\(x^3+4y^2-2xy+x^2+8y^3\)
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\(HPT\Leftrightarrow\left\{{}\begin{matrix}x^3-2xy^2+y\left(x^2-8y^2\right)=0\\x^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-2y\right)\left(x^2+xy+4y^2\right)=0\\x^2-8y^2=-4\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x=2y\\x^2+xy+4y^2=0\end{matrix}\right.\\x^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y\\x^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=2y\\\left(2y\right)^2-8y^2=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2;y=1\\x=-2;y=-1\end{matrix}\right.\).
a) ( x+2y)(x^2-2xy+4y^2)-8y^3+27=0
x^3+(2y)^3 - (2y)^3 + 3^3 = 0
x^3+3^3 = 0
(x+3)(x^2-3x+9)= 0
=> x+3=0 hoặc x^2 - 3x + 9 = 0
x+3 = 0 => x=-3
x^2-3x+9= 0=> x^2-2x.3/2+9/4-9/4+9=0
(x^2-2x.3/2+9/4)+(-9/4+9)=0
(x-3/2)^2 + 25/4=0
(x-3/2)^2 =-25/4
Vì (x-3/2)^2 >= 0 mà -25/4<0 nên k tìm đc x tỏa mãn đk đề bài
Vậy x=-3
b) x^3 + x^2 - 2x - 8 = 0
(x^3-8)+(x^2-2x)=0
(x-2)(x^2+2x+4)+x(x-2)=0
(x-2)(x^2+2x+4+x)=0
(x-2)(x^2+3x+4)=0
=>x-2=0 hoặc x^2+3x+4=0
x-2=0=>x=2
x^2+3x+4=0
x^2+2x.3/2+9/4-9/4+4=0
(x^2+2x.3/2+9/4)+(-9/4+4)=0
(x+3/2)^2+15/4=0
(x+3/2)^2=-15/4
Vì (x+3/2)^2>=0 mà-15/4<0 nên k tìm đc x thỏa mãn đk đề bài
Vậy x=2
Mình chỉ làm đc như này thui b thông cảm
Tick cho mình nha
a) \(x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
\(=\left(x-1\right)^2-\left(2y+1\right)^2=\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
\(=\left(x-2y-3\right)\left(x+2y\right)\)
b) \(x^2-4x^2y^2+y^2+2xy=\left(x^2+2xy+y^2\right)-4x^2y^2\)
\(=\left(x+y\right)^2-4x^2y^2=\left(x+y-2xy\right)\left(x+y+2xy\right)\)
c) \(x^6-x^4+2x^3+2x^2=\left(x^6+2x^3+1\right)-\left(x^4-2x^2+1\right)\)
\(=\left(x^3+1\right)^2-\left(x^2-1\right)^2=\left(x^3+1-x^2+1\right)\left(x^3+1+x^2-1\right)=x^2\left(x^3-x^2+2\right)\left(x+1\right)\)
d) \(x^3+3x^2+3x+1-8y^3=\left(x+1\right)^3-8y^3=\left(x+1-2y\right)\left(x^2+2x+1+2xy+2y+4y^2\right)\)
\(\left(3x-6y\right)\left(x^2+2xy+4y^2\right)-3\left(x^3-8y^3+12\right)\)
\(=3\left(x-2y\right)\left(x^2+2xy+4y^2\right)-3\left(x^3-8y^3+12\right)\)
\(=3\left(x^3-8y^3\right)-3\left(x^3-8y^3+12\right)\)
=-36
P=\(X^2+2Y^2-2XY+8X+8Y+2017\)
P=\(\dfrac{4X^2+8Y^2-8XY+32Y+32X+8068}{4}\)
P=\(\dfrac{(\sqrt{3}X)^2-2.\sqrt{3}X.\dfrac{4}{\sqrt{3}}Y+\left(\dfrac{4}{\sqrt{3}}Y\right)^2-\left(\dfrac{4}{\sqrt{3}}Y\right)^2+8Y^2+X^2+32X+32Y+8068}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+\dfrac{8}{3}Y^2+32X+32Y+8068}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+X^2+2.X.16+16^2+(\dfrac{2\sqrt{2}}{\sqrt{3}}Y)^2+2.\dfrac{2\sqrt{2}}{\sqrt{3}}Y.4\sqrt{6}+\left(4\sqrt{6}\right)^2+7716}{4}\)
P=\(\dfrac{\left(\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y\right)^2+\left(X+16\right)^2+\left(\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}\right)^2}{4}+1929\ge1929\forall X\in R\)
DẤU = XẢY RA \(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{3}X-\dfrac{4}{\sqrt{3}}Y=0\\X+16=0\\\dfrac{2\sqrt{2}}{\sqrt{3}}Y+4\sqrt{6}=0\end{matrix}\right.\)
\(C=3.\left(x^2-8y^3-15\right)-3\left(x-2y\right)\left(x^2+2xy+4y^2\right)\)
\(=3x^2-24y^3-45-3\left[x\left(x^2+2xy+4y^2\right)-2y\left(x^2+2xy+4y^2\right)\right]\)
\(=3x^2-24y^3-45-3\left[\left(x^3+2x^2y+4xy^2\right)-\left(2x^2y+4xy^2+8y^3\right)\right]\)
\(=3x^2-24y^3-45-3\left(x^3+2x^2y+4xy^2-2x^2y-4xy^2-8y^3\right)\)
\(=3x^2-24y^3-45-3\left(x^3-8y^3\right)\)
\(=3x^2-24y^3-45-3x^3+24y^3\)
\(=3x^2-3x^3-45\)
a) x^4 - x^3 - x + 1
= x^3 ( x - 1 ) - ( x- 1 )
= ( x^3 - 1 )(x - 1)
= ( x- 1 )^2 (x^2 + x + 1 )
a)x4-x3-x+1
=x3(x-1)-(x-1)
=(x-1)(x3-1)
=(x-1)(x-1)(x2+x+1)
=(x-1)2(x2+x+1)
b)5x2-4x+20xy-8y
(sai đề)
(x-2y)(x2+2xy+4y2)+8y3
=x3-(2y)3+(2y)3
=x3
=>Giá trị của biểu thức không phụ giá trị của biến y
\(\left(x-2y\right)\left(x^2+2xy+4y^2\right)+8y^3\)
\(=x^3-8y^3+8y^3=x^3\)
Vậy giá trị biểu thức không phụ thuộc biến y
12+13+14+15+16+100+267548384543=