\(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}.Cminh:\frac{x^3+y^3+z^3}{y^3+z^3+t^3}=\frac{x}{t}\)
cho các số dương x,y,z,t . Chứng minh: \(\frac{40}{3}\le\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\)
\(VP=\frac{x}{y+z+t}+\frac{y}{z+t+x}+\frac{z}{t+x+y}+\frac{t}{x+y+z}+\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}=\left(\frac{x}{y+z+t}+\frac{y+z+t}{9x}\right)+\left(\frac{y}{z+t+x}+\frac{z+t+x}{9y}\right)+\left(\frac{z}{t+x+y}+\frac{t+x+y}{9z}\right)+\left(\frac{t}{x+y+z}+\frac{x+y+z}{9t}\right)+\frac{8}{9}\left(\frac{y+z+t}{x}+\frac{z+t+x}{y}+\frac{t+x+y}{z}+\frac{x+y+z}{t}\right)\)\(\ge8\sqrt[8]{\frac{x}{y+z+t}.\frac{y}{z+t+x}.\frac{z}{t+x+y}.\frac{t}{x+y+z}.\frac{y+z+t}{9x}.\frac{z+t+x}{9y}.\frac{t+x+y}{9z}.\frac{x+y+z}{9t}}+\frac{8}{9}\left(\frac{y}{x}+\frac{z}{x}+\frac{t}{x}+\frac{z}{y}+\frac{t}{y}+\frac{x}{y}+\frac{t}{z}+\frac{x}{z}+\frac{y}{z}+\frac{x}{t}+\frac{y}{t}+\frac{z}{t}\right)\)\(\ge\frac{8}{3}+\frac{8}{9}.12\sqrt[12]{\frac{y}{x}.\frac{z}{x}.\frac{t}{x}.\frac{z}{y}.\frac{t}{y}.\frac{x}{y}.\frac{t}{z}.\frac{x}{z}.\frac{y}{z}.\frac{x}{t}.\frac{y}{t}.\frac{z}{t}}=\frac{8}{3}+\frac{8}{9}.12=\frac{40}{3}=VT\left(đpcm\right)\)
Đẳng thức xảy ra khi x = y = z = t > 0
\(Cho\) \(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}\) . Chứng minh rằng \(\frac{x^3+y^3+z^3}{y^3+z^3+t^3}\)\(=\frac{x}{t}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{x}{y}=\frac{y}{z}=\frac{z}{t}=\frac{x+y+z}{y+z+t}\)
Vì \(\frac{x^3+y^3+z^3}{y^3+z^3+t^3}\Leftrightarrow\left(\frac{x+y+z}{y+z+t}\right)^3\)
\(\Rightarrow\left(\frac{x+y+z}{y+z+t}\right)^3=\frac{x+y+z}{y+z+t}.\frac{x+y+z}{y+z+t}.\frac{x+y+z}{y+z+t}=\frac{x}{y}.\frac{y}{z}.\frac{z}{t}=\frac{x}{t}\) (đpcm)
\(Tìm\) \(\frac{t}{y}\) :
\(a)\)\(\frac{t}{x}=\frac{4}{3};\frac{y}{z}=\frac{3}{2};\frac{z}{x}=\frac{1}{6}\)
\(b)\frac{t}{x}=\frac{4}{3};\frac{y}{z}=\frac{2}{3};\frac{z}{x}=\frac{1}{6}\)
\(c)\frac{t}{x}=\frac{3}{4};\frac{y}{z}=\frac{3}{2};\frac{z}{x}=\frac{1}{6}\)
thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
\(
\frac{2.x+2.y+3}{z}=\frac{3.x+3.x+1}{y}=\frac{y+z+2}{x}=\frac{6}{x+y+z}
\)
Tìm x,y,z
Câu 1: 13 + 23 +33 + ... + 1003 = ?
Câu 2: A= 1x2 + 2x3 + 3x4+ ...+ 277x278 = ?
Câu 3: B= 1x2x3 + 2x3x4 +...+ 111x112x113 = ?
Câu 4: Cho \(\frac{x}{y+z+t}=\frac{y}{x+z+t}=\frac{z}{x+y+t}=\frac{t}{x+y+z}\) .Giá trị dương của \(\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
Câu 5: 1x4 + 2x5 + ... + 277x280 =?
Các bạn nhớ ghi cách làm và đáp án ra nhé! Cảm ơn các bạn nhìu!!!
Cho x,y,z,t > 0. Chứng minh rằng:
\(\frac{3}{4}<\frac{x}{x+y+z}+\frac{y}{y+z+t}+\frac{z}{z+t+x}+\frac{t}{t+x+y}<\frac{5}{2}\)
đặt A=x/x+y+z +y/y+z+t +z/z+t+x +t/t+x+y
ta có x/x+y+z>x/x+y+z+t
y/y+z+t>y/x+y+z+t
z/z+t+x>z/z+t+x+y
t/t+x+y>t/x+t+y+z
=>A>x/x+y+t+z +t/x+y+t+z +z/x+y+t+z +y/x+t+y+z=x+y+z+t/x+y+z+t=1>3/4 (1)
*)y/y+z+t<y+x/y+z+t+x
x/x+y+z<x+t/x+y+z+t
z/z+t+x<z+y/x+y+z+t
t/t+x+y<t+z/t+x+y+z
=>A<y+x/x+y+z+t +x+t/x+y+z+t +z+y/x+y+z+t +t+z/x+y+z+t
=y+x+x+t+z+y+t+z/x+y+z+t=2(x+y+z+t)/x+y+z+t=2<5/2 (2)
từ (1) và (2) =>3/4<A<5/2
=>
Ta có:
\(\frac{x}{x+y+z+t}+\frac{y}{x+y+z+t}+\frac{z}{x+y+z+t}+\frac{t}{x+y+z+t}
Ai giúp đc cho 3 tick!!!!!!!!
Cho \(P=\frac{x+y}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+z}=\frac{t}{x+y+z}\)
Tìm x, y, z
a. Tìm x,y,z biết :
\(\frac{x}{5}\)=\(\frac{y}{3};\frac{y}{5}=\frac{z}{7}\)và 5x + y- 2x = 28
b. Tìm 3 số x,y,z biết:
\(\frac{x}{2}=\frac{y}{3};\frac{y}{5}=\frac{z}{4}\)và x-y+z= -49
c. Tìm x,y,z biết:
\(\frac{x}{3}=\frac{y}{4};\frac{y}{5}=\frac{z}{7}\)và 2x + 3y - z = 186
Giải hộ mình với ạ!
b)
Ta có: \(\frac{x}{2}=\frac{y}{3}=>\frac{x}{10}=\frac{y}{15}.\)
\(\frac{y}{5}=\frac{z}{4}=>\frac{y}{15}=\frac{z}{12}.\)
=> \(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}\) và \(x-y+z=-49.\)
Áp dụng tính chất dãy tỉ số bằng nhau ta được:
\(\frac{x}{10}=\frac{y}{15}=\frac{z}{12}=\frac{x-y+z}{10-15+12}=\frac{-49}{7}=-7.\)
\(\left\{{}\begin{matrix}\frac{x}{10}=-7=>x=\left(-7\right).10=-70\\\frac{y}{15}=-7=>y=\left(-7\right).15=-105\\\frac{z}{12}=-7=>z=\left(-7\right).12=-84\end{matrix}\right.\)
Vậy \(\left(x;y;z\right)=\left(-70;-105;-84\right).\)
Chúc bạn học tốt!
a) Ta có: \(\frac{x}{5}\)= \(\frac{y}{3}\) =>\(\frac{x}{25}\)= \(\frac{y}{15}\)
\(\frac{y}{5}\)= \(\frac{z}{7}\) => \(\frac{y}{15}\)= \(\frac{z}{21}\)
=> \(\frac{x}{25}\)= \(\frac{y}{15}\)= \(\frac{z}{21}\)=> \(\frac{5x}{125}\)= \(\frac{y}{15}\)= \(\frac{2z}{42}\)
Áp dụng tính chất dãy tỉ số = nhau
Ta có: \(\frac{5x}{125}\)= \(\frac{y}{15}\)= \(\frac{2z}{42}\)= \(\frac{5x+y-2z}{125+15-42}\)= \(\frac{28}{98}\)= \(\frac{2}{7}\)
Vậy x = \(\frac{50}{7}\)
y = \(\frac{30}{7}\)
z = 6
Bạn xem lại ý sau sao lại có 2 chữ x mà ko có z nhé!
b) Ta có: \(\frac{x}{2}\)= \(\frac{y}{3}\)=> \(\frac{x}{10}\)= \(\frac{y}{15}\)
\(\frac{y}{5}\)= \(\frac{z}{4}\)=> \(\frac{y}{15}\)= \(\frac{z}{12}\)
=> \(\frac{x}{10}\)= \(\frac{y}{15}\)= \(\frac{z}{12}\)
Áp dụng tính chất dãy tỉ số = nhau
Ta có: \(\frac{x}{10}\)= \(\frac{y}{15}\)= \(\frac{z}{12}\)= \(\frac{x-y+z}{10-15+12}\)= \(\frac{-49}{7}\)= -7
Vậy x = -70
y = -105
z = -84
c) Ta có: \(\frac{x}{3}\)= \(\frac{y}{4}\)=> \(\frac{x}{15}\)= \(\frac{y}{20}\)
\(\frac{y}{5}\)= \(\frac{z}{7}\)=> \(\frac{y}{20}\)= \(\frac{z}{28}\)
=> \(\frac{x}{15}\)= \(\frac{y}{20}\)= \(\frac{z}{28}\)= \(\frac{2x}{30}\)= \(\frac{3y}{60}\)= \(\frac{z}{28}\)
Áp dụng tính chất dãy tỉ số = nhau
Ta có: \(\frac{2x}{30}\)= \(\frac{3y}{60}\)= \(\frac{z}{28}\)= \(\frac{2x+3y-z}{30+60-28}\)= \(\frac{186}{62}\)= 3
Vậy x = 45
y = 60
z = 84
a) Ta có : \(\left\{{}\begin{matrix}\frac{x}{5}=\frac{y}{3}\Rightarrow\frac{x}{25}=\frac{y}{15}\\\frac{y}{5}=\frac{z}{7}\Rightarrow\frac{y}{15}=\frac{z}{21}\end{matrix}\right.\Rightarrow}\frac{x}{25}=\frac{y}{15}=\frac{z}{21}\)