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sửa đổi
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$\;$
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d/ c2 $x^2+4x-2xy-4y+y^2$$=x^2+y^2+4-2xy+4x-4y-4$$=(x-y+2)^2-4$$=(x-y)(x-y-4)$
d/ c2 $x^2+4x-2xy-4y+y^2$$=x^2+y^2+4-2xy+4x-4y-4$$=(x-y+2)^2-4$$=(x-y)(x-y+4)$
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sửa đổi
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$\;$
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d/ c1 $x^2+4x-2xy-4y+y^2$$=(x^2-2xy+y^2)+(4x-4y)$$=(x-y)^2+4(x-y)$$=(x-y)(x-y-4)$
d/ c1 $x^2+4x-2xy-4y+y^2$$=(x^2-2xy+y^2)+(4x-4y)$$=(x-y)^2+4(x-y)$$=(x-y)(x-y+4)$
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giải đáp
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chứng minh
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Áp dụng bất đẳng thức Cô si cho 4 số dương$a^4+a^4+a^4+1\geq4\sqrt[4]{a^4.a^4.a^4.1}=4a^3$ $b^4+b^4+b^4+1 \geq4b^3$ $c^4+c^4+c^4+1 \geq4c^3$
$d^4+d^4+d^4+1 \geq4d^3$
=> $3VT+4\geq 4VP$ (1) Ta lại có $a^3+1+1\geq3a$ $b^3+1+1\geq3b$ $c^3+1+1\geq3c$ $d^3+1+1\geq3d$ => $VP+8>=3(a+b+c+d)=12$ => $VP\geq4$ (2) Từ (1) và (2) $3VT+VP\geq4VP$ hay $VT>=VP$
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sửa đổi
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chứng minh
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chứng minh cho a,b,c,d >0 và a+b+c+d=4cmr a^{4}+ b^{4}+ c^{4}+ d^{4} \geq a^{3} +b^{3} +c^{3} +d^{3}
chứng minh cho $a,b,c,d >0 $ và $a+b+c+d=4 $cmr $a^{4}+ b^{4}+ c^{4}+ d^{4} \geq a^{3} +b^{3} +c^{3} +d^{3} $
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sửa đổi
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Giải phương trình:
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MicrosoftInternetExplorer4
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$7x^2 + 7x = \sqrt{1/28(4x+9)}$
$7x^2 +7x = (\sqrt{4x+9})/(2\sqrt{7})$
$7x(x+1) = (\sqrt{4x+9})/(2\sqrt{7})$
$\sqrt{4x+9 = 14\sqrt{7}x^2 +14\sqrt{7}x$
$(\sqrt{7})\sqrt{4x+9}=98x(x+1)$
$x= 5/(7\sqrt{2}) -3/7 hoặc x= -4/7 - (\sqrt{23/2})/7$
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MicrosoftInternetExplorer4
/* Style Definitions */
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{mso-style-name:"Table Normal";
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$7x^2 + 7x = \sqrt{1/28(4x+9)}$
$7x^2 +7x = (\sqrt{4x+9})/(2\sqrt{7})$
$7x(x+1) = (\sqrt{4x+9})/(2\sqrt{7})$
$\sqrt{4x+9} = 14\sqrt{7}x^2 +14\sqrt{7}x$
$(\sqrt{7})\sqrt{4x+9}=98x(x+1)$
$x= 5/(7\sqrt{2}) -3/7 hoặc x= -4/7 - (\sqrt{23/2})/7$
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