\(\frac{2}{3}\sqrt{3}\)- \(\frac{1}{4}\sqrt{18}\)+ \(\frac{2}{5}\sqrt{2}-\frac{1}{4}\sqrt{12}\)

= \(\frac{2}{3}\sqrt{3}-\frac{3}{4}\sqrt{2}+\frac{2}{5}\sqrt{2}-\frac{2}{4}\sqrt{3}\)

= \(\sqrt{3}\left(\frac{2}{3}-\frac{1}{2}\right)\)- \(\sqrt{2}\left(\frac{3}{4}-\frac{2}{5}\right)\)

= \(\frac{\sqrt{3}}{6}\)- \(\frac{7}{20}\sqrt{2}\)

kq ra hơi kì

#mã mã#

\(\left(a+b+c\right)^2+12=4\left(a+b+c\right)\)\(+2\left(ab+bc+ac\right)=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2bc+2ac+12-4\left(a+b+c\right)-2\left(ab+bc+ac\right)=0\)

\(\Rightarrow a^2+b^2+c^2+2\left(ab+ac+bc\right)-2\left(ab+bc+ac\right)-4\left(a+b+c\right)+12=0\)

\(\Rightarrow a^2+b^2+c^2-4a-4b-4c+12=0\)

\(\Rightarrow\left(a^2-4a+4\right)+\left(b^2-4b+4\right)+\left(c^2-4c+4\right)=0\)

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

Ta co: \(\left(a-2\right)^2\ge0\forall a\)

\(\left(b-2\right)^2\ge0\forall b\)

\(\left(c-2\right)^2\ge0\forall c\)

\(\Rightarrow\left(a-2\right)^2+\left(b-2\right)^2+\left(c-2\right)^2=0\Leftrightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}\Leftrightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}\Leftrightarrow}a=b=c=2}\left(\right)\)

(đpcm) 

Mình nghĩ thế này nhé bạn! 

(a + b + c )² + 12 = 4 (a + b +c ) + 2(ab + bc +ac)

  \(\Leftrightarrow\)a² + b² + c² + 2ab + 2bc + 2ac + 12 = 4a + 4b + 4c + 2ab + 2ac + 2bc

\(\Leftrightarrow\) a² + b² + c² - 4a - 4b -4c +12 = 0

\(\Leftrightarrow\)a² - 4a + 4 + b² - 4b + 4 + c² - 4c + 4 =0

\(\Leftrightarrow\)( a -2 )² + (b-2)² + (c-2)² = 0

ta có (a-2 )² \(\ge0\forall a\)

          (b - 2 )² \(\ge0\forall b\)

            (c - 2 )² \(\ge0\forall c\)

mà (a-2)² + (b-2)² + (c-2)² = 0

\(\Rightarrow\hept{\begin{cases}\left(a-2\right)^2=0\\\left(b-2\right)^2=0\\\left(c-2\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}a-2=0\\b-2=0\\c-2=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a=2\\b=2\\c=2\end{cases}\left(đpcm\right)}\)

vậy................... khi a=b = c =2

#mã mã#

chăm vl

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ab+ac}{abc}=2\)

\(\frac{bc+ab+ac}{a+b+c}=2\Leftrightarrow bc+ab+ac=2\left(a+b+c\right)\)

\(\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4=\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}\)( * )

Để \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=2\)thì \(2\left(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}\right)=2\Leftrightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=1\)

\(\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{a^2bc+bac^2+ab^2c}{\left(abc\right)^2}=\frac{abc\left(a+b+c\right)}{\left(abc\right)^2}=\frac{a+b+c}{abc}\)

mà a + b + c = abc \(\Rightarrow\frac{1}{bc}+\frac{1}{ab}+\frac{1}{ac}=\frac{abc}{abc}=1\Leftrightarrow\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\)

thay \(\frac{2}{bc}+\frac{2}{ab}+\frac{2}{ac}=2\) vào ( * ) ta được \(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2=2\left(đpcm\right)\)

\(\text{Ta có: }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=\frac{bc.ac+ab.ac+ab.bc}{ab.bc.ac}\)

\(=\frac{abc.\left(a+b+c\right)}{a^2b^2c^2}=\frac{a+b+c}{abc}=1\left(\text{vì }a+b+c=abc\right)\)

\(\text{Lại có: }\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=2\Leftrightarrow\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=4\)

\(\Leftrightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=4-2.\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=2\text{ vì }\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}=1\text{ từ}\left(1\right)\)

Vậy ...

 lấy sách toán lớp 6 tập 1 tập 2 mở từng bài ra mà chép vào 

k có thì lên tìm sách lớp 6 tập 1 nó cho đọc online

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