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Ta chứng minh BĐT sau với các số dương:
\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\)
Thật vậy, BĐT tương đương: \(\dfrac{x+y}{xy}\ge\dfrac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow x^2-2xy+y^2\ge0\Leftrightarrow\left(x-y\right)^2\ge0\) (luôn đúng)
Áp dụng:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\) ; \(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\) ; \(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\)
Cộng vế với vế:
\(2\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge\dfrac{4}{a+b}+\dfrac{4}{b+c}+\dfrac{4}{c+a}\)
\(\Leftrightarrow\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{2}{a+b}+\dfrac{2}{b+c}+\dfrac{2}{c+a}\)
b.
Ta có:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow\dfrac{3}{a}+\dfrac{3}{b}\ge\dfrac{12}{a+b}\) (1)
\(\dfrac{1}{b}+\dfrac{1}{c}\ge\dfrac{4}{b+c}\Rightarrow\dfrac{2}{b}+\dfrac{2}{c}\ge\dfrac{8}{b+c}\) (2)
\(\dfrac{1}{c}+\dfrac{1}{a}\ge\dfrac{4}{c+a}\) (3)
Cộng vế với vế (1); (2) và (3):
\(\dfrac{4}{a}+\dfrac{5}{b}+\dfrac{3}{c}\ge4\left(\dfrac{3}{a+b}+\dfrac{2}{b+c}+\dfrac{1}{c+a}\right)\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
Lời giải:
a) Áp dụng BĐT Cô-si cho các số dương:
$a^3+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}a$
$b^3+\frac{1}{8}+\frac{1}{8}\geq \frac{3}{4}b$
$\Rightarrow a^3+b^3+\frac{1}{2}\geq \frac{3}{4}(a+b)=\frac{3}{4}$
$\Rightarrow a^3+b^3\geq \frac{1}{4}$ (đpcm)
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
b) Áp dụng BĐT Cauchy-Schwarz:
\(\frac{1}{a^3+b^3}+\frac{3}{ab}=\frac{1}{a^2-ab+b^2}+\frac{1}{ab}+\frac{1}{ab}+\frac{1}{ab}\geq \frac{(1+1+1+1)^2}{a^2-ab+b^2+ab+ab+ab}\)
\(=\frac{16}{(a+b)^2}=16\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=\frac{1}{2}$
\(\frac{x-3}{5}-\frac{2x-1}{10}=\frac{x+1}{2}+\frac{1}{4}\)
\(< =>\frac{\left(x-3\right).4}{20}-\frac{\left(2x-1\right).2}{20}=\frac{\left(x+1\right).10}{20}+\frac{5}{20}\)
\(< =>4x-12-4x+2=10x+10+5\)
\(< =>10x=-10-10-5=-25\)
\(< =>x=-\frac{25}{10}=-\frac{5}{2}\)
\(\frac{x+3}{2}-\frac{2x-1}{3}-1=\frac{x+5}{5}\)
\(< =>\frac{\left(x+3\right).15}{30}-\frac{\left(2x-1\right).10}{30}-\frac{30}{30}=\frac{\left(x+5\right).5}{30}\)\(< =>15x+45-20x+10-30=5x+25\)
\(< =>-5x+25=5x+25< =>10x=0< =>x=0\)
b) \(\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-x-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(\frac{x+1}{3x}-\left(x+1\right)\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{x+1}.\left(x+1\right)\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-2\left(\frac{1}{3x}-1\right)\right]:\frac{x-1}{x}\)
\(=\left[\frac{2}{3x}-\frac{2}{3x}+2\right]:\frac{x-1}{x}\)
\(=2.\frac{x}{x-1}=\frac{2x}{x-1}\left(đpcm\right)\)
a) \(\left(\frac{9}{x^3-9x}+\frac{1}{x+3}\right):\left(\frac{x-3}{x^2+3x}-\frac{x}{3x+9}\right)\)
\(=\left(\frac{9}{x\left(x^2-9\right)}+\frac{1}{x+3}\right):\left(\frac{x-3}{x\left(x+3\right)}-\frac{x}{3\left(x+3\right)}\right)\)
\(=\left(\frac{9}{x\left(x+3\right)\left(x-3\right)}+\frac{x^2-3x}{x\left(x+3\right)\left(x-3\right)}\right)\)
\(:\left(\frac{3x-9}{3x\left(x+3\right)}-\frac{x^2}{3x\left(x+3\right)}\right)\)
\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}:\frac{-x^2+3x-9}{3x\left(x+3\right)}\)
\(=\frac{x^2-3x+9}{x\left(x+3\right)\left(x-3\right)}.\frac{3x\left(x+3\right)}{-x^2+3x-9}\)
\(=\frac{x^2-3x+9}{x-3}.\frac{3}{x^2+3x-9}\)
\(=\frac{x^2-3x+9}{3-x}.\frac{3}{x^2-3x+9}\)
\(=\frac{3}{3-x}\left(đpcm\right)\)
d: =>4x+6=15x-12
=>4x-15x=-12-6=-18
=>-11x=-18
hay x=18/11
e: =>\(45x+27=12+24x\)
=>21x=-15
hay x=-5/7
f: =>35x-5=96-6x
=>41x=101
hay x=101/41
g: =>3(x-3)=90-5(1-2x)
=>3x-9=90-5+10x
=>3x-9=10x+85
=>-7x=94
hay x=-94/7
a) 1 + 3 + 32 + 33 + ... + 311
= (1 + 3 + 32 + 33) + ... + (38 + 39 + 310 + 311)
= 40 + ... + 38.(1 + 3 + 32 + 33)
= 40 + ... + 38. 40
= (1 + ... + 38) . 40 \(⋮\)40
b) Ta có: B = \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
=> B = \(\frac{1}{2.2}+\frac{1}{3.3}+\frac{1}{4.4}+...+\frac{1}{100.100}\)< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
=> B < \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=> B <\(1-\left(\frac{1}{2}-\frac{1}{2}\right)-\left(\frac{1}{3}-\frac{1}{3}\right)-...-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)
=> B < \(1-\frac{1}{100}\)
=> B < 1