tinh
|x-34|^335+|y-12|^34=0
Tính:
\(\left|x-34\right|^{335}+\left|y-12\right|^{34}=0\)
\(\left[x-34\right]^{335}+\left|y-12\right|^{34}=0\)
Để GTBT bằng 0 thì :
\(\left\{\begin{matrix}x-34=0\\y-12=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=34\\y=12\end{matrix}\right.\)
Vậy \(x=34;y=12\) thì GTBT bằng 0.
Vì \(\left|x-34\right|^{335}\ge0;\left|y-12\right|^{34}\ge0\)
\(\Rightarrow\left|x-34\right|^{335}+\left|y-12\right|^{34}\ge0\)
Dấu "=" xảy ra khi \(\left[\begin{matrix}\left|x-34\right|^{335}=0\\\left|x-12\right|^{34}=0\end{matrix}\right.\Rightarrow\left[\begin{matrix}x=34\\y=12\end{matrix}\right.\)
Bài 1:Tìm x,y biết:
a, |x+5|+(3y -4)^2012=0
b,(2x-1)^2+|2y-x|-8=12 - 5.2^2
Answer:
a, \(\left|x+5\right|+\left(3y-4\right)^{2012}=0\)
Ta có: \(\hept{\begin{cases}\left|x+5\right|\ge0\forall x\\\left(3y-4\right)^{2012}\ge0\forall y\end{cases}}\)
\(\Rightarrow\left|x+5\right|+\left(3y-4\right)^{2012}=0\)
\(\Rightarrow\hept{\begin{cases}x+5=0\\3y-4=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=\frac{4}{3}\end{cases}}\)
b, \(\left(2x-1\right)^2+\left|2y-x\right|-8=12-5.2^2\)
\(\Rightarrow\left(2x-1\right)^2+\left|2y-x\right|=12-5.2^2+8\)
\(\Rightarrow\left(2x-1\right)^2+\left|2y-x\right|=0\)
\(\Rightarrow\hept{\begin{cases}2x-1=0\\2y-x=0\end{cases}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\2y-\frac{1}{2}=0\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{1}{2}\\y=\frac{1}{4}\end{cases}}}\)
Tìm x,y,z thuộc Q
a, |x+195 |+|y+18901975 |+|z+2004|
b, |x+92 |+|y+43 |+|z+72 |≤0
c,|x+34 |+|y−15 |+|x+y+z|=0
d, |x+34 |+|y−25 |+|z+12 |≤0
a) Vì : \(\left|x+\frac{19}{5}\right|\ge0\forall x\in R\)
\(\left|y+\frac{1890}{1975}\right|\ge0\forall y\in R\)
\(\left|z-2004\right|\ge0\forall z\in R\)
\(\Rightarrow\left|x+\frac{19}{5}\right|+\left|y+\frac{1890}{1975}\right|+\left|z-2004\right|\ge0\forall x,y,z\in R\)
Dấu''='' xảy ra khi và chỉ khi \(\hept{\begin{cases}x=\frac{-19}{5}\\y=\frac{-1890}{1975}\\z=2004\end{cases}}\)
b,\(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)
Ta có:\(\left|x+\frac{9}{2}\right|\ge0\forall x\)
\( \left|y+\frac{4}{3}\right|\ge0\forall y\)
\(\left|z+\frac{7}{2}\right|\ge0\forall z\)
\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\ge0\forall x,y,z\)
Mà \(\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|\le0\)
\(\Rightarrow\left|x+\frac{9}{2}\right|+\left|y+\frac{4}{3}\right|+\left|z+\frac{7}{2}\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{9}{2}\right|=0\\\left|y+\frac{4}{3}\right|=0\\\left|z+\frac{7}{2}\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+\frac{9}{2}=0\\y+\frac{4}{3}=0\\z+\frac{7}{2}=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{9}{2}\\y=-\frac{4}{3}\\z=-\frac{7}{2}\end{cases}}}\)
c,\(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
Ta có:\(\left|x+\frac{3}{4}\right|\ge0\forall x\)
\(\left|y-\frac{1}{5}\right|\ge0\forall y\)
\(\left|x+y+z\right|\ge0\forall x,y,z\)
\(\Rightarrow\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|\ge0\forall x,y,z\)
Mà \(\left|x+\frac{3}{4}\right|+\left|y-\frac{1}{5}\right|+\left|x+y+z\right|=0\)
\(\Rightarrow\hept{\begin{cases}\left|x+\frac{3}{4}\right|=0\\\left|y-\frac{1}{5}\right|=0\\\left|x+y+z\right|=0\end{cases}\Rightarrow\hept{\begin{cases}x+\frac{3}{4}=0\\y-\frac{1}{5}=0\\x+y+z=0\end{cases}\Rightarrow}\hept{\begin{cases}x=-\frac{3}{4}\\y=\frac{1}{5}\\z=\frac{11}{20}\end{cases}}}\)
Tính giá trị biểu thức
a) x + − 34 , x = − 12
b) − 103 + y , y = − 217
c) z + − 34 + − 103 , z = − 4
34+6231=???
335+78=??
Giả sử x =a/m ; y = b/m( a, b, m ∈ Z, b # 0) và x < y. Hãy chứng tỏ rằng nếu chọn z =(a+b)/2m thì ta có x < z < y
Bạn tham khảo hình ảnh!
Không thấy ib nhé :v
Cre : Hoidap247
Theo đề bài ta có x = \(\frac{a}{m}\), y = \(\frac{b}{m}\) (a, b, m ∈ Z, b # 0)
Vì x < y nên ta suy ra a < b
Ta có: x = \(\frac{2a}{2m}\), y = \(\frac{2b}{2m}\); z = \(\frac{\left(a+b\right)}{2m}\)
Vì a < b => a + a < a + b => 2a < a + b
Do 2a < a + b nên x < z (1)
Vì a < b => a + b < b + b => a + b < 2b
Do a + b < 2b nên z < y (2)
Từ (1) và (2) ta suy ra x < z < y
Theo đề bài ta có x = a/m, y = b/m (a, b, m ∈ Z, b # 0)
Vì x < y nên ta suy ra a < b
Ta có: x = 2a/2m, y = 2b/2m; z = (a+b)/2m
Vì a < b => a + a < a + b => 2a < a + b
Do 2a < a + b nên x < z (1)
Vì a < b => a + b < b + b => a + b < 2b
Do a + b < 2b nên z < y (2)
Từ (1) và (2) ta suy ra x < z < y
Tìm x ; y Z
a, | x + 3y | + | y - 12 | = 0
b, | x - 3 | + | y + 4 | = 1
c , ( 3x + 1 ) \(^2\)+ | y - 5 | = 1
lm nhanh và đầy đủ giùm mk nha
{\displaystyle \in }
tìm x ; y Z , biết :
a, | x + 3y | + | y - 12 | = 0
b, | x - 3 | + | y + 4 | = 1
c, ( 3x + 1 ) \(^2\)+ | y - 5 | = 1
lm nhanh và đầy đủ giùm mk nha
{\displaystyle \in }
Tìm x, biết
a) x : − 2 1 15 + 3 1 2 = − 3 4 b) − 5 8 − x : 3 5 6 + 7 3 4 = − 2
c) x − 3 4 − 1 4 = 0 d) 3 4 : 2 4 9 − − 3 x + 2 2 3 = 3 4
e) x − 1 3 = 2 − 3 x f) 3 5 7 x − 1 5 7 x − 1 3 = 2 3
g) − 3 4 3 x − 1 = − 27 64 h) 4 5 2 x + 5 = 256 625
i) x − 2 15 3 = 8 125 k) x + 3 5 x + 3 2 = 64 27
m) x − 1 x + 5 = 6 7 n) 4 13 . 6 5 + 4 13 . 2 5 . 2 x + 1 2 = 10 13