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Ta có: \(\frac{x+y-3}{z}=\frac{y+z+1}{x}=\frac{z+x+2}{y}=\frac{1}{x+y+z}\)
\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=x+y+z\)
TH1: \(x+y+z=0\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=\frac{x+y+z}{x+y-3+y+z+1+z+x+2}\)
\(=\frac{x+y+z}{x+y+y+z+z+x}=\frac{x+y+z}{2\left(x+y+z\right)}=\frac{1}{2}\)
\(\Rightarrow x+y+z=\frac{1}{2}\)
\(\Rightarrow x+y=\frac{1}{2}-z\)
\(y+z=\frac{1}{2}-x\)
\(z+x=\frac{1}{2}-y\)
Thay \(x+y-3=\frac{1}{2}-z-3\)
\(\Rightarrow\frac{z}{\frac{1}{2}-z+3}=\frac{1}{2}\)
\(\Rightarrow2z=\frac{1}{2}-z-3\)
\(\Rightarrow2z+z=\frac{1}{2}-3\)
\(\Rightarrow3z=-\frac{5}{2}\Rightarrow z=-\frac{5}{6}\)
Thay \(y+z+1=\frac{1}{2}-x+1\)
\(\Rightarrow\frac{x}{\frac{1}{2}-x+1}=\frac{1}{2}\)
\(\Rightarrow2x=\frac{1}{2}-x+1\)
\(\Rightarrow2x+x=\frac{1}{2}+1\)
\(\Rightarrow3x=\frac{3}{2}\Rightarrow x=\frac{1}{2}\)
Thay \(z+x+2=\frac{1}{2}-y+2\)
\(\Rightarrow\frac{y}{\frac{1}{2}-y+2}=\frac{1}{2}\)
\(\Rightarrow2y=\frac{1}{2}-y+2\)
\(\Rightarrow2y+y=\frac{1}{2}+2\)
\(\Rightarrow3y=\frac{5}{2}\Rightarrow y=\frac{5}{6}\)
Ta có: \(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)
\(=\left(\frac{1}{2}+\frac{5}{6}+-\frac{5}{6}-\frac{3}{2}\right)^{2019}\)
\(=\left[\left(\frac{1}{2}-\frac{3}{2}\right)+\left(-\frac{5}{6}+\frac{5}{6}\right)\right]^{2019}\)
\(=\left(-1\right)^{2019}=-1\)
TH2: x + y + z = 0
\(\Rightarrow\frac{z}{x+y-3}=\frac{x}{y+z+1}=\frac{y}{z+x+2}=0\)
\(\Rightarrow x=y=z=0\)
\(A=\left(x+y+z-\frac{3}{2}\right)^{2019}\)
\(=\left(0-\frac{3}{2}\right)^{2019}=\left(-\frac{3}{2}\right)^{2019}\)
Ah! Mk nhầm chút. TH1 là khác 0 nhé!!!!!!
Ta có:
\(\frac{xy}{x+y}=\frac{yz}{y+z}=\frac{zx}{z+x}\rightarrow\frac{x+y}{xy}=\frac{y+z}{yz}=\frac{z+x}{zx}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}=\frac{1}{y}+\frac{1}{z}=\frac{1}{z}+\frac{1}{x}\Rightarrow\frac{1}{x}=\frac{1}{y}=\frac{1}{z}\Rightarrow x=y=z\)
Thay tất cả giá trị x,y,z vào M ta được:
\(M=\frac{2020x^3+2020y^3+2020z^3}{x^3+y^3+z^3}+\frac{2021x^5+2021y^5}{x^5+y^5}\)
\(\Rightarrow M=\frac{2020\left(x^3+y^3+z^3\right)}{x^3+y^3+z^3}+\frac{2021\left(x^5+y^5\right)}{x^5+y^5}\)
\(\Rightarrow M=2020+2021=4041\)
\(\frac{x+5}{4}=\frac{y+3}{3}=\frac{z+1}{2}\), áp dụng tính chất dãy tỉ số bằng nhau
\(\frac{x+5}{4}=\frac{y+3}{3}=\frac{z+1}{2}=\frac{x+5+y+3+z+1}{4+3+2}=\frac{11}{9}\)
rồi tính x,y,z và cho vào M nhé
4y = 3z => z = 4/3.y
\(\frac{x+y+z}{x+y-z}\)\(=\frac{\frac{y}{2}+y+\frac{4}{3}.y}{\frac{y}{2}+y-\frac{4}{3}.y}\)\(=\frac{y.\left(\frac{1}{2}+1+\frac{4}{3}\right)}{y.\left(\frac{1}{2}+1-\frac{4}{3}\right)}\)\(=\frac{\frac{17}{6}}{\frac{1}{6}}=\frac{17.6}{6}=17\)
Ta có: \(x=\frac{y}{2};\frac{y}{3}=\frac{z}{4}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{6};\frac{y}{6}=\frac{z}{8}\)
\(\Rightarrow\frac{x}{3}=\frac{y}{6}=\frac{z}{8}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{x}{3}=\frac{y}{6}=\frac{z}{8}=\frac{x+y+z}{3+6+8}=\frac{x+y-z}{3+6-8}\)
\(\Rightarrow\frac{x+y+z}{3+6+8}=\frac{x+y-z}{3+6-8}\)
\(\Rightarrow\frac{x+y+z}{17}=\frac{x+y-z}{1}\)
\(\Rightarrow\frac{x+y+z}{x+y-z}=\frac{17}{1}=17\)
Vậy .................................................
làm lần lượt nhá,dài dòng quá khó coi.ahihihi!
\(\frac{1-\frac{1}{\sqrt{49}}+\frac{1}{49}-\frac{1}{7\left(\sqrt{7}\right)^2}}{\frac{\sqrt{64}}{2}-\frac{4}{7}+\left(\frac{2}{7}\right)^2-\frac{4}{343}}=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4-\frac{4}{7}+\frac{4}{49}-\frac{4}{343}}\)
\(=\frac{1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}}{4\left(1-\frac{1}{7}+\frac{1}{49}-\frac{1}{343}\right)}=\frac{1}{4}\)
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=\frac{2\left(x-1\right)+3\left(y-2\right)-\left(z-3\right)}{2.2+3.3-4}=\frac{2x+3y-z-5}{9}=\frac{45}{9}=5=\frac{x-1+y-2+z-3}{2+3+4}=\frac{x+y+z-6}{9}\)
=> x+y+z - 6 =9.5
=>x+y+z =45+6 =51
Đặt: \(\frac{x}{2}=\frac{y}{3}=\frac{z}{4}=k\)
\(\Rightarrow\)\(x=2k;\)\(y=3k;\)\(z=4k\)
Ta có: \(P=\frac{y+z-x}{x-y+z}\)
\(=\frac{3k+4k-2k}{2k-3k+4k}\)
\(=\frac{5k}{3k}=\frac{5}{3}\)
P/s: tham khảo nha