\(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\Rightarrow A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=?\)
Đay là 1 bài violympic lớp 7: Cho tỉ lệ thức: \(\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}\). Khi đó \(x+y=kz\). Vậy k=?
Cách 1: vì \(x+y=kz\) nên \(\frac{kz}{z}=\frac{y+z}{x}=\frac{x+z}{y}=k\)\(\Rightarrow\frac{y+z-x-z}{x-y}=k\Rightarrow\frac{y-x}{x-y}=z\Rightarrow k=-1\)
Cách 2: \(\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}=\frac{2\left(x+y+z\right)}{x+y+z}=2\Rightarrow x+y=2z\Rightarrow k=2\)
Mình thấy 2 cách đều đúng mà trong violympic chỉ có 1 ô nên mình ko biết chọn phương án nào. Các bạn giúp mình chỉ ra lỗi sai của 1 cách nhé. Mình xin cảm ơn
à nhầm ở dòng 3 cáii\(\frac{y-x}{x-y}=k\) chứ ko phải như trên đâu nha
<=>\(\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y+y+z+x+z}{z+x+y}=\frac{2.\left(x+y+z\right)}{x+y+z}=2\)
<=>\(\frac{x+y}{z}=\frac{y+z}{x}=\frac{x+z}{y}=\frac{x+y+y+z+x+z}{z+x+y}=\frac{2.\left(x+y+z\right)}{x+y+z}2\)
=>\(\frac{x+y}{z}=2=>x+y=2z\)
Mà \(x+y=kz=>k=2\)
vậy k=2
CHO a,b,c>0 thỏa mãn: \(a^2b^2+b^2c^2+c^2a^2\ge a^2+b^2+c^2\)
CMR: \(\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(a^2+c^2\right)}\ge\frac{\sqrt{3}}{2}\)
ĐẶT \(A=\frac{a^2b^2}{c^3\left(a^2+b^2\right)}+\frac{b^2c^2}{a^3\left(b^2+c^2\right)}+\frac{c^2a^2}{b^3\left(c^2+a^2\right)}\)
ĐẶT:\(\frac{1}{a}=x,\frac{1}{y}=b,\frac{1}{z}=c\)
\(\Rightarrow x^2+y^2+z^2\ge1\)
\(\Rightarrow A=\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{z^2+y^2}\)
TA CÓ:
\(x\left(y^2+z^2\right)=\frac{1}{\sqrt{2}}\sqrt{2x^2\left(y^2+z^2\right)\left(y^2+z^2\right)}\le\frac{1}{\sqrt{2}}\sqrt{\frac{\left(2x^2+2y^2+2z^2\right)^3}{27}}=\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)TƯƠNG TỰ:
\(y\left(x^2+z^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2},z\left(x^2+y^2\right)\le\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}\)LẠI CÓ:
\(A=\frac{x^3}{y^2+z^2}+\frac{y^3}{x^2+z^2}+\frac{z^3}{x^2+y^2}=\frac{x^4}{x\left(y^2+z^2\right)}+\frac{y^4}{y\left(x^2+z^2\right)}+\frac{z^4}{z\left(x^2+y^2\right)}\ge\frac{\left(x^2+y^2+z^2\right)^2}{x\left(y^2+z^2\right)+y\left(x^2+z^2\right)+z\left(x^2+y^2\right)}\ge\frac{1}{3.\frac{2}{3\sqrt{3}}\left(x^2+y^2+z^2\right)\sqrt{x^2+y^2+z^2}}
\)\(\ge\frac{\sqrt{3}}{2}\sqrt{x^2+y^2+z^2}\ge\frac{\sqrt{3}}{2}\)
DẤU BẰNG XẢY RA\(\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\Rightarrow DPCM\)
tại tui trả lời bài này cho 1 bạn ở trên facebook nên phải chụp màn hình lại nên làm v á
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,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Tính:a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
b) Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\) . Tính \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
a) \(A=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}+\frac{\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{2\left(y-z\right)\left(z-x\right)+2\left(x-y\right)\left(z-x\right)+2\left(x-y\right)\left(y-z\right)+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
\(=\frac{\left[\left(x-y\right)+\left(y-z\right)+\left(z-x\right)\right]^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y+y-z+z-x\right)^2}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=0\)
Áp dụng: \(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac\)
b)Ta có: \(\frac{x^2}{y+z}+x=\frac{x^2+x\left(y+z\right)}{y+z}=\frac{x^2+xy+xz}{y+z}=\frac{x\left(x+y+z\right)}{y+z}\)
Tương tự: \(\frac{y^2}{x+z}+y=\frac{y^2+xy+zy}{x+z}=\frac{y\left(x+y+z\right)}{x+z}\)
\(\frac{z^2}{x+y}+z=\frac{z^2+xz+zy}{x+y}=\frac{z\left(x+y+z\right)}{x+y}\)
Suy ra: \(A+\left(x+y+z\right)\)
\(=\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{z+x}+\frac{z\left(x+y+z\right)}{x+y}+\left(x+y+z\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}+1\right)\)
\(=2.\left(x+y+z\right)\)
Nên \(A=2.\left(x+y+z\right)-\left(x+y+z\right)=x+y+z\)
Mình có sai chỗ nào không nhỉ?
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{1}{x+y+z}\)
b) \(\frac{x}{y+z+1}=\frac{y}{x+z+2}=\frac{z}{x+y-2}=x+y+z\)
Tìm x ;y;z
a) \(\frac{y+z+1}{x}=\frac{x+z+2}{y}=\frac{x+y-3}{z}=\frac{y+z+1+x+z+2+x+y-3}{x+y+z}=2\)
\(\Rightarrow x+y+z=\frac{1}{2}\)(do 1/(x+y+z)=2)
\(\Rightarrow y+z=\frac{1}{2}-x;z+x=\frac{1}{2}-y;x+y=\frac{1}{2}-z\)
Thay vào lần lượt ta có:
\(\frac{\frac{1}{2}-x+1}{x}=2\)\(\Rightarrow x=\frac{1}{2}\)
\(\frac{\frac{1}{2}-y+2}{y}=2\)\(\Rightarrow y=\frac{5}{6}\)
\(\frac{\frac{1}{2}-z-3}{z}=2\)\(\Rightarrow z=-\frac{5}{6}\)
\(x+y+z=1,\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
tính giá trị của biểu thức A=\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
chị QA
ta có đề bài <=>
\(\frac{x^2}{y}-2x+y+\frac{y^2}{z}-2y+z+\frac{z^2}{x}-2z+x+\left(x+y+z\right)-\left(x-y\right)^2-\left(y-z\right)^2-\left(z-x\right)^2\)
=\(\frac{\left(x-y\right)^2}{y}-\left(x-y\right)^2+...+\left(x+y+z\right)\)
=\(\left(x-y\right)^2\left(\frac{1}{y}-1\right)+....+\left(x+y+z\right)\)
mà \(\sqrt{x}+\sqrt{y}+\sqrt{z}=1\Rightarrow x,y,z\in\left[0;1\right]\)
=> \(\frac{1}{y}-y>0\)
=> \(A\ge x+y+z\ge\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=\frac{1}{3}\)
CHo \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\)
Tính : \(A=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
\(\Rightarrow\left(\frac{x}{x+y}+\frac{y}{z+x}+\frac{z}{x+y}\right)\cdot\left(x+y+z\right)=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{y^2}{z+x}+\frac{xy}{z+x}+\frac{yz}{z+x}+\frac{z^2}{x+y}+\frac{xz}{x+y}+\frac{yz}{x+y}=x+y+z\)
Rồi bạn cộng 2 phân thức 2,3 5,6 8,9 lại thì được
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{z+x}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
Cho \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\). Tính \(A=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\)
Dễ thấy \(x+y+z\ne0\)
Ta có :
\(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}\right)+\left(x+y+z\right)\left(\frac{y}{x+z}\right)+\left(x+y+z\right)\left(\frac{z}{x+y}\right)\)
\(=\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z\)
\(=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+x+y+z\)
Mà \(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)\)
\(=\left(x+y+z\right).1=x+y+z\)
=> \(x+y+z=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}+x+y+z\)
=> \(0=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\)
Làm xong thấy bài tui làm hới ... @@
P/s : Nếu cần c/m x+y+x khác không thùy công 3 zô :3