ca nha co ai dang on thi giup to may bai nha
\(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+y}\)
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Anh có cách khác nè :
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-z\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{yz\left(x-y+z-x\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(yz-xy\right)-\left(z-x\right)\left(zx-yz\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{y\left(x-y\right)\left(z-x\right)-z\left(x-y\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{1}{xyz}\)
\(\frac{1}{x\left(x-y\right)\left(x-z\right)}+\frac{1}{y\left(y-x\right)\left(y-z\right)}+\frac{1}{z\left(z-x\right)\left(z-y\right)}\)
\(=\frac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-y^2z+yz^2-z^2x+zx^2-x^2y+xy^2}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{-y^2\left(z-x\right)-zx\left(z-x\right)+y\left(z^2-x^2\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(z-x\right)\left(yz+xy-y^2-zx\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(z-y\right)\left[y\left(x-y\right)-z\left(x-y\right)\right]}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{xyz\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{1}{xyz}\)
8:50 gửi--> 9:30 đi
=> bạn phải nhắn tin may ra có kết quả mong đợi
bài 4 : Ta có : \(\frac{1+2y}{18}=\frac{1+4y}{24}\left(1\right)\)
\(\Rightarrow24+48y=18+72y
\)
\(\Rightarrow y=\frac{1}{4}\)
\(\frac{1+4y}{24}=\frac{1+6y}{6x}\left(2\right)\)
Thay y = \(\frac{1}{4}\) vào (2) ta được x = 5 (thõa mãn )
Do x < y
=> \(\frac{a}{m}< \frac{b}{m}\)
=> \(\frac{a}{m}+\frac{a}{m}< \frac{a}{m}+\frac{b}{m}< \frac{b}{m}+\frac{b}{m}\)
=> \(\frac{2a}{m}< \frac{a+b}{m}< \frac{2b}{m}\)
=> \(\frac{a}{m}< \frac{a+b}{m}:2< \frac{b}{m}\)
=> \(\frac{a}{m}< \frac{a+b}{2m}< \frac{b}{m}\)
=> x < z < y
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)
Do đó :
\(\frac{y+z-x}{x}=1\)\(\Rightarrow\)\(2x=y+z\)
\(\frac{z+x-y}{y}=1\)\(\Rightarrow\)\(2y=x+z\)
\(\frac{x+y-z}{z}=1\)\(\Rightarrow\)\(2z=x+y\)
Suy ra :
\(P=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{x}.\frac{y+z}{z}.\frac{x+z}{x}=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)
Vậy \(P=8\)
Đề hơi sai