matrix}\right.\)

e. \(\left\{{}\begin{matrix}4x-3y=5\\5x+3y=13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}9x=18\\4x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

a) \(\begin{cases} 3x -y=5\\ 4x +2y=10 \end{cases} \)

    \(\begin{cases} 12x - 4y= 20\\ 12x +6y= 30 \end{cases} \)

    \(\begin{cases} -10y=-10\\ 3x-y=5 \end{cases} \)

    \(\begin{cases} y=1\\ 3x-1=5 \end{cases} \)

    \(\begin{cases} y=1\\ 3x=6 \end{cases} \)

    \(\begin{cases} y=1\\ x=2 \end{cases} \)

Hpt có nghiệm duy nhất: {1;2}

b)\(\begin{cases} 5x +2y=9\\ x+5y=11 \end{cases} \)

  \(\begin{cases} 5x+2y=9\\ 5x+25y=55 \end{cases} \)

  \(\begin{cases} -23y=-46\\ x+5y=11 \end{cases} \)

  \(\begin{cases} y=2\\ x+ 5*2=11 \end{cases} \)

  \(\begin{cases} y=2\\ x+10=11 \end{cases} \)

Hpt có nghiệm duy nhất:{1;2}

c)\(\begin{cases} 3x+y=10\\ 4x-3y=9 \end{cases} \)

  \(\begin{cases} 12x+4y=40\\ 12x-9y=27 \end{cases} \)

 \(\begin{cases} 13y=13\\ 3x+y=10 \end{cases} \)

 \(\begin{cases} y=1\\ 3x+1=10 \end{cases} \)

 \(\begin{cases} y=1\\ 3x=9 \end{cases} \)

hpt có nghiệm duy nhất:{1;3}

d)\(\begin{cases} 4x+3y=22\\ 5x+3y=26 \end{cases} \)

  \(\begin{cases} 20x+15y=110\\ 20x+12y=104 \end{cases} \)

 \(\begin{cases} 3y=6\\ 4x+3y=22 \end{cases} \)

 \(\begin{cases} y=2\\ 4x+3*2=22 \end{cases} \)

 \(\begin{cases} y=2\\ 4x+6=22 \end{cases} \)

hệ phương trình có nghiệm duy nhất:{2;4}

e)\(\begin{cases} 4x-3y=5\\ 5x+3y=13 \end{cases} \)

  \(\begin{cases} 20x-15y=25\\ 20x+12y=52 \end{cases} \)

 \(\begin{cases} -27y=-27\\ 4x-3y=5 \end{cases} \)

  \(\begin{cases} y=1\\ 4x-3*1=5 \end{cases} \)

   \(\begin{cases} y=1\\ 4x-3=5 \end{cases} \)

Hệ phương trình có nghiệm duy nhất là:{1;2}

ta có 4 x 3 y 2   –   8 x 2 y 3   =   4 x 2 y 2 . x   –   4 x 2 y 2 . 2 y   =   4 x 2 y 2 ( x   –   2 y )    

Vậy 4x³y² – 8x²y³ = 4x²y²(x – 2y)      

Đáp án cần chọn là: C

bấm đúng cho mik đi 

a: \(=\left(-\dfrac{6}{2}\right)\cdot\dfrac{x^3}{x}\cdot\dfrac{y^2}{y^2}=-3x^2\)

b: \(=\left(-\dfrac{1}{4}:\dfrac{1}{2}\right)\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^3}{y^2}=-\dfrac{1}{2}xy\)

c: \(=\dfrac{8}{4}\cdot\dfrac{x^4}{x^3}\cdot\dfrac{y^5}{y^4}=2xy\)

\(a,-6x^3y^2:2xy^2=-3x^2\)

\(b,-\dfrac{1}{4}x^4y^3:\dfrac{1}{2}x^3y^2=-\dfrac{1}{2}xy\)

\(c,8x^4y^5:4x^3y^4=2xy\)

#Urushi

\(\frac{4x-3}{3}=\frac{3y+1}{7}=\frac{4x+3y-2}{5y}\)

\(=\frac{4x-3+3y+1-\left(4x+3y-2\right)}{3+7-5y}\)

\(=\frac{4x-3+3y+1-4x-3y+2}{10-5y}\)

\(=\frac{\left(4x-4x\right)+3y-3y-3+1+2}{10-5y}=0\)

\(\Rightarrow\hept{\begin{cases}4x-3=0\Leftrightarrow x=\frac{3}{4}\\3y+1=0\Leftrightarrow y=-\frac{1}{3}\end{cases}}\)

Vậy \(x=\frac{3}{4};y=-\frac{1}{3}\).

Câu trả lời đúng là :

x = 3/4

y = -1/3

Đáp số : ...

Đặt bthuc = A nhé

ĐKXĐ : \(2x\ne3y\)

\(A=\left[\dfrac{2x\left(4x^2+6xy+9y^2\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{27y^3+36xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}-\dfrac{24xy\left(2x-3y\right)}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{2x\left(2x-3y\right)}{\left(2x-3y\right)}+\dfrac{9y^2+12xy}{\left(2x-3y\right)}\right]\)\(=\left[\dfrac{8x^3+12x^2y+18xy^2-27y^3-36xy^2-48x^2y+72xy^2}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\right]\left[\dfrac{4x^2-6xy+9y^2+12xy}{\left(2x-3y\right)}\right]\)

\(=\dfrac{8x^3-36x^2y+36xy^2-27y^3}{\left(2x-3y\right)\left(4x^2+6xy+9y^2\right)}\cdot\dfrac{4x^2+6xy+9y^2}{2x-3y}\)

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

Với x = 1/3 ; y = -2 (tmđk) thay vào A ta được : A = 2.1/3 - 3.(-2) = 20/3

Ta thấy hai đường thẳng này song song, nên khoảng cách giữa chúng là khoảng cách từ một điểm bất kì từ đường thẳng này tới đường thẳng kia

Chọn điểm \(A\left( {0;4} \right) \in {d_2}\), suy ra \(d\left( {{d_1},{d_2}} \right) = d\left( {A,{d_1}} \right) = \frac{{\left| {4.0 - 3.4 + 2} \right|}}{{\sqrt {{4^2} + {3^2}} }} = 2\)

Vậy khoảng cách giữa hai đường thẳng \({d_1}:4x - 3y + 2 = 0\) và \({d_2}:4x - 3y + 12 = 0\) là 2