a)    5x(x - 2000) - (x - 2000) = 0

tương đương (x - 2000)(5x - 1) = 0

tương đương x = 2000 hoặc x = 1/5

b)   x(x^2 -13) = 0

\(x\left(x-\sqrt{13}\right)\left(x+\sqrt{13}\right)=0\)

tương đương x = 0 hoặ x = \(\sqrt{13}\)hoặc x = \(-\sqrt{13}\)

x . ( 2x - 3 ) = 0 

<=> x = 0 hoặc 2x - 3 = 0

<=> x = 0 hoặc 2x = 3 

<=> x= 0 hoặc x = 3/2 

Vậy x € { 0 ; 3/2 }

 cho 2 cái bằng 0 nhé

x=0 hoặc x=3/2 

h mik nha

Ta có: \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2ab^2c+2bc^2a+2ca^2b\)

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

Vì a + b + c = 0

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

=> ĐPCM

VT :\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2+2\left(ab^2c+a^2bc+abc^2\right)\)

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

\(=a^2b^2+b^2c^2+c^2a^2\)= vp 

=> \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)(đpcm)

http://123link.pro/QfDJ

ĐKXĐ : \(x\ne0\)

Đặt \(x+\frac{1}{x}=y\Rightarrow x^2+\frac{1}{x^2}=y^2-2\)

\(\Rightarrow8\left(x^2+\frac{1}{x^2}\right)-34\left(x+\frac{1}{x}+51\right)=0\)có dạng \(8\left(y^2-2\right)-34y+51=0\)

\(\Leftrightarrow8y^2-34y+35=0\)

\(\Leftrightarrow\left(2y-5\right)\left(4y-7\right)=0\)

\(\Leftrightarrow y\in\left\{\frac{5}{2};\frac{7}{4}\right\}\)

+) Với \(y=\frac{5}{2}\)thì \(x+\frac{1}{x}=\frac{5}{2}\Leftrightarrow2x^2-5x+2=0\)

\(\Leftrightarrow\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}\)

+) Với \(y=\frac{7}{4}\)  thì \(x+\frac{1}{x}=\frac{7}{4}\Leftrightarrow4x^2-7x+4=0\)

\(\Leftrightarrow4\left(x-\frac{7}{8}\right)^2+\frac{15}{16}=0\) ( pt vô ngiệm)

Vậy ... 2 ; 1/2