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Đề là gì nhỉ?

Câu 1 nào em nhỉ?

Ta có: \(\frac{a^2}{a+b}+\frac{b^2}{b+c}+\frac{c^2}{c+a}-\left(\frac{a^2}{a+c}+\frac{b^2}{b+a}+\frac{c^2}{c+b}\right)\)

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

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

=a-b+b-c+c-a

=0

=>\(\frac{2020}{2021}-P=0\)

=>\(P=\frac{2020}{2021}\)

dễ mà

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i would like to be a writer

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

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

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

=>xy+yz+xz=0

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

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

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

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

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

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

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

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

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

I would like to be a wirter

I would like to be a writer

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

=6

    \(\dfrac{5}{9}\) x \(\dfrac{9}{11}\) x \(\dfrac{11}{7}\) x \(\dfrac{6}{5}\) x 7

=  \(\dfrac{5\times7\times9\times11}{5\times7\times9\times11}\) x 6

= 6

\(\overrightarrow{BA}=\left(1;1;2\right)\)

Pt AB có dạng: \(\left\{{}\begin{matrix}x=-2+t\\y=1+t\\z=1+2t\end{matrix}\right.\)

Pt mặt phẳng qua C và vuông góc AB có dạng: 

\(1\left(x-5\right)+y+2z=0\)

\(\Leftrightarrow x+y+2z-5=0\)

Tọa độ H thỏa mãn:

\(\left(-2+t\right)+\left(1+t\right)+2\left(1+2t\right)-5=0\)

\(\Rightarrow t=\dfrac{2}{3}\Rightarrow H\left(-\dfrac{4}{3};\dfrac{5}{3};\dfrac{7}{3}\right)\)