Với a,b,c dương , giải phương trình
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
Cho 3 số dương a , b , c . Giải phương trình : \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
ta co phuong trinh (X+X100/60+X200/60)/3=680
giai pt ta duoc X=340
Cho 3 số dương a,b,c
a, Chứng minh rằng : (a+b+c) . (\(\frac{1}{a}\)+\(\frac{1}{b}+\frac{1}{c}\))\(\ge\)9
b, Giải phương trình : \(\frac{a+b-c}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}\)=1
a)Áp dụng BDT AM-GM ta có:
\(a+b+c\ge3\sqrt[3]{abc}\)
\(\Rightarrow\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{a}\cdot\frac{1}{b}\cdot\frac{1}{c}}=3\sqrt[3]{\frac{1}{abc}}\)
Nhân theo vế ta có:
\(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge3\sqrt[3]{abc}\cdot3\sqrt[3]{\frac{1}{abc}}=9\)
Dấu "=" xảy ra khi \(a=b=c\)
Giải phương trình
a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
b)\(\frac{1}{2}\left(\frac{x-a}{bc}+\frac{x-b}{ca}+\frac{x-c}{ab}\right)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)
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 \(x=a+b+c\)
Giải phương trình: \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
giải phương trình với các tham số a,b,c
\(\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=\frac{3x}{a+b+c}\)
giải phương trình với các tham số a,b,c:
\(\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3\)
\(\Leftrightarrow\dfrac{x-a-b-c}{b+c}+\dfrac{x-b-a-c}{a+c}+\dfrac{x-c-a-b}{a+b}=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a+b+c\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=0\end{matrix}\right.\)
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.
giải và biện luận, chứng minh phương trình sau
\(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
Giải phương trình với ẩn x
a) \(\frac{x+a-1}{a+2}+\frac{x-a}{a-2}+\frac{x-a}{4-a^2}=0\)
b) \(\frac{x-a}{b+c}+\frac{x-b}{a+c}+\frac{x-c}{a+b}=3\)
Mn júp mk vs
Giải phương trình với tham sooa a,b,c
\(\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}+3\)