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The film was good but it is longer than they thought.
b) với mọi a,b,c ϵ R và x,y,z ≥ 0 có : \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\left(1\right)\) Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\) Thật vậy với a,b∈ R và x,y ≥ 0 ta có: \(\frac{a^2}{x}=\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\left(2\right)\) ⇔\(\frac{a^2y}{xy}+\frac{b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\) ⇔\(\frac{a^2y+b^2x}{xy}\ge\frac{\left(a+b\right)^2}{x+y}\) ⇔\(\frac{a^2y+b^2x}{xy}.\left(x+y\right)xy\ge\frac{\left(a+b\right)^2}{x+y}.\left(x+y\right)xy\) ⇔\(\left(a^2y+b^2x\right)\left(x+y\right)\ge\left(a+b\right)^2xy\) ⇔\(a^2xy+b^2x^2+a^2y^2+b^2xy\ge a^2xy+2abxy+b^2xy\) ⇔\(b^2x^2+a^2y^2-2abxy\ge0\) ⇔\(\left(bx-ay\right)^2\ge0\)(luôn đúng ) Áp dụng BĐT (2) có: \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b\right)^2}{x+y}+\frac{c^2}{z}=\frac{\left(a+b+c\right)^2}{x+y+z}\) Dấu ''='' xảy ra ⇔\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}\) Ta có: \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)} \) = \(\frac{1}{a^2}.\frac{1}{ab+ac}+\frac{1}{b^2}.\frac{1}{bc+ac}+\frac{1}{c^2}.\frac{1}{ac+bc}\) =\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\) Áp dụng BĐT (1) ta có: \(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ab}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}++\frac{1}{c}\right)^2}{2\left(ab+bc+ac\right)}\) Mà abc=1⇒\(\left\{{}\begin{matrix}ab=\frac{1}{c}\\bc=\frac{1}{a}\\ac=\frac{1}{b}\end{matrix}\right.\) \(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)}\) \(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\) Có \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\sqrt[3]{\frac{1}{abc}}=3\sqrt[3]{\frac{1}{1}}=3\)( BĐT cosi ) ⇒\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge3\) ⇒\(\frac{\frac{1}{a^2}}{ab+ac}+\frac{\frac{1}{b^2}}{bc+ac}+\frac{\frac{1}{c^2}}{ac+bc}\ge\frac{1}{2}.3=\frac{3}{2}\) Vậy \(\frac{1}{a^3\left(b+c\right)}+\frac{1}{b^3\left(c+a\right)}+\frac{1}{c^3\left(a+b\right)}\ge\frac{3}{2}\) Chúc bạn học tốt !!!
\(\frac{7x-3}{x-1}\)=\(\frac{2}{3}\) <=>( 7x-3).3=2.(x-1) <=> 21x-9= 2x-2 <=> 19x=7 <=>x=\(\frac{7}{19}\)
6. A. re'sort B. ho'tel C. ex'pect D. 'rescue => chọn D 7. A. 'Jungle B. 'luggage C. 'sunbathe D. 'sugar => Không có đáp án
Đề bài là j vậy bạn?
(x-1)(x-3)(x-5)(x-7)-20=0 <=> (x-1)(x-7)(x-3)(x-5)-20=0 <=> (x^2-8x+7)(x^2-8x+15)-20=0 Đặt x^2-8x+7=a => x^2-8x+15= a+8 => a(a+8)-20=0 <=> a^2+8a-20=0 <=>(a^2+8a+16)-36=0 <=> (a+4)^2=36 => \(\left\{{}\begin{matrix}a+4=6\\a+4=-6\end{matrix}\right.\) <=>\(\left\{{}\begin{matrix}a=2\\a=-10\end{matrix}\right.\) *a=2 => x^2-8x+7=2 <=> x^2-8x+5=0 <=>(x^2-8x+16)-11=0 <=>(x-4)^2=11 <=>x-4=√11 <=> x=√11 +4 *a=-10 => x^2-8x+7=-10 <=> x^2-8x+17=0 <=> (x^2-8x+16)+1=0 <=> (x-4)^2=-1 (PT vô nghiệm) Vậy pt có nghiệm x=√11 +4 Chúc bạn học tốt !