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

=>\(x^2+y^2+z^2+2\left(xy+yz+xz\right)=z^2+y^2+x^2\)

=>2(xy+yz+xz)=0

=>xy+yz+xz=0

=>\(\frac{xy+yz+xz}{xyz}=0\)

=>\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\)

=>\(\frac{1}{x}+\frac{1}{y}=-\frac{1}{z}\)

\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}-\frac{3}{xyz}\)

\(=\left(\frac{1}{x}+\frac{1}{y}\right)^3-3\cdot\frac{1}{x}\cdot\frac{1}{y}\left(\frac{1}{x}+\frac{1}{y}\right)+\frac{1}{z^3}-\frac{3}{xyz}\)

\(=\left(\frac{1}{x}+\frac{1}{y}\right)^3+\left(\frac{1}{z}\right)^3-\frac{3}{xy}\cdot\frac{x+y}{xy}-\frac{3}{xyz}\)

\(=\left(-\frac{1}{z}\right)^3+\left(\frac{1}{z}\right)^3-\frac{3\left(x+y\right)}{x^2y^2}-\frac{3}{xyz}\)

\(=\frac{-3\left(x+y\right)}{\left(xy\right)^2}-\frac{3}{xyz}=\frac{-3z\left(x+y\right)-3xy}{\left(xyz\right)^2}=\frac{-3\left(xy+yz+xz\right)}{\left(xyz\right)^2}=0\)

=>\(\frac{1}{x^3}+\frac{1}{y^3}+\frac{1}{z^3}=\frac{3}{xyz}\)

Sửa đề: \(\left(a+b+c\right)^2=a^2+b^2+c^2\)

=>\(a^2+b^2+c^2+2ab+2ac+2bc=a^2+b_{}^2+c^2\)

=>2ab+2ac+2bc=0

=>ab+ac+bc=0

=>bc=-ab-ac; ab=-ac-bc; ac=-ab-bc

\(a^2+2bc\)

\(=a^2+bc+bc=a^2+bc-ab-ac\)

=a(a-b)-c(a-b)

=(a-b)(a-c)

\(b^2+2ac\)

\(=b^2+ac+ac\)

\(=b^2+ac-ab-bc\)

=b(b-a)+c(a-b)

=-b(a-b)+c(a-b)

=(a-b)(c-b)

\(c^2+2ab\)

\(=c^2+ab+ab\)

\(=c^2+ab-ac-bc\)

=c(c-a)-b(c-a)

=(c-a)(c-b)

\(P=\frac{a^2}{a^2+2bc}+\frac{b^2}{b^2+2ac}+\frac{c^2}{c^2+2ab}\)

\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}+\frac{b^2}{\left(a-b\right)\left(c-b\right)}+\frac{c^2}{\left(c-a\right)\left(c-b\right)}\)

\(=\frac{a^2}{\left(a-b\right)\left(a-c\right)}-\frac{b^2}{\left(a-b\right)\left(b-c\right)}+\frac{c^2}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{a^2\left(b-c\right)-b^2\left(a-c\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{a^2b-a^2c-b^2a+b^2c+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(a-b\right)\left(ab-ca-cb+c^2\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=\frac{\left\lbrack a\left(b-c\right)-c\left(b-c\right)\right\rbrack}{\left(a-c\right)\left(b-c\right)}\)

\(=\frac{\left(b-c\right)\left(a-c\right)}{\left(b-c\right)\left(a-c\right)}\)

=1

a: x-4=1

=>x=5

Thay x=5 vào B, ta được:

\(B=\dfrac{5+1}{5-3}=\dfrac{6}{2}=3\)

b: \(A=\dfrac{x}{x-3}-\dfrac{x+1}{x+3}+\dfrac{3x-3}{x-3}\)

\(=\dfrac{x+3x-3}{x-3}-\dfrac{x+1}{x+3}\)

\(=\dfrac{4x-3}{x-3}-\dfrac{x+1}{x+3}\)

\(=\dfrac{\left(4x-3\right)\left(x+3\right)-\left(x-3\right)\left(x+1\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{4x^2+12x-3x-9-\left(x^2-2x-3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{4x^2+9x-9-x^2+2x+3}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{3x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\)

c: \(M=B:A=\dfrac{3x^2+11x-6}{\left(x-3\right)\left(x+3\right)}:\dfrac{x}{x-3}\)

\(=\dfrac{3x^2+11x-6}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x-3}{x}=\dfrac{3x^2+11x-6}{x^2+3x}\)

M=5

=>\(5\left(x^2+3x\right)=3x^2+11x-6\)

=>\(5x^2+15x-3x^2-11x+6=0\)

=>\(2x^2-4x+6=0\)

=>\(x^2-2x+3=0\)

=>\(\left(x-1\right)^2+2=0\)(vô lý)

mình nghĩ là câu hỏi tu từ

Gọi vận tốc của xe máy là x(km/h)

(ĐK: x>0)

Vận tốc của ô tô là x+10(km/h)

Tổng vận tốc của hai xe là 140:2=70(km/h)

Do đó,ta có phương trình:

x+x+10=70

=>2x=60

=>x=30(nhận)

vậy: vận tốc xe máy là 30km/h

vận tốc ô tô là 30+10=40km/h

3 - 4x(25 - 2x) = 8x² + x - 300

3 - 100x + 8x² = 8x² + x - 300

-100x + 8x² - 8x² - x = -300 - 3

-101x = -303

x = -303 : (-101)

x = 3

Vậy S = {3}

ĐKXĐ: \(x\notin\left\{0;3;-3;-\dfrac{3}{2}\right\}\)

\(\dfrac{x^2-6}{x-3}+\dfrac{x^2+3x}{2x+3}\left(\dfrac{x}{x^2-9}-\dfrac{x+3}{x\left(x-3\right)}\right)\)

\(=\dfrac{x^2-6}{x-3}+\dfrac{x\left(x+3\right)}{2x+3}\cdot\left(\dfrac{x}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{x\left(x-3\right)}\right)\)

\(=\dfrac{x^2-6}{x-3}+\dfrac{x\left(x+3\right)}{2x+3}\cdot\dfrac{x^2-\left(x+3\right)^2}{x\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2-6}{x-3}+\dfrac{x^2-x^2-6x-9}{\left(2x+3\right)\left(x-3\right)}\)

\(=\dfrac{x^2-6}{x-3}-\dfrac{3}{x-3}=\dfrac{x^2-9}{x-3}=x+3\)

   Đây là dạng toán nâng cao chuyên đề phép chia đa thức, cấu trúc thi chuyên, thi học sinh giỏi. Hôm nay, Olm.vn sẽ hưỡng dẫn các em giải chi tiết dạng này bằng bezout như sau:

                   Giải:

F(\(x\)) = \(x^3\) + a\(x\) + b ⋮ (\(x\) + 1)(\(x\) + 2)

Theo bezout ta có: F(\(x\)) ⋮ (\(x\) + 1)(\(x\) + 2) khi và chỉ khi: \(\left\{{}\begin{matrix}F\left(-1\right)=0\\F\left(-2\right)=0\end{matrix}\right.\)

                      ⇒\(\left\{{}\begin{matrix}F\left(-1\right)=\left(-1\right)^3+a.\left(-1\right)+b=0\\F\left(-2\right)=\left(-2\right)^3+a.\left(-2\right)+b=0\end{matrix}\right.\)

                       \(\left\{{}\begin{matrix}-1-a+b=0\\-8-2a+b=0\end{matrix}\right.\)

                        \(\left\{{}\begin{matrix}-1-a+b-\left(-8-2a+b\right)=0\\-8-2a+b=0\end{matrix}\right.\)

                         \(\left\{{}\begin{matrix}-1-a+b+8+2a-b=0\\-8-2a+b=0\end{matrix}\right.\)

                         \(\left\{{}\begin{matrix}\left(-1+8\right)+\left(2a-a\right)+\left(b-b\right)=0\\-8-2a+b=0\end{matrix}\right.\)

                          \(\left\{{}\begin{matrix}7+a=0\\-8-2a+b=0\end{matrix}\right.\)

                           \(\left\{{}\begin{matrix}a=-7\\-8-2a+b=0\end{matrix}\right.\)

                             \(\left\{{}\begin{matrix}a=-7\\-8-2.\left(-7\right)+b=0\end{matrix}\right.\)

                              \(\left\{{}\begin{matrix}a=-7\\6+b=0\end{matrix}\right.\)

                               \(\left\{{}\begin{matrix}a=-7\\b=-6\end{matrix}\right.\)

Kết luận: \(x^3\) + a\(x\) + b ⋮ (\(x\) + 1)(\(x\) + 2) ⇔ a = -7; b = - 6

Vậy \(x^3\) + a\(x\) + b = \(x^3\) - 7\(x\) - 6