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HN
1
20 tháng 3 2019
\(\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}=x\left(\frac{x}{y+z}+1-1\right)+y\left(\frac{y}{x+z}+1-1\right)+z\left(\frac{z}{x+y}+1-1\right)\)
\(=x\left(\frac{x+y+z}{y+z}-1\right)+y\left(\frac{x+y+z}{x+z}-1\right)+z\left(\frac{x+y+z}{x+y}-1\right)\)
\(=\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}\right)-\left(x+y+z\right)=0\)
\(M=2019\)
24 tháng 3 2019
\(\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=1\)
\(\Rightarrow\frac{x\left(x+y+z\right)}{y+z}+\frac{y\left(x+y+z\right)}{x+z}+\frac{z\left(x+y+z\right)}{x+y}=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+x+\frac{y^2}{x+z}+y+\frac{z^2}{x+y}+z=x+y+z\)
\(\Rightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
\(\Rightarrow M=2019+0=2019\)
NT
0
LD
13 tháng 4 2019
Vì \(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}=1\)
\(\Rightarrow\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{xy}{z+x}+\frac{zx}{x+y}+\frac{xy}{y+z}+\frac{y^2}{z+x}+\frac{yz}{x+y}+\frac{zx}{y+z}+\frac{yz}{z+x}+\frac{z^2}{x+y}=x+y+z\)
\(\Leftrightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy+yz}{z+x}\right)+\left(\frac{yz+zx}{x+y}\right)+\left(\frac{zx+xy}{y+z}\right)=x+y+z\)
\(\Leftrightarrow\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\frac{y\left(z+x\right)}{z+x}+\frac{z\left(x+y\right)}{x+y}+\frac{x\left(y+z\right)}{y+z}=x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}+x+y+z=x+y+z\)
\(\Leftrightarrow\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
\(\Rightarrow M=2019\)
MA
6 tháng 7 2023
Đặt \(\dfrac{1}{a}=\dfrac{1}{x+y},\dfrac{1}{b}=\dfrac{1}{y+z},\dfrac{1}{c}=\dfrac{1}{z+x}\)
Đề trở thành: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\), tính \(P=\dfrac{bc}{a^2}+\dfrac{ac}{b^2}+\dfrac{ab}{c^2}\)
\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=0\) Tương đương \(ab+bc=-ac\)
\(P=\dfrac{b^3c^3+a^3c^3+a^3b^3}{a^2b^2c^2}=\dfrac{\left(ab+bc\right)\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}=\dfrac{-ac\left(a^2b^2-ab^2c+b^2c^2\right)+a^3c^3}{a^2b^2c^2}\)
\(=\dfrac{a^2c^2-a^2b^2+ab^2c-b^2c^2}{ab^2c}=\dfrac{ac}{b^2}-\dfrac{a}{c}+1-\dfrac{c}{a}\)\(=ac\left(\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\right)-\dfrac{a}{c}+1-\dfrac{c}{a}\) (do \(\dfrac{1}{b}=-\dfrac{1}{a}-\dfrac{1}{c}\) tương đương \(\dfrac{1}{b^2}=\dfrac{1}{a^2}+\dfrac{2}{ac}+\dfrac{1}{c^2}\))
\(=3\)
Vậy P=3
DT
3
1 tháng 11 2020
M+2019=2xy−yz−zx+2020M+2019=2xy−yz−zx+2020
=2xy−yz−zx+x2+y2+z2=2xy−yz−zx+x2+y2+z2
=(x+y−z2)2+3z24≥0=(x+y−z2)2+3z24≥0
⇒Mmin=0⇒Mmin=0 khi ⎧⎩⎨⎪⎪⎪⎪x+y−z2=03z24=0x2+y2+z2=2020{x+y−z2=03z24=0x2+y2+z2=2020
⇔⎧⎩⎨⎪⎪x+y=0z=0x2+y2=2020⇔{x+y=0z=0x2+y2=2020 ⇒⎧⎩⎨⎪⎪x=±1010−−−−√y=−xz=0
VT
2
KB
30 tháng 4 2019
Đặt 1/x = a ; 1/y = b ; 1/z = c
Ta có : \(a+b+c=2;2ab-c^2=4\)
\(a^2+b^2+c^2+2ab+2bc+2ac=2ab-c^2\)
\(\Leftrightarrow a^2+b^2+c^2+2bc+2ac+c^2=0\)
\(\Leftrightarrow\left(a+c\right)^2+\left(b+c\right)^2=0\)
=> a + c = 0 và b + c = 0
=> a = b = -c
\(\Rightarrow\frac{1}{x}=\frac{1}{y}=-\frac{1}{z}\)
Khi đó , ta có : \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=-\frac{2}{z}+\frac{1}{z}=-\frac{1}{z}=2\Rightarrow z=-\frac{1}{2}\)
\(P=\left(x+2y+z\right)^2=4z^2\) \(=4.\left(-\frac{1}{2}\right)^2=1\)
Tham khảo nha
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