Cho \(\frac{a}{b}\)= \(\frac{c}{d}\) CMR:
a) \(\frac{a^2+b^2}{a^2+d^2}\)= \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\left(\frac{a-b}{c-d}\right)^4\)= \(\frac{a^4+b^4}{c^4+d^4}\)
Bài 1:Cho a,b,c,d là các số dương. Chứng minh rằng :
\(\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}+\frac{b^4}{\left(b+c\right)\left(b^2+c^2\right)}+\frac{c^4}{\left(c+d\right)\left(c^2+d^2\right)}+\frac{d^4}{\left(d+a\right)\left(d^2+a^2\right)}\ge\frac{a+b+c+d}{4}\)
Bài 2:Cho \(a>0,b>0,c>0\).\(CM:\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\ge2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)
Bài 3: a) Cho x,y,>0. CMR:\(\frac{x^3}{x^2+xy+y^2}\ge\frac{2x-y}{3}\)
b) Chứng minh rằng\(\Sigma\frac{a^3}{a^2+ab+b^2}\ge\frac{a+b+c}{3}\)
Xét \(\frac{a^3}{a^2+ab+b^2}-\frac{b^3}{a^2+ab+b^2}=\frac{\left(a-b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=a-b\)
Tương tự, ta được: \(\frac{b^3}{b^2+bc+c^2}-\frac{c^3}{b^2+bc+c^2}=b-c\); \(\frac{c^3}{c^2+ca+a^2}-\frac{a^3}{c^2+ca+a^2}=c-a\)
Cộng theo vế của 3 đẳng thức trên, ta được: \(\left(\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\right)\)\(-\left(\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\right)=0\)
\(\Rightarrow\frac{a^3}{a^2+ab+b^2}+\frac{b^3}{b^2+bc+c^2}+\frac{c^3}{c^2+ca+a^2}\)\(=\frac{b^3}{a^2+ab+b^2}+\frac{c^3}{b^2+bc+c^2}+\frac{a^3}{c^2+ca+a^2}\)
Ta đi chứng minh BĐT phụ sau: \(a^2-ab+b^2\ge\frac{1}{3}\left(a^2+ab+b^2\right)\)(*)
Thật vậy: (*)\(\Leftrightarrow\frac{2}{3}\left(a-b\right)^2\ge0\)*đúng*
\(\Rightarrow2LHS=\Sigma_{cyc}\frac{a^3+b^3}{a^2+ab+b^2}=\Sigma_{cyc}\text{ }\frac{\left(a+b\right)\left(a^2-ab+b^2\right)}{a^2+ab+b^2}\)\(\ge\Sigma_{cyc}\text{ }\frac{\frac{1}{3}\left(a+b\right)\left(a^2+ab+b^2\right)}{a^2+ab+b^2}=\frac{1}{3}\text{}\Sigma_{cyc}\left[\left(a+b\right)\right]=\frac{2\left(a+b+c\right)}{3}\)
\(\Rightarrow LHS\ge\frac{a+b+c}{3}=RHS\)(Q.E.D)
Đẳng thức xảy ra khi a = b = c
P/S: Có thể dùng BĐT phụ ở câu 3a để chứng minhxD:
1) ta chứng minh được \(\Sigma\frac{a^4}{\left(a+b\right)\left(a^2+b^2\right)}=\Sigma\frac{b^4}{\left(a+b\right)\left(a^2+b^2\right)}\)
\(VT=\frac{1}{2}\Sigma\frac{a^4+b^4}{\left(a+b\right)\left(a^2+b^2\right)}\ge\frac{1}{4}\Sigma\frac{a^2+b^2}{a+b}\ge\frac{1}{8}\Sigma\left(a+b\right)=\frac{a+b+c+d}{4}\)
bài 2 xem có ghi nhầm ko
3a biến đổi tí là xong
b tuong tự bài 1
1/ Biết \(\frac{a}{b}=\frac{c}{d}\), chứng minh
a) \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\left(\frac{a-d}{c-b}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
2/ Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Chứng minh \(\left(\frac{a+b+c}{b+c+d}\right)^3=\frac{a}{b}\)
3/ Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{d}\)
Chứng minh a=b=c
Mình chỉ làm bài 1a, và bài 3 thôi nhé,còn lại là bạn tự làm nhé
Bài 1:
a, Ta có : \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\)
\(\Rightarrow\left[\frac{a}{b}\right]^2=\left[\frac{c}{d}\right]^2=\left[\frac{a+c}{b+d}\right]^2\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{(a+c)^2}{(b+d)^2}\Rightarrow\frac{a^2+c^2}{b^2+d^2}=\frac{(a+c)^2}{(b+d)^2}\)
Bài 3 : Sửa đề : Cho \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\)
CM : a = b = c
Cách 1 : Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\)
vì \(a+b+c\ne0\)
\(\frac{a}{b}=1\Rightarrow a=b;\frac{b}{c}=1\Rightarrow b=c\)
Do đó : \(a=b=c\).
Cách 2 : Đặt \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=m\), ta có : \(a=bm,b=cm,c=am\)
Do đó : \(a=bm=m(mc)=m\left[m(ma)\right]\)
\(\Rightarrow a=m^3a\Rightarrow m^3=1(a\ne0)\Rightarrow m=1\)
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)
Cách 3 : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}\Rightarrow\frac{a}{b}\cdot\frac{b}{c}\cdot\frac{c}{a}=\left[\frac{a}{b}\right]^3\Rightarrow1=\left[\frac{a}{b}\right]^3\Rightarrow\frac{a}{b}=1\)
Ta có : \(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=1\Rightarrow a=b=c\)
b) \(2< \frac{\left(a+b\right)}{a+b+c}+\frac{\left(b+c\right)}{b+c+d}+\frac{\left(c+d\right)}{c+d+a}+\frac{\left(d+a\right)}{d+a+b}< 4\)
Cho a,b,c,d > 0 CMR :
a)\(A=\frac{\left(a+c\right)}{a+b}+\frac{\left(b+d\right)}{b+c}+\frac{\left(c+a\right)}{c+d}+\frac{\left(d+b\right)}{d+a}4\ge\)
b, \(\frac{a+b}{a+b+c}>\frac{a+b}{a+b+c+d}\); \(\frac{b+c}{b+c+a}>\frac{b+c}{a+b+c+d}\)
\(\frac{c+d}{c+d+a}>\frac{c+d}{a+b+c+d};\frac{d+a}{a+d+b}>\frac{a+d}{a+b+c+d}\)
Cộng các bĐT trên
=> \(B>\frac{2\left(a+b+c+d\right)}{a+b+c+d}=2\)
Ta có Với \(0< \frac{x}{y}< 1\)
=> \(\frac{x}{y}< \frac{x+z}{y+z}\)
Áp dụng ta có
\(B>\frac{a+b+d}{a+b+c+d}+...+\frac{d+a+c}{a+b+c+d}=3\)
Vậy 2<B<3
Cho \(\frac{a}{b}\)=\(\frac{c}{d}\)
CM: A, \(\frac{a^2+b^2}{c^2+d^2}\)=\(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
B, \(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}\)=\(\frac{a^4+b^4}{c^4+d^4}\)
áp dụng phương pháp đặt k đó bạn . có ở trong sách giáo khoa bài hai tỉ số bằng nhau đó
ta có a/b = c/d = k suy ra a=bk, c=dk
suy ra a^2 +b^2 / c^2 + d^2 = (bk)^2 +b^2/(dk)^2 -d^2 =..... bạn rút ra tối giản sẽ được k^2+1/K^2-1(1)
bạn thế vào phân số còn lại rồi rút gọn thành k^2+1/ k^2-1(2)
từ 1 và 2 suy ra .....
câu b tương tự
cho 4 số thực a,b,c,d tm a+b+c+d=4
cmr \(\frac{\left(a+\sqrt{b}\right)^2}{\sqrt{a^2-ab+b^2}}+\frac{\left(b+\sqrt{c}\right)^2}{\sqrt{b^2-bc+c^2}}+\frac{\left(c+\sqrt{d}\right)^2}{\sqrt{c^2-cd+d^2}}+\frac{\left(d+\sqrt{a}\right)^2}{\sqrt{d^2-ad+a^2}}\le16\)
Cho\(\frac{a}{b}\)=\(\frac{c}{d}\) chứng minh
1,\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a.c}{b.d}\)
2,\(\frac{a^2+c^2}{b^2+d^2}\)=\(\frac{a^2-c^2}{b^2-d^2}\)
\(3,\left(a+c\right).\left(b-d\right)=\left(a-c\right).\left(b+d\right)\)
\(4,\left(b+d\right).c=\left(c+c\right).d\)
\(5,\frac{4.a-12.b}{8.a+11.b}=\frac{4.c-12.d}{8.c+11.d}\)
\(6,\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)
\(7,\frac{a^{10}+b^{10}}{\left(a+b\right)^{10}}=\frac{c^{10}+d^{10}}{\left(c+d\right)^{10}}\)
Cho \(\frac{a}{b}\)= \(\frac{c}{d}\)CMR
a) \(\frac{a^2+b^2}{a^2+d^2}\)= \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
b) \(\left(\frac{a-b}{c-d}\right)^4\)= \(\frac{a^4+b^4}{c^4+d^4}\)
a) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
Thay a = bk, c = dk vào \(\frac{a^2+b^2}{c^2+d^2}\) và \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\), ta có:
\(\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2k^2+b^2}{d^2k^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)
\(\frac{\left(bk+b\right)^2}{\left(dk+d\right)^2}=\frac{\left[b\left(k+1\right)\right]^2}{\left[d\left(k+1\right)\right]^2}=\frac{b^2\left(k+1\right)^2}{d^2\left(k+1\right)^2}=\frac{b^2}{d^2}\)
Vì \(\frac{b^2}{d^2}=\frac{b^2}{d^2}\Rightarrow\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
Vậy \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\) với \(\frac{a}{b}=\frac{c}{d}\)
b) Thay a = bk, c = dk vào \(\left(\frac{a-b}{c-d}\right)^4\)và \(\frac{a^4+b^4}{c^4+d^4}\), ta có:
\(\left(\frac{bk-b}{dk-d}\right)^4=\frac{\left(bk-b\right)^4}{\left(dk-d\right)^4}=\frac{\left[b\left(k-1\right)\right]^4}{\left[d\left(k-1\right)\right]^4}=\frac{b^4\left(k-1\right)^4}{d^4\left(k-1\right)^4}=\frac{b^4}{d^4}\)
\(\frac{\left(bk\right)^4+b^4}{\left(dk\right)^4+d^4}=\frac{b^4k^4+b^4}{d^4k^4+d^4}=\frac{b^4\left(k^4+1\right)}{d^4\left(k^4+1\right)}=\frac{b^4}{d^4}\)
Vì \(\frac{b^4}{d^4}=\frac{b^4}{d^4}\Rightarrow\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\)
Vậy \(\left(\frac{a-b}{c-d}\right)^4=\frac{a^4+b^4}{c^4+d^4}\) với \(\frac{a}{b}=\frac{c}{d}\)
1.Chứng minh rằng :
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+b+c+d\)với \(a\ge-1;b\ge-4;c\ge2;d>3\)
2. Chứng minh rằng :
\(\frac{a^2}{b^5}+\frac{b^2}{c^5}+\frac{c^2}{d^5}+\frac{d^2}{a^5}\ge\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)với \(a,b,c,d>0\)
Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)
Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3
\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)
\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)
\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
ta sẽ giết ngươi kí tên dép đờ kiu lờ
Cho tỉ lệ thức \(\frac{a}{b}=\frac{c}{d}\)C/m:
a)\(\frac{a+b}{b}=\frac{c+d}{d}\)
b)\(\frac{â-b}{b}=\frac{c-d}{d}\)
c)\(\frac{a+b}{a}=\frac{c+d}{c}\)
d)\(\frac{a-b}{a}=\frac{c-d}{c}\)
e)\(\frac{a}{a+b}=\frac{c}{c+d}\)
f)\(\frac{a}{a-b}=\frac{c}{c-d}\)
g)\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)
h)\(\frac{\left(a-b\right)^2}{\left(c-d\right)^2}=\frac{a^4+b^4}{c^4+d^4}\)
( áp dụng t/c tỉ lệ thức và dãy tỉ số = nhau)
a,a/b=c/d
<=>a/b+1=c/d+1
<=>a/b+b/b=c/d+d/d
=>a+b/b=c+d/d
b,a/b=c/d
<=>a/b-1=c/d-1
<=>a/b-b/b=c/d-d/d
<=>a-b/b=c-d/d
a,a/b=c/d
=>ad=bc
=>ad+bd=bc+bd
=>d(a+b)=b(c+d)
=>a+b/b=c+d/d
ko bt đ ko