Chứng minh rằng nếu \(\dfrac{a}{b}\)<\(\dfrac{c}{d}\) (b;d > 0) thì \(\dfrac{a}{b}\)<\(\dfrac{a+c}{b+d}\)<\(\dfrac{c}{d}\)
cho a,b,c là các số thực thỏa man: a+\(\dfrac{1}{b}=b+\dfrac{1}{c}=c+\dfrac{1}{a\backslash}\).
a) chứng minh nếu a,b,c đôi một khác nhau thì a2b2c2=1
b) chứng minh rằng nếu a,b,c>0 thì a=b=c
Chứng minh rằng nếu a,b,c \(\ge\)0 và abc=1 thì
\(\dfrac{1}{2+a}+\dfrac{1}{2+b}+\dfrac{1}{2+c}\le1\)
\(\Leftrightarrow\dfrac{\left(a+2\right)\left(b+2\right)+\left(b+2\right)\left(c+2\right)+\left(c+2\right)\left(a+2\right)}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\le1\)
\(\Leftrightarrow\dfrac{ab+bc+ca+4\left(a+b+c\right)+12}{abc+2\left(ab+bc+ca\right)+4\left(a+b+c\right)+8}\le1\)
\(\Leftrightarrow ab+bc+ca+12\le2\left(ab+bc+ca\right)+9\)
\(\Leftrightarrow ab+bc+ca\ge3\)
Hiển nhiên đúng do: \(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}=3\)
Vì abc=1 , ta đặt \(a=\dfrac{x}{y};b=\dfrac{y}{z};c=\dfrac{z}{x}\)
Điều phải chứng minh tương đương với:
\(\dfrac{1}{2+\dfrac{x}{y}}+\dfrac{1}{2+\dfrac{y}{z}}+\dfrac{1}{2+\dfrac{z}{x}}\le1\\ \Leftrightarrow\dfrac{y}{2y+x}+\dfrac{z}{2z+y}+\dfrac{x}{2x+z}\le1\\ \Leftrightarrow\dfrac{2y}{2y+x}+\dfrac{2z}{2z+y}+\dfrac{2x}{2x+z}\le2\\ \Leftrightarrow\dfrac{x}{2y+x}+\dfrac{y}{2z+y}+\dfrac{z}{2x+z}\ge1\left(1\right)\)
Áp dụng bất đẳng thức bunhiacopxki dạng phân thức ta có:
\(\dfrac{x}{2y+x}+\dfrac{y}{2z+x}+\dfrac{z}{2x+z}=\dfrac{x^2}{x^2+2xy}+\dfrac{y^2}{y^2+2zx}+\dfrac{z^2}{z^2+2xy}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
=> bài toán được chứng minh
Dấu bằng xảy ra khi x=y=z=1 <=>a=b=c=1
Cho các số hữu tỉ \(x=\dfrac{a}{b};y=\dfrac{c}{d};z=\dfrac{a+c}{b+d}\left(a,b,c,d\in Z;b>0;d>0\right)\)
Chứng minh rằng nếu x < y thì x < y < z .
Đề bài sai
Ví dụ: với \(a=1;b=2;c=3,d=4\) thì \(x=\dfrac{1}{2}\) ; \(y=\dfrac{3}{4}\) ; \(z=\dfrac{2}{3}\)
Khi đó \(x< y\) nhưng \(z< y\)
\(\text{Vì }\dfrac{a}{b}< \dfrac{c}{d}\text{ nên }ad< bc\left(1\right)\)
\(\text{Xét tích}:a\left(b+d\right)=ab+ad\left(2\right)\)
\(b\left(a+c\right)=ba+bc\left(3\right)\)
\(\text{Từ(1);(2);(3)}\Rightarrow a\left(b+d\right)< b\left(a+c\right)\text{ do đó }\dfrac{a}{b}< \dfrac{a+c}{b+d}\left(4\right)\)
\(\text{Tương tự ta có:}\dfrac{a+c}{b+d}< \dfrac{c}{d}\left(5\right)\)
\(\text{Từ (4);(5) ta được }\dfrac{a}{b}< \dfrac{a+c}{b+d}< \dfrac{c}{d}\)
\(\Rightarrow x< y< z\)
Chứng minh rằng nếu:
(a + b + c + d) (a - b - c + d) = (a - b + c - d) (a + b - c - d)
thì\(\dfrac{a}{c}\)=\(\dfrac{b}{d}\)
(a, b, c, d khác 0)
Ta có: \(\left(a+b+c+d\right)\left(a-b-c+d\right)=\left(a-b+c-d\right)\left(a+b-c-d\right)\)
\(\Leftrightarrow\left(a+d\right)^2-\left(b+c\right)^2=\left(a-d\right)^2-\left(b-c\right)^2\)
\(\Leftrightarrow\left(a+d-a+d\right)\left(a+d+a-d\right)=\left(b+c-b+c\right)\left(b+c+b-c\right)\)
\(\Leftrightarrow2d\cdot2a=2c\cdot2b\)
\(\Leftrightarrow ad=bc\)
hay \(\dfrac{a}{c}=\dfrac{b}{d}\)
Chứng minh rằng: Nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì \(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a-b}{c-d}=\dfrac{a+b}{c+d}\Rightarrow\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\)
Cho phân số \(\dfrac{a}{b}\) . Chứng minh rằng :
Nếu \(\dfrac{a-x}{b-y}=\dfrac{a}{b}\) thì \(\dfrac{x}{y}=\dfrac{a}{b}\)
\(\dfrac{a-x}{b-y}=\dfrac{a}{b}\)
\(\Rightarrow\dfrac{a-x}{a}=\dfrac{b-y}{b}\)
\(\Rightarrow1-\dfrac{x}{a}=1-\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{a}=\dfrac{y}{b}\)
\(\Rightarrow\dfrac{x}{y}=\dfrac{a}{b}\)
Chứng minh rằng nếu \(\dfrac{a}{b}\)=\(\dfrac{b}{d}\) thì \(\dfrac{a^2+b^2}{b^2+d^2}=\dfrac{a}{d}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{d}=k\Leftrightarrow a=bk;b=dk\Leftrightarrow a=bk=dk^2\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{a}{d}=\dfrac{dk^2}{d}=k^2\\\dfrac{a^2+b^2}{b^2+d^2}=\dfrac{d^2k^4+d^2k^2}{d^2k^2+d^2}=\dfrac{d^2k^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=k^2\end{matrix}\right.\\ \LeftrightarrowĐpcm\)
Chứng minh rằng nếu a + c = 2b và 2bd = c.(b + d) với b, d khác 0 thì \(\dfrac{a}{b}=\dfrac{c}{d}\)
Chứng minh rằng nếu a > 0 , b > 0 , c > 0 thì :
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\ge\dfrac{3}{2}\)
Link: https://vn.answers.yahoo.com/question/index?qid=20100612215240AA1bp3C
\(\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}=\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\)
Áp dụng BĐT Cauchy dạng Engel , ta có :
\(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\) ≥ \(\dfrac{\left(a+b+c\right)^2}{2\left(ab+bc+ac\right)}\left(1\right)\)
Lại có : \(a^2+b^2+c^2\text{≥}ab+bc+ac\)
⇔ \(\left(a+b+c\right)^2\text{≥}3\left(ab+bc+ac\right)\left(2\right)\)
⇒ \(\dfrac{a^2}{ab+ac}+\dfrac{b^2}{ab+bc}+\dfrac{c^2}{ac+bc}\text{≥}\dfrac{3\left(ab+bc+ac\right)}{2\left(ab+bc+ac\right)}=\dfrac{3}{2}\)
\("="\text{⇔}a=b=c\)
Chứng minh rằng: nếu \(\dfrac{a}{b}=\dfrac{c}{d}\) thì \(\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
mọi người ơi giúp mik với, ai làm đc mik tick cho
Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{b}{a}=\dfrac{d}{c}\)
\(\Leftrightarrow1+\dfrac{b}{a}=1+\dfrac{d}{c}\)
\(\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Leftrightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\Leftrightarrow\dfrac{a+b}{a}=\dfrac{c+d}{c}\)