Thu gọn: \(\left(x-y+z\right)^2-\left(2x-y+1\right)^2-\left(y-z+2\right)^2+\left(2x-1\right)^2-3\left(2y-3z\right)^2\)
Những câu hỏi liên quan
Thu gọn: \(\left(x-y+z\right)^2-\left(2x-y+1\right)^2-\left(y-z+2\right)^2+\left(2x-1\right)^2-3\left(2y-3z\right)^2\)
hept{begin{cases}3x^2+2y+12zleft(x+2right)3y^2+2z+12xleft(y+2right)3z^2+2x+12yleft(z+2right)end{cases}Leftrightarrowhept{begin{cases}3x^2+2y+12xz+4z3y^2+2z+12xy+4x3z^2+2x+12yz+4yend{cases}}}Cộng 3 vế vào rồi chuyển vế ta được2x^2+2y^2+2z^2-2xy-2yz-2zx+left(x^2+2x+1right)+left(y^2+2y+1right)+left(z^2+2z+1right)0Leftrightarrowleft(x-yright)^2+left(y-zright)^2
+left(z-xright)^2+left(x+1right)^2+left(y+1right)^2+left(z+1right)^20Dễ thấy VP 0Dấu khi x y z -1
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\(\hept{\begin{cases}3x^2+2y+1=2z\left(x+2\right)\\3y^2+2z+1=2x\left(y+2\right)\\3z^2+2x+1=2y\left(z+2\right)\end{cases}\Leftrightarrow\hept{\begin{cases}3x^2+2y+1=2xz+4z\\3y^2+2z+1=2xy+4x\\3z^2+2x+1=2yz+4y\end{cases}}}\)
Cộng 3 vế vào rồi chuyển vế ta được
\(2x^2+2y^2+2z^2-2xy-2yz-2zx+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)+\left(z^2+2z+1\right)=0\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2 +\left(z-x\right)^2+\left(x+1\right)^2+\left(y+1\right)^2+\left(z+1\right)^2=0\)
Dễ thấy VP > 0
Dấu "=" khi x = y = z = -1
Thu gọn biểu thức :
1, \(\left(2x-y\right)^2+2\cdot\left(2x-y\right)\cdot\left(y-x\right)+\left(x-y\right)^2\)
2, \(\left(x-y+z\right)^2+2\cdot\left(x-y+z\right)\cdot\left(y-z\right)+\left(y-z\right)^2\)
1, đa thức đã cho \(\Leftrightarrow\left(2x-y\right)^2-2\left(2x-y\right)\left(x-y\right)+\left(x-y\right)^2=\left[\left(2x-y\right)-\left(x-y\right)\right]^2=\left(2x-y-x+y\right)^2=x^2\)
2, đa thức đã cho \(\Leftrightarrow\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2=\left[\left(x-y+z\right)+\left(y-z\right)\right]^2=\left(x-y+z+y-z\right)^2=x^2\)
--- giải chi tiết lắm rồi đó---
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a, \(\left(2x-y\right)^2+2\left(2x-y\right)\left(y-x\right)+\left(x-y\right)^2\)
\(=4x^2-4xy+y^2+2\left(2xy-2x^2-y^2+xy\right)+x^2-2xy+y^2\)
\(=4x^2-4xy+y^2+4xy-4x^2-2y^2+2xy+x^2-2xy+y^2\)
\(=x^2\)
b, \(\left(x-y+z\right)^2+2\left(x-y+z\right)\left(y-z\right)+\left(y-z\right)^2\)
\(=\left(x-y+z\right)\left[1+2\left(y-z\right)\right]+y^2-2yz+z^2\)
\(=\left(x-y+z\right)\left(1+2y-2z\right)+y^2-2yz+z^2\)
\(=x+2xy-2xz-y-2y^2+2yz+z+2yz-2z^2+y^2-2yz+z^2\)
\(=x-y+z+2xy-2xz+2yz-y^2-z^2\)
Chúc bạn học tốt!!!
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thực hiện phép tính
a,x^3+left[frac{xleft(2y^3-x^3right)}{x^3+y^3}right]^3-left[frac{yleft(2x^3-y^3right)}{x^3+y^3}right]^3
b,frac{frac{xleft(x+yright)}{x-y}+frac{xleft(x+zright)}{x-z}}{1+frac{left(y-zright)^2}{left(x-yright)left(x-zright)}}+frac{frac{yleft(y+zright)}{y-z}+frac{yleft(y+xright)}{y-x}}{1+frac{left(z-xright)^2}{left(y-zright)left(y-xright)}}+frac{frac{zleft(z+xright)}{z-x}+frac{zleft(z+yright)}{z-y}}{1+frac{left(x-yright)^2}{left(z-xright)left(z-yright)}}
c,left[frac{y+z-2x}{fra...
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thực hiện phép tính
a,\(x^3+\left[\frac{x\left(2y^3-x^3\right)}{x^3+y^3}\right]^3-\left[\frac{y\left(2x^3-y^3\right)}{x^3+y^3}\right]^3\)
b,\(\frac{\frac{x\left(x+y\right)}{x-y}+\frac{x\left(x+z\right)}{x-z}}{1+\frac{\left(y-z\right)^2}{\left(x-y\right)\left(x-z\right)}}+\frac{\frac{y\left(y+z\right)}{y-z}+\frac{y\left(y+x\right)}{y-x}}{1+\frac{\left(z-x\right)^2}{\left(y-z\right)\left(y-x\right)}}+\frac{\frac{z\left(z+x\right)}{z-x}+\frac{z\left(z+y\right)}{z-y}}{1+\frac{\left(x-y\right)^2}{\left(z-x\right)\left(z-y\right)}}\)
c,\(\left[\frac{y+z-2x}{\frac{\left(y-z\right)^3}{y^3-z^3}+\frac{\left(x-y\right)\left(x-z\right)}{y^2+yz+z^2}}+\frac{z+x-2y}{\frac{\left(z-x\right)^3}{z^3-x^3}+\frac{\left(y-z\right)\left(y-x\right)}{z^2+xz+x^2}}+\frac{x+y-2z}{\frac{\left(x-y\right)^3}{x^3-y^3}+\frac{\left(z-x\right)\left(z-y\right)}{x^2+xy+y^2}}\right]:\frac{1}{x+y+z}\)
Thu gọn biểu thức :
1, \(\left(x-y-z\right)^2-\left(y+z\right)^2\)
2, \(\left(2x+y\right)^2-4x\cdot\left(2x+y\right)+4x^2\)
3, \(\left(x+y\right)^2-2\cdot\left(x^2-y^2\right)+\left(x-y\right)^2\)
1) \(\left(x-y-z\right)^2-\left(y+z\right)^2=\left(x\right).\left(x-2y-2z\right)=x^2-2yx-2zx\) 2) \(\left(2x+y\right)^2-4x\left(2x+y\right)+4x^2\Leftrightarrow\left(2x+y\right)\left(2x+y-4x\right)+4x^2\)
\(=\left(2x+y\right)\left(y-2x\right)+4x^2=\left(y^2-4x^2\right)+4x^2=y^2-4x^2+4x^2=y^2\)
3) \(\left(x+y\right)^2-2\left(x^2-y^2\right)+\left(x-y\right)^2\)
\(=x^2+2xy+y^2-2x^2+2y^2+x^2-2xy+y^2\)
\(=4y^2=\left(2y\right)^2\)
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Cho x , y , z
\(\left(x-y\right)^2\)+\(\left(y-z\right)^2\)+\(\left(z-x\right)^2\)=\(\left(x+y-2z\right)^2\)+\(\left(y+z-2x\right)^2+\left(z+x-2y\right)^2\)
cmr: x=y=z
Phân tích vế trái ta được: 2(x2 + y2 + z2 − (xy + yz + zx)
Phân tích vế phải ta được: 6(x2 + y2 + z2 − (xy + yz + zx)
Vì VT = VP nên VP - VT=0
→ 4(x2 + y2 + z2 − (xy + yz + zx)) = 0
→2(2 (x2 + y2 + z2 − (xy + yz + zx))) = 0
→2((x − y)2 + (y − z)2 + (z − x)2) = 0
→(x − y)2 + (y − z)2 + (z − x)2 = 0
→(x − y)2 = 0; (y − z)2 = 0; (z − x)2 = 0
→x = y = z
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\(\left\{{}\begin{matrix}3x^2+2y+4=2z\left(x+3\right)\\3y^2+2z+4=2x\left(y+3\right)\\3z^2+2x+4=2y\left(z+3\right)\end{matrix}\right.\)
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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Xem thêm câu trả lời
cho x,y,z là các số thực dương thỏa mãn x+y+z=xyz.CMR
\(\dfrac{x}{1+x^2}+\dfrac{2y}{1+y^2}+\dfrac{3z}{1+z^2}=\dfrac{xyz\left(5x+4y+3z\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)