Chứng minh rằng:
\(\dfrac{a}{b}=\dfrac{c}{d}thì\dfrac{2a+c}{2b+d}=\dfrac{a}{b}\)
Sẽ tick cho các bn!!!
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\left(b,d\ne0\right)\).Chứng minh rằng
\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{2a+b}{2a-b}=\dfrac{2bk+b}{2bk-b}=\dfrac{2k+1}{2k-1}\)
\(\dfrac{2c+d}{2c-d}=\dfrac{2dk+d}{2dk-d}=\dfrac{2k+1}{2k-1}\)
=>\(\dfrac{2a+b}{2a-b}=\dfrac{2c+d}{2c-d}\)
b: \(\dfrac{2a+b}{a-2b}=\dfrac{2bk+b}{bk-2b}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{2k+1}{k-2}\)
=>\(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
Cho \(\dfrac{2a+3c}{2b+3d}\)=\(\dfrac{2a-3c}{2b-3d}\). Chứng minh\(\dfrac{a}{b}\)=\(\dfrac{c}{d}\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c+2a-3c}{2b+3d+2b-3d}=\dfrac{a}{b}\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2a-3c}{2b-3d}=\dfrac{2a+3c-\left(2a-3c\right)}{2b+3d-\left(2b-3d\right)}=\dfrac{c}{d}\)
Suy ra \(\dfrac{a}{b}=\dfrac{c}{d}\)
Cho a,b,c là các số dương, chứng minh rằng
\(\dfrac{2a^2}{2b+c}+\dfrac{2b^2}{2a+c}+\dfrac{c^2}{4a+4b}\ge\dfrac{1}{4}\left(2a+2b+c\right)\)
\(P=\dfrac{4a^2}{4b+2c}+\dfrac{4b^2}{4a+2c}+\dfrac{c^2}{4a+4b}\ge\dfrac{\left(2a+2b+c\right)^2}{8a+8b+4c}\)
\(=\dfrac{\left(2a+2b+c\right)^2}{4\left(2a+2b+c\right)}=\dfrac{1}{4}\left(2a+2b+c\right)\)
Cho số hực dương a,b,c,d, e khác 0 thỏa mãn\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\)
Chứng minh rằng\(\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\)=\(\dfrac{a}{e}\)
Đặt \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}=k\Rightarrow a=bk;b=ck;c=dk;d=ek\)
\(\Rightarrow a=bk=ck^2=dk^3=ek^4;b=ek^3\)
\(\Rightarrow\dfrac{a}{e}=\dfrac{ek^4}{e}=k^4\left(1\right)\)
Ta có \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{e}\Rightarrow\dfrac{a^4}{b^4}=\dfrac{b^4}{c^4}=\dfrac{c^4}{d^4}=\dfrac{d^4}{e^4}=\dfrac{2a^4+3b^4+4c^4+5d^4}{2b^4+3c^4+4d^4+5e^4}\left(2\right)\)
Lại có \(\dfrac{a^4}{b^4}=\left(\dfrac{a}{b}\right)^4=\left(\dfrac{ek^4}{ek^3}\right)^4=k^4\left(3\right)\)
\(\left(1\right)\left(2\right)\left(3\right)\RightarrowĐpcm\)
Chứng minh rằng với a, b, c, d ta đều có: \(\dfrac{ab}{a+b+2c}+\dfrac{bc}{2a+b+c}+\dfrac{ac}{a+2b+c}\le\dfrac{a+b+c}{4}\)
Đề bài sai, BĐT này chỉ đúng với a;b;c dương
Cho các số \(a,b,c,d\) nguyên dương đôi một khác nhau và thỏa mãn: \(\dfrac{2a+b}{a+b}+\dfrac{2b+c}{b+c}+\dfrac{2c+d}{c+d}+\dfrac{2d+a}{d+a}=6\). Chứng minh \(A=abcd\) là số chính phương.
Điều kiện đã cho có thể được viết lại thành \(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+d}+\dfrac{d}{d+a}=2\)
hay \(1-\dfrac{a}{a+b}-\dfrac{b}{b+c}+1-\dfrac{c}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b}{a+b}-\dfrac{b}{b+c}+\dfrac{d}{c+d}-\dfrac{d}{d+a}=0\)
\(\Leftrightarrow\dfrac{b^2+bc-ab-b^2}{\left(a+b\right)\left(b+c\right)}+\dfrac{d^2+da-cd-d^2}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\dfrac{b\left(c-a\right)}{\left(a+b\right)\left(b+c\right)}+\dfrac{d\left(a-c\right)}{\left(c+d\right)\left(d+a\right)}=0\)
\(\Leftrightarrow\left(c-a\right)\left[\dfrac{b}{\left(a+b\right)\left(b+c\right)}-\dfrac{d}{\left(c+d\right)\left(d+a\right)}\right]=0\)
\(\Leftrightarrow\dfrac{b}{\left(a+b\right)\left(b+c\right)}=\dfrac{d}{\left(c+d\right)\left(d+a\right)}\) (do \(c\ne a\))
\(\Leftrightarrow b\left(cd+ca+d^2+da\right)=d\left(ab+ac+b^2+bc\right)\)
\(\Leftrightarrow bcd+abc+bd^2+abd=abd+acd+b^2d+bcd\)
\(\Leftrightarrow abc+bd^2-acd-b^2d=0\)
\(\Leftrightarrow ac\left(b-d\right)-bd\left(b-d\right)=0\)
\(\Leftrightarrow\left(b-d\right)\left(ac-bd\right)=0\)
\(\Leftrightarrow ac=bd\) (do \(b\ne d\))
Do đó \(A=abcd=ac.ac=\left(ac\right)^2\), mà \(a,c\inℕ^∗\) nên A là SCP (đpcm)
Cho a, b, c là các số thực dương. Chứng minh rằng:
\(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}\ge1\)
\(\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b}=\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\)
áp dụng BDT CAUCHY SCHAWRZ
\(=>\dfrac{a^2}{ab+2ac}+\dfrac{b^2}{bc+2ab}+\dfrac{c^2}{ac+2bc}\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ac+2ac+2ab+2bc}\)
\(=\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ac\right)}\ge\dfrac{3\left(ab+bc+ac\right)}{3\left(ab+bc+ac\right)}=1\)
Cho a, b, c là các số dương. Chứng minh rằng:
\(\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ca}{c+a+2b}\le\dfrac{a+b+c}{4}\)
\(\dfrac{bc}{a+b+c+a}\le\dfrac{bc}{4}\cdot\left(\dfrac{1}{a+b}+\dfrac{1}{a+c}\right)\\ \dfrac{ac}{b+c+a+b}\le\dfrac{ac}{4}\cdot\left(\dfrac{1}{b+c}+\dfrac{1}{a+b}\right)\\ \dfrac{ab}{a+c+b+c}\le\dfrac{ab}{4}\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}\right)\\ \Leftrightarrow VT\le\dfrac{1}{a+b}\left(\dfrac{bc}{4}+\dfrac{ac}{4}\right)+\dfrac{1}{a+c}\left(\dfrac{bc}{4}+\dfrac{ab}{4}\right)+\dfrac{1}{b+c}\left(\dfrac{ac}{4}+\dfrac{ab}{4}\right)\\ =\dfrac{1}{a+b}\cdot\dfrac{c\left(a+b\right)}{4}+\dfrac{1}{a+c}\cdot\dfrac{b\left(a+c\right)}{4}+\dfrac{1}{b+c}\cdot\dfrac{a\left(b+c\right)}{4}\\ =\dfrac{c}{4}+\dfrac{b}{4}+\dfrac{a}{4}\\ =\dfrac{a+b+c}{4}\left(đfcm\right)\)
Cho \(\dfrac{a+c}{b+d}=\dfrac{2a-c}{2b-d}\) chứng minh \(\dfrac{a}{b}=\dfrac{c}{d}\)
Giải:
Ta có:
\(\dfrac{a+c}{b+d}=\dfrac{2a-c}{2b-d}\)
\(\Leftrightarrow\dfrac{2a-c}{2b-d}=\dfrac{a+c}{b+d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{2a-c}{2b-d}=\dfrac{a+c}{b+d}=\dfrac{2a-c-\left(a+c\right)}{2b-d-\left(b+d\right)}=\dfrac{2a-c-a-c}{2b-d-b-d}=\dfrac{a}{b}\)
\(\Leftrightarrow\dfrac{a+c}{b+d}=\dfrac{a}{b}=\dfrac{a+c-a}{b+d-b}=\dfrac{c}{d}\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\) (đpcm)