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1.
C/m bổ đề: \(a^3-b^3\ge\frac{1}{4}\left(a^3-b^3\right)\) với \(\forall a,b\in R,a\ge b\)
\(\Leftrightarrow4a^3-4b^3-\left(a^3-3a^2b+3ab^2-b^3\right)\ge0\)
\(\Leftrightarrow3a^3+3a^2b-3ab^2-3b^3\ge0\)
\(\Leftrightarrow3\left(a^2-b^2\right)\left(a+b\right)\ge0\)
\(\Leftrightarrow3\left(a+b\right)^2\left(a-b\right)\ge0\)(đúng)
Theo bài ra: \(a^3-b^3\ge3a-3b-4\)
\(\Leftrightarrow\) Cần c/m: \(\left(a-b\right)^3\ge12a-12b-16\)(1)
Thật vậy:
\(\left(1\right)\)\(\Leftrightarrow\left(a-b\right)^3-12\left(a-b\right)+16\ge0\)
\(\Leftrightarrow\left[\left(a-b\right)^3-8\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a-b\right)+4\right]-12\left(a-b-2\right)\ge0\)
\(\Leftrightarrow\left(a-b-2\right)\left[\left(a-b\right)^2+2\left(a+b\right)-8\right]\ge0\)
\(\Leftrightarrow\left(a-b-2\right)^2\left(a-b+4\right)\ge0\) (đúng với mọi a,b thỏa mãn \(a,b\in R,a\ge b\))
2.
\(BĐT\Leftrightarrow\frac{1}{\frac{a+b}{ab}}+\frac{1}{\frac{c+d}{cd}}\le\frac{1}{\frac{a+b+c+d}{\left(a+c\right)\left(b+d\right)}}\)
\(\Leftrightarrow\frac{ab}{a+b}+\frac{cd}{c+d}\le\frac{\left(a+c\right)\left(b+d\right)}{a+b+c+d}\)
\(\Leftrightarrow\frac{ab\left(c+d\right)+cd\left(a+b\right)}{\left(a+b\right)\left(c+d\right)}\le\)\(\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\frac{abc+abd+acd+bcd}{ac+ad+bc+bd}\le\frac{ab+ad+bc+cd}{a+b+c+d}\)
\(\Leftrightarrow\left(ad+ab+bc+cd\right)\left(ac+ad+bc+bd\right)\ge\)\(\left(a+b+c+d\right)\left(abc+abd+acd+bcd\right)\)
\(\Leftrightarrow\left(ad\right)^2-2abcd+\left(bc\right)^2\ge0\)
\(\Leftrightarrow\left(ad-bc\right)^2\ge0\) (đúng với mọi a,b,c,d>0)
Làm tạm một câu rồi đi chơi, lát làm cho.
4)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT\ge\frac{\left(1+1+1\right)^2}{a^2+b^2+c^2+2ab+2bc+2ca}=\frac{9}{\left(a+b+c\right)^2}\ge\frac{9}{1}=9\)
Dấu "=" xảy ra \(\Leftrightarrow a=b=c=\frac{1}{3}\)
1.
a) 13\(\frac{1}{3}\) : 1\(\frac{1}{3}\) = 26 : (2x - 1)
<=> \(\frac{40}{3}:\frac{4}{3}\) = 13x - 26
<=> 10 + 26 = 13x
<=> 13x = 36
<=> x = \(\frac{36}{13}\)
b) 0,2 : 1\(\frac{1}{5}\) = \(\frac{2}{3}\) : (6x + 7)
<=> \(\frac{1}{5}:\frac{6}{5}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{6}\) = \(\frac{1}{9}x\) : \(\frac{2}{21}\)
<=> \(\frac{1}{9}x\) = \(\frac{2}{21}.\frac{1}{6}\) = \(\frac{1}{63}\)
<=> x = \(\frac{1}{7}\)
c) \(\frac{37-x}{x+13}\) = \(\frac{3}{7}\)
<=> (37 - x) . 7 = 3.(x + 13)
<=> 119 - 7x = 3x + 39
<=> -7x - 3x = 39 - 119
<=> -10x = -80
<=> x = 8
d) \(\frac{x-1}{x+5}=\frac{6}{7}\)
<=> 7(x - 1) = 6(x + 5)
<=> 7x - 7 = 6x + 30
<=> 7x - 6x = 30 + 7
<=> x = 37
e)
2\(\frac{2}{\frac{3}{0,002}}\) = \(\frac{1\frac{1}{9}}{x}\)
<=> \(\frac{1501}{750}\) = \(\frac{10}{9}:x\)
<=> x = \(\frac{10}{9}:\frac{1501}{750}\) = \(\frac{2500}{4503}\)
Bài 2. đề sai
Bài 3.
a) 6,88 : x = \(\frac{12}{27}\)
<=> x = 6,88 : \(\frac{12}{27}\)
<=> x = 15,48
b) 8\(\frac{1}{3}\) : \(11\frac{2}{3}\) = 13 : 2x
<=> \(\frac{25}{3}:\frac{35}{3}\) = 13 : 2x
<=> \(\frac{5}{7}=13:2x\)
<=> 2x = \(13:\frac{5}{7}\) = \(\frac{91}{5}\)
<=> x = 9,1
Bài 1:
a)Áp dụng Bđt Bunhiacopski ta có:
\(3a^2+4b^2\ge\frac{\left(3a+4b\right)^2}{7}=7\)
b)Áp dụng Bđt Bunhiacopski ta có:
\(\left(3a^2+5b^2\right)\left[\left(\frac{2}{\sqrt{3}}\right)^2+\left(-\frac{3}{\sqrt{5}}\right)^2\right]\ge\left(2a-3b\right)^2=49\)
\(\Rightarrow3a^2+5b^2\ge\frac{735}{47}\)
c)Áp dụng Bđt Bunhiacopski ta có:
\(\left(7a^2+11b^2\right)\left[\left(\frac{3}{\sqrt{7}}\right)^2+\left(\frac{5}{\sqrt{11}}\right)^2\right]\ge\left(\frac{3}{\sqrt{7}}\cdot\sqrt{7}a-\frac{5}{\sqrt{11}}\cdot\sqrt{11}b\right)^2=64\)
\(\Rightarrow\frac{274}{77}\left(7a^2+11b^2\right)\ge64\)
\(\Rightarrow7a^2+11b^2\ge\frac{2464}{137}\)
d)Áp dụng Bđt Bunhiacopski ta có:
\(\left(1^2+2^2\right)\left(a^2+b^2\right)\ge\left(a+2b\right)^2=4\)
\(\Rightarrow a^2+b^2\ge\frac{4}{5}\)
Biến đổi như sau $$\dfrac{bc}{6}+\dfrac{ac}{3}+\dfrac{ab}{2}=1 \leftrightarrow \dfrac{b}{2}.\dfrac{c}{3}+\dfrac{c}{3}.\dfrac{a}{1}+\dfrac{a}{1}.\dfrac{b}{2}=1$$
Đặt $(\dfrac{a}{1},\dfrac{b}{2},\dfrac{c}{3})=(x,y,z), x,y,z>0 \rightarrow xy+yz+zx=1$
Mặt khác $$A=\dfrac{1}{a^2+1}+\dfrac{1}{(\dfrac{b}{2})^2+1}+\dfrac{1}{(\dfrac{c}{3})^2+1}=\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}$$
Do đó ta cần tìm max của $$\dfrac{1}{x^2+1}+\dfrac{1}{y^2+1}+\dfrac{1}{z^2+1}$$
Với $$xy+yz+zx=1$$
Thật vậy thay
$$1=xy+yz+zx \rightarrow A=\sum{\dfrac{1}{x^2+xy+yz+zx}}=\sum{\dfrac{1}{(x+y)(y+z)}}=\dfrac{(x+y)+(y+z)+(z+x)}{(x+y)(y+z)(z+x)}=\dfrac{2(x+y+z)}{(x+y)(y+z)(z+x)}$$
Áp dụng bdt $(x+y)(y+z)(z+x)\geq \dfrac{8}{9}(x+y+z)(xy+yz+xz)$
Suy ra $A\le \dfrac{2(x+y+z)}{\dfrac{8}{9}(x+y+z)(xy+xz+zx)}$ thay $xy+yz+zx=1 \rightarrow A\le \dfrac{9}{4}$
Dấu $= \leftrightarrow x=y=z=\sqrt{\dfrac{1}{3}} \rightarrow a=..., b=...,c=...$ Làm tiếp hộ mình
1)ĐK:\(x\in\left[-3;\frac{6}{5}\right]\)
pt\(\Leftrightarrow3\left(x^2-x+2\right)-3\left[\sqrt{6-5x}-\left(x-2\right)\right]+\left[3\sqrt{x+3}-\left(x+5\right)\right]=0\)
\(\Leftrightarrow\left(x^2-x+2\right)\left(\frac{3}{\sqrt{6-5x}+x-2}+\frac{1}{3\sqrt{x+3}+x+5}+3\right)=0\)
\(\Leftrightarrow x^2\)-x+2=0(do(...)>0)
\(\Leftrightarrow x=-2\)hoặc \(x=1\)(t/m)
ÁD BĐT Bunhiacopxki:
\(\left(a+b+c\right)\left[\frac{a}{\left(b+c\right)^2}+\frac{b}{\left(c+a\right)^2}+\frac{c}{\left(a+b\right)^2}\right]\ge\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\right)^2\)
Lại có:\(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}=\left(a+b+c\right)\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\)
\(=\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)-3\ge\frac{9}{2\left(a+b+c\right)}-3=\frac{3}{2}\)
\(\Rightarrow VT\ge\left(\frac{3}{2}\right)^2\)=\(\frac{9}{4}\)(đpcm)
Dấu''='' xảy ra\(\Leftrightarrow a=b=c=\frac{1}{3}\)
