cho |x| = |y|
và x < 0, y > 0
Tính : 1. x + y 2. 1/x + 1/y
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
x-y-z=0
=>x=y+z và y=x-z và z=x-y
B=(1-z/x)(1-x/y)(1+y/z)+2023
\(=\dfrac{x-z}{x}\cdot\dfrac{y-x}{y}\cdot\dfrac{y+z}{z}+2023\)
\(=\dfrac{y}{x}\cdot\dfrac{-z}{y}\cdot\dfrac{x}{z}+2023=2023-1=2022\)
Ta có: \(x-y-z=0\)
\(\Rightarrow x-y=z\)
\(x-z=y\)
\(y+z=x\)
\(\Rightarrow B=\left(1-\dfrac{z}{x}\right)\left(1-\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\)
\(=\dfrac{x-z}{x}.\dfrac{-\left(y-x\right)}{y}.\dfrac{z+y}{z}\)
\(=\dfrac{y}{x}.-\dfrac{z}{y}.\dfrac{z}{x}=-1\)
\(\Rightarrow B=-1\)
2x2 + 2y2 + 3xy - x + y + 1 = 0
2x2 + 2y2 + 4xy - xy - x + y + 1 = 0
(2x2 + 2y2 + 4xy) + (-xy - x) + (y + 1) = 0
2(x + y)2 - x(y + 1) + (y + 1) = 0
2(x + y)2 + (y + 1)(1 - x) = 0
Do (x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 \(\ge0\)
\(\Rightarrow\) 2(x + y)2 + (y + 1)(1 - x) = 0 \(\Leftrightarrow\) (y + 1)(1 - x) = 0
\(\Rightarrow y+1=0;1-x=0\)
*) y + 1 = 0
y = -1
*) 1 - x = 0
x = 1
Với x = 1; y = -1, ta có:
B = [1 + (-1)]2018 + (1 - 2)2018 + (-1 - 1)2018
= 1 + 22018
3x^2+3y^2+4xy-2x+2y+2=0
=>2x^2+4xy+2y^2+x^2-2x+1+y^2+2y+1=0
=>x=1 và y=-1
M=(1-1)^2017+(1-2)^2018+(-1+1)^2015=1
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) \(B=\dfrac{1}{x^{2020}}+\dfrac{1}{y^{2020}}+\dfrac{1}{z^{2020}}=\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}+\dfrac{1}{\left(-1\right)^{2020}}=\dfrac{1}{1}+\dfrac{1}{1}+\dfrac{1}{1}=3\)
\(5x2+5y2+8xy-2x+2y+2=0\)
(=) \((4x^2 + 8xy + 4y^2) + (x^2 - 2x +1) + (y^2 + 2y +1) = 0 \)
(=) \(4(x+y)^2 + (x-1)^2 + (y+1)^2 = 0 \)
Ta có \(\begin{cases} 4(x+y)^2 ≥ 0 \\ (x-1)^2 ≥ 0 \\ (y+1)^2 ≥ 0 \end{cases} \)
=> \(4(x+y)^2 + (x-1)^2 + (y+1)^2 ≥ 0 \)
Vậy để \(4(x+y)^2 + (x-1)^2 + (y+1)^2 = 0 \)
(=) \(\begin{cases} 4(x+y)^2 = 0 \\ (x-1)^2 = 0 \\ (y+1)^2 = 0 \end{cases} \)
(=) \(\begin{cases} x = -y \\ x = 1 \\ y = -1 \end{cases} \)
(=) \(\begin{cases} x = 1 \\ y = -1 \end{cases} \)
Vậy \(M=(x+y)^{2015}+(x-2)^{2016}+(y+1)^{2017} M=(1-1)^{2015} + (1-2)^{2016} + (-1+1)^{2017} M=0^{2015} + (-1)^{2016} +0^{2017} M= 1 \)Vậy M = 1
\(x+y+z=0\\ \Rightarrow\left\{{}\begin{matrix}x=-y-z\\y=-z-x\\z=-x-y\end{matrix}\right.\)
\(\dfrac{xy}{x^2+y^2-z^2}+\dfrac{yz}{y^2+z^2-x^2}+\dfrac{zx}{z^2+x^2-y^2}\)
\(=\dfrac{xy}{x^2+y^2-\left(-x-y\right)^2}+\dfrac{yz}{y^2+z^2-\left(-y-z\right)^2}+\dfrac{zx}{z^2+x^2-\left(-z-x\right)^2}\)
\(=\dfrac{xy}{x^2+y^2-\left(x+y\right)^2}+\dfrac{yz}{y^2+z^2-\left(y+z\right)^2}+\dfrac{zx}{z^2+x^2-\left(z+x\right)^2}\)
\(=\dfrac{xy}{x^2+y^2-x^2-2xy-y^2}+\dfrac{yz}{y^2+z^2-y^2-2yz-z^2}+\dfrac{zx}{z^2+x^2-z^2-2zx-x^2}\)
\(=\dfrac{xy}{-2xy}+\dfrac{yz}{-2yz}+\dfrac{zx}{-2zx}\)
\(=-\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{2}\)
\(=-\dfrac{3}{2}\)
Ta có : x < 0 <=>|x| = -x (1)
y > 0 <=> |y|= y (2)
Mà |x| = |y| (3)
Từ (1)(2)(3) => -x = y hay y = -x
1) Với y = -x
=> x + y = x + (-x) = 0
2) Với y = -x
=> \(\frac{1}{x}\)+ \(\frac{1}{y}\)\(=\frac{1}{x}+\frac{1}{-x}=\frac{1}{x}+\frac{-1}{x}=0\)( Vì x \(\ne\)0)