Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
( x - 1 )2018 + ( y + 3 )2020 + ( z - 5 )2022 = 0
Ta thấy : ( x - 1 )2018 \(\ge0\) ; ( y + 3 )2020 \(\ge0\) ; ( z - 5 )2022 \(\ge0\)
\(\Rightarrow\left(x-1\right)^{2018}+\left(y+3\right)^{2020}+\left(z-5\right)^{2022}\ge0\)
Theo đề,ta có : \(\left(x-1\right)^{2018}=\left(y+3\right)^{2020}=\left(z-5\right)^{2022}=0\)
+) \(\left(x-1\right)^{2018}=0\Rightarrow x-1=0\Rightarrow x=1\)
+) \(\left(y+3\right)^{2020}=0\Rightarrow y+3=0\Rightarrow y=-3\)
=) \(\left(z-5\right)^{2022}=0\Rightarrow z-5=0\Rightarrow z=5\)
Vậy : x = 1 ; y = -3 ; z = 5
\(\text{Ta có:}\)
\(\hept{\begin{cases}\left(x-1\right)^{2018}\ge0\\\left(y+3\right)^{2020}\ge0\\\left(z-5\right)^{2022}\ge0\end{cases}}\text{mà:}\left(x-1\right)^{2018}+\left(y-2\right)^{2020}+\left(z-3\right)^{2022}=0\text{ nên:}\)
\(\hept{\begin{cases}\left(x-1\right)^{2018}=0\\\left(y+3\right)^{2018}=0\\\left(z-5\right)^{2018}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=1\\y=-3\\z=5\end{cases}}\)
bạn tự kết luận
Ta có \(M^2=y\left(x-z\right)-z\left(x-y\right)\)
\(=xy-yz-xz+yz\)
\(=xy-xz\)
\(=x\left(y-z\right)=-20\left(-5\right)=100\)
\(M^2=100\Rightarrow\orbr{\begin{cases}M=10\\M=-10\end{cases}}\)
Ta có : \(\left\{{}\begin{matrix}x-y=-9\\y-z=-10\\z+x=11\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-9\\z=10+y\\z+x=11\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-9\\z=10+y\\y-9+10+y=11\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=y-9\\z=10+y\\2y+1=11\end{matrix}\right.\)
=> \(\left\{{}\begin{matrix}x=5-9=-4\\z=10+5=15\\y=5\end{matrix}\right.\)
Vậy các giá trị x; y; z lần lượt là -4; 5; 15 .
Ta có:
\(\left\{{}\begin{matrix}x-y=-9\\y-z=-10\end{matrix}\right.\)\(\Rightarrow\left(x-y\right)+\left(y-z\right)=\left(-9\right)+\left(-10\right)\)
\(\Rightarrow x-z=-19\) ; \(z+x=11\)
\(\Rightarrow\left(x-z\right)+\left(z+x\right)=-19+11\)
\(\Rightarrow2x=-8\Rightarrow x=-4\)
\(\Rightarrow y=5;z=15\)
Có \(VT\)ko âm với mọi \(x,y,z\in Z\)
\(\Rightarrow\hept{\begin{cases}\left|x\right|=0\\\left|y\right|=0\\\left|z\right|=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=0\\y=0\\z=0\end{cases}}\)
Vậy x = 0 ; y = 0 ; z = 0
\(xy+12=x+y\)
<=> \(xy-x-y+12=0\)
<=> \(x\left(y-1\right)-\left(y-1\right)+11=0\)
<=> \(\left(x-1\right)\left(y-1\right)=-11\)
làm nốt nha
\(x\left(y+1\right)=2y+5\)
<=> \(x\left(y+1\right)-2y-5=0\)
<=> \(x\left(y+1\right)-2\left(y+1\right)-3=0\)
<=> \(\left(x-2\right)\left(y+1\right)=3\)
làm nốt nha
bạn nen biet cach giai ban cong tung ve ta co;
x-y =8
y-z=10
x+z=12
cong; 2x=30
x=15
y=7
z=-3
x+y+z=19
