Cho a/b=c/d . CMR a/a-b = c/c-d , a/3a - b = c/3c - d
a,cho 3a-b/3a+b=3c-d/3c+d cmr a/b=c/d
b,cho a/b=c/d cmr:b^2+d^2/a^2+c^2=bd/ac
a) Ta có\(\frac{3a-b}{3a+b}=\frac{3c-d}{3c+d}\)
=> (3a - b)(3c + d) = (3a + b)(3c - d)
=> 9ac + 3ad - 3bc - bd = 9ac - 3ad + 3bc - bd
=> 3ad - 3bc = -3ad + 3bc
=> 3ad + 3ad = 3bc + 3bc
=> 6ad = 6bc
=> ad = bc
=> \(\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)
b) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Khi đó \(\frac{b^2+d^2}{a^2+c^2}=\frac{b^2+d^2}{\left(bk\right)^2+\left(dk\right)^2}=\frac{b^2+d^2}{d^2k^2+d^2k^2}=\frac{b^2+d^2}{k^2\left(b^2+d^2\right)}=\frac{1}{k^2}\)(1);
\(\frac{bd}{ac}=\frac{bd}{bkdk}=\frac{1}{k^2}\left(2\right)\)
Từ (1)(2) => \(\frac{b^2+d^2}{a^2+c^2}=\frac{bd}{ac}\)(đpcm)
Cho a/b=c/d CMR
a, 3a/b=3c/d
b, a+b/b=c+b/a
c, 3a+b/b= 3c+d/d
a ta có a/b=c/d=>ac=bd.nhân cả 2 vế vs 3 ta được 3ac=3bd=>3a/b=3c/d
c từ ý a có 3a/b=3c/d=>3a/b+1=3c/d +1(cộng cả hai vế vs 1).sau đó quy đồng được 3a+b/b=3c+d/d
còn ý b thì hình như bạn chép sai r thì phải,đề bài đúng chắc là như thế nầy a+b/b=c+a/a.nếu đề bài như thế thì sẽ giải giông ý c bạn nha!^^
cho a/b=c/d CMR 2a+b/3a-b=2c+d/3c-a
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)
Áp dụng t/c của DTSBN , ta có :
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2a+b}{2c+d}=\dfrac{3a-b}{3c-d}\\ \Rightarrow\dfrac{2a+b}{2c+d}=\dfrac{3a-b}{3c-d}\\ \Rightarrow\dfrac{2a+b}{3a-b}=\dfrac{2a+d}{3c-a}\left(đpcm\right)\)
cho a/b = c/d, cmr: \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow c=na,d=nb\)
Thay vào \(\dfrac{c}{3c+d}\), ta có
\(\dfrac{c}{3c+d}=\dfrac{na}{3na+nb}\)\(=\dfrac{na}{n\left(3a+b\right)}=\dfrac{na:n}{n\left(3a+b\right):n}=\dfrac{a}{3a+b}\)
FUCK MY LIFE!!!
cho a/b = c/d .Chứng minh
a) 3a-c/3b-d = 2a+3c/2b+3d
b) 3a-b/3a+d = 3c-a/3c+d
c) a^2 - b^2/c^2-d^2 = 2ab + b^2/2cd + d^2
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\dfrac{3a-c}{3b-d}=\dfrac{3bk-dk}{3b-d}=k\)
\(\dfrac{2a+3c}{2b+3d}=\dfrac{2bk+3dk}{2b+3d}=k\)
Do đó: \(\dfrac{3a-c}{3b-d}=\dfrac{2a+3c}{2b+3d}\)
c: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\dfrac{b^2}{d^2}\)
\(\dfrac{2ab+b^2}{2cd+d^2}=\dfrac{2\cdot bk\cdot b+b^2}{2\cdot dk\cdot d+d^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{2ab+b^2}{2cd+d^2}\)
Cho a,b,c,d khác của a/b - c/d
CMR: 20-4b/3a = 2c-4d/3c
Cho a,b,c,d khác a/b - c/d
CMR: 20-4b/3a = 2c- 4d/3c
Cmr nếu a/b=c/d thì
a. a+b/a-b=c+d/c-d
b. (a+b)^2/(a-b)^2=(c+d)^2/(c-d)^2
c. 2a+5b/3a-4b=2c+5d/3c-4d
Cho a, b, c, d > 0. CMR \(\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\ge\dfrac{2}{3}\)
Áp dụng BĐT Cauchy-Schwarz dạng Engel ta có:
\(VT=\dfrac{a}{b+2c+3d}+\dfrac{b}{c+2d+3a}+\dfrac{c}{d+2a+3b}+\dfrac{d}{a+2b+3c}\)
\(=\dfrac{a^2}{ab+2ac+3ad}+\dfrac{b^2}{bc+2bd+3ab}+\dfrac{c^2}{cd+2ac+3bc}+\dfrac{d^2}{ad+2bd+3cd}\)
\(\ge\dfrac{\left(a+b+c+d\right)^2}{4\left(ab+ad+bc+bd+ca+cd\right)}\ge\dfrac{\left(a+b+c+d\right)^2}{\dfrac{3}{2}\left(a+b+c+d\right)^2}=\dfrac{2}{3}\)
*Chứng minh \(4\left(ab+ad+bc+bd+ca+cd\right)\le\dfrac{3}{2}\left(a+b+c+d\right)^2\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-d\right)^2+\left(b-c\right)^2+\left(b-d\right)^2+\left(a-c\right)^2+\left(c-d\right)^2\ge0\)
Cho\(\frac{a}{b}=\frac{c}{d}\).CMR:\(\frac{a}{3a+b}=\frac{c}{3c+d}\)