\(a+b+c=0\Rightarrow\left(a+b\right)=-c\Rightarrow\left(a+b\right)^2=\left(-c\right)^2\Leftrightarrow a^2+b^2+2ab=c^2\)

\(\Rightarrow a^2+b^2-c^2=-2ab\)

Tương tự: \(b^2+c^2-a^2=-2bc;\text{ }c^2+a^2-b^2=-2ca\)

\(\Rightarrow VT=-\frac{1}{2bc}-\frac{1}{2ca}-\frac{1}{2ab}=-\frac{1}{2}\left(\frac{a+b+c}{abc}\right)=0=VP\text{ (đpcm)}\)

b là trung bình cộng của a và c \(\Rightarrow b=\frac{a+c}{2}\)

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

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

\(\left(5a-3b+8c\right)\left(5a-3b-8c\right)=\left(5a-3b\right)^2-64c^2=25a^2-2.5.3ab+9b^2-16\left(a^2-b^2\right)\)

\(=9a^2-2.3.5ab+25b^2=\left(3a-5b\right)^2\)

\(0=\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zx\right)=x^2+y^2+z^2+2.0\)

\(\Rightarrow x^2+y^2+z^2=0\Rightarrow x=y=z=0\)

\(B=\left(-1\right)^{2007}+0^{2008}+1^{2009}=0\)

Cách 1: 

\(+\left(a^2+b^2+c^2\right)^2=a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(+0=\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ca\right)=14+2\left(ab+bc+ca\right)\)

\(\Rightarrow ab+bc+ca=-7\)

\(+\left(-7\right)^2=\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab.bc+bc.ca+ca.ab\right)\)

\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=a^2b^2+b^2c^2+c^2a^2+2abc.0\)

\(\Rightarrow a^2b^2+b^2c^2+c^2a^2=49\)

Từ các điều trên suy ra:

\(14^2=a^4+b^4+c^4+2.49\)

\(\Rightarrow a^4+b^4+c^4=14^2-2.49=98\)

Cách 2:

\(+a+b+c=0\Rightarrow a+b=-c\)

\(+14=a^2+b^2+c^2=a^2+b^2+\left(-a-b\right)^2=a^2+b^2+a^2+b^2+2ab=2\left(a^2+b^2+ab\right)\)

\(\Rightarrow a^2+b^2+ab=7\)

\(+a^4+b^4+c^4=a^4+b^4+\left[-\left(a+b\right)\right]^4=\left(a^2+b^2\right)^2-2a^2b^2+\left(a^2+b^2+2ab\right)^2\)

\(=\left(a^2+b^2\right)^2-2a^2b^2+\left(a^2+b^2\right)^2+4\left(a^2+b^2\right).ab+4a^2b^2\)

\(=2\left(a^2+b^2\right)^2+4\left(a^2+b^2\right).ab+2a^2b^2\)

\(=2\left(a^2+b^2+ab\right)^2\)

\(=2.7^2=98\)

Bình phương lên biến đổi tương đương là ra (chắc vậy).

Hình như thiếu x+y+z = ko ?

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{1}{x+y+z}\)

\(\Leftrightarrow\left(x+y+z\right)\left(xy+yz+zx\right)=xyz\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)=0\)

\(\Leftrightarrow x=-y\text{ hoặc }y=-z\text{ hoặc }z=-x\)

\(+\text{Nếu }x=-y\text{ thì }x^8=\left(-y\right)^8=y^8\Rightarrow x^8-y^8=0\Rightarrow M=\frac{3}{4}\)

\(+\text{Nếu }y=-z\text{ thì }y^9=\left(-z\right)^9=-z^9\Rightarrow y^9+z^9=0\Rightarrow M=\frac{3}{4}\)

\(+\text{Nếu }z=-x\text{ thì }z^{10}=\left(-x\right)^{10}=x^{10}\Rightarrow z^{10}-x^{10}=0\Rightarrow M=\frac{3}{4}\)

\(\text{Vậy M}=\frac{3}{4}.\)

ui trui, trieu dang gioi zay ma con hoi , la thiek

cây ba bù là cu ba mày chứ gì ngu gì mà trả lời