Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Đề bài sai
Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)
Khi đó \(x< y\) nhưng \(z< y\)
\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)
\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)
\(b\left(a+c\right)=ba+bc\left(3\right)\)
\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)
\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)
\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
\(\Rightarrow x< y< z\)
Ta có : \(2\left(x+y\right)=5\left(y+z\right)=3\left(x+z\right)\)
=> \(\dfrac{2\left(x+y\right)}{30}=\dfrac{5\left(y+z\right)}{30}=\dfrac{3\left(x+z\right)}{30}\)
=> \(\dfrac{x+y}{15}=\dfrac{y+z}{6}=\dfrac{x+z}{10}\)
Áp dụng t/c dãy tỉ số bằng nhau:
\(\dfrac{x+y}{15}=\dfrac{y+z}{6}=\dfrac{x+z}{10}=\dfrac{x+z-y-z}{10-6}=\dfrac{x+y-x-z}{15-10}=\dfrac{x-y}{4}=\dfrac{y-z}{5}\left(ĐPCM\right)\)
a) Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
=> ad = bc
Ta có : (a + 2c)(b + d)
= a(b + d) + 2c(b + d)
= ab + ad + 2cb + 2cd (1)
Ta có : (a + c)(b + 2d)
= a(b + 2d) + c(b + 2b)
= ab + a2d + cb + c2b
= ab + c2d + ad + c2b (Vì ad = cd) (2)
Từ (1),(2) => (a + 2c)(b + d) = (a + c)(b + 2d) (ĐPCM)
Sửa đề bài : P = \(\dfrac{x+y}{z+t}+\dfrac{y+z}{t+x}+\dfrac{z+t}{x+y}+\dfrac{t+x}{y+z}\)
Ta có : \(\dfrac{x}{y+z+t}=\dfrac{y}{z+t+x}=\dfrac{z}{t+x+y}=\dfrac{t}{x+y+z}\)
=> \(\dfrac{y+z+t}{x}=\dfrac{z+t+x}{y}=\dfrac{t+x+y}{z}=\dfrac{x+y+z}{t}\)
=> \(\dfrac{y+z+t}{x}+1=\dfrac{z+t+x}{y}+1=\dfrac{t+x+y}{z}+1=\dfrac{x+y+z}{t}+1\)=> \(\dfrac{y+z+t+x}{x}=\dfrac{z+t+x+y}{y}=\dfrac{t+x+y+z}{z}=\dfrac{x+y+z+t}{t}\)TH1: x + y + z + t # 0
=> x = y = z = t
Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
P = \(\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}+\dfrac{x+x}{x+x}\)
P = 1 + 1 + 1 + 1 = 4
TH2 : x + y + z + t = 0
=> x + y = -(z + t)
y + z = -(t + x)
z + t = -(x + y)
t + x = -(y + z)
Ta có : P = \(\dfrac{x+y}{z+t}=\dfrac{y+z}{t+x}=\dfrac{z+t}{x+y}=\dfrac{t+x}{y+z}\)
P = \(\dfrac{-\left(z+t\right)}{z+t}=\dfrac{-\left(t+x\right)}{t+x}=\dfrac{-\left(x+y\right)}{x+y}=\dfrac{-\left(y+z\right)}{y+z}\)
P = (-1) + (-1) + (-1) + (-1)
P = -4
Vậy ...
\(\dfrac{x-y}{x+y}=\dfrac{z-x}{z+x}\\ \Rightarrow\left(x-y\right)\left(z+x\right)=\left(x+y\right)\left(z-x\right)\\ \Rightarrow xz+x^2-yz-yx=xz-x^2+yz-yx\\ \Rightarrow xz-xz+x^2+x^2=yz+yz-yx+yx\\ \Rightarrow2x^2=2yz\\ \Rightarrow x^2=yz\)