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Sửa đề: \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}\le0\)(1)
Ta có: \(\left|3x-5\right|\ge0;\left(2y+5\right)^{2018}\ge0;\left(4z-3\right)^{2020}\ge0.\)mọi x,y, z.
=> \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}\ge0\)với mọi x, y,z.
Như vậy (1) chỉ xảy ra trường hợp: \(\left|3x-5\right|+(2y+5)^{2018}+\left(4z-3\right)^{2020}=0\)
<=> \(\hept{\begin{cases}3x-5=0\\2y+5=0\\4z-3=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=\frac{5}{3}\\y=-\frac{5}{2}\\z=\frac{3}{4}\end{cases}}\)
Vậy...
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a) Ta có: \(A=\left(-\dfrac{3}{4}x^2y^5z^3\right)\cdot\left(\dfrac{5}{3}x^3y^4z^2\right)\)
\(=\left(\dfrac{-3}{4}\cdot\dfrac{5}{3}\right)\cdot\left(x^2\cdot x^3\right)\cdot\left(y^5\cdot y^4\right)\cdot\left(z^3\cdot z^2\right)\)
\(=\dfrac{-5}{4}x^5y^9z^5\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Ta có : \(\frac{5z-6y}{4}=\frac{6x-4z}{5}=\frac{4y-5x}{6}=\frac{20z-24y}{16}=\frac{30x-20z}{25}=\frac{24y-30x}{36}\)
\(=\frac{20z-24y+30x-20z+24y-30x}{16+25+36}=0\)
\(\Rightarrow\hept{\begin{cases}\frac{5z-6y}{4}=0\\\frac{6x-4z}{5}=0\end{cases}\Leftrightarrow\hept{\begin{cases}5z-6y=0\\6x-4z=0\end{cases}}\Leftrightarrow\hept{\begin{cases}5z=6y\\6x=4z\end{cases}}\Leftrightarrow\hept{\begin{cases}\frac{y}{5}=\frac{z}{6}\\\frac{z}{6}=\frac{x}{4}\end{cases}}}\)
\(\Leftrightarrow\frac{y}{5}=\frac{z}{6}=\frac{x}{4}=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}\)
Áp dụng tính chất dãy tỉ số bằng nhau , ta có :
\(\frac{y}{5}=\frac{z}{6}=\frac{x}{4}=\frac{3x}{12}=\frac{2y}{10}=\frac{5z}{30}=\frac{3x-2y+5z}{12-10+30}=\frac{96}{32}=3\)
\(\Rightarrow\hept{\begin{cases}x=3.4=12\\y=3.5=15\\z=3.6=18\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1) ADTCDTSBN, ta có:
\(\frac{x}{3}=\frac{y}{4}=\frac{z}{5}\)= \(\frac{2x^2+2y^2-3z^2}{18+32-75}=\frac{-100}{-25}\)= 4
* \(\frac{x}{3}=4\)=> x = 3 . 4 = 12
- \(\frac{y}{4}=4\)=> y = 4 . 4 = 16
* \(\frac{z}{5}=4\)=> z = 5 . 4 = 20
Vậy x = 12
y = 16
z = 20
![](https://rs.olm.vn/images/avt/0.png?1311)
\(E=\left(2x-5\right)^{10}-12\ge-12\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{5}{2}\)
Vậy \(E_{min}=-12\Leftrightarrow x=\dfrac{5}{2}\)
\(F=\left(x+5\right)^8+\left|x+5\right|+22\ge22\)
Dấu "=" xảy ra \(\Leftrightarrow x=-5\)
Vậy \(F_{min}=22\Leftrightarrow x=-5\)
\(G=17-\left|3x-2\right|\)
Dấu "=" xảy ra \(x=\dfrac{2}{3}\)
Vậy \(G_{max}=17\Leftrightarrow x=\dfrac{2}{3}\)
\(K=17-\left|3x-2\right|-\left(2-3x\right)^{2020}\le17\)
Dấu "=" xảy ra \(\Leftrightarrow x=\dfrac{2}{3}\)
Vậy \(K_{max}=17\Leftrightarrow x=\dfrac{2}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1 :
Vì \(\sqrt{3x+2y+z}\ge0\forall x;y;z\)
\(\left|y-\frac{1}{2}\right|\ge0\forall y\)
\(\left(z-2\right)^2\ge0\forall z\)
\(\Rightarrow A\ge2018\forall x;y;z\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}3x+2y+z=0\\y-\frac{1}{2}=0\\z-2=0\end{cases}\Leftrightarrow\hept{\begin{cases}3x+2\cdot\frac{1}{2}+2=0\\y=\frac{1}{2}\\z=2\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-1\\y=\frac{1}{2}\\z=2\end{cases}}}\)
Vậy........
Bài 2 :
Lý luận tương tự câu 1) ta có :
\(\hept{\begin{cases}x-1=0\\y+1=0\\x+y+z=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=1\\y=-1\\1-1+z=0\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=-1\\z=0\end{cases}}}\)
Thay x; y; z vào P ta có :
\(P=1^{2018}+\left(-1\right)^{2019}+0^{2020}\)
\(P=1-1+0\)
\(P=0\)