a,

\(\left|x+\dfrac{9}{2}\right|\ge0\forall x\\ \left|y+\dfrac{4}{3}\right|\ge0\forall y\\ \left|z+\dfrac{7}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{9}{2}=0\\y+\dfrac{4}{3}=0\\z+\dfrac{7}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-9}{2}\\y=\dfrac{-4}{3}\\z=\dfrac{-7}{2}\end{matrix}\right.\)

Vậy \(x=\dfrac{-9}{2};y=\dfrac{-4}{3};z=\dfrac{-7}{2}\)

d,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{1}{5}\right|\ge0\forall y\\ \left|x+y+z\right|\ge0\forall x,y,z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{1}{5}=0\\x+y+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-3}{4}+\dfrac{1}{5}+z=0\end{matrix}\right.\\\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\\dfrac{-11}{20}+z=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{1}{5}\\z=\dfrac{11}{20}\end{matrix}\right.\)

Bạn mới hỏi ở dưới rồi :v

b,

\(\left|x+\dfrac{3}{4}\right|\ge0\forall x\\ \left|y-\dfrac{2}{5}\right|\ge0\forall y\\ \left|z+\dfrac{1}{2}\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\forall x,y,z\\ \)

Mà \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\\ \Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{2}{5}\right|=0\\\left|z+\dfrac{1}{2}\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{3}{4}=0\\y-\dfrac{2}{5}=0\\z+\dfrac{1}{2}=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-3}{4}\\y=\dfrac{2}{5}\\z=\dfrac{-1}{2}\end{matrix}\right.\)

Vậy ...

c,

\(\left|x+\dfrac{19}{5}\right|\ge0\forall x\\ \left|y+\dfrac{1890}{1975}\right|\ge0\forall y\\ \left|z-2004\right|\ge0\forall z\\ \Rightarrow\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\)

Mà

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\\ \Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\\\left|y+\dfrac{1890}{1975}\right|=0\\\left|z-2004\right|=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x+\dfrac{19}{5}=0\\y+\dfrac{1890}{1975}=0\\z-2004=0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x=\dfrac{-19}{5}\\y=\dfrac{-1890}{1975}=\dfrac{-378}{395}\\z=2004\end{matrix}\right. \)

Vậy ...

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\forall x;y;z\)

mà A\(\le0\)

​​\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\)​ phải bằng 0 đê thỏa mãn điều kiện

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\\\left|y+\dfrac{4}{3}\right|=0\\\left|z+\dfrac{7}{2}\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{9}{2}\\y=-\dfrac{4}{3}\\z=-\dfrac{7}{2}\end{matrix}\right.\)

Vậy....

b;c)I hệt câu a nên làm tương tự nhá

d)

Hơi tắt nhá

a) Đặt \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=B\)

B=\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\\\left|y-\dfrac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-\dfrac{3}{4}\\y=\dfrac{1}{5}\\x+y+z=0\end{matrix}\right.\)

Thay ra ta tính đc :\(z=-\dfrac{11}{20}\)

Vậy....

\(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|\ge0\\\left|y+\dfrac{4}{3}\right|\ge0\\\left|z+\dfrac{7}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\ge0\\\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{9}{2}\right|=0\Rightarrow x=-\dfrac{9}{2}\\\left|y+\dfrac{4}{3}\right|=0\Rightarrow y=-\dfrac{4}{3}\\\left|z+\dfrac{7}{2}\right|=0\Rightarrow z=-\dfrac{7}{2}\end{matrix}\right.\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\\left|y-\dfrac{2}{5}\right|\ge0\\\left|z+\dfrac{1}{2}\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\)

\(\Rightarrow\left[{}\begin{matrix}\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\ge0\\\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|\le0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{2}{5}\right|+\left|z+\dfrac{1}{2}\right|=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}\right|=0\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}\right|=0\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2004\right|=0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|\ge0\\ \left|y+\dfrac{1980}{1975}\right|\ge0\\\left|z-2004\right|\ge0\end{matrix}\right.\)

\(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1980}{1975}\right|+\left|z-2004\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{19}{5}\right|=0\Rightarrow x=-\dfrac{19}{5}\\ \left|y+\dfrac{1980}{1975}\right|=0\Rightarrow y=-\dfrac{1980}{1975}\\\left|z-2004\right|=0\Rightarrow z=2004\end{matrix}\right.\)

\(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|=0\)

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|\ge0\\ \left|y-\dfrac{1}{5}\right|\ge0\\\left|x+y+z\right|\ge0\end{matrix}\right.\)

\(\Rightarrow\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\ge0\)

Dấu "=" xảy ra khi:

\(\left\{{}\begin{matrix}\left|x+\dfrac{3}{4}\right|=0\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{1}{5}\right|=0\Rightarrow y=\dfrac{1}{5}\\\left|x+y+z\right|=0\Rightarrow z+-\dfrac{11}{20}=0\Rightarrow z=\dfrac{11}{20}\end{matrix}\right.\)

Đặt \(\left|x+\dfrac{9}{2}\right|+\left|y+\dfrac{4}{3}\right|+\left|z+\dfrac{7}{2}\right|=A\)

\(\Rightarrow A\ge0\)

Mà ĐK đề là \(A\le0\)

\(\Rightarrow A=0\)

\(\left[{}\begin{matrix}\left|x+\dfrac{3}{4}=0\right|\Rightarrow x=-\dfrac{3}{4}\\\left|y-\dfrac{2}{5}=0\right|\Rightarrow y=\dfrac{2}{5}\\\left|z+\dfrac{1}{2}=0\right|\Rightarrow z=-\dfrac{1}{2}\end{matrix}\right.\)

Các câu còn lại tương tự nhé

\(\frac{x}{x+1}+\frac{y}{y+1}+\frac{z}{z+1}=1-\frac{1}{x+1}+1-\frac{1}{y+1}+1-\frac{1}{z+1}=3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)

vì \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}>=\frac{9}{x+1+y+1+z+1}=\frac{9}{1+3}=\frac{9}{4}\)(bđt svacxo)

\(\Rightarrow3-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)< =3-\frac{9}{4}=\frac{3}{4}\)

dấu = xảy ra khi x=y=z=\(\frac{1}{3}\)

no biet

dự đoán điểm rơi : x = y = z > 0 

dùng Cô si :bớt. Ta làm như sau 

Giải :    Đặt P = \(\frac{x^3}{y\left(z+2x\right)}+\frac{y^3}{z\left(x+2y\right)}+\frac{z^3}{x\left(y+2z\right)}.\)áp dụng bất đẳng thức Cô si cho 3 số dương.

ta có :  \(\frac{x^3}{y\left(z+2x\right)}+\frac{y}{3}+\frac{z+2x}{9}\ge3\sqrt[3]{\frac{x^3.y.\left(z+2x\right)}{y.\left(z+2x\right).3.9}}=x.\left(1\right)..\)

chứng minh tương tự ta có :

\(\frac{y^3}{z.\left(x+2y\right)}+\frac{z}{3}+\frac{x+2y}{9}\ge y\left(2\right).\)\(\frac{z^3}{x.\left(y+2z\right)}+\frac{x}{3}+\frac{y+2z}{9}\ge z.\left(3\right).\)

Cộng vế với vế các bất đẳng thức (1) , (2) và (3) ta được :  

                                 \(P+\frac{2}{3}.\left(x+y+z\right)\ge x+y+z\)

                =>    \(P\ge\frac{x+y+z}{3}.\) đấu " = " xẩy ra khi x = y = z  > 0  ( đpcm ) 

Ta chứng minh \(\frac{x^4+y^4}{x^2+y^2}\ge\frac{\frac{\left(x^2+y^2\right)^2}{2}}{x^2+y^2}=\frac{x^2+y^2}{2}\)

Tương tự và cộng lại

\(\Rightarrow VT\ge x^2+y^2+z^2\ge xy+xz+yz=3\)

chứng minh kiểu j vậy bạn ? , Chỉ mình rõ hơn được không ? 

Đặt \(a=2x+y+z;b=2y+z+x;c=2z+x+y\)

\( \implies\) \(a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)\) 

\( \implies\) \(a+b+c=4x+4y+4z\)

\( \implies\) \(x+y+z=\frac{a+b+c}{4}\) 

+)Ta có : \(a=2x+y+z\)

\(\iff\) \(a=x+\left(x+y+z\right)\)

\(\iff\) \(a-\left(x+y+z\right)=x\)

\(\iff\) \(a-\frac{a+b+c}{4}=x\)

\(\iff\) \(x=\frac{3a-b-c}{4}\)

+)Ta có :\(b=2y+z+x\)

\(\iff\) \(b=y+\left(y+z+x\right)\)

\(\iff\)\(b-\left(y+z+x\right)=y\)

\(\iff\) \(b-\frac{a+b+c}{4}=y\)

\(\iff\)\(y=\frac{3b-c-a}{4}\)

+)Ta có :\(c=2z+x+y\)

\(\iff\) \(c=z+\left(z+x+y\right)\)

\(\iff\) \(c-\left(z+x+y\right)=z\)

\(\iff\) \(c-\frac{a+b+c}{4}=z\)

\(\iff\)\(z=\frac{3c-a-b}{4}\)

​​​\( \implies\)​ \(\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}\) 

 \(=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}\)

 \(=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)\)

 \(=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)\)

 \(=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\)

Áp dụng bất đẳng thức ( BĐT Cosi ) : \(m+n\)\( \geq\)\(2\sqrt{mn}\) \(\left(m;n>0\right)\)ta được : 

\(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{a}.\frac{a}{b}}\) = 2 \( \implies\) \(\frac{b}{a}+\frac{a}{b}\) \( \geq\) 2 

\(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 \(\sqrt{\frac{c}{a}.\frac{a}{c}}\) = 2 \( \implies\) \(\frac{c}{a}+\frac{a}{c}\) \( \geq\) 2 

\(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 \(\sqrt{\frac{b}{c}.\frac{c}{b}}\) = 2 \( \implies\) \(\frac{b}{c}+\frac{c}{b}\) \( \geq\) 2 

\( \implies\) \(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\) \( \geq\) 2 + 2 + 2 

\( \implies\) ​​​\(\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\)​ \( \geq\) 6 

\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{6}{4}\)

\( \implies\) \(\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \( \geq\) \(\frac{3}{2}\)

\( \implies\) \(-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(-\frac{3}{2}\)

\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{9}{4}-\frac{3}{2}\)

\( \implies\) \(\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]\) \(\leq\) \(\frac{3}{4}\) 

Dấu " = " xảy ra khi a = b = c hay x = y = z 

https://chat.lazi.vn/upload/images/2020/03/file_xot1584973258.jpg

Chúc bạn hok tốt!!!