Question

Mn giúp mk với

B1: Tìm x,y biết:

a) \(\dfrac{x}{5}=\dfrac{y}{4}\)và x² -y² =1

b) \(|2x-y+\dfrac{1}{2}|+\left(x+y-\dfrac{3}{2}\right)^2=0\)

Answer 1

a) Áp dụng t/c dãy tỉ số bằng nhau :

\(\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{25}=\dfrac{1}{9}\Rightarrow x^2=\dfrac{25}{9}\Rightarrow x=\pm\dfrac{5}{3}\\\dfrac{y^2}{16}=\dfrac{1}{9}\Rightarrow y^2=\dfrac{16}{9}\Rightarrow y=\pm\dfrac{4}{3}\end{matrix}\right.\)

Vậy............

Answer 2

Đặt \(\dfrac{x}{5}=\dfrac{y}{4}=t\Rightarrow x=5t;y=4t\)(*)

Thay (*) vào \(x^2-y^2=1\) ta được:

\(\left(5t\right)^2\cdot\left(4t\right)^2=1\)

\(\Rightarrow5^2\cdot t^2-4^2\cdot t^2=1\cdot4\)

\(\Rightarrow t^2\left(25-16\right)=1\)

\(\Rightarrow t^2\cdot9=1\)

\(\Rightarrow t^2=\dfrac{1}{9}\)

\(\Rightarrow t=\dfrac{1}{3}\) hoặc \(t=\dfrac{-1}{3}\)

\(\Rightarrow x=5t;y=4t\)(tự tính nhá )

b) \(\left|2x-y+\dfrac{1}{2}\right|+\left(x+y-\dfrac{3}{2}\right)^2=0\) khi \(\left|2x-y+\dfrac{1}{2}\right|=\left(x+y-\dfrac{3}{2}\right)=0\) hoặc \(\left|2x-y+\dfrac{1}{2}\right|\) và\(\left(x+y-\dfrac{3}{2}\right)^2\) là 2 số đối nhau

mà \(\left|2x-y+\dfrac{1}{2}\right|\) và \(\left(x+y-\dfrac{3}{2}\right)^2\) đều lớn hơn hoặc bằng 0 nên không thể là 2 số đối nhau

\(\Rightarrow\left|2x-y+\dfrac{1}{2}\right|=\left(x+y-\dfrac{3}{2}\right)^2=0\)

\(\left|2x-y+\dfrac{1}{2}\right|=0\Rightarrow2x-y+\dfrac{1}{2}=0\)

\(\Rightarrow2x-y=-\dfrac{1}{2}\)

\(\Rightarrow y=2x-\left(-\dfrac{1}{2}\right)=2x+\dfrac{1}{2}\) (1)

\(\left(x+y-\dfrac{3}{2}\right)^2=0\Rightarrow x+y-\dfrac{3}{2}=0\)

\(\Rightarrow x+y=\dfrac{3}{2}\) (2)

Thay (1) vào (2) ta được:

\(x+2x+\dfrac{1}{2}=\dfrac{3}{2}\)

\(\Rightarrow3x=\dfrac{3}{2}-\dfrac{1}{2}=\dfrac{2}{2}=1\)

\(\Rightarrow x=\dfrac{1}{3}\)

\(\Rightarrow y=2x+\dfrac{1}{2}=2\cdot\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{7}{6}\)

Answer 3

a, \(\dfrac{x}{5}=\dfrac{y}{4}\)và x²-y²= 1
Ta có:\(\dfrac{x}{2}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{4}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{4-16}=\dfrac{-1}{12}=-0,083\\ \Rightarrow\dfrac{x}{2}=-0,083\Rightarrow x=-0,083\cdot2=-0,166\\ \Rightarrow\dfrac{y}{4}=-0,083\Rightarrow y=-0,083\cdot4=-0,332\)

Answer 4

chết mình bị nhầm r sr bạn nha

Answer 5

a)\(\dfrac{x}{5}\) = \(\dfrac{y}{4}\)

-->\(\dfrac{x^2}{25}\) =\(\dfrac{y^2}{16}\) =\(\dfrac{x^2-y^2}{25-16}\) =\(\dfrac{1}{9}\) (Vì x² - y²=1)

--> \(\dfrac{x^2}{25}\)=\(\dfrac{1}{9}\) -->x²=\(\dfrac{1}{9}\) .25=\(\dfrac{25}{9}\)=( \(\dfrac{-5}{3}\))²=( \(\dfrac{5}{3}\))²

\(\dfrac{y^2}{16}\)=\(\dfrac{1}{9}\)--> y²=\(\dfrac{1}{9}\) . 16=\(\dfrac{16}{9}\)=( \(\dfrac{-4}{3}\))² =( \(\dfrac{4}{3}\))²
Vậy x=\(\dfrac{5}{3}\) và y= \(\dfrac{4}{3}\) hoặc x=\(\dfrac{-5}{3}\) và y =\(\dfrac{-4}{3}\)

Answer 6

\(a,\dfrac{x}{5}=\dfrac{y}{4}\)và x²-y²=1
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{x}{5}=\dfrac{y}{4}\Rightarrow\dfrac{x^2}{25}=\dfrac{y^2}{16}=\dfrac{x^2-y^2}{25-16}=\dfrac{1}{9}=0,1 \\ \Rightarrow\dfrac{x}{5}=0,1\Rightarrow x=0,1\cdot5=0,5\\ \Rightarrow\dfrac{y}{4}=0,1\Rightarrow y=0,1\cdot4=0,4.\)

Answer 7

mình sai nữa r