Cho \(\frac{a}{b}=\frac{c}{d}\).CMR :
a/ \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\)
b/\(\frac{2a+3c}{2d+3d}=\frac{2a-3c}{2b-3d}\)
c/\(\frac{a^2+c^2}{b^2+d^2}=\frac{ac^2}{bd}\)
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a/ do \(\frac{a}{b}\) = \(\frac{c}{d}\) = \(\frac{a+c}{b+d}\)=\(\frac{a-c}{b-d}\)(điều phải suy ra)
bạn viết sai đề bài b nhé phân số đầu là \(\frac{2a+3c}{2b+3d}\)
b/ đặt \(\frac{a}{b}\)= \(\frac{c}{d}\) là K
a=Kb;c=Kd
ta có:\(\frac{2a+3c}{2b+3d}\)= \(\frac{2Kb+3Kd}{2b+3d}\) = \(\frac{k\left(2b+3d\right)}{2b+3d}\) = K (1)
\(\frac{2a-3c}{2b-3d}\) = \(\frac{2Kb-3Kd}{2b-3d}\) = \(\frac{k\left(2b-3d\right)}{2b-3d}\) =K (2)
từ (!) và (2) suy ra \(\frac{2a+3c}{2b+3d}\) = \(\frac{2a-3c}{2b-3d}\)
Đặt a/b=c/d=k
=>a=bk; c=dk
a: \(\left(a+c\right)\cdot\left(b-d\right)=\left(bk+dk\right)\left(b-d\right)=k\left(b^2-d^2\right)\)
\(\left(a-c\right)\left(b+d\right)=\left(bk-dk\right)\left(b+d\right)=k\left(b^2-d^2\right)\)
Do đó: \(\left(a+c\right)\left(b-d\right)=\left(a-c\right)\left(b+d\right)\)
b: \(\left(2a+3c\right)\left(2b-3d\right)=\left(2bk+3dk\right)\left(2b-3d\right)=k\left(4b^2-9d^2\right)\)
\(\left(2a-3c\right)\left(2b+3d\right)=\left(2bk-3dk\right)\left(2b+3d\right)=k\left(4b^2-9d^2\right)\)
Do đó: \(\left(2a+3c\right)\left(2b-3d\right)=\left(2a-3c\right)\left(2b+3d\right)\)
1.
\(P=\frac{a^4}{abc}+\frac{b^4}{abc}+\frac{c^4}{abc}\ge\frac{\left(a^2+b^2+c^2\right)^2}{3abc}=\frac{\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\left(a+b+c\right)}{3abc\left(a+b+c\right)}\)
\(P\ge\frac{\left(a^2+b^2+c^2\right).3\sqrt[3]{a^2b^2c^2}.3\sqrt[3]{abc}}{3abc\left(a+b+c\right)}=\frac{3\left(a^2+b^2+c^2\right)}{a+b+c}\)
Dấu "=" khi \(a=b=c\)
2.
\(P=\sum\frac{a^2}{ab+2ac+3ad}\ge\frac{\left(a+b+c+d\right)^2}{4\left(ab+ac+ad+bc+bd+cd\right)}\ge\frac{\left(a+b+c+d\right)^2}{4.\frac{3}{8}\left(a+b+c+d\right)^2}=\frac{2}{3}\)
Dấu "=" khi \(a=b=c=d\)
Áp dụng dãy tỉ số bằng nhau ta có:
\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C+2A-3C}{2B+3D+2B-3D}=\frac{4A}{4B}=\frac{A}{B}\left(1\right)\)\(\frac{2A+3C}{2B+3D}=\frac{2A-3C}{2B-3D}=\frac{2A+3C-2A+3C}{2B+3D-2B+3D}=\frac{6C}{6D}=\frac{C}{D}\left(2\right)\)
Từ (1) và (2) suy ra : \(\frac{A}{B}=\frac{C}{D}\)
Giải :
Từ đảng thức : \(\frac{2a+3c}{2b+3d}=\frac{2a-3c}{2b-3d}\)
\(\Rightarrow\left(2a+3c\right).\left(2b-3d\right)=\left(2b+3d\right).\left(2a-3c\right)\)
\(\Rightarrow4ab-6ad+6bc-9cd=4ab-6bc+6ad-9cd\)
\(\Rightarrow\left(4ab-6ad+6bc-9cd\right)-\left(4ab-6bc+6ad-9cd\right)=0\)
\(\Rightarrow4ab-6ad+6bc-9cd-4ab+6bc-6ad+9cd=0\)
\(\Rightarrow\left(4ab-4ab\right)-\left(6ad+6ad\right)+\left(6bc+6bc\right)-\left(9cd-9cd\right)=0\)
\(\Rightarrow-12ad+12bc=0\)
\(\Rightarrow12bc=12ad\)
\(\Rightarrow bc=ad\)
\(\Rightarrow\frac{a}{b}=\frac{c}{d}\left(\text{đpcm}\right)\)
Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\begin{cases}a=kb\\c=kd\end{cases}\)
a) => \(\frac{2a+c}{2b+d}=\frac{2kb+kd}{2b+d}=\frac{k\left(2b+d\right)}{2b+d}=k\) (1)
\(\frac{2a-3c}{2b-3d}=\frac{2kb-3kd}{2b-3d}=\frac{k\left(2b-3d\right)}{2b-3d}=k\) (2)
Từ (1) và (2) => \(\frac{2a+c}{2b+d}=\frac{2a-3c}{2b-3d}\)
b) => \(\frac{ab}{cd}=\frac{kbb}{kdd}=\frac{b^2}{d^2}\) (1)
\(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(kb\right)^2+b^2}{\left(kd\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\) (2)
Từ (1) và (2) => \(\frac{ab}{cd}=\frac{a^2+b^2}{c^2+d^2}\)
đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
=>\(\frac{2a-3c}{2b-3d}=\frac{2bk-3dk}{2b-3d}=\frac{k.\left(2b-3d\right)}{2b-3d}=k\)
suy ra:\(\frac{a}{b}=\frac{c}{d}=\frac{2a-3c}{2b-3d}\)( vì cùng = k)