Tìm x thuộc Z biết
a) xy + 4y + 3x = 28
b)2xy – 3y+ 3x = 7
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Áp dụng t/c dãy tỉ số bằng nhau:
a.
\(\dfrac{x}{3}=\dfrac{y}{5}=\dfrac{2x}{6}=\dfrac{4y}{20}=\dfrac{2x+4y}{6+20}=\dfrac{28}{26}=\dfrac{14}{13}\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.\dfrac{14}{13}=\dfrac{52}{13}\\y=5.\dfrac{14}{13}=\dfrac{70}{13}\end{matrix}\right.\)
(Em có nhầm đề 26 thành 28 ko nhỉ, số xấu quá)
b.
\(4x=5y\Rightarrow\dfrac{x}{5}=\dfrac{y}{4}=\dfrac{3x}{15}=\dfrac{-2y}{-8}=\dfrac{3x-2y}{15-8}=\dfrac{35}{7}=5\)
\(\Rightarrow\left\{{}\begin{matrix}x=5.5=25\\y=4.2=20\end{matrix}\right.\)
c.
\(\dfrac{x}{-3}=\dfrac{y}{-7}=\dfrac{2x}{-6}=\dfrac{4y}{-28}=\dfrac{2x+4y}{-6-28}=\dfrac{68}{-34}=-2\)
\(\Rightarrow\left\{{}\begin{matrix}x=-3.\left(-2\right)=6\\y=-7.\left(-2\right)=14\end{matrix}\right.\)
d.
\(\dfrac{x}{2}=\dfrac{y}{-3}=\dfrac{z}{4}=\dfrac{4x}{8}=\dfrac{-3y}{9}=\dfrac{-2z}{-8}=\dfrac{4x-3y-2z}{8+9-8}=\dfrac{16}{9}\)
\(\Rightarrow\left\{{}\begin{matrix}x=2.\dfrac{16}{9}=\dfrac{32}{9}\\y=-3.\dfrac{16}{9}=-\dfrac{48}{9}\\z=4.\dfrac{16}{9}=\dfrac{64}{9}\end{matrix}\right.\)
Bài 2:
a: \(3x^2-3xy=3x\left(x-y\right)\)
b: \(x^2-4y^2=\left(x-2y\right)\left(x+2y\right)\)
c: \(3x-3y+xy-y^2=\left(x-y\right)\left(3+y\right)\)
d: \(x^2-y^2+2y-1=\left(x-y+1\right)\left(x+y-1\right)\)
A) \(-4x2xy^2+3x^2.\frac{1}{3}y+\left(-5\right)xy.\frac{1}{5}xy=-8x^2y^2+x^2y+\left(-x^2y^2\right)=-9x^2y^2+x^2y\)
B) \(\frac{4}{3}x^4y^7-3x^4y^7=\frac{-5}{3}x^4y^7\)
C) \(\frac{2}{3}x^3y^4+3x^3y^4=3\frac{2}{3}x^3y^4\)
CHÚC BN HỌC TỐT!!!
b: 2xy-3y+3x=7
=>y(2x-3)+3x-4,5=7-4,5
=>2y(x-1,5)+3(x-1,5)=2,5
=>(x-1,5)(2y+3)=2,5
=>(2x-3)(2y+3)=5
=>(2x-3;2y+3)∈{(1;5);(5;1);(-1;-5);(-5;-1)}
=>(2x;2y)∈{(4;2);(8;-2);(2;-8);(-2;-4)}
=>(x;y)∈{(2;1);(4;-1);(1;-4);(-1;-2)}
a: xy+4y+3x=28
=>y(x+4)+3x+12=28+12
=>(x+4)(y+3)=40
=>(x+4;y+3)∈{(1;40);(40;1);(-1;-40);(-40;-1);(2;20);(20;2);(-2;-20);(-20;-2);(4;10);(10;4);(-4;-10);(-10;-4);(5;8);(8;5);(-5;-8);(-8;-5)}
=>(x;y)∈{(-3;37);(36;-2);(-5;-43);(-44;-4);(-2;17);(16;-1);(-6;-23);(-24;-5);(0;7);(6;1);(-8;-13);(-14;-7);(1;5);(4;2);(-9;-11);(-12;-8)}