Tìm y
2020-y/8*10
Cho biết các số x,y,z thỏa mãn :
x2+2y+1=0
y2+2z+1=0
z2+2x+1=0
Tính giá trị biểu thức:
a) A = x2020 + y2020+z2020
b) B=\(\dfrac{1}{x^{2022}}+\dfrac{1}{y^{2022}}+\dfrac{1}{z^{2022}}\)
Ta có: \(\left\{{}\begin{matrix}x^2+2y+1=0\\y^2+2z+1=0\\z^2+2x+1=0\end{matrix}\right.\)
\(\Rightarrow x^2+2y+1+y^2+2z+1+z^2+2x+1=0\)
\(\Rightarrow\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
\(\Rightarrow x=y=z=-1\)(do \(\left(x+1\right)^2,\left(y+1\right)^2,\left(z+1\right)^2\ge0\forall x,y,z\))
a) \(A=x^{2020}+y^{2020}+z^{2020}=\left(-1\right)^{2020}+\left(-1\right)^{2020}+\left(-1\right)^{2020}=1+1+1=3\)
b) \(B=\dfrac{1}{x^{2020}}+\dfrac{1}{y^{2020}}+\dfrac{1}{z^{2020}}=\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}=3\)
bài 9 tìm y
y x 8 = y + 10
y x 10 + 9 = y x 3 + 3
a, y x 8 = y + 10
=> 8y=y+10
=> 8y-y=10
=> 7y=10
=> y=10/7
b, y x 10 + 9 = y x 3 + 3
=> 10y+9=3y+3
=>10y+9-3y=3
=> 10y-3y+9=3
=> 10y-3y=3-9
=> 7y=-6
=> y=-6/7
:)
y= 6/7 nhaaaaaaa!
Tìm y :2020-y:8*10=1000
2020-y:8×10=1000
y:8×10=2020-1000
y:8×10=1020
y:8=1020:10
y:8=102
y=102×8
y=816
Vậy y=816
y = 816
tk cho mk nha !! Chúc bạn năm mới vui vẻ gặp nhiều may mắn
Thank you very much
Tìm x,y
1/x + y/2= 5/8
x/10-1/y=3/10
tìm y biết 2020 - y/ 8*10=1000
Tìm số nguyên x biết:
(2x-10)^2+(8-y)^10-10=-10
\(\left(2x-10\right)^2+\left(8-y\right)^{10}-10=-10\)
\(\Rightarrow\left(2x-10\right)^2+\left(8-y\right)^{10}=0\)
Vì \(\hept{\begin{cases}\left(2x-10\right)^2\ge0\\\left(8-y\right)^{10}\ge0\end{cases}\Rightarrow\left(2x-10\right)^2+\left(8-y\right)^{10}\ge0}\)
Mà \(\left(2x-10\right)^2+\left(8-y\right)^{10}=0\)
\(\Rightarrow\hept{\begin{cases}\left(2x-10\right)^2=0\\\left(8-y\right)^{10}=0\end{cases}\Rightarrow\hept{\begin{cases}2x-10=0\\8-y=0\end{cases}\Rightarrow}\hept{\begin{cases}x=5\\y=8\end{cases}}}\)
\(\left(2x-10\right)^2+\left(8-y\right)^{10}-10=-10\)
\(\Leftrightarrow\left(2x-10\right)^2+\left(8-y\right)^{10}=0\)
Mà \(\hept{\begin{cases}\left(2x-10\right)^2\ge0\\\left(8-y\right)^{10}\ge0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}\left(2x-10\right)^2=0\\\left(8-y\right)^{10}=0\end{cases}}\)
\(\Leftrightarrow\hept{\begin{cases}2x-10=0\\8-y=0\end{cases}}\) \(\Leftrightarrow\hept{\begin{cases}x=5\\y=8\end{cases}}\)
Vậy ..
Tìm y,biết:
y*1,15+y:10-y*0,25=8
Tìm y biết: 2020 - y : 8 x 10 = 1000
Tìm y biết:
2020 - y : 8 x 10 = 1000
y : 8 x 10 = 2000 - 1000
y : 8 x 10 = 1000
y : 8 = 1000 : 10
y : 8 = 100
y = 100 x 8
y = 800
Vậy y = 800
\(2020-y:8\times10=1000\)
\(2020-y:80=1000\)
\(y:80=2020-1000\)
\(y:80=1020\)
\(y=1020\times80\)
\(y=81600\)
2020-y:8=1000:10
2020-y:8=100
2020-y=100x8
2020-y=800
y=2020-800=1220
6/4 - 8/10 - y=4/6 . Tìm y?