Tìm x,y \(\in\) Z biết :
a) 7(x - 2004)2 = 23 - y2
b) (1 + 2x).(2 - y) = -12
c) (1 + 3x).(1 + y) = 6
d) \(\left|x+5\right|+2.\left|3-y\right|=0\)
e) (x - 2)2 + 2005(1 + y)4 = 0
g) (2x2 +1).(2y + 1) = -27
Những câu hỏi liên quan
Bài 1:
a) a)left(x^2+yright)left(y^2+xright)left(x-yright)^2 x,yin Z
b)x^2left(y+3right)yz^2 x,y,zin Z_+
c)xleft(x+1right)left(x+7right)left(x+8right)y^2 x,yin Z_+
d)x^4+x^2-y^2+y+100 x,yin Z
e)x^3-y^3-2y^2-3y-10 x,yin Z_+
f)x^4-y^4+z^4+2x^2y^2+3x^2+4z^2+10 x,y,zin Z
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Bài 1:
a) \(a)\left(x^2+y\right)\left(y^2+x\right)=\left(x-y\right)^2\) \(x,y\in Z\)
\(b)x^2\left(y+3\right)=yz^2\) \(x,y,z\in Z_+\)
\(c)x\left(x+1\right)\left(x+7\right)\left(x+8\right)=y^2\) \(x,y\in Z_+\)
\(d)x^4+x^2-y^2+y+10=0\) \(x,y\in Z\)
\(e)x^3-y^3-2y^2-3y-1=0\) \(x,y\in Z_+\)
\(f)x^4-y^4+z^4+2x^2y^2+3x^2+4z^2+1=0\) \(x,y,z\in Z\)
tìm x và y
a) \(\left(x-1\right)^2+\left(y+3\right)^2=0\)
b) \(2\left(x-5\right)^4+5\left|2y-7\right|^5=0\)
c) \(3\left(x-2y\right)^{2004}+4\left|y+\frac{1}{2}\right|=0\)
d) \(\left|x+3y-1\right|+\left(2y-\frac{1}{2}\right)^{2000}=0\)
a. x=1 y= -3
b. x=5 y=7/2
c. x= -1 y= -1/2
d. x=1/4 y= 1/4
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a) x = 1
y = -3
b) x = 5
y = 7/2
c) x = -1
y = -1/2
d) x = 1/4
y = 1/4
nha bn
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Tìm x, y, z:
a) left|x+dfrac{19}{5}right|+left|y+dfrac{1890}{1975}right|+left|z-2005right|0
b) left|x+dfrac{3}{4}right|+left|y-dfrac{1}{5}right|+left|x+y+zright| 0
c) dfrac{16}{2^x}1
d) left(2x-1right)^3-27
e) left(x-2right)^21
f) left(x+dfrac{1}{2}right)^2dfrac{4}{25}
g) left(x-1right)^2left(x-1right)^6
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Tìm x, y, z:
a) \(\left|x+\dfrac{19}{5}\right|+\left|y+\dfrac{1890}{1975}\right|+\left|z-2005\right|=0\)
b) \(\left|x+\dfrac{3}{4}\right|+\left|y-\dfrac{1}{5}\right|+\left|x+y+z\right|\) = 0
c) \(\dfrac{16}{2^x}=1\)
d) \(\left(2x-1\right)^3=-27\)
e) \(\left(x-2\right)^2=1\)
f) \(\left(x+\dfrac{1}{2}\right)^2=\dfrac{4}{25}\)
g) \(\left(x-1\right)^2=\left(x-1\right)^6\)
a)
Ta thấy \(\left\{\begin{matrix} |x+\frac{19}{5}|\geq 0\\ |y+\frac{1890}{1975}|\geq 0\\ |z-2005|\geq 0\end{matrix}\right., \forall x,y,z\in\mathbb{Z}\)
\(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|\geq 0\)
Do đó, để \(|x+\frac{19}{5}|+|y+\frac{1890}{1975}|+|z-2005|=0\) thì :
\(\left\{\begin{matrix} |x+\frac{19}{5}|= 0\\ |y+\frac{1890}{1975}|= 0\\ |z-2005|=0\end{matrix}\right.\Rightarrow x=\frac{-19}{5}; y=\frac{-1890}{1975}; z=2005\)
b) Giống phần a, vì trị tuyệt đối của một số luôn không âm nên để tổng các trị tuyệt đối bằng $0$ thì:
\(\left\{\begin{matrix} |x+\frac{3}{4}|=0\\ |y-\frac{1}{5}|=0\\ |x+y+z|=0\end{matrix}\right.\Rightarrow \left\{\begin{matrix} x=-\frac{3}{4}\\ y=\frac{1}{5}\\ z=-(x+y)=\frac{11}{20}\end{matrix}\right.\)
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c) \(\frac{16}{2^x}=1\Rightarrow 16=2^x\)
\(\Leftrightarrow 2^4=2^x\Rightarrow x=4\)
d) \((2x-1)^3=-27=(-3)^3\)
\(\Rightarrow 2x-1=-3\)
\(\Rightarrow 2x=-2\Rightarrow x=-1\)
e) \((x-2)^2=1=1^2=(-1)^2\)
\(\Rightarrow \left[\begin{matrix} x-2=1\\ x-2=-1\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=3\\ x=1\end{matrix}\right.\)
f) \((x+\frac{1}{2})^2=\frac{4}{25}=(\frac{2}{5})^2=(\frac{-2}{5})^2\)
\(\Rightarrow \left[\begin{matrix} x+\frac{1}{2}=\frac{2}{5}\\ x+\frac{1}{2}=-\frac{2}{5}\end{matrix}\right.\Rightarrow \left[\begin{matrix} x=\frac{-1}{10}\\ x=\frac{-9}{10}\end{matrix}\right.\)
g) \((x-1)^2=(x-1)^6\)
\(\Leftrightarrow (x-1)^6-(x-1)^2=0\)
\(\Leftrightarrow (x-1)^2[(x-1)^4-1]=0\)
\(\Rightarrow \left[\begin{matrix} (x-1)^2=0\\ (x-1)^4=1=(-1)^4=1^4\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x-1=-1\\ x-1=1\end{matrix}\right.\end{matrix}\right.\)
\(\Rightarrow \left[\begin{matrix} x=1\\ \left[\begin{matrix} x=0\\ x=2\end{matrix}\right.\end{matrix}\right.\)
Vậy \(x=\left\{0;1;2\right\}\)
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a \(\left(x-1\right)^2-\left(y+1\right)^2=0\)
\(x+3y-5=0\)
b \(xy-2x-y+2=0\)
3x+y=8
c \(\left(x+y\right)^2-4\left(x+y\right)=12\)
\(\left(x-y\right)^2-2\left(x-y\right)=3\)
d \(2x-y=1\)
\(2x^2+xy-y^2-3y=-1\)
a.
\(\left\{{}\begin{matrix}\left(x-1\right)^2-\left(y+1\right)^2=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1-y-1\right)\left(x-1+y+1\right)=0\\x+3y-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y-2\right)\left(x+y\right)=0\\x+3y-5=0\end{matrix}\right.\)
TH1: \(\left\{{}\begin{matrix}x-y-2=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{11}{4}\\y=\dfrac{3}{4}\end{matrix}\right.\)
TH2: \(\left\{{}\begin{matrix}x+y=0\\x+3y-5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-\dfrac{5}{2}\\y=\dfrac{5}{2}\end{matrix}\right.\)
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b.
\(\left\{{}\begin{matrix}xy-2x-y+2=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(y-2\right)-\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)\left(y-2\right)=0\\3x+y=8\end{matrix}\right.\)
TH1:
\(\left\{{}\begin{matrix}x-1=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=5\end{matrix}\right.\)
TH2:
\(\left\{{}\begin{matrix}y-2=0\\3x+y=8\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
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c.
\(\left\{{}\begin{matrix}\left(x+y\right)^2-4\left(x+y\right)-12=0\\\left(x-y\right)^2-2\left(x-y\right)=3\end{matrix}\right.\)
Xét pt:
\(\left(x+y\right)^2-4\left(x+y\right)-12=0\)
\(\Leftrightarrow\left(x+y+2\right)\left(x+y-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+y+2=0\\x+y-6=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}y=-x-2\\y=6-x\end{matrix}\right.\)
TH1: \(y=-x-2\) thế vào \(\left(x-y\right)^2-2\left(x-y\right)=3\)
\(\Rightarrow\left(2x+2\right)^2-2\left(2x+2\right)=3\)
\(\Leftrightarrow4x^2+4x-3=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\Rightarrow y=-\dfrac{5}{2}\\x=-\dfrac{3}{2}\Rightarrow y=-\dfrac{1}{2}\end{matrix}\right.\)
TH2: \(y=6-x\) thế vào...
\(\left(2x-6\right)^2-2\left(2x-6\right)=3\)
\(\Leftrightarrow4x^2-28x+45=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\Rightarrow y=\dfrac{7}{2}\\y=\dfrac{9}{2}\Rightarrow y=\dfrac{3}{2}\end{matrix}\right.\)
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Bài 2: Tìm x, y biết :
a) \(\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)
b) \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2=0\)
Ta có : \(\left(3x-\frac{y}{5}\right)^2\ge0;\left(2y+\frac{3}{7}\right)^2\ge0\)
\(=>\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2\ge0\)
Mà \(\left(3x-\frac{y}{5}\right)^2+\left(2y+\frac{3}{7}\right)^2=0\)nên dấu "=" xảy ra
\(< =>\hept{\begin{cases}3x-\frac{y}{5}=0\\2y+\frac{3}{7}=0\end{cases}}< =>\hept{\begin{cases}3x-\frac{y}{5}=0\\y=-\frac{3}{14}\end{cases}}\)
\(< =>\hept{\begin{cases}x=-\frac{1}{70}\\y=-\frac{3}{14}\end{cases}}\)
Ta có : \(\left(x+y-\frac{1}{4}\right)^2\ge0;\left(x-y+\frac{1}{5}\right)^2\ge0\)
Cộng theo vế ta được : \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2\ge0\)
Mà \(\left(x+y-\frac{1}{4}\right)^2+\left(x-y+\frac{1}{5}\right)^2=0\)nên dấu "=" xảy ra
\(< =>\hept{\begin{cases}y+x=\frac{1}{4}\\y-x=\frac{1}{5}\end{cases}}< =>\hept{\begin{cases}y=\frac{9}{40}\\x=\frac{1}{40}\end{cases}}\)
Giải hệa) left{{}begin{matrix}x^2left(y^2+1right)+2yleft(x^2+x+1right)3left(x^2+xright)left(y^2+yright)1end{matrix}right.b) left{{}begin{matrix}left(6x+5right)sqrt{2x+1}-2y-3y^30y+sqrt{x}sqrt{2x^2+4x-23}end{matrix}right.Giải bất ptdfrac{9}{left|x-5right|-3}geleft|x-2right|
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Giải hệ
a) \(\left\{{}\begin{matrix}x^2\left(y^2+1\right)+2y\left(x^2+x+1\right)=3\\\left(x^2+x\right)\left(y^2+y\right)=1\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\left(6x+5\right)\sqrt{2x+1}-2y-3y^3=0\\y+\sqrt{x}=\sqrt{2x^2+4x-23}\end{matrix}\right.\)
Giải bất pt
\(\dfrac{9}{\left|x-5\right|-3}\ge\left|x-2\right|\)
Tìm Min, Max :
a)A x + y + 1 biết x^2+2xy+3left(x+yright)+2y^2+20
b)B x + y + 1 biết x^2+2xy+7left(x+yright)+2y^2+100
c)C x^2+y^2 biết x^2left(x^2+y^2-3right)+left(y^2-4right)^21
d)D x + y biết x^2+2y^2+2xy+3x+3y-40
e)E x^2+y^2 biết left(x^2-y^2+1right)^2+4x^2y^2-x^2-y^20
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Tìm Min, Max :
a)A = x + y + 1 biết \(x^2+2xy+3\left(x+y\right)+2y^2+2=0\)
b)B = x + y + 1 biết \(x^2+2xy+7\left(x+y\right)+2y^2+10=0\)
c)C = \(x^2+y^2\) biết \(x^2\left(x^2+y^2-3\right)+\left(y^2-4\right)^2=1\)
d)D = x + y biết \(x^2+2y^2+2xy+3x+3y-4=0\)
e)E = \(x^2+y^2\) biết \(\left(x^2-y^2+1\right)^2+4x^2y^2-x^2-y^2=0\)
Giải hệ phương trình:
a)left{{}begin{matrix}dfrac{3x+2}{x-1}-dfrac{3y-1}{y+2}0dfrac{2}{x-1}+dfrac{3}{y+2}1end{matrix}right.
b)left{{}begin{matrix}dfrac{4x-5}{x+1}+dfrac{2y-3}{y-5}8dfrac{3}{x+1}-dfrac{2}{y-5}-1end{matrix}right.
c)left{{}begin{matrix}dfrac{x+y-2}{x+1}+dfrac{3-x}{y+1}dfrac{5}{4}dfrac{3left(x+y-2right)}{x+1}-dfrac{5-x+2y}{y+1}dfrac{3}{4}end{matrix}right.
d)left{{}begin{matrix}dfrac{x-y+1}{x-3}+dfrac{x+1}{y-3}dfrac{-7}{2}dfrac{2left(x-y+1right)}{x-3}-dfrac{x+y-2}{y-3}-dfrac{9}{2}...
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Giải hệ phương trình:
a)\(\left\{{}\begin{matrix}\dfrac{3x+2}{x-1}-\dfrac{3y-1}{y+2}=0\\\dfrac{2}{x-1}+\dfrac{3}{y+2}=1\end{matrix}\right.\)
b)\(\left\{{}\begin{matrix}\dfrac{4x-5}{x+1}+\dfrac{2y-3}{y-5}=8\\\dfrac{3}{x+1}-\dfrac{2}{y-5}=-1\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}\dfrac{x+y-2}{x+1}+\dfrac{3-x}{y+1}=\dfrac{5}{4}\\\dfrac{3\left(x+y-2\right)}{x+1}-\dfrac{5-x+2y}{y+1}=\dfrac{3}{4}\end{matrix}\right.\)
d)\(\left\{{}\begin{matrix}\dfrac{x-y+1}{x-3}+\dfrac{x+1}{y-3}=\dfrac{-7}{2}\\\dfrac{2\left(x-y+1\right)}{x-3}-\dfrac{x+y-2}{y-3}=-\dfrac{9}{2}\end{matrix}\right.\)
e)\(\left\{{}\begin{matrix}x^2-y^2+2y=1\\\left(x+y\right)^2-2x-2y=0\end{matrix}\right.\)
f)\(\left\{{}\begin{matrix}4x^2+y^2-4xy=4\\x^2+y^2-2\left(xy+8\right)=0\end{matrix}\right.\)
28 tháng 10 2021 lúc 15:39
1.Cho x+y+z0. CMR:a) 5left(x^3+y^3+z^3right)left(x^2+y^2+z^2right)6left(x^5+y^5+z^5right)b) x^7+y^7+z^77xyzleft(x^2y^2+y^2z^2+z^2x^2right)c) 10left(x^7+y^7+z^7right)7left(x^2+y^2+z^2right)left(x^5+y^5+z^5right)d) 2left(x^5+y^5+z^5right)5xyzleft(x^2+y^2+z^2right)2. Tìm n∈ N để biểu thức sau là số nguyên tố a) An^3-4n^2-4n-1b) Bn^3-6n^2+9n-2c) Cn^{1975}+n^{1973}+1
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1.Cho x+y+z=0. CMR:
a) \(5\left(x^3+y^3+z^3\right)\left(x^2+y^2+z^2\right)=6\left(x^5+y^5+z^5\right)\)
b) \(x^7+y^7+z^7=7xyz\left(x^2y^2+y^2z^2+z^2x^2\right)\)
c) \(10\left(x^7+y^7+z^7\right)=7\left(x^2+y^2+z^2\right)\left(x^5+y^5+z^5\right)\)
d) \(2\left(x^5+y^5+z^5\right)=5xyz\left(x^2+y^2+z^2\right)\)
2. Tìm n∈ N để biểu thức sau là số nguyên tố
a) \(A=n^3-4n^2-4n-1\)
b) \(B=n^3-6n^2+9n-2\)
c) \(C=n^{1975}+n^{1973}+1\)
Vì bài dài nên mình sẽ tách ra nhé.
1a. Ta có:
$x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)=-2(xy+yz+xz)$
$x^3+y^3+z^3=(x+y+z)^3-3(x+y)(y+z)(x+z)=-3(x+y)(y+z)(x+z)$
$=-3(-z)(-x)(-y)=3xyz$
$\Rightarrow \text{VT}=-30xyz(xy+yz+xz)(1)$
------------------------
$x^5+y^5=(x^2+y^2)(x^3+y^3)-x^2y^2(x+y)$
$=[(x+y)^2-2xy][(x+y)^3-3xy(x+y)]-x^2y^2(x+y)$
$=(z^2-2xy)(-z^3+3xyz)+x^2y^2z$
$=-z^5+3xyz^3+2xyz^3-6x^2y^2z+x^2y^2z$
$=-z^5+5xyz^3-5x^2y^2z$
$\Rightarrow 6(x^5+y^5+z^5)=6(5xyz^3-5x^2y^2z)$
$=30xyz(z^2-xy)=30xyz[z(-x-y)-xy]=-30xyz(xy+yz+xz)(2)$
Từ $(1);(2)$ ta có đpcm.
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1b.
$x^4+y^4=(x^2+y^2)^2-2x^2y^2=[(x+y)^2-2xy]^2-2x^2y^2$
$=(z^2-2xy)^2-2x^2y^2=z^4+2x^2y^2-4xyz^2$
$x^3+y^3=(x+y)^3-3xy(x+y)=-z^3+3xyz$
Do đó:
$x^7+y^7=(x^4+y^4)(x^3+y^3)-x^3y^3(x+y)$
$=(z^4+2x^2y^2-4xyz^2)(-z^3+3xyz)+x^3y^3z$
$=7x^3y^3z-14x^2y^2z^3+7xyz^5-z^7$
$\Rightarrow \text{VT}=7x^3y^3z-14x^2y^2z^3+7xyz^5$
$=7xyz(x^2y^2-2xyz^2+z^4)$
$=7xyz(xy-z^2)$
$=7xyz[xy+z(x+y)]^2=7xyz(xy+yz+xz)^2$
$=7xyz[x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)]$
$=7xyz(x^2y^2+y^2z^2+z^2x^2)$ (đpcm)
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1c. Sử dụng kq phần a,b:
\(10(x^7+y^7+z^7)=70xyz(xy+yz+xz)^2\)
\(=-35xyz(xy+yz+xz).-2(xy+yz+xz)=-35xyz(x+y+z)(x^2+y^2+z^2)\)
\(=\frac{7}{6}.-30xyz(xy+yz+xz)(x^2+y^2+z^2)=\frac{7}{6}.6(x^5+y^5+z^5).(x^2+y^2+z^2)\)
\(=7(x^5+y^5+z^5)(x^2+y^2+z^5)\)
(đpcm)
1d. Áp dụng kq phần a
$6(x^5+y^5+z^5)=-30xyz(xy+y+xz)=15xyz.-2(xy+yz+xz)=15xyz(x^2+y^2+z^2)$
$\Rightarrow 2(x^5+y^5+z^5)=5xyz(x^2+y^2+z^2)$ (đpcm)
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