Question

Answer 1

a. 

Áp dụng TCDTSBN:

$\frac{x}{2}=\frac{y}{5}=\frac{x+y}{2+5}=\frac{-21}{7}=-3$

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

Khi đó:

$A=5x+4y=5(-6)+4(-15)=-90$

Answer 2

b.

$\frac{x}{-3}=\frac{y}{5}$

$\Rightarrow (\frac{x}{-3})^2=(\frac{y}{5})^2=\frac{xy}{3(-5)}=\frac{-135}{-15}=9$

Suy ra có các TH sau:

TH1: $\frac{x}{-3}=3; \frac{y}{5}=3\Rightarrow x=-9; y=15$

$\Rightarrow B=3(-9)+5.15=48$

TH2: $\frac{x}{-3}=-3; \frac{y}{5}=-3\Rightarrow x=9; y=-15$

$\Rightarrow B=-48$

TH3: $\frac{x}{-3}=3; \frac{y}{5}=-3\Rightarrow x=-9; y=-15\Rightarrow xy=135\neq -135$ (loại) 

TH4:  $\frac{x}{-3}=-3; \frac{y}{5}=3\Rightarrow x=9; y=15\Rightarrow xy=135\neq -135$ (loại) 

Answer 3

c.

$\frac{x}{2}=\frac{y}{5}; \frac{y}{3}=\frac{z}{2}$

$\Rightarrow \frac{x}{6}=\frac{y}{15}=\frac{z}{10}$

Áp dụng TCDTSBN:

$ \frac{x}{6}=\frac{y}{15}=\frac{z}{10}=\frac{2x}{12}=\frac{3y}{45}=\frac{4z}{40}=\frac{2x+3y-4z}{12+45-40}=\frac{-34}{17}=-2$

$\Rightarrow x=(-2).6=-12; y=-2.15=-30; z=-2.10=-20$

Answer 4

d.

Đặt $\frac{a}{6}=\frac{b}{-4}=t\Rightarrow a=6t; b=-4t$

Khi đó:

$a^2-b^2=5$

$\Rightarrow (6t)^2-(-4t)^2=5$

$\Rightarrow 36t^2-16t^2=5$

$\Rightarrow 20t^2=5\Rightarrow t^2=\frac{1}{4}\Rightarrow t=\pm \frac{1}{2}$

Nếu $t=\frac{1}{2}$ thì $a=6t=3; b=-4t=-2$

Nếu $t=\frac{-1}{2}$ thì $a=6t=-3; b=-4t=2$