tinh gia tri cua bieu thuc:
\(y=\frac{x+y}{z}+\frac{x+z}{y}+\frac{y+z}{x}\)neu \(\frac{1}{x}+\frac{1}{z}+\frac{1}{y}=0\)
Cho x,y,z khác 0 và x-y-z=0 . tinh gia tri cua bieu thuc \(\left(1-\frac{z}{x}\right)\left(1-\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\)
cho \(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)
tinh gia tri cua bieu thuc B=\(\frac{x+y-z}{x+2y-z}\)
1) A= \(\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
b) Cho 3 so x,y,z la 3 so khac 0 thoa man dieu kien :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}\)
Hay tinh gia tri bieu thuc:\(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)\)
Bài 1 :
Ta có :
\(A=\frac{\frac{3}{4}-\frac{3}{11}+\frac{3}{13}}{\frac{5}{7}-\frac{5}{11}+\frac{5}{13}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{4}-\frac{5}{6}+\frac{5}{8}}\)
\(A=\frac{3\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}}{\frac{5}{2}\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}\right)}\)
\(A=\frac{3}{5}+\frac{1}{\frac{5}{2}}\)
\(A=\frac{3}{5}+\frac{2}{5}\)
\(A=1\)
\(b)\) Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{y+z-x}{x}=\frac{z+x-y}{y}=\frac{x+y-z}{z}=\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{2\left(x+y+z\right)}{x+y+z}=2\)
Đo đó :
\(\frac{y+z-x}{x}=2\)\(\Rightarrow\)\(y+z=3x\)\(\left(1\right)\)
\(\frac{z+x-y}{y}=2\)\(\Rightarrow\)\(x+z=3y\)\(\left(2\right)\)
\(\frac{x+y-z}{z}=2\)\(\Rightarrow\)\(x+y=3z\)\(\left(3\right)\)
Lại có : \(B=\left(1+\frac{x}{y}\right)\left(1+\frac{y}{z}\right)\left(1+\frac{z}{x}\right)=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\)
Thay (1), (2) và (3) vào \(B=\frac{x+y}{y}.\frac{y+z}{z}.\frac{x+z}{x}\) ta được :
\(B=\frac{2z}{y}.\frac{2x}{z}.\frac{2y}{x}=\frac{8xyz}{xyz}=8\)
Vậy \(B=8\)
Chúc bạn học tốt ~
bạn phùng minh quân câu 1 a tại sao lại rút gọn được \(\frac{3.\left(\frac{1}{4}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{7}-\frac{1}{11}+\frac{1}{13}\right)}=\frac{3}{5}\) vậy nó không cùng nhân tử mà
câu b \(\frac{y+z-x+z+x-y+x+y-z}{x+y+z}=\frac{\left(y-y+y\right)+\left(-x+x+x\right)+\left(z+z-z\right)}{x+y+z}=\frac{x+y+z}{x+y+z}=1\)sao lại ra bằng 2
(mình chỉ góp ý thôi nha tại mình làm thấy nó sai sai)
Cho \(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{t+x+y}=\frac{t}{x+y+z}\)
Tinh gia tri cua da thuc\(P=\frac{x+y}{z+t}+\frac{y+z}{t+x}+\frac{z+t}{x+y}+\frac{t+x}{y+z}\)
\(choP=\frac{1}{x+y+z}.\frac{1}{xy+yz+zx}.\left[\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right]\left[\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right]\)
chung minh rang gia tri bieu thuc P luon luon duong voi moi x,y,z khac 0
cho x,y,z la cac so thuc thoa x+y+z=0, x+1>0, y+1>0, z+1>0. tim GTLN cua P=\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+4}\)
cho x,y,z,t la cac so duong. tim GTNN cua A=\(\frac{x-t}{t+y}+\frac{t-y}{y+z}+\frac{y-z}{z+x}+\frac{z-x}{x+t}\)
cho biet xyz=1.tinh gia tri cua A=\(\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{xz+z+1}\)
\(A=\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1}\)
\(A=\frac{xz}{xyz+xz+z}+\frac{yxz}{yz.xz+xyz+xz}+\frac{z}{zx+z+1}\) Thay xyz=1 vào ta được:
\(A=\frac{xz}{xz+z+1}+\frac{1}{z+1+xz}+\frac{z}{zx+z+1}\)
\(A=\frac{zx+z+1}{zx+z+1}=1\)
=> A=1
1. Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1.\)
Tinh \(A=\frac{x^2+y^2-z^2}{y+z}+\frac{-x^2+y^2+z^2}{z+x}+\frac{x^2-y^2+z^2}{x+y}\)
2. Cho a,b,c>0 va ab+bc+ca=1. Tinh gia tri \(A=a+b-\sqrt{\frac{\left(1+a^2\right)\left(1+b^2\right)}{\left(1+c^2\right)}}\)
1) A = \(\frac{x^2+\left(y-z\right)\left(y+z\right)}{y+z}+\frac{y^2+\left(z-x\right)\left(z+x\right)}{z+x}+\frac{\left(x-y\right)\left(x+y\right)+z^2}{x+y}\)
A = \(\frac{x^2}{y+z}+\left(y-z\right)+\frac{y^2}{z+x}+\left(z-x\right)+\left(x-y\right)+\frac{z^2}{x+y}\)
A = \(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
Nhân cả hai vế của \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) với x ta được:
\(\frac{x^2}{y+z}+\frac{yx}{z+x}+\frac{zx}{x+y}=x\)
Tương tự, ta nhân hai vế với y; z rồi cộng từng vế 2 đẳng thức với nhau ta được:
\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy}{z+x}+\frac{yz}{z+x}\right)+\left(\frac{xy}{y+z}+\frac{xz}{y+z}\right)+\left(\frac{zx}{x+y}+\frac{yz}{x+y}\right)=x+y+z\)
=> A + \(\frac{\left(x+z\right)y}{z+x}+\frac{\left(y+z\right)x}{y+z}+\frac{z\left(x+y\right)}{x+y}\) = x+ y + z
=> A + y + x + z = x + y + z
=> A = 0
Vậy A = 0
cho cac so x,y,z va x+y+z khac 0 thoa man dieu kien
\(\frac{x+2y}{x+2y-z}+\frac{y+2z}{y+2z-x}+\frac{z+2x}{z+2x-+y}\)
tinh gt bieu thuc \(T=\frac{x^2+y^2}{xy}+\frac{y^2+z^2}{yz}+\frac{z^2+x^2}{zx}\)