x^4 + x^3 + 5x^2 + 8x + 2 - m x^2 - x + 5 x^2 + 2x + 2 x^4 - x^3 + 5x^2 2x^3 + 8x 2x^3 - 2x^2 + 10x 2x^2 - 2x + 2 2x^2 - 2x + 10 -8 - m

Để \(P⋮Q\)<=> -8 - m = 0

<=> m = -8

Bài làm

P = x⁴ + x³ + 5x² + 8x + 2 - m

Q = x² - x + 5 

Gọi H là thương trong phép chia P cho Q

Ta có : P bậc 4 , Q bậc 2 => H bậc 2

=> H có dạng x² + ax + b

Khi đó : P chia hết cho Q <=> P = Q.H

<=> x⁴ + x³ + 5x² + 8x + 2 - m = ( x² - x + 5 )( x² + ax + b )

<=> x⁴ + x³ + 5x² + 8x + 2 - m = x⁴ + ax³ + bx² - x³ - ax² - bx + 5x² + 5ax + 5b

<=> x⁴ + x³ + 5x² + 8x + 2 - m = x⁴ + ( a - 1 )x³ + ( b - a + 5 )x² + ( 5a - b )x + 5b

Đồng nhất hệ số ta có :

\(\hept{\begin{cases}a-1=1\\b-a+5=5\\5a-b=8\end{cases}};5b=2-m\)

=> \(\hept{\begin{cases}a=b=2\\m=-8\end{cases}}\)

Vậy m = -8

a) \(A=2.5\sqrt{2}-5\sqrt{4}-2.5\sqrt{2}=10\sqrt{2}-10-10\sqrt{2}=-10\)

b) đkxđ: \(x\ge1\)

\(\sqrt{x-1}=3\Leftrightarrow x-1=9\Leftrightarrow x=10\)

P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

= \(1-\frac{1}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

= \(1-\left(\frac{1}{x+3}+\frac{1}{x-2}+\frac{5}{\left(x-2\right)\left(x+3\right)}\right)\)

\(=1-\left(\frac{x+3+x-2+5}{\left(x-2\right)\left(x+3\right)}\right)=1-\frac{2x+6}{\left(x-2\right)\left(x+3\right)}=1-\frac{2\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(=1-\frac{2}{x-2}\)

Khi đó P = \(1-\frac{2}{x-2}\)

Khi P = -3/4

=> \(1-\frac{2}{x-2}=-\frac{3}{4}\)

=> \(\frac{2}{x-2}=\frac{7}{4}\)

=> 7(x - 2) = 2.4

=> 7(x - 2) = 8

=> x - 2 = 8/7

=> x = \(\frac{22}{7}\)

Vậy khi x = 22/7 thì P = -3/4

\(P=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}+\frac{1}{2-x}\)

\(=\frac{x+2}{x+3}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{1}{x-2}\)

\(=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x-2\right)\left(x+3\right)}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-12-x}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x-4}{x-2}\)

Neu P = -3/4 thi : 

\(\frac{x-4}{x-2}=-\frac{3}{4}\Leftrightarrow4x-16=-3x+6\Leftrightarrow7x=22\Leftrightarrow x=\frac{22}{7}\)

\(x^2-2xy+2y^2-x+8=\left(x-y-\frac{1}{2}\right)^2+\left(y-\frac{1}{2}\right)^2+\frac{15}{2}\ge\frac{15}{2}\)

Dấu "=" xảy ra khi \(x=1,y=\frac{1}{2}\)