Chứng minh rằng:\(\left(2x^2-y\right)\left(2y^2-x\right)+\left(x+y\right)\left(2x^2+2y^2\right)=\left(2xy+x\right)\left(2xy+y\right)\)
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SGK Kết nối tri thức với cuộc sống - Trang 47
21 tháng 7 2023 lúc 15:42
Rút gọn các biểu thức:
a) \(\left( {2x - 5y} \right)\left( {2x + 5y} \right) + {\left( {2x + 5y} \right)^2}\)
b) \(\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\)
a)
\(\begin{array}{l}\left( {2x - 5y} \right)\left( {2x + 5y} \right) + {\left( {2x + 5y} \right)^2}\\ = \left( {2x + 5y} \right)\left( {2x - 5y + 2x + 5y} \right)\\ = \left( {2x + 5y} \right).4x\\ = 2x.4x + 5y.4x\\ = 8{x^2} + 20xy\end{array}\)
b)
\(\begin{array}{l}\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\ = {x^3} + {\left( {2y} \right)^3} + {\left( {2x} \right)^3} - {y^3}\\ = {x^3} + 8{y^3} + 8{x^3} - {y^3}\\ = \left( {{x^3} + 8{x^3}} \right) + \left( {8{y^3} - {y^3}} \right)\\ = 9{x^3} + 7{y^3}\end{array}\)
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giải hệ phương trình
a) left{{}begin{matrix}sqrt{2x^2+2y^2}+sqrt{frac{4}{3}left(x^2+xy+y^2right)}2left(x+yright)sqrt{3x+1}+sqrt{5x+4}3xy-y+3end{matrix}right.
b) left{{}begin{matrix}sqrt{5x^2+2xy+2y^2}+sqrt{2x^2+2xy+5y^2}3left(x+yright)sqrt{x+2y+1}+2sqrt[3]{12x+7y+8}2xy+x+5end{matrix}right.
c)left{{}begin{matrix}x^2+xy+x+30left(x+1right)^2+3left(y+1right)+2left(xy-sqrt{x^2y+2y}right)0end{matrix}right.
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giải hệ phương trình
a) \(\left\{{}\begin{matrix}\sqrt{2x^2+2y^2}+\sqrt{\frac{4}{3}\left(x^2+xy+y^2\right)}=2\left(x+y\right)\\\sqrt{3x+1}+\sqrt{5x+4}=3xy-y+3\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\\\sqrt{x+2y+1}+2\sqrt[3]{12x+7y+8}=2xy+x+5\end{matrix}\right.\)
c)\(\left\{{}\begin{matrix}x^2+xy+x+3=0\\\left(x+1\right)^2+3\left(y+1\right)+2\left(xy-\sqrt{x^2y+2y}\right)=0\end{matrix}\right.\)
b)\(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}=3\left(x+y\right)\)
\(\Rightarrow\left(\sqrt{5x^2+2xy+2y^2}+\sqrt{2x^2+2xy+5y^2}\right)^2=\left(3\left(x+y\right)\right)^2\)
\(\Leftrightarrow\sqrt{\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)}=x^2+7xy+y^2\)
\(\Rightarrow\left(5x^2+2xy+2y^2\right)\left(2x^2+2xy+5y^2\right)=\left(x^2+7xy+y^2\right)^2\)
\(\Leftrightarrow9\left(x-y\right)^2\left(x+y\right)^2=0\)\(\Leftrightarrow\left[{}\begin{matrix}x=y\\x=-y\end{matrix}\right.\)
\(\rightarrow\left(x;y\right)\in\left\{\left(0;0\right),\left(1;1\right)\right\}\)
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caau a) binh phuong len ra no x=y tuong tu
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c)
ĐK $y \geqslant 0$
Hệ đã cho tương đương với
$\left\{\begin{matrix} 2x^2+2xy+2x+6=0\\ (x+1)^2+3(y+1)+2xy=2\sqrt{y(x^2+2)} \end{matrix}\right.$
Trừ từng vế $2$ phương trình ta được
$x^2+2+2\sqrt{y(x^2+2)}-3y=0$
$\Leftrightarrow (\sqrt{x^2+2}-\sqrt{y})(\sqrt{x^2+2}+3\sqrt{y})=0$
$\Leftrightarrow x^2+2=y$
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giải giúp mik bt này vs mn!
1)left{{}begin{matrix}2x^2+y^2+x3left(xy+1right)+2ydfrac{2}{3+sqrt{2x-y}}+dfrac{2}{3+sqrt{4-5x}}dfrac{9}{2x-y+9}end{matrix}right.
2)left{{}begin{matrix}left(x+3y+1right)sqrt{2xy+2y}yleft(3x+4y+3right)left(sqrt{x+3}-sqrt{2y-2}right)left(x-3+sqrt{x^2+x+2y-4}right)4end{matrix}right.
3)left{{}begin{matrix}x-dfrac{1}{x}y-dfrac{1}{y}2yx^3+1end{matrix}right.
4)left{{}begin{matrix}sqrt{2x-3}left(y^2+2011right)left(5-yright)+sqrt{y}yleft(y-x+2right)3x+3end{matrix}right.
5...
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giải giúp mik bt này vs mn!
1)\(\left\{{}\begin{matrix}2x^2+y^2+x=3\left(xy+1\right)+2y\\\dfrac{2}{3+\sqrt{2x-y}}+\dfrac{2}{3+\sqrt{4-5x}}=\dfrac{9}{2x-y+9}\end{matrix}\right.\)
2)\(\left\{{}\begin{matrix}\left(x+3y+1\right)\sqrt{2xy+2y}=y\left(3x+4y+3\right)\\\left(\sqrt{x+3}-\sqrt{2y-2}\right)\left(x-3+\sqrt{x^2+x+2y-4}\right)=4\end{matrix}\right.\)
3)\(\left\{{}\begin{matrix}x-\dfrac{1}{x}=y-\dfrac{1}{y}\\2y=x^3+1\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}\sqrt{2x-3}=\left(y^2+2011\right)\left(5-y\right)+\sqrt{y}\\y\left(y-x+2\right)=3x+3\end{matrix}\right.\)
5)\(\left\{{}\begin{matrix}x^3+2x^2=x^2y+2xy\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14=x-2}\end{matrix}\right.\)
5,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x\left(x+y\right)\left(x+2\right)=0\\2\sqrt{x^2-2y-1}+\sqrt[3]{y^3-14}=x-2\end{matrix}\right.\)
Thay từng TH rồi làm nha bạn
3,\(hpt\Leftrightarrow\left\{{}\begin{matrix}x-y=\frac{1}{x}-\frac{1}{y}=\frac{y-x}{xy}\\2y=x^3+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x-y\right)\left(1+\frac{1}{xy}\right)=0\\2y=x^3+1\end{matrix}\right.\)
thay nhá
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Bài 1:ĐKXĐ: \(2x\ge y;4\ge5x;2x-y+9\ge0\)\(\Rightarrow2x\ge y;x\le\frac{4}{5}\Rightarrow y\le\frac{8}{5}\)
PT(1) \(\Leftrightarrow\left(x-y-1\right)\left(2x-y+3\right)=0\)
+) Với y = x - 1 thay vào pt (2):
\(\frac{2}{3+\sqrt{x+1}}+\frac{2}{3+\sqrt{4-5x}}=\frac{9}{x+10}\) (ĐK: \(-1\le x\le\frac{4}{5}\))
Anh quy đồng lên đê, chắc cần vài con trâu đó:))
+) Với y = 2x + 3...
Giải các hệ phương trình :a) left{{}begin{matrix}left(x-3right)left(2y+5right)left(2x+7right)left(y-1right)left(4x+1right)left(3y-6right)left(6x-1right)left(2y+3right)end{matrix}right.;b) left{{}begin{matrix}left(x+yright)left(x-1right)left(x-yright)left(x+1right)left(2xyright)left(y-xright)left(y+1right)left(y+xright)left(y-2right)-2xyend{matrix}right..
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Giải các hệ phương trình :
a) \(\left\{{}\begin{matrix}\left(x-3\right)\left(2y+5\right)=\left(2x+7\right)\left(y-1\right)\\\left(4x+1\right)\left(3y-6\right)=\left(6x-1\right)\left(2y+3\right)\end{matrix}\right.\);
b) \(\left\{{}\begin{matrix}\left(x+y\right)\left(x-1\right)=\left(x-y\right)\left(x+1\right)\left(2xy\right)\\\left(y-x\right)\left(y+1\right)=\left(y+x\right)\left(y-2\right)-2xy\end{matrix}\right.\).
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
\(\frac{x^2}{\left(x-y\right)^2\left(x+y\right)}-\frac{2xy^2}{x^4-2x^2y^2+y^4}+\frac{y^2}{\left(x^2-y^2\right)\left(z+y\right)}\)
Giải PT và HPT:
1)left{{}begin{matrix}xy+x+y3frac{1}{x^2+2x}+frac{1}{y^2+2y}frac{2}{3}end{matrix}right.
2)left(sqrt{x+4}-2right)left(sqrt{4-x}+2right)2x
3)left{{}begin{matrix}xyleft(x+yright)29xyleft(3x-yright)+626x^3-2y^3end{matrix}right.
4)left{{}begin{matrix}x^2-2xy+x-2y+30y^2-x^2+2xy+2x-20end{matrix}right.
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Giải PT và HPT:
1)\(\left\{{}\begin{matrix}xy+x+y=3\\\frac{1}{x^2+2x}+\frac{1}{y^2+2y}=\frac{2}{3}\end{matrix}\right.\)
2)\(\left(\sqrt{x+4}-2\right)\left(\sqrt{4-x}+2\right)=2x\)
3)\(\left\{{}\begin{matrix}xy\left(x+y\right)=2\\9xy\left(3x-y\right)+6=26x^3-2y^3\end{matrix}\right.\)
4)\(\left\{{}\begin{matrix}x^2-2xy+x-2y+3=0\\y^2-x^2+2xy+2x-2=0\end{matrix}\right.\)
26 tháng 10 2021 lúc 17:19
Ghpt:
a) \(\left\{{}\begin{matrix}x^2+2y^2=2x-2xy+1\\3x^2+2xy-y^2=2x-y+5\end{matrix}\right.\)
b) \(\left\{{}\begin{matrix}4xy+4x^2+4y^2+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=3\end{matrix}\right.\)
a) rút gọn biểu thức: \(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)
b) tìm x biết: \(x\left(3x+2\right)+\left(x+1\right)^2-\left(2x-5\right)\left(2x+5\right)=12\)
a)A=x3+x2y+y2x+y3+2x2y+2xy2
=x3+3x2y+3xy2+y3
A=(x+y)3
b)=3x2+2x+(x2+2x+1)-(4x2-25)=12
3x2+2x+x2+2x+1-4x2+25=12
4x+26=12
= >4x=6/13
= >x=6,5
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