a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)

\(\Leftrightarrow\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4\left(a+b+c\right)}{a+b+c}=0\)

\(\Leftrightarrow\left(x-a-b-x\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)

b)đề bài như trên

\(\Leftrightarrow\left(\frac{x-a-b-c}{bc}\right)+\left(\frac{x-b}{ca}-\frac{1}{a}-\frac{1}{c}\right)+\left(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b}\right)=0\)

\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)

\(a,\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)

\(a,\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)

\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}-\frac{4a+4b+4c-4x}{a+b+c}=0\)

\(\Leftrightarrow\left(a+b+c-x\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\right)=0\)

\(\Leftrightarrow a+b+c-x=0\)Do \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\ne0\)

\(\Leftrightarrow x=a+b+c\)

Vậy phương trình có nghiệm \(x=a+b+c\)

pt <=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x+a+b+c}{a+b+c}=5\) (Cộng 4 vào mỗi vế)

   <=> \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x+a+b+c-5\left(a+b+c\right)}{a+b+c}=0\)

   <=>  \(\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4a-4b-4c}{a+b+c}=0\)

   <=>  \(\left(a+b+c-x\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}\right)=0\)

Áp dụng bất đẳng thức Cauchy - Schwarz dạng engel, ta có :

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{\left(1+1+1\right)^2}{a+b+c}=\frac{9}{a+b+c}>\frac{4}{a+b+c}\)

=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-\frac{4}{a+b+c}>0\)

Vậy phương trình trên có nghiệm là 

x = a + b + c 

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x = a + b + c

\(\Leftrightarrow\left(\frac{x-b-c}{a}-1\right)+\left(\frac{x-c-a}{b}-1\right)+\left(\frac{x-a-b}{c}-1\right)=0\\ \)

\(\Leftrightarrow\left(x-p\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=0\)

=> x=p=(a+b+c)

sao lại là p

Ký hiệu viết cho thôi

​

moi hok lop 6

ms hok lóp 7

ta co phuong trinh (X+X100/60+X200/60)/3=680 

giai pt ta duoc X=340

sao tr loi ben duoi lai nhay len ten vay troi

\(\Leftrightarrow\dfrac{x-a-b-c}{b+c}+\dfrac{x-b-a-c}{a+c}+\dfrac{x-c-a-b}{a+b}=0\)

\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=a+b+c\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=0\end{matrix}\right.\)

Xét \(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=0\)

\(\Leftrightarrow\dfrac{\left(a+b\right)\left(b+c\right)+\left(b+c\right)\left(c+a\right)+\left(a+b\right)\left(a+c\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}=0\)ĐK: \(\left\{{}\begin{matrix}a\ne-b\\b\ne-c\\c\ne-a\end{matrix}\right.\)

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

\(\Leftrightarrow a^2+b^2+c^2+3\left(ab+bc+ca\right)=0\)

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

\(\Leftrightarrow\left\{{}\begin{matrix}a+b+c=0\\ab+bc+ca=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\ab-\left(a+b\right)b-\left(a+b\right)a=0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}c=-\left(a+b\right)\\ab+a^2+b^2=0\end{matrix}\right.\)\(\Leftrightarrow a=b=c=0\)

Vậy với x=a+b+c hoặc a=b=c=0 thì pt thỏa mãn.

bạn có thể cho mình cách gải được k

cách giải là đi hỏi thầy,cô