\(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}+\frac{1}{x}\)
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3 tháng 4 2020
a) \(\frac{a+b-x}{c}+\frac{b+c-x}{a}+\frac{c+a-x}{b}+\frac{4x}{a+b+c}=1\)
\(\Leftrightarrow\frac{a+b-x}{c}+1+\frac{b+c-x}{a}+1+\frac{c+a-x}{b}+1+\frac{4x}{a+b+c}-4=0\)
\(\Leftrightarrow\frac{a+b+c-x}{c}+\frac{a+b+c-x}{a}+\frac{a+b+c-x}{b}+\frac{4x-4\left(a+b+c\right)}{a+b+c}=0\)
\(\Leftrightarrow\left(x-a-b-x\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
3 tháng 4 2020
b)đề bài như trên
\(\Leftrightarrow\left(\frac{x-a-b-c}{bc}\right)+\left(\frac{x-b}{ca}-\frac{1}{a}-\frac{1}{c}\right)+\left(\frac{x-c}{ab}-\frac{1}{a}-\frac{1}{b}\right)=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\frac{1}{bc}+\frac{1}{ca}+\frac{1}{ab}\right)=0\)
19 tháng 7 2018
a)\(\frac{1}{a+b-x}\)=\(\frac{1}{a}\)+\(\frac{1}{b}\)-\(\frac{1}{x}\)\(\Leftrightarrow\)\(\frac{1}{a+b-x}\)+\(\frac{1}{x}\)=\(\frac{a+b}{ab}\)\(\Leftrightarrow\)\(\frac{x+a+b-x}{x\left(a+b-x\right)}\)=\(\frac{a+b}{ab}\)
\(\Leftrightarrow\)\(\frac{a+b}{xa+xb-x^2}\)=\(\frac{a+b}{ab}\)\(\Leftrightarrow\)\(xa+xb-x^2\)=\(ab\)\(\Leftrightarrow\)\(xa+xb-x^2-ab\)=\(0\)
\(\Leftrightarrow\)\(a\left(x-b\right)-x\left(x-b\right)=0\)\(\Leftrightarrow\)\(\left(x-b\right)\left(a-x\right)=0\)\(\Leftrightarrow\)\(x=b;x=a\)
b) \(\Leftrightarrow\)\(\frac{1}{\left(x+a-1\right)\left(x+a+1\right)}+\frac{1}{\left(x+a+1\right)\left(x-a+1\right)}\)=\(\frac{1}{\left(x-a-1\right)\left(x+a+1\right)}+\frac{1}{\left(x-a+1\right)\left(x+a-1\right)}\)\(\Leftrightarrow\)\(\frac{1}{\left(x+a-1\right)\left(x+a+1\right)}-\frac{1}{\left(x-a-1\right)\left(x+a+1\right)}\)=\(\frac{1}{\left(x-a+1\right)\left(x+a-1\right)}-\frac{1}{\left(x+a+1\right)\left(x-a+1\right)}\)\(\Leftrightarrow\)\(\frac{1}{\left(x+a+1\right)}\left(\frac{1}{x+a-1}-\frac{1}{x-a-1}\right)\)=\(\frac{1}{x-a+1}\left(\frac{1}{x+a-1}-\frac{1}{x+a+1}\right)\)\(\Leftrightarrow\)\(\frac{1}{x+a+1}.\frac{-2a}{\left(x+a-1\right)\left(x-a-1\right)}=\frac{1}{x-a+1}.\frac{2}{\left(x+a-1\right)\left(x+a+1\right)}\)(Quy dong phan so ttrong dau ngoac)
\(\Leftrightarrow\)\(\frac{-2a}{x-a-1}=\frac{2}{x-a+1}\)\(\Leftrightarrow\)\(-2a\left(x-a+1\right)=2\left(x-a-1\right)\)\(\Leftrightarrow\)\(-ax+a^2-a=x-a-1\)\(\Leftrightarrow\)\(-ax-x+a^2-1=0\)\(\Leftrightarrow\)\(\left(a+1\right)\left(-x+a-1\right)=0\)
neu a+1=0 thi phuong trinh co vo so nghiem, neu a+1\(\ne\)0 thi x=a-1
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 7 2020
a/ \(\Leftrightarrow\frac{x+y}{xy}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2+y^2-2xy\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Vậy BĐT đã cho đúng
b/ \(\frac{a}{a+b^2}=\frac{a}{a\left(a+b+c\right)+b^2}=\frac{a}{a^2+b^2+a\left(b+c\right)}\le\frac{a}{2ab+a\left(b+c\right)}=\frac{1}{b+b+b+c}\)
\(\Rightarrow\frac{a}{a+b^2}=\frac{1}{b+b+b+c}\le\frac{1}{16}\left(\frac{1}{b}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{16}\left(\frac{3}{b}+\frac{1}{c}\right)\)
Tương tự: \(\frac{b}{b+c^2}\le\frac{1}{16}\left(\frac{3}{c}+\frac{1}{a}\right)\) ; \(\frac{c}{c+a^2}\le\frac{1}{16}\left(\frac{3}{a}+\frac{1}{c}\right)\)
Cộng vế với vế:
\(VT\le\frac{1}{16}\left(\frac{4}{a}+\frac{4}{b}+\frac{4}{c}\right)=\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Dấu "=" xảy ra khi \(a=b=c=\frac{1}{3}\)
10 tháng 10 2016
a)Có: \(\frac{a}{x}+\frac{b}{x-1}+\frac{c}{x+1}=\frac{a\left(x-1\right)\left(x+1\right)+bx\left(x+1\right)+cx\left(x-1\right)}{x\left(x-1\right)\left(x+1\right)}\)
\(\frac{a\left(x^2-1\right)+bx^2+bx+cx^2+cx}{x\left(x^2-1\right)}=\frac{ax^{2\:}-a+bx^2+bx+cx^2-cx}{x^3-x}\)
\(=\frac{\left(a+b+c\right)x^2+\left(b-c\right)x-a}{x^3-x}\)
Do đó: \(\frac{6x^2-x-1}{x^3-x}=\frac{\left(a+b+c\right)x^2+\left(b-c\right)x-a}{x^3-x}\)
Đồng nhất hai phân thức trên ta được:
\(\begin{cases}a+b+c=6\\b-c=-1\\a=1\end{cases}\)\(\Leftrightarrow\begin{cases}a=1\\b=2\\c=3\end{cases}\)
Phần b tương tự
PA
25 tháng 2 2017
\(\frac{1}{a+b-x}=\frac{1}{a}+\frac{1}{b}-\frac{1}{x}\) (ĐKXĐ: x \(\ne\) 0 và x \(\ne\) a + b)
<=> \(\frac{1}{a+b-x}+\frac{1}{x}-\frac{1}{a}-\frac{1}{b}=0\)
<=> \(\frac{x}{x\left(a+b-x\right)}+\frac{a+b-x}{x\left(a+b-x\right)}-\frac{b}{ab}-\frac{a}{ab}\)
<=> \(\frac{a+b}{x\left(a+b-x\right)}-\frac{a+b}{ab}=0\)
<=> \(\left(a+b\right)\left(\frac{1}{x\left(a+b-x\right)}-\frac{1}{ab}\right)=0\)
* Nếu a = - b thì tập nghiệm cuả pt là S = R
* Nếu a \(\ne\) b thì \(\frac{1}{x\left(a+b-x\right)}-\frac{1}{ab}=0\)
<=> \(\frac{ab}{abx\left(a+b-x\right)}-\frac{x\left(a+b-x\right)}{abx\left(a+b-x\right)}=0\)
<=> \(\frac{ab-\text{ax}-bx+x^2}{abx\left(a+b-x\right)}=0\)
<=> \(\frac{b\left(a-x\right)-x\left(a-x\right)}{abx\left(a+b-x\right)}=0\)
<=> \(\frac{\left(a-x\right)\left(b-x\right)}{abx\left(a+b-x\right)}=0\)
<=> \(\left[\begin{matrix}a-x=0\\b-x=0\end{matrix}\right.\)
<=> \(\left[\begin{matrix}x=a\\x=b\end{matrix}\right.\)
Vậy tập nghiệm của pt là S = {a ; b}
PA
25 tháng 2 2017
\(\frac{x+1}{x^2+x+1}-\frac{x-1}{x^2-x+1}=\frac{3}{x\left(x^4+x^2+1\right)}\) (ĐKXĐ: x \(\ne\) 0
<=> \(\frac{x\left(x+1\right)\left(x^2-x+1\right)}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}-\frac{x\left(x-1\right)\left(x^2+x+1\right)}{x\left(x^2-x+1\right)\left(x^2+x+1\right)}=\frac{3}{x\left(x^2+x+1\right)\left(x^2-x+1\right)}\)
=> \(\left(x^4+x\right)-\left(x^4-x\right)=3\)
<=> \(2x-3=0\)
<=> \(x=\frac{3}{2}\) (nhận)
Vậy S = {1,5}