N
nhocngo976
bài này lâu rồi mà chưa ai làm
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N
nhockthongay_girlkute
0
01263812493
Giải pt ngộ 
[tex] \huge \blue b^4-3b^3+2b^2+6b+2=0[/TEX]
===========......===========@-)@-)@-)
Lâu wa' hok ai post bài, em cho 1 bài
[tex]\huge \blue 2x^3-x^2+5x-3=0[/tex]
Típ!!!
[tex] \huge \blue x^8 +3x^4+1=0 [/tex]
[tex] \huge \blue b^4-3b^3+2b^2+6b+2=0[/TEX]
===========......===========@-)@-)@-)
Lâu wa' hok ai post bài, em cho 1 bài
[tex]\huge \blue 2x^3-x^2+5x-3=0[/tex]
Típ!!!
[tex] \huge \blue x^8 +3x^4+1=0 [/tex]
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N
nhocngo976
giải hệ
[TEX]\left{\begin {y+xy^2=6x^2 \\ 1+x^2y^2=5x^2[/TEX]
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[TEX]\left{\begin {y+xy^2=6x^2 \\ 1+x^2y^2=5x^2[/TEX]
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nhockthongay_girlkute
x=0; y=0 ko là nghiêm
[TEX]HPT\Leftrightarrow\left{\frac{y}{x^2}+\frac{y^2}{x}=6\\{\frac{1}{x^2}+y^2=5[/TEX]
[TEX]\Leftrightarrow\ \left{\frac{y}{x}(\frac{1}{x}+y)=6\\{(\frac{1}{x}+y)^2-\frac{2y}{x}=5[/TEX]
đến đây chắc là dc
N
nhockthongay_girlkute
R
rua_it
[TEX]\sqrt{1-x^2}+\sqrt[4]{x^2+x-1}+\sqrt[6]{1-x}-1=0[/TEX]
[TEX]\left{ a= \sqrt{1-x^2} \\ b= \sqrt[4]{x^2+x-1} \\ c = \sqrt[6]{1-x}[/TEX]
[TEX]\Rightarrow \left{ a +b+c = 1(1) \\ a^2 + b^4 + c^6 = 1 [/TEX]
[TEX]Do\ a,b,c \ge 0 \ nen\ tu\ (1)\ \Rightarrow a,b,c \le 1 \Rightarrow 0\le a,b,c \le 1 \Rightarrow \left\{ a \le a^2 \\ b \le b^4 \\ c \le c^6 \right. \Rightarrow a+b+c \le a^2+b^4+c^6 =1 [/TEX]
[TEX]Dau\ "="\ \Leftrightarrow .........[/TEX]
N
nhocngo976
típ
1, [TEX]\left{\begin x^2-y\sqrt{xy}=31 \\ y^2-x\sqrt{xy}=72[/TEX]
2[TEX]\left{\begin x^3-8x=y^3+2y \\ x^2-3=3(y^2+1)[/TEX]
1, [TEX]\left{\begin x^2-y\sqrt{xy}=31 \\ y^2-x\sqrt{xy}=72[/TEX]
2[TEX]\left{\begin x^3-8x=y^3+2y \\ x^2-3=3(y^2+1)[/TEX]
D
duynhana1
Nhân chéo đặt [TEX]\left{ a = \sqrt{x} \\ b =\sqrt{y} [/TEX]
[TEX]72(a^4-ab^3) = 31 (b^4 - a^3b) [/TEX]
[TEX]\Leftrightarrow \left{ a^3 - b^3 = 0 \\ 72a = -31b [/TEX]
Đơn giản.
[TEX]6 (x^3- y^3)=(8x+ 2y)(x^2- 3y^2)[/TEX]
N
nhocngo976
3,[TEX]\left{ \begin{8x^3y^3+27=18 \\ 4x^2y+6x=y^2[/TEX]
4,[TEX] x+\|y| +x^3+2x^2\|y| +2xy^2 \|y^3| =0 \tex{and} x \|y|=-2[/TEX]
4,[TEX] x+\|y| +x^3+2x^2\|y| +2xy^2 \|y^3| =0 \tex{and} x \|y|=-2[/TEX]
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D
duynhan1
[TEX]y = 0 \Rightarrow x = 0 [/TEX] không là nghiệm của hệ.
[TEX]* y \not= 0 .\ Ta\ co:\ [/TEX]
[TEX]\left{ (2xy)^3 + 27 = 18 \\ (2xy)^2 + 6xy = y^3 [/TEX]
[TEX]\Leftrightarrow \left{ x+ |y| + (x+|y| )^3 - x|y| (x + |y|) =0 \\ x |y| = -2 [/TEX]
N
nguyentung2510
[TEX]\left{ 2x + x^2 y=y \\ 2y + y^2 z=z \\ 2z + z^2 x=x [/TEX]
[TEX]\left{ 6x(y^2 +z^2 ) =13yz \\ 6y(x^2 +z^2 ) =5xz \\ 6z(x^2 +y^2 ) =5xy [/TEX]
[TEX]\left{ 6x(y^2 +z^2 ) =13yz \\ 6y(x^2 +z^2 ) =5xz \\ 6z(x^2 +y^2 ) =5xy [/TEX]
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nhocngo976
\Leftrightarrow[TEX]\left{\begin{ y=\frac{2x}{1-x^2} \\ z=\frac{2y}{1-y^2} \\ x=\frac{2z}{1-z^2}[/TEX]
đặt [TEX]x=tana [/TEX]
\Rightarrow[TEX]\left{y=tan2a \\z=tan4a \\x=tan8a[/TEX]
\Rightarrow[TEX]tana=tan8a[/TEX]
N
nhocngo976
+, x=y=z=0 là nghiệm hệ
xét x, y, z #0
hệ \Leftrightarrow[TEX]\left{\begin{ \frac{6xy}{z}+\frac{6xz}{y}=13 \\ \frac{6xy}{z}+\frac{6yz}{x}=5 \\ \frac{6zx}{y}+\frac{6zy}{x}=5[/TEX]
dặt [TEX]\frac{xy}{z}=a, \frac{yz}{x}=b, \frac{zx}{y}=c[/TEX]
D
duynhana1
0
01263812493
M
minhkhac_94
Dễ thấy pt chỉ có nghiệm duy nhất2. [TEX]\huge \blue 4x^3+3x=2 [/TEX]
Đặt
[TEX]x = \frac{1}{2}(d - \frac{1}{d})\\[/TEX]
[TEX]pt \Rightarrow \frac{1}{2}({d^3} - \frac{1}{{{d^3}}}) = 2\\[/TEX]
[TEX]\Leftrightarrow {d^6} - 4{d^3} - 1 = 0\\[/TEX]
[TEX] \Leftrightarrow \left[ \begin{array}{l}{d^3} = 2 + \sqrt 5 \\{d^3} = 2 - \sqrt 5 \end{array} \right. \Rightarrow d = \sqrt[3]{{2 + \sqrt 5 }}\\[/TEX]
[TEX] \Rightarrow x = x = \frac{1}{2}(\sqrt[3]{{2 + \sqrt 5 }} - \frac{1}{{\sqrt[3]{{2 + \sqrt 5 }}}})[/TEX]
tex chán
N
nhocngo976
2, [TEX]\left{\begin{ sinxcosy=sinz \\cosxsiny=cosz \\2003x+2004y=2005z^2[/TEX]
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K
keropik
N
nhocngo976
Ko biết có trùng ko
2, từ hệ ta có x, y dương
[TEX]\frac{x+y}{2} \geq \sqrt{xy}[/TEX] \Leftrightarrow[TEX]-\sqrt{x+y}{2} \leq -\sqrt{xy}[/TEX]
\Leftrightarrow[TEX]\frac{x+y}{2} \leq x+y-sqrt{xy}=3 [/TEX]\Leftrightarrow[TEX]x+y \leq 6[/TEX]
bình phương (1) ta dc [TEX]x+y+2+2\sqrt{xy+x+y+1}[/TEX] \leq[TEX]6+2+2\sqrt{(\frac{6}{2})^2+6+1}=16[/TEX]
dấu = khi x=y=3