Cho a,b \(\ge0\) và \(a^2+b^2=1\) . Tìm GTNN của biểu thức \(A=\left(1+a\:\right)\left(1+\frac{1}{b}\right)+\left(1+b\right)\left(1+\frac{1}{a}\right)\)
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\(\frac{2x-1}{3x^2+7x+2}+\frac{3}{9x^2+15x+4}-\frac{2x+7}{3x^2-5x-12}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{2x-1}{\left(3x+1\right)\left(x+2\right)}+\frac{3}{\left(3x+1\right)\left(3x+4\right)}-\frac{2x+7}{\left(4x+3\right)\left(x-3\right)}=\frac{5}{\left(x+2\right)}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{3x+1}+\frac{1}{3x+1}-\frac{1}{3x+4}+\frac{1}{3x+4}-\frac{1}{x-3}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{1}{x+2}-\frac{1}{x-3}=\frac{5}{x+2}\)
\(\Leftrightarrow\frac{x-3-x-2}{\left(x+2\right)\left(x-3\right)}=\frac{5\left(x-3\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow5x-3=-5\)
\(\Leftrightarrow x=-\frac{2}{5}\)
Chúc bạn học tốt !!!
a) (x - 3)(x2 + 3x + 9) + x(x + 2)(2 - x) = 1
=> x3 - 27 + x(4 - x2) = 1
=> x3 - 27 + 4x - x3 = 1
=> 4x - 27 = 1
=> 4x = 1 + 27
=> 4x = 28
=> x = 28 : 4 = 7
b) (x + 1)3 - (x - 1)3 - 6(x - 1)2 = -10
=>(x + 1 - x + 1)[(x + 1)2 + (x + 1)(x - 1) + (x - 1)2] - 6(x2 - 2x + 1) = -10
=> 2(x2 + 2x + 1 + x2 - 1 + x2 - 2x + 1) - 6x2 + 12x - 6 = -10
=> 2(3x2 + 1) - 6x2 + 12x - 6 = -10
=> 6x2 + 2 - 6x2 + 12x - 6 = -10
=> 12x - 4 = -10
=> 12x = -10 + 4
=> 12x = -6
=> x = -6 : 12 = -1/2
Ta có:
a) A = x2 + 6x + 10 = (x2 + 6x + 9) + 1 = (x + 3)2 + 1 \(\ge\)1 \(\forall\)x
Dấu "=" xảy ra <=> x + 3 = 0 <=> x = -3
Vậy MinA = 1 <=> x = -3
b) B = 4x2 - 12x + 13 = 4(x2 - 3x + 9/4) + 4 = 4(x - 3/2)2 + 4 \(\ge\)4 \(\forall\)x
Dấu "=" xảy ra <=> x - 3/2 = 0 <=> x = 3/2
Vậy MinB = 4 <=> x = 3/2
A=-3.(1-(2/3)2)(1-(2/5)2)...(1-(2/11)2)=-3.(1-2/3)(1+2/3)(1-2/5)(1+2/5)...(1-2/11)(1+2/11)=-3.\(\frac{1}{3}\).\(\frac{5}{3}\).\(\frac{3}{5}\).\(\frac{7}{5}\)...\(\frac{9}{11}.\frac{13}{11}\)
= -\(\frac{13}{11}\)
\(A=\frac{\left(1+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right).....\left(51^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)....\left(52^4+\frac{1}{4}\right)}\)
\(=\frac{\left(1+1+\frac{1}{2}\right)\left(1-1+\frac{1}{2}\right)....\left(11^2-11+\frac{1}{2}\right)}{\left(2+2^2+\frac{1}{2}\right)\left(2^2-2+\frac{1}{2}\right)....\left(12^2-12+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}\left(1.2+\frac{1}{2}\right)\left(2.3+\frac{1}{2}\right)....\left(11.12+\frac{1}{2}\right)}{\left(2.3+\frac{1}{2}\right)\left(3.4+\frac{1}{2}\right)....\left(12.13+\frac{1}{2}\right)}\)
\(=\frac{\frac{1}{2}}{12.13+\frac{1}{2}}\)
\(=\frac{1}{313}\)
Chúc bạn học tốt !!!
m^3 - m = (m^2-1)m = (m-1)(m+1)m là tích 3 stn liên tiếp -> chia hết cho 6
Áp dụng BĐT AM - GM ta có ;
\(A=\left(a+1\right)\left(1+\frac{1}{b}\right)+\left(b+1\right)\left(1+\frac{1}{a}\right)\)
\(=\frac{a}{b}+\frac{b}{a}+a+\frac{1}{a}+b+\frac{1}{b}+2\)
\(=\frac{a}{b}+\frac{b}{a}+\left(a+\frac{1}{2a}\right)+\left(b+\frac{1}{2b}\right)+\frac{1}{2a}+\frac{1}{2b}+2\)
\(\ge2\sqrt{\frac{a}{b}.\frac{b}{a}}+2\sqrt{a.\frac{1}{2a}}+2\sqrt{b.\frac{1}{2b}}+2\sqrt{\frac{1}{2a}.\frac{1}{2b}}+2\)
\(=4+2\sqrt{2}+\frac{1}{\sqrt{ab}}\ge4+2\sqrt{2}+\frac{1}{\frac{\sqrt{2\left(a^2+b^2\right)}}{2}}\)
\(=4+3\sqrt{2}\)
Dấu " = " xảy ra khi \(a=b=\frac{1}{\sqrt{2}}\)
Chúc bạn học tốt !!!