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giả sử :c^2>a^2>b^2 khi đó ta có :
\(\frac{b^2+c^2}{a^2+3}+\frac{c^2-a^2}{b^2+4^2}+\frac{a^2-b^2}{c^2+5}\le\frac{b^2+c^2}{b^2+3}+\frac{c^2-a^2}{b^2+3}+\frac{a^2-b^2}{b^2+3}=\frac{2c^2}{b^2+3}\le\frac{2}{3}.c^2\)
Như vậy ta có :\(a^2+b^2+c^2\le\frac{2}{3}.c^2\). Điều này xảy ra khi a=b=c
chuc bn hk tốt!
Đặt \(\dfrac{a}{3}=\dfrac{b}{4}=\dfrac{c}{5}=k\Rightarrow a=3k;b=4k;c=5k\)
\(\Rightarrow4\left(a-b\right)\left(b-c\right)=4\left(3k-4k\right)\left(4k-5k\right)\)
\(=4.\left[\left(3-4\right).k\right].\left[\left(4-5\right).k\right]\)
\(=4.\left[-k\right].\left[-k\right]=4k^2\left(1\right)\)
\(\Rightarrow\left(a-c\right)^2=\left(3k-5k\right)^2=\left[\left(3-5\right).k\right]^2=\left[-2k\right]^2=4k^2\left(2\right)\)
Từ \(\left(1\right),\left(2\right)\Rightarrow4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\)
Vậy \(4\left(a-b\right)\left(b-c\right)=\left(a-c\right)^2\left(dpcm\right)\)
Câu 1
Ta có : \(\frac{a}{b}=\frac{c}{d}=>\left(\frac{a}{b}+1\right)=\left(\frac{c}{d}+1\right)\left(=\right)\frac{a+b}{b}=\frac{c+d}{d}\)
=> ĐPCM
Câu 2
Ta có \(\frac{a}{b}=\frac{c}{d}=>\frac{b}{a}=\frac{d}{c}=>\left(\frac{b}{a}+1\right)=\left(\frac{d}{c}+1\right)\left(=\right)\frac{b+a}{a}=\frac{d+c}{c}=>\frac{a}{b+a}=\frac{c}{d+c}\)
=> ĐPCM
Câu 3
Câu 3
Ta có \(\frac{a+b}{a-b}=\frac{c+d}{c-d}\)(=) (a+b).(c-d)=(a-b).(c+d)(=)ac-ad+bc-bd=ac+ad-bc-bd(=)-ad+bc=ad-bc(=) bc+bc=ad+ad(=)2bc=2ad(=)bc=ad=> \(\frac{a}{b}=\frac{c}{d}\)
=> ĐPCM
Câu 4
Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
\(=>\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
Ta có \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\left(1\right)\)
Lại có \(\frac{a^2+c^2}{b^2+d^2}=\frac{b^2k^2+c^2k^2}{b^2+d^2}=\frac{k^2.\left(b^2+d^2\right)}{b^2+d^2}=k^2\left(2\right)\)
Từ (1) và (2) => ĐPCM
1: Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>\(a=b\cdot k;c=d\cdot k\)
\(\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\)
\(\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\)
Do đó: \(\dfrac{a}{a+b}=\dfrac{c}{c+d}\)
2: \(\dfrac{2a+b}{a-2b}=\dfrac{2\cdot bk+b}{bk-2b}=\dfrac{b\left(2k+1\right)}{b\left(k-2\right)}=\dfrac{2k+1}{k-2}\)
\(\dfrac{2c+d}{c-2d}=\dfrac{2dk+d}{dk-2d}=\dfrac{d\left(2k+1\right)}{d\left(k-2\right)}=\dfrac{2k+1}{k-2}\)
Do đó: \(\dfrac{2a+b}{a-2b}=\dfrac{2c+d}{c-2d}\)
3: \(\dfrac{a+b}{a-b}=\dfrac{bk+b}{bk-b}=\dfrac{b\left(k+1\right)}{b\cdot\left(k-1\right)}=\dfrac{k+1}{k-1}\)
\(\dfrac{c+d}{c-d}=\dfrac{dk+d}{dk-d}=\dfrac{d\left(k+1\right)}{d\left(k-1\right)}=\dfrac{k+1}{k-1}\)
Do đó: \(\dfrac{a+b}{a-b}=\dfrac{c+d}{c-d}\)
4: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5\cdot bk+3b}{5dk+3d}=\dfrac{b\left(5k+3\right)}{d\left(5k+3\right)}=\dfrac{b}{d}\)
\(\dfrac{5a-3b}{5c-3d}=\dfrac{5\cdot bk-3b}{5\cdot dk-3d}=\dfrac{b\left(5k-3\right)}{d\left(5k-3\right)}=\dfrac{b}{d}\)
Do đó: \(\dfrac{5a+3b}{5c+3d}=\dfrac{5a-3b}{5c-3d}\)