Mình thiếu nhé. Câu b chứng minh p(ở câu a) < t

a, \(p=\frac{x+y}{y+z}=\frac{\frac{a}{m}+\frac{b}{m}}{\frac{b}{m}+\frac{a+b}{m}}=\frac{\frac{a+b}{m}}{\frac{a+b^2}{m}}=\frac{a+b}{a+b^2}\)

\(\frac{\frac{1}{4}+\frac{1}{2}}{\frac{1}{2}+\frac{3}{4}}=\frac{\frac{1}{4}+\frac{2}{4}}{\frac{2}{4}+\frac{1+2}{4}}=\frac{1+2}{1+2^2}=\frac{3}{5}\)

Hok tốt !!!!!!!!!

a, xy+2x-y=5

=> x(y+2)-y-2=3

=>x(y+2)-(y+2)=3

=>(x-1)(y+2)=3

=>\(\hept{\begin{cases}x-1=3\Rightarrow x=4\\y+2=1\Rightarrow y=-1\end{cases}}\); \(\hept{\begin{cases}x-1=1\Rightarrow x=2\\y+2=3\Rightarrow y=1\end{cases}}\)

=>\(\hept{\begin{cases}x-1=-1\Rightarrow x=0\\y+2=-3\Rightarrow y=-5\end{cases}}\); \(\hept{\begin{cases}x-1=-3\Rightarrow x=-2\\y+2=-1\Rightarrow y=-3\end{cases}}\)

vậy (x;y)\(\in\)(4,-1);(2,1);(0,-5);(-2.-3)

từ\(\frac{2bz-3cy}{a}\)=\(\frac{3cx-az}{2b}=\frac{ay-2bx}{3c}\)

=>\(\frac{2abz-3acy}{a}\)=\(\frac{6bcx-2abz}{2b}\)=\(\frac{3cay-6cbx}{3c}\)

=\(\frac{2abz-3acy+6bcx-2abz+3cay-6cbx}{2a+4b+6c}\)=0

=>\(\frac{2bz-3cy}{a}=0\)=>2bz=3cy=>\(\frac{z}{3c}\)=\(\frac{y}{2b}\)(1)

=>\(\frac{3cx-az}{2b}\)=0 =>3cx=az =>\(\frac{x}{a}\)=\(\frac{z}{3c}\)(2)

=>\(\frac{ay-2bx}{3c}=0\)=>ay=2bx =>\(\frac{y}{2b}\)=\(\frac{x}{a}\)(3)

Từ (1),(2) và (3) suy ra\(\frac{x}{a}=\frac{y}{2b}=\frac{z}{3c}\)đpcm

Bài 1

Cho a , b , c > 0 . CM : \(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge\frac{a+b}{b+c}+\frac{b+c}{a+b}\left(1\right)\)

\(\Leftrightarrow\left(a+b\right)^2+\left(b+c\right)^2+\left(a+b\right)\left(b+c\right)\le\frac{a\left(a+b\right)\left(b+c\right)}{b}+\frac{b\left(a+b\right)\left(b+c\right)}{c}+\frac{c\left(a+b\right)\left(b+c\right)}{a}\)

\(=\frac{a^2c}{b}+a^2+ab+ac+\frac{b^2\left(a+b\right)}{c}+b^2+ab+c^2+bc+\frac{cb\left(b+c\right)}{a}\)

Mặt khác : \(\left(a+b\right)^2+\left(b+c\right)^2+\left(a+b\right)\left(b+c\right)=a^2+ac+c^2+3b^2+3ab+3bc\)

Do đó ta cần chứng minh :

\(\frac{a^2c}{b}+\frac{b^2\left(a+b\right)}{c}+\frac{cb\left(b+c\right)}{a}\ge2b^2+2bc+ab\left(2\right)\)

\(VT=\frac{a^2c}{b}+\frac{b^2\left(a+b\right)}{c}+\frac{cb\left(b+c\right)}{a}=\frac{1}{2}\left(\frac{a^2c}{b}+\frac{b^3}{c}\right)+\frac{1}{2}\left(\frac{a^2c}{b}+\frac{c^2b}{a}\right)+\frac{1}{2}\left(\frac{b^3}{c}+\frac{c^2b}{a}\right)+b^2\left(\frac{c}{a}+\frac{a}{c}\right)\)

\(\ge ab+\sqrt{ac^3}+\sqrt{\frac{b^4c}{a}}+2b^2\ge ab+2bc+2b^2=VP\)

Dấu " = " xảy ra khi a=b=c

Bài 2 :

Vì x , y , z > 0 ta có :

Áp dụng BĐT Cô - si đối với 2 số dương \(\frac{x^2}{y+z}\) và \(\frac{y+z}{4}\)

ta được :

\(\frac{x^2}{y+z}+\frac{y+z}{4}\ge2\sqrt{\frac{x^2}{y+z}.\frac{y+z}{4}}=2.\frac{x}{2}=x\left(1\right)\) .

Tương tự ta cũng có :
\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge y\left(2\right);\frac{z^2}{x+y}+\frac{x+y}{4}\ge z\left(3\right)\)

Cộng theo vế (1) , (2) và (3) ta được :
\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\frac{x+y+z}{2}\ge x+y+z\Rightarrow P\ge\left(x+xy+z\right)-\frac{x+y+z}{2}=1\)

Dấu " = " xảy ra \(\Leftrightarrow x=y=z=\frac{2}{3}\)

Vậy \(P=1\Leftrightarrow x=y=z=\frac{2}{3}\)

Ta có \(\frac{2a+b+c}{b+c}=\frac{2b+c+a}{c+a}=\frac{2c+a+b}{a+b}\Rightarrow\frac{2a}{b+c}+1=\frac{2b}{a+c}+1=\frac{2c}{a+b}+1\)

=> \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\Rightarrow\frac{a}{b+c}+\frac{b}{a+c}+\frac{c}{a+b}=\frac{3}{2}\)

^_^ 

Bài 1: Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\)

\(\Rightarrow\hept{\begin{cases}a=2016k\\b=2017k\\c=2018k\end{cases}}\).Thay vào M,ta có:

 \(M=4\left(2016k-2017k\right)\left(2017k-2018k\right)-\left(2018k-2016k\right)^2\)

\(=4.\left(-1k\right)\left(-1k\right)-\left(2k\right)^2\)

\(=4k^2-4k^2=0\)

\(\frac{x}{y+z+t}=\frac{y}{z+t+x}=\frac{z}{y+x+t}=\frac{t}{x+y+z}=\frac{x+y+z+t}{2\left(x+y+z+t\right)}=\frac{1}{2}\)

=>2x=y+z+t

2y=x+z+t

2z+x+y+t

2t=x+y+z

=>x+y=2(z+t)(1)

y+z=2(x+t)(2)

z+t=2(x+y)(3)

t+x=2(y+z)(4)

Thay 1;2;3 và 4 vào P

=>P=2+2+2+2=8

bài 2 tương tự

ác mộng sai rồi

Bài 1: Theo đề : \(2ab+6bc+2ac=7abc\) \(;a,b,c>0\)

Chia cả 2 vế cho \(abc>0\Rightarrow\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)

Đặt: \(\hept{\begin{cases}x=\frac{1}{a}\\y=\frac{1}{b}\\z=\frac{1}{c}\end{cases}}\Rightarrow\hept{\begin{cases}x,y,z>0\\2z+6x+2y=7\end{cases}}\)

Khi đó: \(M=\frac{4ab}{a+2b}+\frac{9ac}{a+4c}+\frac{4bc}{b+c}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)

\(\Rightarrow M=\frac{4}{2x+y}+2x+y+\frac{9}{4x+z}+4x+z+\frac{4}{y+z}+y+z-\left(2x+y+4x+z+y+z\right)\)

\(=\left(\frac{2}{\sqrt{x+2y}}-\sqrt{x+2y}\right)^2+\left(\frac{3}{\sqrt{4x+z}}-\sqrt{4x+z}\right)^2+\left(\frac{2}{\sqrt{y+z}}-\sqrt{y+z}\right)^2+17\ge17\)

Khi: \(\hept{\begin{cases}x=\frac{1}{2}\\y=z=1\end{cases}}\Rightarrow M=17\)

\(Min_M=17\Leftrightarrow a=2;b=1;c=1\)

ミ★๖ۣۜBăηɠ ๖ۣۜBăηɠ ★彡 chém bài khó nhất rồi nên em xin mạn phép chém bài dễ ạ.

2/\(VT=\Sigma_{cyc}\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}=\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\)

\(\ge\Sigma_{cyc}\frac{\left(y+z\right)\left(2x+y+z\right)}{\frac{\left(2x+y+z\right)^2}{4}}=\Sigma_{cyc}\frac{4\left(y+z\right)}{2x+y+z}=\Sigma_{cyc}\frac{2\left(y+z-2x\right)}{2x+y+z}+6\)

\(=\Sigma_{cyc}\left(\frac{2\left(x+y+z\right)\left(y+z-2x\right)}{2x+y+z}-\frac{3}{2}\left(y+z-2x\right)\right)+6\)

\(=\Sigma_{cyc}\frac{\left(y+z-2x\right)^2}{2\left(2x+y+z\right)}+6\ge6\)

a ) Đặt A = \(\frac{-a+b+c}{2a}+\frac{a-b+c}{2b}+\frac{a+b-c}{2c}=\frac{1}{2}\left(-1+\frac{b}{a}+\frac{c}{a}+\frac{a}{b}-1+\frac{c}{b}+\frac{a}{c}+\frac{b}{c}-1\right)\)

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

Do a ; b ; c > 0 , áp dụng BĐT Cô - si cho các cặp số dương , ta có :

\(A\ge\frac{1}{2}\left[2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{\frac{b}{c}.\frac{c}{b}}+2\sqrt{\frac{a}{c}.\frac{c}{a}}-3\right]=\frac{1}{2}\left(2+2+2-3\right)=\frac{3}{2}\)

Dấu " = " xảy ra \(\Leftrightarrow a=b=c\)

b ) \(P=\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{x+y}=\frac{x^2}{xy+xz}+\frac{y^2}{xy+yz}+\frac{z^2}{xz+yz}\ge\frac{\left(x+y+z\right)^2}{2\left(xy+yz+xz\right)}\ge\frac{3\left(xy+yz+xz\right)}{2\left(xy+yz+xz\right)}=\frac{3}{2}\)

( áp dụng BĐT Cauchy - Schwarz )

Dấu " = " xảy ra \(\Leftrightarrow x=y=z\)