2.Chứng minh nếu a/b=c/d thi
a)\(\frac{a+b}{b}\)=\(\frac{c+d}{d}\)
b)\(\frac{2a+3b}{2c+3d}\)=\(\frac{5a-2b}{5c-2d}\)
cho \(\frac{a}{b}=\frac{c}{d}\)(b,d khác 0)
\(\frac{2a+b}{2a-b}=\frac{2c+d}{2c-d}\)
\(\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)
a.a/b=c/d=>.a/c=b/d=>2a/2c=b/d
ap dung tính chất dãy tỉ sồ bàng nhau ya có
2a/2c=b/d=2a+b/2c+d=2a-b/2c-d
=>2a+b/2a-b=2c+d/2c-d
b.a/b=c/d=>a/c=b/d=>5a/5c=3b/3d=3a/3c=2b/2d
áp dụng tính chat dãy ti số bang nhau ta co
5a/5c=3b/3d=3a/3c=2b/2d=5a-3b/5c-3d=3a+2b/3c+2d
5a-3b/3a+2b=5c-3d/3c+2d
bạn bấm vào đây cho mình nhé !CMR:từ tỉ lệ thức $\frac{a}{b}=\frac{c}{d}$ab =cd ta suy ra được $\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}$5a−3b3a+2b =5c−3d3c+2d
dat a/b=c/d=k a=kb ;c=kd Xet 5a+3b/5a-3b=5kb+3b/5kb-3b= b(5k+3)/b(5k-3)=5k+3/5k-3 (1) Xet 5c+3d/5c-3d=5kd+3d/5kd-3d= d(5k+3)/d(5k-3)= 5k+3/5k-3 (2) Tu (1) va (2) Suy ra 5a+3b/5a-3b =5c+3d/5c-3d
Chứng minh:\(\frac{5a+3b}{5c+3d}=\frac{2a-3b}{2c-3d}\) Cho a/b=c/d.
Đặt
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
\(VT:\frac{5a+3b}{5c+3d}=\frac{5bk+3b}{5dk+3d}=\frac{b\cdot\left(5k+3\right)}{d\cdot\left(5k+3\right)}=\frac{b}{d}\)
\(VP:\frac{2a-3b}{2c-3d}=\frac{2bk-3b}{2dk-3d}=\frac{b\cdot\left(2k-3\right)}{d\cdot\left(2k-3\right)}=\frac{b}{d}\)
Vì \(\frac{b}{d}=\frac{b}{5}\Rightarrow\frac{5a+3b}{5c+3d}=\frac{2a-3b}{2c-3d}\)
Vậy \(\frac{5a+3b}{5c+3d}=\frac{2a-3b}{2c-3d}\left(đpcm\right)\)
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
\(\Rightarrow\frac{5a}{5c}=\frac{3b}{3d}=\frac{2a}{2c}\)
Áp dụng tc của dãy tỉ số bằng nhau ta có:
\(\frac{5a}{5c}=\frac{3b}{3d}=\frac{2a}{2c}=\frac{5a+3b}{5c+3d}=\frac{2a-3b}{2a-3c}\)
Vậy \(\frac{5a+3b}{5c+3d}=\frac{2a-3b}{2a-3c}\left(đpcm\right)\)
\(Cho\) \(\frac{a}{b}=\frac{c}{d}\)
\(CMR:\)\(a,\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)
\(b,\frac{2a+b}{a-2b}=\frac{2c+d}{c-2d}\)
a) \(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\\\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{3a}{3c}=\frac{2b}{2d}=\frac{3a+2b}{3c+2d}\end{cases}}\)
\(\Rightarrow\frac{5a-3b}{5c-3d}=\frac{3a+2b}{3c+2d}\)
\(\Rightarrow\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)
b) Chứng minh tương tự
a, \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b, \(\frac{a^2.b^2}{c^2.d^2}=\frac{a^4+b^4-2a^2b^2}{c^4+d^4-2c^2d^2}\)
a, a/b=c/d
<=>a/c=b/d
<=>2a/2c=3b/3d=2a+3b/2c+3d=2a-3b/2c-3d
<=>2a+3b/2a-3b=2c+3d/2c-3d(đpcm)
c/m nếu a/b=c/d
a)\(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)
b)\(\frac{5a^3+3b^6}{11a^2-8b^2}=\frac{5c+3cd}{11c^2-8d^2}\)
Chứng minh rằng nếu \(\frac{a}{b}=\frac{c}{d}\)
a, \(\frac{5a+9c}{5b+9d}=\frac{2a}{2b}\)
b, \(\frac{5a+3b}{3a-3b}=\frac{5c+3d}{5c-3d}\)
giúp mik với ạ
\(\frac{a}{b}=\frac{c}{d}=\frac{2a}{2b}=\frac{5a}{5b}=\frac{9c}{9d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2a}{2b}=\frac{5a}{5b}=\frac{9c}{9d}=\frac{5a+9c}{5b+9d}\)
đpcm
b) bạn xem lại đề nhé
a, Theo tính chất dãy tỉ số bằng nhau ta có :
\(\hept{\begin{cases}\frac{a}{b}=\frac{c}{d}=\frac{2a}{2b}\\\frac{a}{b}=\frac{c}{d}=\frac{5a}{5b}=\frac{9c}{9d}=\frac{5a+9c}{5b+9d}\end{cases}}\)
\(\Rightarrow\frac{5a+9c}{5b+9d}=\frac{2a}{2b}\) ( đpcm )
b, Sai đề nha là \(\frac{5a+3b}{5a-3b}\)
Ta có : \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)
Theo tính chất dãy tỉ số bằng nhau ta có :
\(\hept{\begin{cases}\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a+3b}{5c+3d}\\\frac{a}{c}=\frac{b}{d}=\frac{5a}{5c}=\frac{3b}{3d}=\frac{5a-3b}{5c-3d}\end{cases}}\)
\(\Rightarrow\frac{5a+3b}{5c+3d}=\frac{5a-3b}{5c-3d}\)
\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=>a=bk,c=dk\)
\(=>\frac{5a+9c}{5b+9d}=\frac{5bk+9dk}{5b+9d}=\frac{k.\left(5b+9d\right)}{5b+9d}=k\)
\(\frac{2a}{2b}=\frac{2bk}{2b}=k\)
\(=>\frac{5a+9c}{5b+9d}=\frac{2a}{2b}\)
Cho a, b, c, d là các số thực dương. Chứng minh :
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
Cho tỉ lệ thức : a/b = c/d chứng minh rằng :
a) A - B /2a = C - D / 2c ; A + B / B = C+ D /D
b) 5a - 3b / 3a+2b = 5c - 3d / 3c+2d
Cho \(a;b;c;d>0\)chứng minh
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\ge\frac{2}{3}\)
cre: dự vào đề tóan quốc tế mỹ
\(\text{Σ}\frac{a}{b+2c+3d}=\text{Σ}\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{6\left(ab+bc+cd+ad\right)}\)
\(=\frac{\left(a+b\right)^2+\left(c+d\right)^2+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}=\frac{a^2+c^2+b^2+d^2+2ab+2cd+2\left(a+b\right)\left(c+d\right)}{6\left(ab+bc+cd+ad\right)}\)
\(\ge\frac{4\left(ab+bc+cd+ad\right)}{6\left(ab+bc+cd+ad\right)}=\frac{2}{3}\)
Dấu = xảy ra khi a=b=c=d
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{a}{b+2c+3d}+\frac{b}{c+2d+3a}+\frac{c}{d+2a+3b}+\frac{d}{a+2b+3c}\)
\(=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
\(\ge\frac{\left(a+b+c+d\right)^2}{4.\left(ab+ad+bc+bd+ca+cd\right)}\)\(\ge\frac{\left(a+b+c+d\right)^2}{\frac{3}{2}.\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=d\)
\(VT=\frac{a^2}{ab+2ac+3ad}+\frac{b^2}{bc+2bd+3ab}+\frac{c^2}{cd+2ac+3bc}+\frac{d^2}{ad+2bd+3cd}\)
Áp dụng BĐT Svac-xơ cho 3 số dương ta được :
\(VT\ge\frac{\left(a+b+c+d\right)^2}{4ab+4ac+4ad+4bc+4bd+4cd}\)
Áp dụng BĐT phụ \(x^2+y^2\ge2xy\) ta được :
\(a^2+b^2\ge2ab;a^2+c^2\ge2ac;a^2+d^2\ge2ad\)
\(b^2+c^2\ge2bc;b^2+d^2\ge2bd;c^2+d^2\ge2cd\)
\(\Rightarrow3\left(a^2+b^2+c^2+d^2\right)\ge2\left(ab+ac+ad+bc+bd+cd\right)\)
Ta lại có : \(\left(a+b+c+d\right)^4=a^2+b^2+c^2+d^2+2ab+2ac+2ad+2bc+2bd+2cd\)
\(\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{3}\)
\(\Rightarrow VT\ge\frac{\left(a+b+c+d\right)^4}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{8\left(ab+ac+ad+bc+bd+cd\right)}{12\left(ab+ac+ad+bc+bd+cd\right)}=\frac{2}{3}\)
Dấu "=" xảy ra khi \(a=b=c=d\)