\(\left(-1\frac{1}{2}\right)\left(-1\frac{1}{3}\right)\left(-1\frac{1}{4}\right)...\left(-1\frac{1}{2003}\right)\left(-1\frac{1}{2004}\right)\)

\(=-\frac{3}{2}.\frac{4}{3}.\frac{5}{4}.....\frac{2004}{2003}.\frac{2005}{2004}\)

\(=-\frac{3.4.5.....2004.2005}{2.3.4.....2003.2004}=\frac{-2005}{2}\)

Do \(a^3+b^3+c^3=3abc\)
\(\Rightarrow a^3+b^3+c^3-3abc=0\)
\(\Rightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ac\right)=0\)
Mà \(a+b+c\ne0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ac=0\)
\(\Rightarrow a^2+b^2+c^2=ab+bc+ac\)
Khi đó:
\(\left(a+b+c\right)^2=a^2+b^2+c^2+2\left(ab+bc+ac\right)\)\(=\left(a^2+b^2+c^2\right)+2\left(a^2+b^2+c^2\right)=3\left(a^2+b^2+c^2\right)\)
Vậy: \(N=\frac{a^2+b^2+c^2}{\left(a+b+c\right)^2}=\frac{a^2+b^2+c^2}{3\left(a^2+b^2+c^2\right)}=\frac{1}{3}\)

\(\left(4\frac{46}{65}+x\right).1\frac{1}{12}=5,75\)

\(\Rightarrow\frac{306}{65}+x.\frac{13}{12}=\frac{23}{4}\)

\(\Rightarrow\frac{51}{10}+\frac{13}{12}x=\frac{23}{4}\)

\(\Rightarrow306x=65x=345\)

\(\Rightarrow65x=39\)

\(\Rightarrow x=\frac{3}{5}\)

b, \(\frac{5}{4}-\left(\frac{3}{2}x+0,5\right)=1\frac{1}{4}\)

\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-0,5=\frac{5}{4}\)

\(\Rightarrow\frac{5}{4}-\frac{3}{2}x-\frac{1}{2}=\frac{5}{4}\)

\(\Rightarrow\frac{3}{4}-\frac{3}{2}x=\frac{5}{4}\)

\(\Rightarrow3-6x=5\)

\(\Rightarrow-6x=2\)

\(\Rightarrow x=-\frac{1}{3}\)

đề học sinh giỏi hồi chiều ak!!!!!!!!! khó v:

a, => |5/3.x| = 1/6

=> 5/3.x = -1/6 hoặc 5/3.x = 1/6

=> x = -1/10 hoặc x = 1/10

Tk mk nha

Trả lời

Hình như b viết thiếu đề hay sao ý

Ng ta ko cho 3a^2+3b^2 bằng bao nhiêu ag

Ta có

3a^2+3b^2=10ab

3a^2-10ab+3b^2=0

3a^2-9ab-ab+3b^2=0

3a(a-3b)-b(a-3b)=0

(a-3b)(3a-b)=0

=>a-3b=0=>a=3b

=>3a-b=0=>3a=b

thay vào biểu thức

P=a-b/a+b=3b-b/3b+b=2b/4b=1/2

vậy P=1/2

Có:

\(a^3+b^3+c^3=3abc\\\Leftrightarrow a^3+b^3+c^3-3abc=0\\\Leftrightarrow (a+b)^3+c^3-3ab(a+b)-3abc=0\\\Leftrightarrow (a+b+c)^3-3(a+b)c(a+b+c)-3ab(a+b+c)=0\\\Leftrightarrow (a+b+c)[(a+b+c)^2-3(a+b)c-3ab]=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2+2ab+2bc+2ac-3ac-3bc-3ab)=0\\\Leftrightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=0\\\Leftrightarrow a^2+b^2+c^2-ab-bc-ac=0(vì.a+b+c\ne0)\\\Leftrightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ac=0\\\Leftrightarrow (a^2-2ab+b^2)+(b^2-2bc+c^2)+(a^2-2ac+c^2)=0\\\Leftrightarrow (a-b)^2+(b-c)^2+(a-c)^2=0\)

Ta thấy: \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\forall a,b\\\left(b-c\right)^2\ge0\forall b,c\\\left(a-c\right)^2\ge0\forall a,c\end{matrix}\right.\)

\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\forall a,b,c\)

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

nên: \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\a-c=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a=b\\b=c\\a=c\end{matrix}\right.\Leftrightarrow a=b=c\)

Thay \(a=b=c\) vào \(A\), ta được:

\(A=\dfrac{\left(2016+\dfrac{a}{a}\right)+\left(2016+\dfrac{b}{b}\right)+\left(2016+\dfrac{c}{c}\right)}{2017^3}\left(a,b,c\ne0\right)\)

\(=\dfrac{2016+1+2016+1+2016+1}{2017^3}\)

\(=\dfrac{2016\cdot3+1\cdot3}{2017^3}\)

\(=\dfrac{3\cdot\left(2016+1\right)}{2017^3}\)

\(=\dfrac{3}{2017^2}\)

Vậy: ...

Ok , mình sẽ làm !

Ta có :

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

\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}\)

\(\Rightarrow\frac{a+b}{c}-1+1=\frac{b+c}{a}-1+1=\frac{c+a}{b}-1+1\)

\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}\left(1\right)\)

+) Trường hợp 1 : \(a+b+c=0\)

\(\Rightarrow\hept{\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}}\)

Ta có :

 \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-a}{a}.\frac{-c}{c}.\frac{-b}{b}\)

\(\Leftrightarrow P=-1.\left(-1\right).\left(-1\right)=-1\)

+) Trường hợp 2 : \(a+b+c\ne0\)

Áp dụng tính chất của dãy tỉ số bằng nhau cho ( 1 ) , ta có :

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

\(\Rightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)

Ta lại có :

 \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)

\(\Leftrightarrow P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{c+b}{b}\)

\(\Leftrightarrow P=2.2.2=8\)

Vậy....................

Đề sai nhé bạn ! Bạn kiểm tra lại!

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