Bài 1:Tìm x,y biết
x/x+y+5=y/x+y-5=x+y
Những câu hỏi liên quan
tìm quan hệ x và y biết
x\(^4\)=y\(^4\)
x\(^5\)=y\(^5\)
Ta có:
\(x^4=y^4\)
\(\Rightarrow x^4-y^4=0\)
\(\Rightarrow\left(x^2\right)^2-\left(y^2\right)^2=0\)
\(\Rightarrow\left(x^2-y^2\right)\left(x^2+y^2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-y^2=0\\x^2+y^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x-y=0\\x+y=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
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Ta có:
\(x^5=y^5\)
\(\Rightarrow x^5-y^5=0\)
\(\Rightarrow x-y=0\)
\(\Rightarrow x=y\)
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tìm các số hữu tỉx,y,z biếtx(x+y+z=-5;y(x+y+z)=9;z(x+y+z)=5
tìm x,y thuộc z biết
x/7=9/y và x>y
\(\Leftrightarrow xy=63\)
\(\Leftrightarrow\left(x,y\right)\in\left\{\left(1;63\right);\left(3;21\right);\left(7;9\right);\left(-63;-1\right);\left(-21;-3\right);\left(-9;-7\right)\right\}\)
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tìm x,y biết
x^3+y^3/6=x^3-2y^3 và x^6 .y^6
P=(1+x/y)(1+y/z)(1+x/z).biếtx+y+z=0,x,y,z khác 0
Xem chi tiết
Ta có: \(x+y+z=0\)
nên \(\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\)
Ta có: \(P=\left(1+\dfrac{x}{y}\right)\left(1+\dfrac{y}{z}\right)\left(1+\dfrac{z}{x}\right)\)
\(=\dfrac{x+y}{y}\cdot\dfrac{y+z}{z}\cdot\dfrac{x+z}{x}\)
\(=\dfrac{-z}{y}\cdot\dfrac{-x}{z}\cdot\dfrac{-y}{x}\)
\(=\dfrac{-\left(x\cdot y\cdot z\right)}{x\cdot y\cdot z}=-1\)
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Tìm các cặp số nguyên (x,y) biết
x/3-1/y+1=1/6
\(\dfrac{x}{3}-\dfrac{1}{y+1}=\dfrac{1}{6}\)
=>\(\dfrac{xy+x-3}{3\left(y+1\right)}=\dfrac{1}{6}\)
=>\(2\left(xy+x-3\right)=1\)
=>2xy+2x-6=1
=>2xy+2x=7
=>2x(y+1)=7
=>x(y+1)=7/2
mà x,y nguyên
nên \(\left(x,y\right)\in\varnothing\)
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1/y+1=x/3-1/6
1/y+1=2x/6-1/6
1/y+1= 2x-1/6
=> 1.6=(y+1).(2x-1)
ta có bảng
y+1 6 1
Past lives couldn't ever hold me down Lost love is sweeter when it's finally found I've got the strangest feeling This isn't our first time around Past lives couldn't ever come between us Sometimes the dreamers finally wake up Don't wake me I'm not dreaming Don't wake me I'm not dreaming All my past lives they got nothing on me Golden eagle you're the one and only flying high Through the cities in the sky I'll take you way back, countless centuries Don't you remember that you were meant to be My Queen of Hearts, meant to be my love Through all of my lives I'd never thought I'd wait so long for you The timing is right The stars are aligned So save that heart for me 'Cause girl you know that you're my destiny (d-destiny) Swear to the moon, the stars, the sons and the daughters Our love is deeper than the oceans of water I need you now, I've waited oh so long (Gimme love) I need you now, I've waited oh so long Passing seasons, empty bottles of wine My ancient kingdom came crashing down without you Baby child, I'm lost without your love Diamond sparrow, my moonlit majesty You know I need you, come flying back to me Through all of my lives I'd never thought I'd wait so long for you The timing is right The stars are aligned So save that heart for me 'Cause girl you know that you're my destiny (d-destiny) Respect to the moon, the stars, their sons and their daughters Our love is deeper than the oceans of water Save that heart for me And girl I'll give you everything you need (everything you need) Here's to our past lives, our mothers and fathers Our love is deeper than the oceans of water I need you now, I've waited oh so long, yeah (Gimme love) I need you now, I've waited oh so long I need you now, I've waited oh so long, yeah (Gimme love) I need you now, I've waited oh so long So save that heart for me 'Cause girl you know that you're my destiny (d-destiny) Respect to the moon, the stars, their sons and their daughters Our love is deeper than the oceans of water Save that heart for me And girl I'll give you everything you need (everything you need) Here's to our past lives, our mothers and fathers Our love is deeper than the oceans of wat2x-1 1 6 ...
y 5 0
x 1 7/2
loại
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tìm 2 số x và y biết
x phần 3 = x cộng y = 20
\(\dfrac{x}{3}=x+y=20\Rightarrow x=60\Rightarrow60+y=20\Rightarrow y=-40\)
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Ta có:
\(\dfrac{x}{3}=20\)
\(\Rightarrow\)\(x=60\)
Lại có:
\(x+y=20\)
\(\Rightarrow\)\(y=20-60\)
\(\Rightarrow\)\(y=-40\)
Vây x = 60 và y = - 40
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\(\dfrac{x}{3}=x+y\)
\(\Leftrightarrow x-\dfrac{1}{3}x=-y\)
\(\Leftrightarrow y=-\dfrac{2}{3}x\)
Ta có: x+y=20
\(\Leftrightarrow x\cdot\dfrac{1}{3}=20\)
hay x=60
=> y=40
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Bài 1 : Tìm x ,y,z biết:
a, 3/x-1 = 4/y-2 = 5/z-3 và x+y+z = 18
b, 3/x-1 = 4/y-2 = 5/z-3 và x.y.z = 192
Bài 2 : Tìm x,y,z biết : x^3+y^3/6 = x^3-2y^3/4 và x^6.y^6 = 64
Bài 3 : Tìm x,y,z biết :x+4/6 = 3y-1/8 = 3y-x-5/x
Bài 4 :Tìm x,y,z biết : x+y+2005/z = y+z-2006 = z+x+1/y = 2/x+y+z
bài 1 : a,ta có 3/x-1 =4/y-2=5/z-3 => x-1/3=y-2/4=z-3/5
áp dụng .... => x-1+y-2+z-3 / 3+4+5 = x+y+z-1-2-3/3+4+5 = 12/12=1
do x-1/3 = 1 => x-1 = 3 => x= 4 ( tìm y,z tương tự
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Bài 1:
a) Ta có: 3/x - 1 = 4/y - 2 = 5/z - 3 => x - 1/3 = y - 2/4 = z - 3/5 áp dụng ... =>x - 1 + y - 2 + z - 3/3 + 4 + 5 = x + y + z - 1 - 2 - 3/3 + 4 + 5 = 12/12 = 1 do x - 1/3 = 1 => x - 1 = 3 => x = 4 ( tìm y, z tương tự )
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Tính A=x(x+2)+y(y-2)-2xy+37, biếtx-y=7
Tính A=x^3+y^3 biết x+y=2 và x^2+y^2=10
Tính gt biểu thức P=(1+x/y)(1+y/z)(1+x/z),cho x+y+z=0,x , y , z khác 0
Xem chi tiết
a) Ta có: \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)
\(=x^2+2x+y^2-2y-2xy+37\)
\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)
\(=\left(x-y\right)^2+2\left(x-y\right)+37\)
\(=\left(x-y\right)\left(x-y+2\right)+37\)(1)
Thay x-y=7 vào biểu thức (1), ta được:
\(A=7\cdot\left(7+2\right)+37=7\cdot9+37=100\)
Vậy: Khi x-y=7 thì A=100
b) Ta có: \(x+y=2\)
\(\Leftrightarrow\left(x+y\right)^2=4\)
\(\Leftrightarrow x^2+y^2+2xy=4\)
\(\Leftrightarrow2xy+10=4\)
\(\Leftrightarrow2xy=-6\)
\(\Leftrightarrow xy=-3\)
Ta có: \(A=x^3+y^3\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)\)(2)
Thay x+y=2; \(x^2+y^2=10\) và xy=-3 vào biểu thức (2), ta được:
\(A=2\cdot\left(10+3\right)=2\cdot13=26\)
Vậy: Khi x+y=2 và \(x^2+y^2=10\) thì A=26
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\(\Rightarrow A=x^2+2x+y^2-2y-2xy+37=x^2-2xy+y^2+2\left(x-y\right)+37=\left(x-y\right)^2+2\left(x-y\right)+37=7^2+2\cdot7+37=100\)
\(\Rightarrow A=x^3+y^3=\left(x+y\right)\left(x^2+y^2-xy\right)=\left(x+y\right)\left[x^2+y^2-\dfrac{\left(x+y\right)^2-\left(x^2+y^2\right)}{2}\right]=2\cdot\left[10+3\right]=2\cdot13=26\) \(\Rightarrow\left\{{}\begin{matrix}x+y=-z\\x+z=-y\\y+z=-x\end{matrix}\right.\) \(\Rightarrow P=\left(\dfrac{x+y}{y}\right)\left(\dfrac{y+z}{z}\right)\left(\dfrac{x+z}{x}\right)=-\dfrac{z}{y}\cdot\dfrac{-x}{z}\cdot-\dfrac{y}{x}=-1\)
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