mk k hiểu câu hỏi

Bài 1:A=4x4+7x2y2+3y4+5y2=4x2(x2+y2)+3y2(x2+y2)+5y2=20x2+15y2+5y2=20(x2+y2)=100.

A=4x⁴+7x²y²+3y⁴+5y²

=4x²(x²+y²)+3y²(x²+y²)+5y²

=20x²+15y²+5y² 

=20x²+(15+5)y²

=20(x²+y²)=100

Cộng 2 biểu thức lại thành 3P

Tìm ra P

Thay P vào P-Q tìm được Q

P = 3x² - 2x + 3y² - 2y + 6xy +2018

P = 3(x² + y² + 2xy) - 2(x + y) + 2018

P = 3[(x + y)² - 2xy + 2xy] -2.5 + 2018

P = 3[ 5² +0] - 10 + 2018

P = 3.25 + 2008

P = 75 + 2008

P = 2083

Chắc đề bài là \(Q=\dfrac{3}{9x^2+6xy+y^2}+\dfrac{3}{3x^2+6xy+2y^2}\)

Từ giả thiết ta có:

\(2x^3+2xy^2+xy^2+y^3=2\left(x^2+y^2\right)\)

\(\Leftrightarrow2x\left(x^2+y^2\right)+y\left(x^2+y^2\right)=2\left(x^2+y^2\right)\)

\(\Leftrightarrow2x+y=2\)

Do đó:

\(Q=3\left(\dfrac{1}{9x^2+6xy+y^2}+\dfrac{1}{3x^2+6xy+2y^2}\right)\)

\(Q\ge\dfrac{3.4}{12x^2+12xy+3y^2}=\dfrac{4}{\left(2x+y\right)^2}=1\)

\(Q_{min}=1\) khi \(\left\{{}\begin{matrix}2x+y=2\\9x^2+6xy+y^2=3x^2+6xy+2y^2\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=\sqrt{6}-2\\y=6-2\sqrt{6}\end{matrix}\right.\)

\(a,P=3x^2-2x+3y^2-2y+6xy-100\)

\(P=3\left(x^2+y^2\right)-\left[2\left(x+y\right)\right]+6xy-100\)

\(P=3\left(x^2+y^2+2xy-2xy\right)-2.5+6xy-100\)

\(P=3\left(x+y\right)^2-6xy-10+6xy-100\)

\(P=3.25-10-100\)

\(P=-35\)

\(b,Q=x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10\)

\(Q=\left(x+y\right)\left(x^2-xy+y^2\right)-2\left(x^2+y^2+2xy-2xy\right)+3xy.5-4xy+3.5+10\)\(Q=5.\left(x^2+y^2+2xy-3xy\right)-2\left(x+y\right)^2+4xy+15xy-4xy+25\)

\(Q=5.5-15xy-2.25+15xy+25\)

\(Q=25-50+25=0\)

a) P= 3x² -2x + 3y²-2y + 6xy -100

= (3x²+ 3y² + 6xy) - 2(x+y) -100

=3(x² + y² +2xy) - 2(x+y) -100

=3(x+y)² - 2(x+y) -100

=3 . 5² -2 .5 -100

=35

b) Q=x³ + y³ -2x² -2y² + 3xy (x+y) -4xy + 3(x+y) + 10

=(x³ +y³) + 3xy (x+y) + 3(x+y) -4xy -2x² -2y² + 10

=(x+y) (x² -xy +y² ) + 3xy (x+y) + 3 (x+y) - 2 (2xy + x² +y² ) + 10

=(x+y) (x² -xy +y² + 3xy ) + 3(x+y) -2 (2xy + x² + y² ) + 10

=(x+y) (x² +2xy +y² ) + 3(x+y) - 2(x+y)² + 10

= (x+y)³ + 3(x+y) - 2 (x+y)² + 10

=5³ + 3.5 -2. 5²+ 10

=100

a)

\(P=3x^2-2x+3y^2-2y+6xy-100\)

\(P=\left(3x^2+3xy\right)+\left(3y^2+3xy\right)-\left(2x+2y\right)-100\)

\(P=3x\left(x+y\right)+3y\left(x+y\right)-2\left(x+y\right)-100\)

\(P=\left(x+y\right)\left(3x+3y-2\right)-100\)

Vì x + y = 5 (1)

=> 3(x + y) = 15

=> 3x + 3y = 15 (2)

Thay (1), (2) vào P ta được:

\(P=5\left(15-2\right)-100\)

\(P=5.13-100\)

\(P=-35\)

Câu a : \(\left(x^2-3x\right)^2+2x^2\left(3x-5\right)=2\)

\(\Leftrightarrow x^4-6x^3+9x^2+6x^3-10x^2-2=0\)

\(\Leftrightarrow x^4-x^2-2=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x^2-2\right)=0\)

\(\Leftrightarrow x^2-2=0\) ( do \(x^2+1>0\) )

\(\Leftrightarrow x=\pm2\)