2xy - x - 4y = 10 

=> 2xy - x - 4y + 2 = 10 +2

=> x(2y-1) - 2(2y -1)= 12 

=> (x-2)(2y-1) = 12 

ta có 12 = 1 x 12 = 3 x 4 = 2 x 6 

ta xét 6TH 

Th1 x-1 = 1 

và 2y-1 = 12 

=> x= 2 

và y = 13/2 

Th khác tự xét

2y(x-2)-x+2=12

(x-2)(2y-1)=12

2y-1={-3,-1,1,3} =>y={-1,0,1,2}

x-2={-4,-12,12,4}=>x={2,-10,14,6}

(x,y)=(2,-1);(-10,0);(14,1);(6,2)

dễ quá

\(a,\Leftrightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\\ \Leftrightarrow24y=24\Leftrightarrow y=1\\ b,\Leftrightarrow y^3+9y^2+27y+27-y^3-3y^2-3y-1=56\\ \Leftrightarrow6y^2+24y-30=0\\ \Leftrightarrow y^2+4y-5=0\\ \Leftrightarrow\left(y-1\right)\left(y+5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\)

a) \(\Leftrightarrow y^3-6y^2+12y-8-y^3+27+6y^2+12y+6=49\)

\(\Leftrightarrow24y=24\Leftrightarrow y=1\)

b) \(\Leftrightarrow y^3+9y^2+27y+27-y^3-3y^2-3y-1=56\)

\(\Leftrightarrow6y^2+24y-30=0\)

\(\Leftrightarrow6\left(y-1\right)\left(y+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}y=1\\y=-5\end{matrix}\right.\)

a) (y-2)³-(y-3)(y²+3y+9)+6(y+1)²=49
    \(y^3-6y^2+12y-8-y^3+27+6\left(y^2+2y+1\right)=49\)
    \(-6y^2+12y+25+6y^2+12y+6=49\)
     \(24y+31=49\)
     24y=18
         y=0,75

a)  2x² - 98 = 0

     2x²        = 0 + 98

     2x²        = 98

       x²        = 98 : 2

       x²         = 49

       x          = \(\sqrt{49}\)

=>   x   = 7

Ta có : 2x² - 98 = 0

=> 2(x² - 49) = 0

Mà : 2 > 0

Nên x² - 49 = 0

=> x² = 49

=> x² = -7;7

2x²-98=0

=>2(x²-49)=0

=>x²-49=0

x²=49+0

x²=7²

x=7

Lấy P(x) - Q(x) -2x^2 thì ra G(x) nhé 

Thanks bạn

Ta có:

\(15x^4y^4-M=10x^2y^4+6x^2y^4\)

\(\Leftrightarrow M=15x^4y^4-\left(10x^2y^4+6x^2y^2\right)\)

\(\Leftrightarrow M=15x^4y^4-16x^2y^4\)

Thay \(x=-\dfrac{1}{2};x=2\) vào M ta có:

\(M=15\cdot\left(-\dfrac{1}{2}\right)^4\cdot2^4-16\cdot\left(-\dfrac{1}{2}\right)^2\cdot2^4=-49\)

\(\dfrac{x}{-3}=\dfrac{y}{5}\)⇒\(\dfrac{x}{-6}=\dfrac{y}{10}\)

\(\dfrac{y}{2}=\dfrac{z}{7}\)⇒\(\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

⇒\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}\)

Áp dụng tính chất dãy tỉ số bằng nhau, ta có:

\(\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{z}{35}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\)

⇒\(\left\{{}\begin{matrix}x=-6.-6=36\\y=-6.10=-60\\z=-6.35=-210\end{matrix}\right.\)

\(a,\dfrac{x}{-3}=\dfrac{y}{5}\Rightarrow\dfrac{x}{-6}=\dfrac{y}{10};\dfrac{y}{2}=\dfrac{z}{7}\Rightarrow\dfrac{y}{10}=\dfrac{z}{35}\\ \Rightarrow\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-6}=\dfrac{y}{10}=\dfrac{z}{35}=\dfrac{2x}{-12}=\dfrac{3y}{30}=\dfrac{2x-3y+z}{-12-30+35}=\dfrac{42}{-7}=-6\\ \Rightarrow\left\{{}\begin{matrix}x=36\\y=-60\\z=-210\end{matrix}\right.\)

\(b,6x=4y=z\Rightarrow\dfrac{6x}{12}=\dfrac{4y}{12}=\dfrac{z}{12}\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{12}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{2x-3y+z}{4-9+12}=\dfrac{42}{7}=6\\ \Rightarrow\left\{{}\begin{matrix}x=12\\y=18\\z=72\end{matrix}\right.\)

\(c,x=-2y\Rightarrow\dfrac{x}{-2}=y\Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}\\ 7y=2z\Rightarrow\dfrac{y}{2}=\dfrac{z}{7}\\ \Rightarrow\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}\)

Áp dụng t/c dtsbn:

\(\dfrac{x}{-4}=\dfrac{y}{2}=\dfrac{z}{7}=\dfrac{2x}{-8}=\dfrac{3y}{6}=\dfrac{2x-3y+z}{-8+6+7}=\dfrac{42}{5}\\ \Rightarrow\left\{{}\begin{matrix}x=-\dfrac{168}{5}\\y=\dfrac{84}{5}\\z=\dfrac{294}{5}\end{matrix}\right.\)

Bạn xem lại đề giúp mình nhé

đây là đề thi của tớ nên ko sai đâu

Kho vay ban

Vâng, có ai giúp được k ạ =(((

Để mk nghĩ đã