Cho x=y+1. Chứng minh rằng:
a)\(x^3-y^3-3xy=1\)
b)\(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)=x^{16}-y^{16}\)
Rút gọn biểu thức :
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
b) P=\(\left(5x-1\right)+2\left(1-5x\right)\left(4+5x\right)+\left(5x+4\right)^2\)
c) Q=\(\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
d) P = \(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(a,2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(=2x^2+2y^2+x^2+2xy+y^2+x^2-2xy+y^2=3\left(x^2+y^2\right)\)\(b,\left(5x-1\right)+2\left(1-5x\right)\left(4x+5\right)+\left(5x+4\right)\)\(=\left[\left(5x-1\right)-\left(5x+4\right)\right]^2=25\)
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)\(P=12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(2P=\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(2P=5^{32}-1\Rightarrow P=\dfrac{5^{32}-1}{2}\)
a) \(2\left(x-y\right)\left(x+y\right)+\left(x+y\right)^2+\left(x-y\right)^2\)
\(\Leftrightarrow2\left(x^2-y^2\right)+x^2+2xy+y^2+x^2-2xy+y^2\)
\(\Leftrightarrow2x^2-2y^2+x^2+2xy+y^2+x^2-2xy+y^2\)
\(\Leftrightarrow4x^2\)
Rút gọn biểu thức với x - y = 1
\(\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x^{16}+y^{16}\right)\)
Ta có \(x-y=1\)
\(=>x+y=\left(x+y\right).\left(x-y\right)\)
\(A=\left(x+y\right).\left(x-y\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^2-y^2\right).\left(x^2+y^2\right).\left(x^4+y^4\right)\)
\(A=\left(x^4-y^4\right).\left(x^4+y^4\right)\)
\(A=x^8-y^8\)
= \(-\left[\left(x-y\right)\left(x^2-y^2\right)\left(x^4-y^4\right)\left(x^8-y^8\right)\left(x^{16}-y^{16}\right)\right]\)
= \(-\left[\left(x-y\right)\left(x-y\right)^2\left(x-y\right)^4\left(x-y\right)^8\left(x-y\right)^{16}\right]\)
= \(-\left(1\cdot1^2\cdot1^4\cdot1^8\cdot1^{16}\right)\)
= -1
rut gon bieu thuc
\(Q=\left(x-y\right)^3+\left(y+x\right)^3+\left(y-x\right)^3-3xy\left(x+y\right)\)
\(P=12\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
Giải hệ phương trình
a, \(\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)=8\\x\left(x+1\right)+y\left(y+1\right)+xy=17\end{matrix}\right.\)
b, \(\left\{{}\begin{matrix}x+y+xy=3\\x^2+y^2=2\end{matrix}\right.\)
c, \(\left\{{}\begin{matrix}x^2+y^2+xy=4\\x-y-3xy=16\end{matrix}\right.\)
GIÚP MÌNH NHÉ
\(\left\{{}\begin{matrix}x+y+3xy=-3\\xy+1=0\end{matrix}\right.\)
___
\(\left\{{}\begin{matrix}x^2-y^2=16\\x+y=8\end{matrix}\right.\)
Câu 1:
Từ $xy+1=0\Leftrightarrow xy=-1$
Thay vào PT(1): $x+y=-3-3xy=-3-3(-1)=0$
$\Leftrightarrow x=-y$. Thay vào đk $xy=-1$ thì:
$(-y)y=-1$
$\Leftrightarrow y^2=1\Leftrightarrow y=\pm 1$
Với $y=1$ thì $x=-y=-1$
Với $y=-1$ thì $x=-y=1$
Vậy $(x,y)=(1,-1), (-1,1)$
Câu 2:
$x^2-y^2=16$
$\Leftrightarrow (x-y)(x+y)=16$
$\Leftrightarrow 8(x-y)=16$
$\Leftrightarrow x-y=2$
Kết hợp với $x+y=8$ thì:
$(x-y)+(x+y)=2+8$
$\Leftrightarrow 2x=10$
$\Leftrightarrow x=5$
$y=8-x=8-5=3$
Vậy.............
\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\left(1+x^2\right)^2\left(1+\dfrac{1}{y^4}\right)=8\\\left(1+y^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\end{matrix}\right.\)
Em cảm ơn ạ !!!
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: \(\dfrac{1}{4}+y^2=\dfrac{1}{2}\Rightarrow y=\pm\dfrac{1}{2}\)
\(\left\{{}\begin{matrix}x^2+y^2=\dfrac{1}{2}\\x^3+3xy^2=\dfrac{1}{2}\end{matrix}\right.\)
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
b.
Trừ vế cho vế:
\(\left(1+x^2\right)^2\left(1+\dfrac{1}{y^4}\right)-\left(1+y^2\right)^2\left(1+\dfrac{1}{x^4}\right)=0\)
\(\Leftrightarrow\left(1+x^2\right)^2-\left(1+y^2\right)^2+\left(\dfrac{1+x^2}{y^2}\right)^2-\left(\dfrac{1+y^2}{x^2}\right)^2=0\)
\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2+2\right)+\left(\dfrac{x^4+x^2-y^4-y^2}{x^2y^2}\right)\left(\dfrac{1+x^2}{y^2}+\dfrac{1+y^2}{x^2}\right)=0\)
\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2+2\right)+\left(\dfrac{\left(x^2-y^2\right)\left(x^2+y^2+1\right)}{x^2y^2}\right)\left(\dfrac{x^2+1}{y^2}+\dfrac{y^2+1}{x^2}\right)=0\)
\(\Leftrightarrow\left(x^2-y^2\right)\left(x^2+y^2+2+\left(\dfrac{x^2+y^2+1}{x^2y^2}\right)\left(\dfrac{x^2+1}{y^2}+\dfrac{y^2+1}{x^2}\right)\right)=0\)
\(\Leftrightarrow x^2=y^2\) (ngoặc to hiển nhiên dương)
Thế vào pt đầu:
\(\left(1+x^2\right)^2\left(1+\dfrac{1}{x^4}\right)=8\)
Ta có: \(\left(1+x^2\right)^2\left(1+\dfrac{1}{x^4}\right)\ge4x^2.2\sqrt{1.\dfrac{1}{x^4}}=8\)
Đẳng thức xảy ra khi và chỉ khi \(x^2=1\)
Vậy nghiệm của hệ là \(x^2=y^2=1\Rightarrow x;y\)
hệ phương trình
1, \(\left\{{}\begin{matrix}\frac{1}{x+y}+\frac{1}{x-y}=\frac{5}{8}\\\frac{1}{x+y}-\frac{1}{x-y}=-\frac{3}{8}\end{matrix}\right.\)
2, \(\left\{{}\begin{matrix}\frac{4}{2x-3y}+\frac{5}{3x+y}=2\\\frac{3}{3x+y}-\frac{5}{2x-3y}=21\end{matrix}\right.\)
3, \(\left\{{}\begin{matrix}\frac{7}{x-y+2}+\frac{5}{x+y-1}=\frac{9}{2}\\\frac{3}{x-y+2}+\frac{2}{x+y-1}=4\end{matrix}\right.\)
4, \(\left\{{}\begin{matrix}\frac{3}{x}+\frac{5}{y}=-\frac{3}{2}\\\frac{5}{x}-\frac{2}{y}=\frac{8}{3}\end{matrix}\right.\)
5 , \(\left\{{}\begin{matrix}\frac{2}{x+y-1}-\frac{4}{x-y+1}=-\frac{14}{5}\\\frac{3}{x+y-1}+\frac{2}{x-y+1}=-\frac{13}{5}\end{matrix}\right.\)
6 , \(\left\{{}\frac{\frac{2x-3}{2y-5}=\frac{3x+1}{3y-4}}{2\left(x-3\right)-3\left(y+20=-16\right)}}\)
7\(\left\{{}\begin{matrix}\left(x+3\right)\left(y+5\right)=\left(x+1\right)\left(y+8\right)\\\left(2x-3\right)\left(5y+7\right)=2\left(5x-6\right)\left(y+1\right)\end{matrix}\right.\)
1)Sắp xếp theo thứ tự tăng dần
\(\left(\dfrac{-2}{5}\right)^3;\left(0,3\right)^2;\left(\dfrac{-3}{4}\right)^3;\left(-1,2\right)^2\)
2)Tìm x và y sao cho
\(\left(\dfrac{x}{y}\right)=\left(\dfrac{x}{y}\right)^2\)
3)Chứng minh rằng
a)\(16^5+2^{15}\) chia hết cho 33
b)\(333^{555}+555^{333}\) chia hết cho 37
4)Tìm x và y biết
a)\(\left(x-3\right)^2+\left(y+2\right)^2=0\)
b)\(\left(x-12+y\right)^{200}+\left(x-4-y\right)^{200}=0\)
5)Tính giá trị của x
a)\(8^x>16\)
b)\(\left(\dfrac{2}{3}\right)^x=\dfrac{4}{9}\)
c)\(\left(\dfrac{1}{2}\right)^{x-1}=8\)
d)\(3^x< 16\)
e)\(3^x< 27\)
Help me mai 5/8 6h mik đi học rồi
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:
\(\left\{{}\begin{matrix}\left(x-12+y\right)^{200}=0\\\left(x-y-4\right)^{200}=0\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x-12+y=0\Rightarrow x+y=12\\x-y-4=0\Rightarrow x-y=4\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\left(x+y\right)+\left(x-y\right)=12+4\Rightarrow x+y+x-y=16\Rightarrow2x=16\Rightarrow x=8\\y=8-4=4\end{matrix}\right.\)
Chứng minh giá trị của mỗi biểu thức sau không phụ thuộc vào giá trị của biến:
a) \(A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\)
b) \(B = \left( {x - 2y} \right)\left( {{x^2} + 2{\rm{x}}y + 4{y^2}} \right) - \left( {{x^3} - 8{y^3} + 10} \right)\)
c) \(C = 4{\left( {x + 1} \right)^2} + {\left( {2{\rm{x}} - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) - 4{\rm{x}}\)
a)
\(\begin{array}{l}A = 0,2\left( {5{\rm{x}} - 1} \right) - \dfrac{1}{2}\left( {\dfrac{2}{3}x + 4} \right) + \dfrac{2}{3}\left( {3 - x} \right)\\A = x - 0,2 - \dfrac{1}{3}x - 2 + 2 - \dfrac{2}{3}x\\ = \left( {x - \dfrac{1}{3}x - \dfrac{2}{3}x} \right) + \left( {\dfrac{{ - 1}}{2} - 2 + 2} \right)\\ = - \dfrac{1}{2}\end{array}\)
Vậy \(A = - \dfrac{1}{2}\) không phụ thuộc vào biến x
b)
\(\begin{array}{l}B = \left( {x - 2y} \right)\left( {{x^2} + 2{\rm{x}}y + 4{y^2}} \right) - \left( {{x^3} - 8{y^3} + 10} \right)\\B = \left[ {x - {{\left( {2y} \right)}^3}} \right] - {x^3} + 8{y^3} - 10\\B = {x^3} - 8{y^3} - {x^3} + 8{y^3} - 10 = - 10\end{array}\)
Vậy B = -10 không phụ thuộc vào biến x, y.
c)
\(\begin{array}{l}C = 4{\left( {x + 1} \right)^2} + {\left( {2{\rm{x}} - 1} \right)^2} - 8\left( {x - 1} \right)\left( {x + 1} \right) - 4{\rm{x}}\\{\rm{C = 4}}\left( {{x^2} + 2{\rm{x}} + 1} \right) + \left( {4{{\rm{x}}^2} - 4{\rm{x}} + 1} \right) - 8\left( {{x^2} - 1} \right) - 4{\rm{x}}\\C = 4{{\rm{x}}^2} + 8{\rm{x}} + 4 + 4{{\rm{x}}^2} - 4{\rm{x}} + 1 - 8{{\rm{x}}^2} + 8 - 4{\rm{x}}\\C = \left( {4{{\rm{x}}^2} + 4{{\rm{x}}^2} - 8{{\rm{x}}^2}} \right) + \left( {8{\rm{x}} - 4{\rm{x}} - 4{\rm{x}}} \right) + \left( {4 + 1 + 8} \right)\\C = 13\end{array}\)
Vậy C = 13 không phụ thuộc vào biến x