giải phương trình với các tham số a,b,c:
\(\frac{x-a}{b+c}+\frac{x-b}{c+a}+\frac{x-c}{a+b}=3\)
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SC
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TT
3 tháng 3 2020
a) ĐKXĐ : \(x\ne\pm a\).
Với \(a=-3\) khi đó ta có pt :
\(A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left(-9+1\right)}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{\left(x-3\right)\left(x+3\right)-\left(x+3\right)\left(-3-x\right)}{\left(-3-x\right)\left(-3+x\right)}+\frac{24}{\left(-3-x\right)\left(-3+x\right)}=0\)
\(\Rightarrow x^2-9-\left(-3x-x^2-9-3x\right)+24=0\)
\(\Leftrightarrow2x^2+6x+24=0\)
\(\Leftrightarrow x^2+3x+12=0\) ( vô nghiệm )
Phần b) tương tự.
3 tháng 3 2020
\(A=\frac{x+a}{a-x}-\frac{x-a}{a+x}=\frac{a\left(3x+1\right)}{a^2-x^2}\)
\(=\frac{x+a}{a-x}+\frac{x-a}{a+x}=\frac{a\left(3+1\right)}{\left(a-x\right)\left(a+x\right)}\)
\(=\frac{\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)}{\left(a-x\right)\left(a+1\right)}=\frac{a\left(3a+1\right)}{\left(a+x\right)\left(a-x\right)}\)
\(\Leftrightarrow\left(x+a\right)^2+\left(x-a\right)\left(a-x\right)=a\left(3a+1\right)\)
\(\Leftrightarrow x^2+2ax+a^2-ax-x^2-a^2+ax=3a^2+a\)
\(\Leftrightarrow2ax=3a^2+a\)
\(\Leftrightarrow x=\frac{3a^2+a}{2a}\left(a\ne0\right)\)
a) Khi x=-3 => \(x=\frac{3\cdot\left(-3\right)^2-3}{2\left(-3\right)}=-13\)
b) a=1
\(\Leftrightarrow x=\frac{3\cdot1^2+1}{2\cdot1}=2\)
MN
4 tháng 3 2020
a) \(ĐKXĐ:x\ne\pm3\)
Với a = -3
\(\Leftrightarrow A=\frac{x-3}{-3-x}-\frac{x+3}{-3+x}=\frac{-3\left[3.\left(-3\right)+1\right]}{\left(-3\right)^2-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}=\frac{24}{9-x^2}\)
\(\Leftrightarrow\frac{3-x}{x+3}-\frac{x+3}{x-3}+\frac{24}{x^2-9}=0\)
\(\Leftrightarrow\frac{-\left(x-3\right)^2-\left(x+3\right)^2+24}{x^2-9}=0\)
\(\Leftrightarrow-x^2+6x-9-x^2-6x-9+24=0\)
\(\Leftrightarrow-2x^2+6=0\)
\(\Leftrightarrow x^2=3\)
\(\Leftrightarrow x=\pm\sqrt{3}\)(tm)
Vậy với \(a=-3\Leftrightarrow x\in\left\{\sqrt{3};-\sqrt{3}\right\}\)
b) \(ĐKXĐ:x\ne\pm1\)
Với a = 1
\(\Leftrightarrow A=\frac{x+1}{1-x}-\frac{x-1}{1+x}=\frac{3+1}{1-x^2}\)
\(\Leftrightarrow\frac{x+1}{1-x}-\frac{x-1}{1+x}+\frac{4}{x^2-1}=0\)
\(\Leftrightarrow\frac{-\left(x+1\right)^2-\left(x-1\right)^2+4}{x^2-1}=0\)
\(\Leftrightarrow-x^2-2x-1-x^2+2x-1+4=0\)
\(\Leftrightarrow-2x^2+2=0\)
\(\Leftrightarrow x^2=1\)
\(\Leftrightarrow x=\pm1\)(ktm)
Vậy với \(a=1\Leftrightarrow x\in\varnothing\)
c) \(ĐKXĐ:a\ne\pm\frac{1}{2}\)
Thay \(x=\frac{1}{2}\)vào phương trình, ta đươc :
\(A=\frac{\frac{1}{2}+a}{a-\frac{1}{2}}-\frac{\frac{1}{2}-a}{a+\frac{1}{2}}=\frac{a\left(3a+1\right)}{a^2-\frac{1}{4}}\)
\(\Leftrightarrow\frac{a+\frac{1}{2}}{a-\frac{1}{2}}+\frac{a-\frac{1}{2}}{a+\frac{1}{2}}-\frac{3a^2+a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow\frac{\left(a+\frac{1}{2}\right)^2+\left(a-\frac{1}{2}\right)^2-3a^2-a}{a^2-\frac{1}{4}}=0\)
\(\Leftrightarrow a^2+a+\frac{1}{4}+a^2-a+\frac{1}{4}-3a^2-a=0\)
\(\Leftrightarrow-a^2-a+\frac{1}{2}=0\)
\(\Leftrightarrow a^2+a-\frac{1}{2}=0\)
\(\Leftrightarrow\left(a+\frac{1}{2}\right)^2-\frac{3}{4}=0\)
\(\Leftrightarrow\orbr{\begin{cases}a=\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{\sqrt{3}-1}{2}\\a=-\frac{\sqrt{3}}{2}-\frac{1}{2}=\frac{-\sqrt{3}-1}{2}\end{cases}}\)(TM)
Vậy với \(x=\frac{1}{2}\Leftrightarrow a\in\left\{\frac{\sqrt{3}-1}{2};\frac{-\sqrt{3}-1}{2}\right\}\)
RR
16 tháng 5 2018
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
27 tháng 3 2020
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\(\Leftrightarrow\dfrac{x-a-b-c}{b+c}+\dfrac{x-b-a-c}{a+c}+\dfrac{x-c-a-b}{a+b}=0\)
\(\Leftrightarrow\left(x-a-b-c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=a+b+c\\\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}=0\end{matrix}\right.\)
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.