\(D=x^2+y^2+x-6y+5\)

\(D=\left(x^2+x+\frac{1}{2}^2\right)+\left(y^2-6y+9\right)-\frac{17}{4}\)

\(D=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\le-\frac{17}{4}\)

dấu "=" xảy ra khi và chỉ khi \(\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}\)

\(< =>MIN:D=-\frac{17}{4}\)

Trả lời:

\(D=x^2+y^2+x-6y+5=x^2+y^2+x-6y+\frac{1}{4}+9-\frac{17}{4}\)

\(=\left(x^2+x+\frac{1}{4}\right)+\left(y^2-6y+9\right)-\frac{17}{4}\)

\(=\left(x+\frac{1}{2}\right)^2+\left(y-3\right)^2-\frac{17}{4}\ge-\frac{17}{4}\forall x;y\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}x+\frac{1}{2}=0\\y-3=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{2}\\y=3\end{cases}}}\)

Vậy GTNN của D = - 17/4 khi x = -1/2; y = 3

Trả lời:

\(C=x^2+x-2=x^2+2x\frac{1}{2}+\frac{1}{4}-\frac{9}{4}=\left(x+\frac{1}{2}\right)^2-\frac{9}{4}\ge-\frac{9}{4}\forall x\)

Dấu "=" xảy ra khi \(x+\frac{1}{2}=0\Leftrightarrow x=-\frac{1}{2}\)

Vậy GTNN của C = - 9/4 khi x = - 1/2

các bạn trả lời giúp mik vs

\(\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2bc+2ac.\)\(\Rightarrow VT=\left(a^2+2ab+b^2\right)+\left(a^2+2ac+c^2\right)+\left(b^2+2bc+c^2\right).\)

\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(a+c\right)^2\left(đpcm\right)\)

Trả lời:

2x ( 2x - 1 )² - 3 ( x + 3 )( x - 3 ) - 4x ( x + 1 )²

= 2x ( 4x² - 4x + 1 ) - 3 ( x² - 9 ) - 4x ( x² + 2x + 1 )

= 8x³ - 8x² + 2x - 3x² + 27 - 4x³ - 8x² - 4x

= 4x³ - 19x² - 2x + 27

Ta có (n² + 3n - 1)(n + 2) - n³ + 2 

= n³ + 3n² - n + 2n² + 6n - 2 - n³ + 2

= 5n² + 5n = 5n(n + 1) \(⋮5\)

a) Ta có: (x + 1)(x + 3)(x + 5)(x + 7) + 15 = 0

<=> (x² + 8x + 7)(x² + 8x + 15) + 15 = 0

<=> (x² + 8x + 7)² + 8(x² + 8x + 7) + 15 = 0

<=> (x² + 8x +7 )²  + 3(x² + 8x + 7) + 5(x² + 8x + 7) + 15 = 0

<=> (x² + 8x + 7 + 3)(x² + 8x  + 7 +5) = 0

<=> (x² + 8x + 10)(x² + 8x + 12) = 0

<=> \(\orbr{\begin{cases}x^2+8x+10=0\\x^2+8x+12=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+4\right)^2-6=0\\x^2+6x+2x+12=0\end{cases}}\)

<=> \(\orbr{\begin{cases}\left(x+4\right)^2=6\left(1\right)\\\left(x+6\right)\left(x+2\right)=0\left(2\right)\end{cases}}\)

Giải (1) <=> \(\orbr{\begin{cases}x+4=\sqrt{6}\\x+4=-\sqrt{6}\end{cases}}\) <=> \(\orbr{\begin{cases}x=\sqrt{6}-4\\x=-\sqrt{6}-4\end{cases}}\)

Giải (2) <=> \(\orbr{\begin{cases}x=-6\\x=-2\end{cases}}\)

b) Ta có: (x² + x)(x² + x + 1) = 6

<=> (x² + x)² + (x²  + x) - 6 = 0

<=> (x² + x)² + 3(x² + x) - 2(x² + x) - 6 = 0

<=> (x² + x + 3)(x² + x - 2) = 0

<=> x² + 2x - x - 2 = 0 (vì x²  + x + 3 = (x + 1/2)^2 + 11/4 > 0)

<=> (x + 2)(x - 1) = 0

<=> \(\orbr{\begin{cases}x=-2\\x=1\end{cases}}\)

A=x(x-1)+(x+y)(y-x)

=x²-x+y²-x²

=y²-x

#H

Trả lời:

A = x ( x - 1 ) + ( x + y ) ( y - x )

= x² - x + y² - x²

= y² - x