cho ba số dương a,b,c thỏa mãn \(\dfrac{a^{ }}{a+b}+\dfrac{b^2}{b+c}+\dfrac{c^2}{c+a}=\dfrac{1015}{2016}\)
tính giá trị biểu thức \(\dfrac{a^2}{a+c}+\dfrac{b^2}{a+b}+\dfrac{c^2}{c+b}\)
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ta có:\(\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=1\)
=>\(\left[\dfrac{a}{c+b}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\right].\left(a+b+c\right)=a+b+c\)
=>\(\dfrac{a^2}{c+b}+\dfrac{ab}{a+c}+\dfrac{ac}{a+b}+\dfrac{b^2}{a+c}+\dfrac{ba}{c+d}+\dfrac{bc}{a+b}+\dfrac{ca}{c+d}+\dfrac{cb}{a+c}+\dfrac{c^2}{a+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+\dfrac{b\left(a+c\right)}{a+c}+\dfrac{c\left(a+b\right)}{a+b}+\dfrac{a\left(b+c\right)}{c+b}=a+b+c\)=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}+a+b+c=a+b+c\)
=>\(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=0\)
chúc bạn học tốt ^ ^
Ta có \(\dfrac{a}{2009}\)=\(\dfrac{b}{2010}\)=\(\dfrac{c}{2011}\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{c-a}{2011-2009}=\dfrac{c-a}{2}\left(1\right)\)
\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{a-b}{2009-2010}=\dfrac{a-b}{-1}\)(2)\(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=\dfrac{b-c}{2010-2011}=\dfrac{b-c}{-1}\left(3\right)\)
Từ (1),(2),(3) \(_{\Rightarrow}\)\(\dfrac{c-a}{2}=\dfrac{a-b}{-1}=\dfrac{b-c}{-1}\Rightarrow\dfrac{\left(a-c\right)^{ }2}{2^{ }2}=\dfrac{\left(a-b\right)}{-1}\times\dfrac{\left(b-c\right)}{-1}\)
\(\Rightarrow\dfrac{\left(a-c\right)^2}{4}=\dfrac{\left(a-b\right)\times\left(b-c\right)}{1}\Rightarrow4\left(a-b\right).\left(b-c\right)=\left(a-c\right)^2\)
\(\Rightarrow M=4\left(a-b\right).\left(a-c\right)-\left(c-a\right)^2=0\)
Vậy M = 0
đặt \(\dfrac{a}{2009}=\dfrac{b}{2010}=\dfrac{c}{2011}=k\) ta có:
\(\Rightarrow a=2009k\left(1\right)\\ \Rightarrow b=2010k\left(2\right)\\ \Rightarrow c=2011k\left(3\right)\)
thay 1;2;3 vào M ta có:
\(M=4\left(2009k-2010k\right)\left(2010k-2011k\right)-\left(2011k-2009k\right)^2\\ \Rightarrow M=4.\left(-k\right)\left(-k\right)-\left(2k\right)^2\\ \Rightarrow M=4k^2-\left(2k\right)^2\\ \Rightarrow M=\left(2k\right)^2-\left(2k\right)^2\\ \Rightarrow M=0\)Vậy M = 0
Tu \(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
Hay \(\dfrac{1}{a}=\dfrac{1}{b}=\dfrac{1}{c}\Leftrightarrow a=b=c\)
Thay vao M ta co: \(M=\dfrac{a\cdot a+a\cdot a+a\cdot a}{a^2+a^2+a^2}=\dfrac{2019}{2019}=\dfrac{2018}{2018}=\dfrac{2017}{2017}=\dfrac{2016}{2015+1}=1\)
=>\(\dfrac{10a+b}{10b+c}=\dfrac{b}{c}\)
=>10ac+bc=10b^2+bc
=>ac=b^2
=>a/b=b/c=k
=>a=bk; b=ck
=>a=ck^2; b=ck
\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{c^2k^4+c^2k^2}{c^2k^2+c^2}=k^2\)
\(\dfrac{a}{c}=\dfrac{ck^2}{c}=k^2\)
=>\(\dfrac{a^2+b^2}{b^2+c^2}=\dfrac{a}{c}\)
a, ĐKXĐ: x≠±3
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\dfrac{-1}{x^2}\)
b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:
\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4
c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x2>0 (Đúng với mọi x)
Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)