C/m:
\(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)
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C/m:
\(x^4+y^4+\left(x+y\right)^4=2\left(x^2+xy+y^2\right)^2\)
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379+920=
Sử dụng khai triển nhị thức Newton:
\(\left(x+y\right)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4\)
\(\Rightarrow x^4+y^4+\left(x+y\right)^4=2x^4+4x^3y+6x^2y^2+4xy^3+2x^4=2\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)\)
Khai triển:\(2\left(x^2+xy+y^2\right)^2=2\left(x^4+x^2y^2+y^4+2x^3y+2xy^3+2x^2y^2\right)=2\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)\)
\(\Rightarrowđpcm\)
Giải hệ phương trình
left{{}begin{matrix}x^2+y^2+xy13x^4+y^4+x^2y^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^2-xy13left(x^2+y^2right)^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^213+xyleft[left(x+yright)^2-2xyright]^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}left(x+yright)^2-xy13left(13-xyright)^2-left(xyright)^291end{matrix}right.
Leftrightarrowleft{{}begin{matrix}xy3left(x+yright)^216end{matri...
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Giải hệ phương trình
\(\left\{{}\begin{matrix}x^2+y^2+xy=13\\x^4+y^4+x^2y^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(x^2+y^2\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2=13+xy\\\left[\left(x+y\right)^2-2xy\right]^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-xy=13\\\left(13-xy\right)^2-\left(xy\right)^2=91\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xy=3\\\left(x+y\right)^2=16\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\) hoặc x+y = -4
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x+y=4\\xy=3\end{matrix}\right.\\\left\{{}\begin{matrix}x+y=-4\\xy=3\end{matrix}\right.\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\\\left\{{}\begin{matrix}x=-3\\y=-1\end{matrix}\right.\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=3\\y=1\end{matrix}\right.\)hoặc \(\left\{{}\begin{matrix}x=-1\\y=-3\end{matrix}\right.\)
Mọi người có thể giải thích từ dấu tương đương thứ 3 xuống 4. tại sao lại như vậy k?
Bài 2. (1 điểm) Thu gọn các đa thức sau:
a) $\left( 4{{x}^{4}}-8{{x}^{2}}{{y}^{2}}+12{{x}^{5}}y \right) \, : \, \left( -4{{x}^{2}} \right);$
b) ${{x}^{2}}\left( x-{{y}^{2}} \right)-xy\left( 1-xy \right)-{{x}^{3}}.$
a) \(\left(4x^4-8x^2y^2+12x^5y\right):\left(-4x^2\right)\)
\(=4x^4:-4x^2-8x^2y^2:-4x^2+12x^4y:-4x^2\)
\(=-x^2+2y^2-3x^2y\)
b) \(x^2\left(x-y^2\right)-xy\left(1-xy\right)-x^3\)
\(=x^3-x^2y^2-xy+x^2y^2-x^3\)
\(=-xy\)
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a)(4x4 - 8x2y2 +12x5y): ( -4x2)
=-x2+2y2-3x3y
b)
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a) ( 4x4-8x2y2+12x5y ) : (-4x2)
= 4x4 : (-4x2) - 8x2y2 : (-4x2) + 12x5y : (-4x2)
= -x2 + 2y2 - 3x3y
b) x2 (x-y2) - xy(1-xy) - x3
= x3 - x2y2 - xy + x2y2 - x3
= -xy
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1.tìm x
a) left(8-5xright)left(x+2right)+4left(x-2right)left(x+1right)+2left(x-2right)left(x+2right)
b) 4left(x-1right)left(x+5right)-left(x+2right)left(x+5right)3left(x-1right)left(x+2right)
2. CMR
a) left(x-yright)left(x^4+x^3y+x^2y^2+xy^3+y^4right)x^5-y^5
b)left(x+yright)left(x^4-x^3y+x^2y^2-xy^3+y^4right)x^5+y^5
c)left(x+aright)left(x+bright)x^2+left(a+bright)x+ab
giúp mik nha
chiều nay nộp r
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1.tìm x
a) \(\left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)\)
b) \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
2. CMR
a) \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
b)\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
c)\(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
giúp mik nha
chiều nay nộp r
2. CMR:
a. \(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5-y^5\)
Ta có: VT=\(\left(x-y\right)\left(x^4+x^3y+x^2y^2+xy^3+y^4\right)=x^5+x^4y+x^3y^2+x^2y^3+xy^4-x^4y-x^3y^2-x^2y^3-xy^4-y^5=x^5-y^5=VP\)=> đpcm.
b. \(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5+y^5\)
Ta có: VT=\(\left(x+y\right)\left(x^4-x^3y+x^2y^2-xy^3+y^4\right)=x^5-x^4y+x^3y^2-x^2y^3+xy^4+x^4y-x^3y^2+x^2y^3-xy^4+y^5=x^5+y^5=VP\)
=> đpcm.
c. \(\left(x+a\right)\left(x+b\right)=x^2+\left(a+b\right)x+ab\)
\(\Leftrightarrow x^2+bx+ax+ab=x^2+ax+bx+ab\) (đúng)
=> đpcm.
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1.
b. \(4\left(x-1\right)\left(x+5\right)-\left(x+2\right)\left(x+5\right)=3\left(x-1\right)\left(x+2\right)\)
\(\Leftrightarrow4\left(x^2+5x-x-5\right)-\left(x^2+5x+2x+10\right)=3\left(x^2+2x-x-2\right)\)
\(\Leftrightarrow4x^2+20x-4x-20-x^2-5x-2x-10=3x^2+6x-3x-6\)
\(\Leftrightarrow4x^2+20x-4x-x^2-5x-2x-3x^2-6x+3x=20+10-6\)
\(\Leftrightarrow6x=24\)
\(\Leftrightarrow x=4\)
Vậy ....
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\(a\text{)}.\: \left(8-5x\right)\left(x+2\right)+4\left(x-2\right)\left(x+1\right)+2\left(x-2\right)\left(x+2\right)=0\\ \Leftrightarrow8x-5x^2+16-10x+4x^2-4x-8+2x^2-8=0\\ \Leftrightarrow x^2-6x=0\Leftrightarrow x\left(x-6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
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a) \(\left(x+a\right)\left(x+2a\right)\left(x+3a\right)\left(x+4a\right)+a^4\)
b)\(\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
c) A= \(2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)^2-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
Giải các hệ
left{{}begin{matrix}sqrt{x+y}+sqrt{2x+y+2}73x+2y23end{matrix}right.
left{{}begin{matrix}x^2+y+x^3y+xy^2+xyfrac{-5}{4}x^4+y^2+xyleft(1+2xright)frac{-5}{4}end{matrix}right.
left{{}begin{matrix}left(x^2+1right)+yleft(x+yright)7yleft(x^2+1right)left(x+y-2right)-yend{matrix}right.
left{{}begin{matrix}xleft(x+y+1right)3left(x+yright)^2-frac{5}{x^2}-1end{matrix}right.
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Giải các hệ
\(\left\{{}\begin{matrix}\sqrt{x+y}+\sqrt{2x+y+2}=7\\3x+2y=23\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x^2+y+x^3y+xy^2+xy=\frac{-5}{4}\\x^4+y^2+xy\left(1+2x\right)=\frac{-5}{4}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(x^2+1\right)+y\left(x+y\right)=7y\\\left(x^2+1\right)\left(x+y-2\right)=-y\end{matrix}\right.\)
\(\left\{{}\begin{matrix}x\left(x+y+1\right)=3\\\left(x+y\right)^2-\frac{5}{x^2}=-1\end{matrix}\right.\)
Giải hệ pta)\(\left\{{}\begin{matrix}x^3+xy^2+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+2y^2+xy+2x-4=0\end{matrix}\right.\)b) \(\left\{{}\begin{matrix}x^4+x^2y^2-y^2=y^3+x^2y+x^2\\5\left(x\sqrt{y+7}+\left(x+1\right)\sqrt{2x+y+8}\right)=13\left(2x...
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a.
\(\Leftrightarrow\left\{{}\begin{matrix}x\left(x^2+y^2\right)+\left(x^2+y^2-4\right)\left(y+2\right)=0\\x^2+y^2+\left(x+y-2\right)\left(y+2\right)=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x^2+y^2-4\right)\left(y+2\right)=-x\left(x^2+y^2\right)\\-\left(x^2+y^2\right)=\left(x+y-2\right)\left(y+2\right)\end{matrix}\right.\)
\(\Rightarrow\left(x^2+y^2-4\right)\left(y+2\right)=x\left(x+y-2\right)\left(y+2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y+2=0\left(\text{không thỏa mãn}\right)\\x^2+y^2-4=x\left(x+y-2\right)\end{matrix}\right.\)
\(\Rightarrow x^2+y^2-4=x^2+x\left(y-2\right)\)
\(\Leftrightarrow\left(y+2\right)\left(y-2\right)=x\left(y-2\right)\)
\(\Leftrightarrow\left[{}\begin{matrix}y=2\\x=y+2\end{matrix}\right.\)
Thế vào pt dưới:
\(\Rightarrow\left[{}\begin{matrix}x^2+8+2x+2x-4=0\\\left(y+2\right)^2+2y^2+y\left(y+2\right)+2\left(y+2\right)-4=0\end{matrix}\right.\)
\(\Leftrightarrow...\)
Câu b chắc chắn đề sai, nhìn 2 vế pt đầu đều có \(x^2\) thì chúng sẽ rút gọn, không ai cho đề như thế hết
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1,left{{}begin{matrix}x^2+xy-3x+y0x^4+3x^2y-5x^2+y^20end{matrix}right.
2, left{{}begin{matrix}left(2x-1right)^2+4left(y-1right)^222xyleft(x-1right)left(y-2right)1end{matrix}right.
3, left{{}begin{matrix}left(x^2+y^2right)left(x+y+1right)25left(y+1right)x^2+xy+2y^2+x-8y9end{matrix}right.
4,left{{}begin{matrix}5x^2y-4xy^2+3y^2-2left(x+yright)0xyleft(x^2+y^2right)+2left(x+yright)^2end{matrix}right.
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1,\(\left\{{}\begin{matrix}x^2+xy-3x+y=0\\x^4+3x^2y-5x^2+y^2=0\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\left(2x-1\right)^2+4\left(y-1\right)^2=22\\xy\left(x-1\right)\left(y-2\right)=1\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\left(x^2+y^2\right)\left(x+y+1\right)=25\left(y+1\right)\\x^2+xy+2y^2+x-8y=9\end{matrix}\right.\)
4,\(\left\{{}\begin{matrix}5x^2y-4xy^2+3y^2-2\left(x+y\right)=0\\xy\left(x^2+y^2\right)+2=\left(x+y\right)^2\end{matrix}\right.\)
Phân tích đa thức thành nhân tử :
1) \(A=\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(xy+yz+zx\right)^2\)
2)\(B=2\left(x^4+y^4+z^4\right)-\left(x^2+y^2+z^2\right)-2\left(x^2+y^2+z^2\right)\left(x+y+z\right)^2+\left(x+y+z\right)^4\)
3)\(\left(a+b+c\right)^3-4\left(a^3+b^3+c^3\right)-12abc\)