Cho x=y+1. Chứng minh rằng:
a)\(x^3-y^3-3xy=1\)
b)\(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)=x^{16}-y^{16}\)
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TN
6 tháng 6 2017
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
6 tháng 6 2017
c)\(Q=\left(x-y\right)^3+\left(x+y\right)^3+\left(x-y\right)^3-3xy\left(x+y\right)\)
\(=x^3-3x^2y+3xy^2-y^3+x^3+3x^2y+3xy^2+y^3-x^3+3x^2y-3xy^2+y^3-3xy^2-3x^2y\)
\(=x^3+y^3\)
d)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
2 tháng 10 2017
Ta có \(x-y=1\)
\(=>x+y=\left(x+y\right).\left(x-y\right)\)
\(A=\left(x+y\right).\left(x-y\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^2-y^2\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^4-y^4\right).\left(x^4+y^4\right)\)
\(A=x^8-y^8\)
C
18 tháng 9 2019
= \(-\left[\left(x-y\right)\left(x^2-y^2\right)\left(x^4-y^4\right)\left(x^8-y^8\right)\left(x^{16}-y^{16}\right)\right]\)
= \(-\left[\left(x-y\right)\left(x-y\right)^2\left(x-y\right)^4\left(x-y\right)^8\left(x-y\right)^{16}\right]\)
= \(-\left(1\cdot1^2\cdot1^4\cdot1^8\cdot1^{16}\right)\)
= -1
NV
Nguyễn Việt Lâm
Giáo viên
27 tháng 7 2021
a.
\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}y^2=\dfrac{1}{2}-x^2\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow x^3+3x\left(\dfrac{1}{2}-x^2\right)=\dfrac{1}{2}\)
\(\Leftrightarrow4x^3-3x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(2x-1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{1}{2}\end{matrix}\right.\)
- Với \(x=-1\) thế vào pt đầu: \(1+y^2=\dfrac{1}{2}\Rightarrow y^2=-\dfrac{1}{2}\) (vô nghiệm)
- Với \(x=\dfrac{1}{2}\) thế vào pt đầu: \(\dfrac{1}{4}+y^2=\dfrac{1}{2}\Rightarrow y=\pm\dfrac{1}{2}\)
HN
27 tháng 7 2021
\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
Dễ thấy x = 0 không phải nghiệm ta nhân tử mẫu phương trình đầu cho 3x thì được
\(\Leftrightarrow\left\{{}\begin{matrix}3x^3+3xy^2=\dfrac{3x}{2}\left(1\right)\\x^3+3xy^2=\dfrac{1}{2}\left(2\right)\end{matrix}\right.\)
Lấy (1) - (2) thì đơn giản rồi ha
5 tháng 8 2017
2) \(\dfrac{x}{y}=\left(\dfrac{x}{y}\right)^2\)
\(\Rightarrow\left(\dfrac{x}{y}\right)^2-\dfrac{x}{y}=0\)
\(\Rightarrow\dfrac{x}{y}\left(\dfrac{x}{y}-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{y}=0\Rightarrow x=0;y\in R\\\dfrac{x}{y}-1=0\Rightarrow\dfrac{x}{y}=1\Rightarrow x=y\end{matrix}\right.\)
3) \(16^5+2^{15}=\left(2^4\right)^5+2^{15}=2^{20}+2^{15}=2^{15}.2^5+2^{15}.1=2^{15}.33⋮33\rightarrowđpcm\)
4)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
\(\left\{{}\begin{matrix}\left(x-3\right)^2\ge0\\\left(y+2\right)^2\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-3\right)^2+\left(y+2\right)^2\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-3\right)^2=0\Rightarrow x-3=0\Rightarrow x=3\\\left(y+2\right)^2=0\Rightarrow y+2=0\Rightarrow y=-2\end{matrix}\right.\)
\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}\ge0\\\left(x-4-y\right)^{200}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(x-12+y\right)^{200}+\left(x-y-4\right)^{200}\ge0\)
Dấu "=" xảy ra khi:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)
a. Do \(x=y-1\Rightarrow x-y=1\)
Ta có:
\(A=x^3-y^3-3xy=\left(x-y\right)^3+3xy\left(x-y\right)-3xy=1^3+3xy.1-3xy=1\left(đpcm\right)\)
b. \(B=\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\)
(Do \(x-y=1\))
(Bạn áp dụng hằng đẳng thức \(x^2-y^2=\left(x-y\right)\left(x+y\right)\)vào bài toán)
Kết quả, \(B=x^{16}-y^{16}\left(đpcm\right)\)
a)\(x=y+1\Rightarrow x-y=1\Rightarrow\left(x-y\right)^3=1\)
Hay x3- 3xy(x-y) - y3=1 => x3- y3 -3xy =1
b) 1.(x+y)(x2+y2)(x4+y4)(x8+y8) = (x-y)(x+y)......................=(x2-y2)(x2+y2)..........=(x4-y4)(x4+y4)......=(x8-y8)(x8+y8) =x16-y16