cho \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}.tínhP=\dfrac{a+b}{2c}+\dfrac{b+c}{3a}+\dfrac{c+a}{4b}\)
Cho a+b+c+d ≠ 0 thỏa mãn:
\(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính P = \(\dfrac{2a+5b}{3c+4d}+\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Cho a+b+c+d ≠ 0 và \(\dfrac{a}{b+c+d}=\dfrac{b}{a+c+d}=\dfrac{c}{b+a+d}=\dfrac{d}{c+b+a}\)
Tính giá trị biểu thức:
P = \(\dfrac{2a+5b}{3c+4d}-\dfrac{2b+5c}{3d+4a}+\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Bài 5: Cho \(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}\) . Tính \(S=\dfrac{a+b}{2c}=\dfrac{b+c}{3a}=\dfrac{c+a}{4b}\) (với \(a,b,c\ne0\)). Lưu ý: áp dụng t/c dãy tỉ số bằng nhau cần điều kiện phân số mẫu khác 0)
cảm ơn nhiều ạaaaaaaaaaaaa ❤
Đề bài \(S=\dfrac{a+b}{2c}+\dfrac{b+c}{3a}+\dfrac{c+a}{4b}\) đúng hơn chứ nhỉ?
ĐKXĐ: \(\left\{{}\begin{matrix}b\ne-c\\c\ne-a\\a\ne-b\end{matrix}\right.\) và \(a,b,c\ne0\)
Áp dụng t/c dtsbn:
\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+c+a+a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\left\{{}\begin{matrix}2a=b+c\\2b=c+a\\2c=a+b\end{matrix}\right.\)
\(\Rightarrow S=\dfrac{a+b}{2c}+\dfrac{b+c}{3a}+\dfrac{c+a}{4b}=\dfrac{2c}{2c}+\dfrac{2a}{3a}+\dfrac{2b}{4b}=1+\dfrac{2}{3}+\dfrac{1}{2}=\dfrac{13}{6}\)
Cho a+b+c+d khác 0 sao cho: \(\dfrac{b+c+d}{a}=\dfrac{a+c+d}{b}=\dfrac{b+a+d}{c}=\dfrac{c+b+a}{d}\)
Hãy tính: M = \(\dfrac{2a+5b}{3c+4d}-\dfrac{2b+5c}{3d+4a}-\dfrac{2c+5d}{3a+4b}+\dfrac{2d+5a}{3c+4b}\)
Tìm a,b,c biết \(\dfrac{3c-4b}{2}=\dfrac{4a-2c}{3}=\dfrac{2b-3a}{4}\) và c+b-a = -30
Bài 7: Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\). Chứng minh rằng ta có các tỉ lệ thức sau( giả thiết các tỉ lệ thức phải chứng minh đều có nghĩa):
a)\(\dfrac{a-b}{a+b}=\dfrac{c-d}{c+d}\) b)\(\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c)\(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\) d)\(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
ai hộ mik vs
a, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-b}{c-d}\Rightarrow\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
b, Áp dụng t/c dtsbn:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a}{2c}=\dfrac{5b}{5d}=\dfrac{3a}{4c}=\dfrac{4b}{4d}=\dfrac{2a+5b}{2c+5d}=\dfrac{3a-4b}{3c-4d}\Rightarrow\dfrac{2a+5b}{3a-4b}=\dfrac{2c+5d}{3c-4d}\)
c, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\left(k-1\right)^2}{d^2\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
d, Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow a=bk;c=dk\)
Ta có \(\dfrac{ac}{bd}=\dfrac{bk\cdot dk}{bd}=k^2\)
\(\dfrac{a^2+c^2}{b^2+d^2}=\dfrac{b^2k^2+d^2k^2}{b^2+d^2}=\dfrac{k^2\left(b^2+d^2\right)}{b^2+d^2}=k^2\)
Do đó \(\dfrac{ac}{bd}=\dfrac{a^2+c^2}{b^2+d^2}\)
cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)
(a,b,c,d khác 0)
chứng tỏ rằng
bài 1: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
bài 2:\(\dfrac{2a+c}{3a-c}=\dfrac{2b+d}{3b-d}\)
bài 3:\(\dfrac{5a-2c}{3a-4c}=\dfrac{5b-2c}{3b-4d}\)
giúp nhanh nha
Bài 1: Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\)
\(\Leftrightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)
\(\dfrac{a}{a+c}=\dfrac{ck}{ck+c}=\dfrac{ck}{c\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{b}{b+d}=\dfrac{dk}{dk+d}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+c}=\dfrac{b}{b+d}\)
a, a,b,c>0. CMR:\(\dfrac{ab}{a+b+2c}+\dfrac{bc}{b+c+2a}+\dfrac{ac}{a+c+2b}\le\dfrac{a+b+c}{4}\)
b, a,b,c>0. CMR:\(\dfrac{ab}{a+3b+2c}+\dfrac{bc}{b+3c+2a}+\dfrac{ac}{c+3a+2b}\le\dfrac{a+b+c}{6}\)
a.
\(\sum\dfrac{ab}{a+c+b+c}\le\dfrac{1}{4}\sum\left(\dfrac{ab}{a+c}+\dfrac{ab}{b+c}\right)=\dfrac{a+b+c}{4}\)
2.
\(\dfrac{ab}{a+3b+2c}=\dfrac{ab}{a+b+2c+2b}\le\dfrac{ab}{9}\left(\dfrac{4}{a+b+2c}+\dfrac{1}{2b}\right)=4.\dfrac{ab}{a+b+2c}+\dfrac{a}{18}\)
Quay lại câu a
\(b,\dfrac{ab}{a+3b+2c}=\left(\dfrac{1}{9}ab\right)\cdot\dfrac{9}{\left(a+c\right)+\left(b+c\right)+2b}\le\left(\dfrac{1}{9}ab\right)\cdot\left(\dfrac{1}{a+c}+\dfrac{1}{b+c}+\dfrac{1}{2b}\right)=\dfrac{1}{9}\cdot\left(\dfrac{ab}{a+b}+\dfrac{ab}{b+c}+\dfrac{a}{2}\right)\)
Cmtt: \(\dfrac{bc}{b+3c+2a}\le\dfrac{1}{9}\cdot\left(\dfrac{bc}{a+b}+\dfrac{bc}{a+b}+\dfrac{b}{2}\right);\dfrac{ca}{c+3a+2b}\le\dfrac{1}{9}\cdot\left(\dfrac{ca}{b+c}+\dfrac{ca}{a+b}+\dfrac{c}{2}\right)\)
\(\Rightarrow VT\le\dfrac{1}{9}\left(\dfrac{bc+ca}{a+b}+\dfrac{ab+ac}{b+c}+\dfrac{ab+bc}{a+c}+\dfrac{a+b+c}{2}\right)\\ \le\dfrac{1}{9}\left(a+b+c+\dfrac{a+b+c}{2}\right)=\dfrac{1}{9}\cdot\dfrac{3}{2}\left(a+b+c\right)=\dfrac{a+b+c}{6}\)
Dấu $"="$ khi $a=b=c$
cho a,b,c dương và \(a^4b^4+b^4c^4+c^4a^4=3a^4b^4c^4\).chứng minh:
\(\dfrac{1}{a^3b+2c^2+1}+\dfrac{1}{b^3c+2a^2+1}+\dfrac{1}{c^3a+2b^2+1}\le\dfrac{3}{4}\)