a: (x-2)(y-3)=5

=>\(\left(x-2\right)\cdot\left(y-3\right)=1\cdot5=5\cdot1=\left(-1\right)\cdot\left(-5\right)=\left(-5\right)\cdot\left(-1\right)\)

=>\(\left(x-2;y-3\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(3;8\right);\left(7;4\right);\left(1;-2\right);\left(-3;2\right)\right\}\)

b: (2x-1)*(y-4)=-11

=>\(\left(2x-1\right)\cdot\left(y-4\right)=1\cdot\left(-11\right)=\left(-11\right)\cdot1=\left(-1\right)\cdot11=11\cdot\left(-1\right)\)

=>\(\left(2x-1;y-4\right)\in\left\{\left(1;-11\right);\left(-11;1\right);\left(-1;11\right);\left(11;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(1;-7\right);\left(-5;5\right);\left(0;15\right);\left(6;3\right)\right\}\)

c: xy-2x+y=3

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

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

=>\(\left(x+1\right)\cdot\left(y-2\right)=1\cdot1=\left(-1\right)\cdot\left(-1\right)\)

=>\(\left(x+1;y-2\right)\in\left\{\left(1;1\right);\left(-1;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(0;3\right);\left(-2;1\right)\right\}\)

a: (x-3)(2y+7)=1

=>(x-3;2y+7)∈{(1;1);(-1;-1)}

=>(x;2y)∈{(4;-6);(2;-8)}

=>(x;y)∈{(4;-3);(2;-4)}

b: (x+1)(y+2)=-3

=>(x+1;y+2)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(0;-5);(-4;-1);(-2;1);(2;-3)}

mà x<y

nên (x;y)∈{(-4;-1);(-2;1)}

c: xy+2x+y=-5

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

=>(x+1)(y+2)=-3

=>(x+1;y+2)∈{(1;-3);(-3;1);(-1;3);(3;-1)}

=>(x;y)∈{(0;-5);(-4;-1);(-2;1);(2;-3)}

a: x/2=-5/y

=>xy=-10

=>\(\left(x,y\right)\in\left\{\left(1;-10\right);\left(-10;1\right);\left(-1;10\right);\left(10;-1\right);\left(2;-5\right);\left(-5;2\right);\left(-2;5\right);\left(5;-2\right)\right\}\)

b: =>xy=12

mà x>y>0

nên \(\left(x,y\right)\in\left\{\left(12;1\right);\left(6;2\right);\left(4;3\right)\right\}\)

c: =>(x-1)(y+1)=3

=>\(\left(x-1;y+1\right)\in\left\{\left(1;3\right);\left(3;1\right);\left(-1;-3\right);\left(-3;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(2;2\right);\left(4;0\right);\left(0;-4\right);\left(-2;-2\right)\right\}\)

d: =>y(x+2)=5

=>\(\left(x+2;y\right)\in\left\{\left(1;5\right);\left(5;1\right);\left(-1;-5\right);\left(-5;-1\right)\right\}\)

=>\(\left(x,y\right)\in\left\{\left(-1;5\right);\left(3;1\right);\left(-3;-5\right);\left(-7;-1\right)\right\}\)

Rxfxtd

Lời giải:

a. Áp dụng TCDTSBN:

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

$\Rightarrow x=-3.2=-6; y=-3.5=-15$

b. Áp dụng TCDTSBN:

$\frac{x}{2}=\frac{y}{3}; \frac{y}{4}=\frac{z}{7}$

$\Rightarrow \frac{x}{8}=\frac{y}{12}=\frac{z}{21}$

$=\frac{2x}{16}=\frac{y}{12}=\frac{z}{21}=\frac{2x-y+z}{16-12+21}=\frac{50}{25}=2$

$\Rightarrow x=8.2=16; y=2.12=24; z=2.21=42$

c.

$\frac{x}{2}=\frac{y}{3}=\frac{z}{4}$

$\Rightarrow \frac{x^2}{4}=\frac{y^2}{9}=\frac{z^2}{16}=\frac{2z^2}{32}$

$=\frac{x^2-y^2+2z^2}{4-9+32}=\frac{108}{27}=4$

$\Rightarrow x^2=4.4=16; y^2=9.4=36; z^2=4.4=16$

Kết hợp với đkxđ suy ra:
$(x,y,z)=(4,6,4); (-4; -6; -4)$

Em cảm ơn ạ

`A)2/3=x/60`

`=>40/60=x/60`

`=>x=40`

`B)-1/2=y/18`

`=>-9/18=y/18`

`=>y=-9`

`C)3/x=y/35=-36/84`

Mà `-36/84=(-3 xx 12)/(7 xx 12)=-3/7`

`=>3/x=-3/7`

`=>x=-7`

`y/35=-3/7=-15/35`

`=>y=-15`

`D)7/x=y/27=-42/54`

Mà `-42/54=(-7 xx 6)/(9 xx 6)=-7/9`

`=>7/x=-7/9`

`=>x=-9`

`y/27=-7/9=-21/27`

`=>y=-21`

Ông viết rõ ràng đc ko?

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

\(\dfrac{x}{10}=\dfrac{y}{15}=\dfrac{z}{12}=\dfrac{x-y+z}{10-15+12}=\dfrac{-49}{7}=-7\)

Do đó: x=-70; y=-135; z=-84

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

Do đó: x=-70; y=-135; z=-84

a: x:y:z=10:3:4

=>\(\frac{x}{10}=\frac{y}{3}=\frac{z}{4}\)

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

\(\frac{x}{10}=\frac{y}{3}=\frac{z}{4}=\frac{x+2y-3z}{10+2\cdot3-3\cdot4}=\frac{-20}{10+6-12}=\frac{-20}{4}=-5\)

=>\(\begin{cases}x=-5\cdot10=-50\\ y=-5\cdot3=-15\\ z=-5\cdot4=-20\end{cases}\)

b: \(\frac{x}{2}=\frac{y}{3}\)

=>\(\frac{x}{10}=\frac{y}{15}\) (1)

\(\frac{y}{5}=\frac{z}{4}\)

=>\(\frac{y}{15}=\frac{z}{12}\) (2)

Từ (1),(2) suy ra \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\)

mà x-y+z=-49

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

\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7\)

=>\(\begin{cases}x=-7\cdot10=-70\\ y=-7\cdot15=-105\\ z=-7\cdot12=-84\end{cases}\)

c: Đặt \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)

=>x=2k; y=3k; z=4k

\(xy+z^2=88\)

=>\(2k\cdot3k+\left(4k\right)^2=88\)

=>\(6k^2+16k^2=88\)

=>\(22k^2=88\)

=>\(k^2=4\)

=>k=2 hoặc k=-2

TH1: k=2

=>\(\begin{cases}x=2\cdot2=4\\ y=3\cdot2=6\\ z=4\cdot2=8\end{cases}\)

TH2: k=-2

=>\(\begin{cases}x=-2\cdot2=-4\\ y=-2\cdot3=-6\\ z=-2\cdot4=-8\end{cases}\)

a: Áp dụng tính chất của DTSBN, ta được:

x/5=y/2=(x-y)/(5-2)=9/3=3

=>x=15; y=6

b: =>(x-3)/12=3/(x-3)

=>(x-3)^2=36

=>(x-9)(x+3)=0

=>x=9 hoặc x=-3

c; x/2=y/3

=>x/10=y/15

y/5=z/4

=>y/15=z/12

=>x/10=y/15=z/12=(x-y-z)/(10-15-12)=-49/-17=49/17

=>x=490/17; y=735/17; z=588/17

A nhé

Thanks