<img alt="" src="https://scontent.fhph1-1.fna.fbcdn.net/v/t1.15752-9/188947531_640784093987848_9184495106730499173_n.png?_nc_cat=102&ccb=1-3&_nc_sid=ae9488&_n...

Câu 5: \(P=\frac{2x+yz}{y+z}+\frac{2y+zx}{z+x}+\frac{2z+xy}{x+y}\left(x,y,z>0\right)\) . Ta có: \(\frac{2x+yz}{y+z}=\frac{x\left(x+y+z\right)+yz}{y+z}\) (vì \(x+y+z=2\) ). \(\Rightarrow\frac{2x+yz}{y+z}=\frac{x\left(x+y\right)+xz+yz}{y+z}=\frac{x\left(x+y\right)+z\left(x+y\right)}{y+z}\) \(=\frac{\left(x+z\right)\left(x+y\right)}{y+z}\) . Chứng minh tương tự, ta được: \(\frac{2y+zx}{z+x}=\frac{\left(x+y\right)\left(y+z\right)}{z+x}\) . Chứng minh tương tự, ta được: \(\frac{2z+xy}{x+y}=\frac{\left(y+z\right)\left(z+x\right)}{x+y}\) . Do đó: \(P=\frac{\left(x+y\right)\left(x+z\right)}{y+z}+\frac{\left(x+y\right)\left(y+z\right)}{z+x}+\frac{\left(z+x\right)\left(y+z\right)}{x+y}\) . Đặt \(x+y=a;y+z=b;z+x=c\left(a,b,c>0\right)\) thì \(a+b+c=2\left(x+y+z\right)=2.2=4\) . Do đó: \(P=\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\) . Vì \(a,b,c>0\) nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được: \(\frac{ac}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ac.ab}{bc}}=2a\) \(\left(1\right)\) . Dấu bằng xảy ra \(\Leftrightarrow\frac{ac}{b}=\frac{ab}{c}\Leftrightarrow\frac{c}{b}=\frac{b}{c}\Leftrightarrow b=c>0\) . Chứng minh tương tự, ta được: \(\frac{ab}{c}+\frac{bc}{a}\ge2b\) \(\left(2\right)\) . Dấu bằng xảy ra \(\Leftrightarrow a=c>0\) . Chứng minh tương tự, ta được: \(\frac{bc}{a}+\frac{ac}{b}\ge2c\) \(\left(3\right)\) . Dấu bằng xảy ra \(\Leftrightarrow a=b>0\) . Từ \(\left(1\right),\left(2\right),\left(3\right)\) , ta được: \(\frac{ac}{b}+\frac{ab}{c}+\frac{ab}{c}+\frac{bc}{a}+\frac{bc}{a}+\frac{ac}{b}\ge2a+2b+2c\) . \(\Leftrightarrow2\left(\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\right)\ge2\left(a+b+c\right)\) . \(\Leftrightarrow\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\ge a+b+c\) . \(\Leftrightarrow P\ge4\) (vì \(a+b+c=4\) ). Dấu bằng xảy ra. \(\Leftrightarrow\hept{\begin{cases}a=b=c\\a,b,c>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=y+z=z+x\\x,y,z>0\end{cases}}\Leftrightarrow x=y=z\) Mà \(x+y+z=2\) nên \(x=y=z=\frac{2}{3}\) . Vậy \(minP=4\Leftrightarrow x=y=z=\frac{2}{3}\) . Câu trả lời: Câu 5: \(P=\frac{2x+yz}{y+z}+\frac{2y+zx}{z+x}+\frac{2z+xy}{x+y}\left(x,y,z>0\right)\) . Ta có: \(\frac{2x+yz}{y+z}=\frac{x\left(x+y+z\right)+yz}{y+z}\) (vì \(x+y+z=2\) ). \(\Rightarrow\frac{2x+yz}{y+z}=\frac{x\left(x+y\right)+xz+yz}{y+z}=\frac{x\left(x+y\right)+z\left(x+y\right)}{y+z}\) \(=\frac{\left(x+z\right)\left(x+y\right)}{y+z}\) . Chứng minh tương tự, ta được: \(\frac{2y+zx}{z+x}=\frac{\left(x+y\right)\left(y+z\right)}{z+x}\) . Chứng minh tương tự, ta được: \(\frac{2z+xy}{x+y}=\frac{\left(y+z\right)\left(z+x\right)}{x+y}\) . Do đó: \(P=\frac{\left(x+y\right)\left(x+z\right)}{y+z}+\frac{\left(x+y\right)\left(y+z\right)}{z+x}+\frac{\left(z+x\right)\left(y+z\right)}{x+y}\) . Đặt \(x+y=a;y+z=b;z+x=c\left(a,b,c>0\right)\) thì \(a+b+c=2\left(x+y+z\right)=2.2=4\) . Do đó: \(P=\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\) . Vì \(a,b,c>0\) nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được: \(\frac{ac}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ac.ab}{bc}}=2a\) \(\left(1\right)\) . Dấu bằng xảy ra \(\Leftrightarrow\frac{ac}{b}=\frac{ab}{c}\Leftrightarrow\frac{c}{b}=\frac{b}{c}\Leftrightarrow b=c>0\) . Chứng minh tương tự, ta được: \(\frac{ab}{c}+\frac{bc}{a}\ge2b\) \(\left(2\right)\) . Dấu bằng xảy ra \(\Leftrightarrow a=c>0\) . Chứng minh tương tự, ta được: \(\frac{bc}{a}+\frac{ac}{b}\ge2c\) \(\left(3\right)\) . Dấu bằng xảy ra \(\Leftrightarrow a=b>0\) . Từ \(\left(1\right),\left(2\right),\left(3\right)\) , ta được: \(\frac{ac}{b}+\frac{ab}{c}+\frac{ab}{c}+\frac{bc}{a}+\frac{bc}{a}+\frac{ac}{b}\ge2a+2b+2c\) . \(\Leftrightarrow2\left(\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\right)\ge2\left(a+b+c\right)\) . \(\Leftrightarrow\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\ge a+b+c\) . \(\Leftrightarrow P\ge4\) (vì \(a+b+c=4\) ). Dấu bằng xảy ra. \(\Leftrightarrow\hept{\begin{cases}a=b=c\\a,b,c>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=y+z=z+x\\x,y,z>0\end{cases}}\Leftrightarrow x=y=z\) Mà \(x+y+z=2\) nên \(x=y=z=\frac{2}{3}\) . Vậy \(minP=4\Leftrightarrow x=y=z=\frac{2}{3}\) .

Câu 3: \(\frac{1}{3a+b}+\frac{2}{a+3b}=\frac{3}{2a+2b}\) . \(\Leftrightarrow\left[\frac{1}{3a+b}-\frac{1}{2\left(a+b\right)}\right]+2\left[\frac{1}{a+3b}-\frac{1}{2\left(a+b\right)}\right]=0\) . \(\Leftrightarrow\left[\frac{2\left(a+b\right)}{2\left(a+b\right)\left(3a+b\right)}-\frac{3a+b}{2\left(a+b\right)\left(3a+b\right)}\right]\) \(+2\left[\frac{2\left(a+b\right)}{2\left(a+b\right)\left(a+3b\right)}-\frac{a+3b}{2\left(a+b\right)\left(a+3b\right)}\right]=0\) . \(\Leftrightarrow\frac{2a+2b-3a-b}{2\left(a+b\right)\left(3a+b\right)}+2.\frac{2a+2b-a-3b}{2\left(a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{b-a}{2\left(a+b\right)\left(3a+b\right)}+\frac{2\left(a-b\right)}{2\left(a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{b-a}{2\left(a+b\right)}\left(\frac{1}{3a+b}-\frac{2}{a+3b}\right)=0\) . Vì \(0< a< b\) nên \(a+b>0;b-a>0\) \(\frac{b-a}{\left(a+b\right)}>0\Rightarrow\frac{b-a}{2\left(a+b\right)}>0\) \(\Rightarrow\frac{b-a}{2\left(a+b\right)}\ne0\) . Lúc đó: \(\frac{1}{3a+b}-\frac{2}{a+3b}=0:\frac{b-a}{2\left(a+b\right)}\) . \(\Leftrightarrow\frac{1}{3a+b}-\frac{2}{a+3b}=0\) . \(\Leftrightarrow\frac{a+3b}{\left(3a+b\right)\left(a+3b\right)}-\frac{2\left(3a+b\right)}{\left(3a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{a+3b-6a-2b}{\left(3a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{b-5a}{\left(3a+b\right)\left(a+3b\right)}=0\) . Vì \(0< a< b\) nên \(3a+b>0;a+3b>0\) \(\Rightarrow\left(3a+b\right)\left(a+3b\right)>0\Rightarrow\left(3a+b\right)\left(a+3b\right)\ne0\) . Do đó: \(\frac{b-5a}{\left(3a+b\right)\left(a+3b\right)}=\frac{0}{\left(3a+b\right)\left(a+3b\right)}\) . \(\Rightarrow b-5a=0\Leftrightarrow b=5a\) (thỏa mãn \(0< a< b\) ). \(M=\frac{3}{3a+b}+\frac{2}{a+3b}-\frac{3}{a+b}\) . Thay \(b=5a\) vào \(M\) , ta được: \(M=\frac{3}{3a+5a}+\frac{2}{a+3.5a}-\frac{3}{a+5a}\) . \(M=\frac{3}{8a}+\frac{2}{16a}-\frac{3}{6a}=\frac{3}{8a}+\frac{1}{8a}-\frac{1}{2a}=\frac{1}{4a}-\frac{1}{2a}\) . \(=\frac{1}{2a}-\frac{1}{2a}=0\) Vậy \(M=0\) . Câu trả lời: Câu 3: \(\frac{1}{3a+b}+\frac{2}{a+3b}=\frac{3}{2a+2b}\) . \(\Leftrightarrow\left[\frac{1}{3a+b}-\frac{1}{2\left(a+b\right)}\right]+2\left[\frac{1}{a+3b}-\frac{1}{2\left(a+b\right)}\right]=0\) . \(\Leftrightarrow\left[\frac{2\left(a+b\right)}{2\left(a+b\right)\left(3a+b\right)}-\frac{3a+b}{2\left(a+b\right)\left(3a+b\right)}\right]\) \(+2\left[\frac{2\left(a+b\right)}{2\left(a+b\right)\left(a+3b\right)}-\frac{a+3b}{2\left(a+b\right)\left(a+3b\right)}\right]=0\) . \(\Leftrightarrow\frac{2a+2b-3a-b}{2\left(a+b\right)\left(3a+b\right)}+2.\frac{2a+2b-a-3b}{2\left(a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{b-a}{2\left(a+b\right)\left(3a+b\right)}+\frac{2\left(a-b\right)}{2\left(a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{b-a}{2\left(a+b\right)}\left(\frac{1}{3a+b}-\frac{2}{a+3b}\right)=0\) . Vì \(0< a< b\) nên \(a+b>0;b-a>0\) \(\frac{b-a}{\left(a+b\right)}>0\Rightarrow\frac{b-a}{2\left(a+b\right)}>0\) \(\Rightarrow\frac{b-a}{2\left(a+b\right)}\ne0\) . Lúc đó: \(\frac{1}{3a+b}-\frac{2}{a+3b}=0:\frac{b-a}{2\left(a+b\right)}\) . \(\Leftrightarrow\frac{1}{3a+b}-\frac{2}{a+3b}=0\) . \(\Leftrightarrow\frac{a+3b}{\left(3a+b\right)\left(a+3b\right)}-\frac{2\left(3a+b\right)}{\left(3a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{a+3b-6a-2b}{\left(3a+b\right)\left(a+3b\right)}=0\) . \(\Leftrightarrow\frac{b-5a}{\left(3a+b\right)\left(a+3b\right)}=0\) . Vì \(0< a< b\) nên \(3a+b>0;a+3b>0\) \(\Rightarrow\left(3a+b\right)\left(a+3b\right)>0\Rightarrow\left(3a+b\right)\left(a+3b\right)\ne0\) . Do đó: \(\frac{b-5a}{\left(3a+b\right)\left(a+3b\right)}=\frac{0}{\left(3a+b\right)\left(a+3b\right)}\) . \(\Rightarrow b-5a=0\Leftrightarrow b=5a\) (thỏa mãn \(0< a< b\) ). \(M=\frac{3}{3a+b}+\frac{2}{a+3b}-\frac{3}{a+b}\) . Thay \(b=5a\) vào \(M\) , ta được: \(M=\frac{3}{3a+5a}+\frac{2}{a+3.5a}-\frac{3}{a+5a}\) . \(M=\frac{3}{8a}+\frac{2}{16a}-\frac{3}{6a}=\frac{3}{8a}+\frac{1}{8a}-\frac{1}{2a}=\frac{1}{4a}-\frac{1}{2a}\) . \(=\frac{1}{2a}-\frac{1}{2a}=0\) Vậy \(M=0\) .

Nguyễn Văn Bình 1

31 tháng 5 2021 - olm

Mọi người giúp mình câu 3,4,5 với!!!!

#Hỏi cộng đồng OLM #Toán lớp 8

Phạm Thành Đông

31 tháng 5 2021

Câu 5:

\(P=\frac{2x+yz}{y+z}+\frac{2y+zx}{z+x}+\frac{2z+xy}{x+y}\left(x,y,z>0\right)\).

Ta có:

\(\frac{2x+yz}{y+z}=\frac{x\left(x+y+z\right)+yz}{y+z}\)(vì \(x+y+z=2\)).

\(\Rightarrow\frac{2x+yz}{y+z}=\frac{x\left(x+y\right)+xz+yz}{y+z}=\frac{x\left(x+y\right)+z\left(x+y\right)}{y+z}\)\(=\frac{\left(x+z\right)\left(x+y\right)}{y+z}\).

Chứng minh tương tự, ta được:

\(\frac{2y+zx}{z+x}=\frac{\left(x+y\right)\left(y+z\right)}{z+x}\).

Chứng minh tương tự, ta được:

\(\frac{2z+xy}{x+y}=\frac{\left(y+z\right)\left(z+x\right)}{x+y}\).

Do đó:

\(P=\frac{\left(x+y\right)\left(x+z\right)}{y+z}+\frac{\left(x+y\right)\left(y+z\right)}{z+x}+\frac{\left(z+x\right)\left(y+z\right)}{x+y}\).

Đặt \(x+y=a;y+z=b;z+x=c\left(a,b,c>0\right)\)thì \(a+b+c=2\left(x+y+z\right)=2.2=4\). Do đó:

\(P=\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\).

Vì \(a,b,c>0\) nên áp dụng bất đẳng thức Cô-si cho 2 số dương, ta được:

\(\frac{ac}{b}+\frac{ab}{c}\ge2\sqrt{\frac{ac.ab}{bc}}=2a\)\(\left(1\right)\).

Dấu bằng xảy ra \(\Leftrightarrow\frac{ac}{b}=\frac{ab}{c}\Leftrightarrow\frac{c}{b}=\frac{b}{c}\Leftrightarrow b=c>0\).

Chứng minh tương tự, ta được:

\(\frac{ab}{c}+\frac{bc}{a}\ge2b\)\(\left(2\right)\).

Dấu bằng xảy ra \(\Leftrightarrow a=c>0\).

Chứng minh tương tự, ta được:
\(\frac{bc}{a}+\frac{ac}{b}\ge2c\)\(\left(3\right)\).

Dấu bằng xảy ra \(\Leftrightarrow a=b>0\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\) , ta được:

\(\frac{ac}{b}+\frac{ab}{c}+\frac{ab}{c}+\frac{bc}{a}+\frac{bc}{a}+\frac{ac}{b}\ge2a+2b+2c\).

\(\Leftrightarrow2\left(\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\right)\ge2\left(a+b+c\right)\).

\(\Leftrightarrow\frac{ac}{b}+\frac{ab}{c}+\frac{bc}{a}\ge a+b+c\).

\(\Leftrightarrow P\ge4\)(vì \(a+b+c=4\)).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}a=b=c\\a,b,c>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x+y=y+z=z+x\\x,y,z>0\end{cases}}\Leftrightarrow x=y=z\)

Mà \(x+y+z=2\)nên \(x=y=z=\frac{2}{3}\).

Vậy \(minP=4\Leftrightarrow x=y=z=\frac{2}{3}\).

Đúng(0)

Phạm Thành Đông

31 tháng 5 2021

Câu 3:

\(\frac{1}{3a+b}+\frac{2}{a+3b}=\frac{3}{2a+2b}\).

\(\Leftrightarrow\left[\frac{1}{3a+b}-\frac{1}{2\left(a+b\right)}\right]+2\left[\frac{1}{a+3b}-\frac{1}{2\left(a+b\right)}\right]=0\).

\(\Leftrightarrow\left[\frac{2\left(a+b\right)}{2\left(a+b\right)\left(3a+b\right)}-\frac{3a+b}{2\left(a+b\right)\left(3a+b\right)}\right]\)\(+2\left[\frac{2\left(a+b\right)}{2\left(a+b\right)\left(a+3b\right)}-\frac{a+3b}{2\left(a+b\right)\left(a+3b\right)}\right]=0\).

\(\Leftrightarrow\frac{2a+2b-3a-b}{2\left(a+b\right)\left(3a+b\right)}+2.\frac{2a+2b-a-3b}{2\left(a+b\right)\left(a+3b\right)}=0\).

\(\Leftrightarrow\frac{b-a}{2\left(a+b\right)\left(3a+b\right)}+\frac{2\left(a-b\right)}{2\left(a+b\right)\left(a+3b\right)}=0\).

\(\Leftrightarrow\frac{b-a}{2\left(a+b\right)}\left(\frac{1}{3a+b}-\frac{2}{a+3b}\right)=0\).

Vì \(0< a< b\)nên \(a+b>0;b-a>0\)\(\frac{b-a}{\left(a+b\right)}>0\Rightarrow\frac{b-a}{2\left(a+b\right)}>0\)\(\Rightarrow\frac{b-a}{2\left(a+b\right)}\ne0\). Lúc đó:

\(\frac{1}{3a+b}-\frac{2}{a+3b}=0:\frac{b-a}{2\left(a+b\right)}\).

\(\Leftrightarrow\frac{1}{3a+b}-\frac{2}{a+3b}=0\).

\(\Leftrightarrow\frac{a+3b}{\left(3a+b\right)\left(a+3b\right)}-\frac{2\left(3a+b\right)}{\left(3a+b\right)\left(a+3b\right)}=0\).

\(\Leftrightarrow\frac{a+3b-6a-2b}{\left(3a+b\right)\left(a+3b\right)}=0\).

\(\Leftrightarrow\frac{b-5a}{\left(3a+b\right)\left(a+3b\right)}=0\).

Vì \(0< a< b\)nên \(3a+b>0;a+3b>0\)\(\Rightarrow\left(3a+b\right)\left(a+3b\right)>0\Rightarrow\left(3a+b\right)\left(a+3b\right)\ne0\).

Do đó:

\(\frac{b-5a}{\left(3a+b\right)\left(a+3b\right)}=\frac{0}{\left(3a+b\right)\left(a+3b\right)}\).

\(\Rightarrow b-5a=0\Leftrightarrow b=5a\)(thỏa mãn \(0< a< b\)).

\(M=\frac{3}{3a+b}+\frac{2}{a+3b}-\frac{3}{a+b}\).

Thay \(b=5a\)vào \(M\), ta được:

\(M=\frac{3}{3a+5a}+\frac{2}{a+3.5a}-\frac{3}{a+5a}\).