3x²+4y²=7xy

<=> 3x²-3xy+4y²-4xy=0

<=> 3x(x-y)-4y(x-y)=0

<=> (3x-4y)(x-y)=0

<=> 3x-4y=0 hoặc x-y=0

<=> 3x=4y hoặc x=y

<=> y = \(\frac{3}{4}\)x hoặc x=y

+) y = \(\frac{3}{4}\)x, ta có:

F = \(\frac{4.\frac{3}{4}x+2x}{5.\frac{3}{4}x-7x}+\)\(\frac{3x-2.\frac{3}{4}x}{10.\frac{3}{4}x-4x}\)

F = \(\frac{5x}{-\frac{13}{4}x}+\frac{\frac{3}{2}x}{\frac{7}{2}x}\)

F = \(-\frac{20}{13}+\frac{3}{7}=-\frac{101}{91}\)

+) x = y, ta có:

F = \(\frac{4x+2x}{5x-7x}+\frac{3x-2x}{10x-4x}\)

F = \(\frac{6x}{-2x}+\frac{1x}{6x}=-3+\frac{1}{6}=-\frac{17}{6}\)

Từ \(3x^2+4y^2=7xy\Rightarrow3x^2+4y^2-7xy=0\)

\(\Rightarrow3x^2-4xy-3xy+4y^2=0\)

\(\Rightarrow x\left(3x-4y\right)-y\left(3x-4y\right)=0\)

\(\Rightarrow\left(x-y\right)\left(3x-4y\right)=0\)\(\Rightarrow\left[\begin{matrix}x=y\\x=\frac{4y}{3}\end{matrix}\right.\)

*)Xét \(x=y\) ta có \(F=\frac{4y+2y}{5y-7y}+\frac{3y-2y}{10y-4y}=\frac{6y}{-2y}+\frac{y}{6y}=-3+\frac{1}{6}=-\frac{17}{6}\)

*)Xét \(x=\frac{4y}{3}\) ta có \(F=\frac{4y+2\cdot\frac{4y}{3}}{5y-7\cdot\frac{4y}{3}}+\frac{3\cdot\frac{4y}{3}-2y}{10y-4\cdot\frac{4y}{3}}=\frac{4y+\frac{8y}{3}}{5y-\frac{28y}{3}}+\frac{4y-2y}{10y-\frac{16y}{3}}=\frac{-20}{13}+\frac{3}{7}=\frac{-101}{91}\)

1) 2x + 2y - x(x+y)

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

= (2 - x)(x + y)

2/ 5x² - 5xy -10x + 10y

= 5x(x - y) - 10(x - y)

= (5x - 10(x - y)

3/ 4x² + 8xy - 3x - 6y

= 4x(x + 2y) - 3(x + 2y)

= (4x - 3)(x + 2y)

1) 2x + 2y - x(x + y) 

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

= (2 - x)(x + y)

2) 5x² - 5xy - 10x + 10y 

= 5x(x - y) - 10(x - y)

= (5x - 10)(x - y)

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

3) 4x² + 8xy - 3x - 6y  

= 4x(x + 2y) - 3(x + 2y)

= (4x - 3)(x + 2y)

4) 2x² + 2y² - x²z + z - y²z - 2 

= 2(x² + y² - z(x² + y²) - (2 - z)

= (2 - z)(x² + y²) - (2 - z)

= (2 - z)(x² + y²)

5) x² + xy - 5x - 5y

= x(x + y) - 5(x + y)

= (x - 5)(x + y)

6) x(2x - 7) - 4x + 14 

= x(2x - 7) - 2(2x - 7) 

= (x - 2)(2x - 7)

7)x² - 3x + xy - 3y  

= x(x + y) - 3(x + y)

= (x - 3)(x + y)

5/ x² + xy - 5x - 5y 

= x(x + y) - 5(x + y)

= (x - 5)(x + y)

6/ x(2x - 7) - 4x + 14

= 2x² - 7x - 4x + 14

= (2x² - 4x) - (7x - 14)

= 2x(x - 2) -7(x - 2)

= (2x - 7)(x - 2)

7/ x² - 3x + xy - 3y

= x(x - 3) + y(x - 3)

= (x + y)(x - 3) 

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

<=> \(\left\{{}\begin{matrix}3x-2y=4\\7x=14\end{matrix}\right.< =>\left\{{}\begin{matrix}x=2\\y=1\end{matrix}\right.\)

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

=> Hệ có vô số nghiệm.

3)\(\left\{{}\begin{matrix}3x-4y=-2\\10x+4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}3x-4y=-2\\13x=26\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

4)\(\left\{{}\begin{matrix}6x+15y=9\\6x-4y=28\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}6x+15y=9\\19y=19\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-1\end{matrix}\right.\)

a, bậc 6 

b, bậc 6 

c, bậc 12 

d, bậc 9 

e, bậc 8 

a)Ta có: \(2x=3y;5y=7z\)và \(x-y-z=-27\)

\(\Rightarrow\frac{x}{3}=\frac{y}{2};\frac{y}{7}=\frac{z}{5}\)và\(x-y-z=-27\)

\(\Rightarrow\frac{x}{21}=\frac{y}{14}=\frac{z}{10}\)và \(x-y-z=-27\)

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

\(\frac{x}{21}=\frac{y}{14}=\frac{z}{10}=\frac{x-y-z}{21-14-10}=\frac{-27}{-3}=9\)

Ta có:\(\frac{x}{21}=9\Rightarrow x=9.21=189\)

          \(\frac{y}{14}=9\Rightarrow y=9.14=126\)

         \(\frac{z}{10}=9\Rightarrow z=9.10=90\)

Vậy:\(x=189;y=126\)và\(z=90\)

b) \(\frac{x}{4}=\frac{y}{5}=\frac{z}{6}\)và\(x^2-2y^2+z^2=18\)

\(\Rightarrow\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}\)và\(x^2-2y^2+z^2=18\)

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

\(\frac{x^2}{16}=\frac{2y^2}{50}=\frac{z^2}{36}=\frac{x^2-2y^2+z^2}{16-50+36}=\frac{18}{2}=9\)

Ta có:\(\frac{x^2}{16}=9\Rightarrow x^2=144\Rightarrow\orbr{\begin{cases}x=12\\x=-12\end{cases}}\)

\(\frac{2y^2}{50}=9\Rightarrow2y^2=450\Rightarrow y^2=225\Rightarrow\orbr{\begin{cases}y=15\\y=-15\end{cases}}\)

\(\frac{z^2}{36}=9\Rightarrow z^2=324\Rightarrow\orbr{\begin{cases}z=18\\z=-18\end{cases}}\)

Vậy: \(x=12;y=15;z=18\)hoặc \(x=-12;y=-15;z=-18\)

c) \(x:y:z=3:8:5\)và\(3x+y-2z=14\)

\(\Rightarrow\frac{x}{3}=\frac{y}{8}=\frac{z}{5}\)và\(3x+y-2z=14\)

\(\Rightarrow\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}\)và \(3x+y-2z=14\)

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

\(\frac{3x}{9}=\frac{y}{8}=\frac{2z}{10}=\frac{3x+y-2z}{9+8-10}=\frac{14}{7}=2\)

Ta có: \(\frac{3x}{9}=2\Rightarrow3x=18\Rightarrow x=6\)

\(\frac{y}{8}=2\Rightarrow y=16\)

\(\frac{2z}{10}=2\Rightarrow2z=20\Rightarrow z=10\)

Vậy:\(x=6;y=16;z=10\)

Ta có :\(15x=10y=6z\Rightarrow\hept{\begin{cases}15x=10y\\10y=6z\end{cases}}\Rightarrow\hept{\begin{cases}3x=2y\\5y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{5}\end{cases}}\Rightarrow\frac{x}{2}=\frac{y}{3}=\frac{z}{5}\)

Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{5}=k\Rightarrow\hept{\begin{cases}x=2k\\y=3k\\z=5k\end{cases}}\)

Khi đó 5x³ + 2y³ - z³ = 31

=> 5(2k)³ + 2(3k)³ - (5k)³ = 31

=> 40k³ + 54k³ - 125k³ = 31

=> -31k³ = 31

=> k³ = -1

=> k = -1

=> x = -2 ; y = -3 ; z = -5

b) Ta có 7x = 14y = 6z =>  \(\hept{\begin{cases}7x=14y\\14y=6z\end{cases}}\Rightarrow\hept{\begin{cases}x=2y\\7y=3z\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{2}=\frac{y}{1}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\hept{\begin{cases}\frac{x}{6}=\frac{y}{3}\\\frac{y}{3}=\frac{z}{7}\end{cases}}\Rightarrow\frac{x}{6}=\frac{y}{3}=\frac{z}{7}\)

Đặt \(\frac{x}{6}=\frac{y}{3}=\frac{z}{7}=k\Rightarrow\hept{\begin{cases}x=6k\\y=3k\\z=7k\end{cases}}\)

Khi đó 2x² - 3y² = 5

<=> 2.(6k)² - 3.(3k)² = 5

=> 72k² - 27k² = 5

=> 45k² = 5

=> k² = 1/9

=> k = \(\pm\frac{1}{3}\)

Nếu k = 1/3 => x = 2 ; y = 1 ; z = 7/3

Nếu k = -1/3 => x = -2 ; y = - 1 ; z = -7/3

Vậy các cặp (x;y;z) thỏa mãn là : (2;1;7/3) ; (-2 ; - 1; -7/3)

c) Ta có : \(3x=8y=5z\Rightarrow\frac{3x}{120}=\frac{8y}{120}=\frac{5z}{120}\Rightarrow\frac{x}{40}=\frac{y}{15}=\frac{z}{24}\)

Đặt \(\frac{x}{40}=\frac{y}{15}=\frac{z}{24}=k\Rightarrow\hept{\begin{cases}x=40k\\y=15k\\z=24k\end{cases}}\)

Khi đó |x - 2y| = 5

<=> |40k - 2.15k| = 5

=>  |10k| = 5

=> \(\orbr{\begin{cases}10k=5\\10k=-5\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{1}{2}\\k=-\frac{1}{2}\end{cases}}\)

Nếu k = 5 => x = 20 ; y = 7,5 ; z = 12

Nếu k = -5 => x = -20 ; y =-7,5 ; z = -12

d) 4x = 5y = 6z => \(\frac{4x}{60}=\frac{5y}{60}=\frac{6z}{60}\Rightarrow\frac{x}{15}=\frac{y}{12}=\frac{z}{10}\)

Đặt \(\frac{x}{15}=\frac{y}{12}=\frac{z}{10}=k\Rightarrow\hept{\begin{cases}x=15k\\y=12k\\z=10k\end{cases}}\)

Khi đó (3x - 2y)² = 16

<=> (3.15k - 2.12k)² = 16

=> (45k -24k)² = 16

=> (21k)² = 16

=> \(\orbr{\begin{cases}21k=4\\21k=-4\end{cases}}\Rightarrow\orbr{\begin{cases}k=\frac{4}{21}\\k=-\frac{4}{21}\end{cases}}\)

Nếu k = 4/21 => x = 20/7 ; y = 16/7 ; z = 40/21

Nếu k = -4/21 => x = -20/7 ; y = -16/7 ; z = -40/21

Ai có cách làm khác không 

khó + lười + nhiều = không làm

Hello

ko thích làm

\(a,\Leftrightarrow\left\{{}\begin{matrix}5x+15y=-10\\5x-4y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}19y=-21\\5x-4y=11\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{21}{19}\\5x-4\left(-\dfrac{21}{19}\right)=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{25}{19}\\y=-\dfrac{21}{19}\end{matrix}\right.\)

\(c,\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\10x-5y=-40\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x+5y=1\\13x=-39\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-3\\y=2\end{matrix}\right.\\ d,\Leftrightarrow\left\{{}\begin{matrix}5x-10y=-30\\5x-3y=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}5x-3y=5\\-7y=-35\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=5\end{matrix}\right.\\ e,\Leftrightarrow\left\{{}\begin{matrix}2\left(x+y\right)+3\left(x-y\right)=4\\2\left(x+y\right)+4\left(x-y\right)=10\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x-y=6\\2\left(x+y\right)+3\cdot6=4\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x-y=6\\x+y=-7\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{1}{2}\\y=-\dfrac{13}{2}\end{matrix}\right.\)

xy+x+y=2

xy+x+y+1=2+1

(xy+x)+(y+1)=3

x(y+1)+(y+1)=3

(x+1)(y+1)=3=1.3=3.1=-1.-3=-3.-1

\(\Rightarrow\left[{}\begin{matrix}x+1=1;y+1=3\\x+1=3;y+1=1\\x+1=-1;y+1=-3\\x+1=-3;y+1=-1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0;y=2\\x=2;y=0\\x=-2;y=-4\\x=-4;y=-2\end{matrix}\right.\)

Vậy:.................

xy+14+2y+7x= -10

\(\Leftrightarrow\)y(x+2)+7(x+2)=-10

\(\Leftrightarrow\)(y+7)(x+2)=-10=1.(-10)=2.(-5)=5.(-2)=10.(-1)