Question

Bài 1 . Cho \(\frac{a}{b}\)=\(\frac{b}{c}=\frac{c}{d}\)

Chứng minh \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{d}\)

Bài 2 . Tìm A biết A = \(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)

Bài 3 . tìm x, y, z biết\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) và x-2y +3z =-10

Giúp mình nha mai mình có tiết kiểm tra rùi

Answer 1

1)Ta có:\(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{a+b+c}{b+c+d}\)

\(\Rightarrow\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{d}=\frac{a}{d}\)(đpcm)

Answer 2

Ta có:A=\(\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{c+a}\)

\(\Rightarrow A=\frac{a}{b+c}=\frac{c}{a+b}=\frac{b}{a+c}=\frac{a+c+b}{b+c+a+b+a+c}\)\(\Rightarrow A=\frac{a+b+c}{2a+2b+2c}=\frac{\left(a+b+c\right)}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy A=\(\frac{1}{2}\)

Answer 3

Đặt \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)=k

\(\Rightarrow x=2k+1;y=3k+2;z=4k+3\)

Mà x-2y+3z=-10

Hay (2k+1)-2\(\cdot\)(3k+2)+3\(\cdot\)(4k+3)=-10

2k+1-6k-4+12k+9=-10

8k+6=-10

8k=(-10)-6

8k=-16

k=(-16):8

k=(-2)

\(\Rightarrow x=\left(-2\right)\cdot2+1=-3\)

\(y=\left(-2\right)\cdot3+2=-4\)

\(z=\left(-2\right)\cdot4+3=-5\)

Answer 4

Bài 2:

Giải:

+) Xét a + b + c = 0 thì

-a = b + c

-b = a + c

-c = a + b

Ta có: \(A=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)

\(\Rightarrow A=\frac{a}{-a}=\frac{b}{-b}=\frac{c}{-c}\)

\(\Rightarrow A=-1\)

+) Xét a + b + c \(\ne\) 0

Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(A=\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)

Vậy \(A=-1\) hoặc \(A=\frac{1}{2}\)