Hỏi đáp Toán lớp 9

Đặt số cần tìm là \(\overline{ab},\left(0\le a,b\le9;a,b\inℕ;a\ne0,a+b=8\right)\)

Số sau khi đổi vị trí là \(\overline{ba}\).

Theo bài ra ta có: \(\overline{ab}-\overline{ba}=18\Leftrightarrow10a+b-\left(10b+a\right)=18\Leftrightarrow9a-9b=18\Leftrightarrow a-b=2\)

\(\Rightarrow a-\left(8-a\right)=2\Leftrightarrow2a=10\Leftrightarrow a=5\Rightarrow b=3\)(thỏa) 

Áp dụng bđt Bunyakovsky dạng phân thức ta có :

\(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4\ge\frac{\left[\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\right]^2}{3}\)(1)

Tiếp tục sử dụng bđt Bunyakovsky dạng phân thức ta có :

\(\left(\frac{a}{a+b}\right)^2+\left(\frac{b}{b+c}\right)^2+\left(\frac{c}{c+a}\right)^2\ge\frac{\left(\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\right)^2}{3}\)(2)

Đặt  \(A=\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\)

Áp dụng bđt Cauchy ta có :

\(\frac{a}{a+b}+\frac{a+b}{4a}\ge2\sqrt{\frac{a}{a+b}\cdot\frac{a+b}{4a}}=1\)

=> \(A+\frac{a+b}{4a}+\frac{b+c}{4b}+\frac{c+a}{4c}\ge3\)

=> \(A+\frac{a}{4a}+\frac{b}{4a}+\frac{b}{4b}+\frac{c}{4b}+\frac{c}{4c}+\frac{a}{4c}\ge3\)

=> \(A+\frac{3}{4}+\frac{b}{4a}+\frac{c}{4b}+\frac{a}{4c}\ge3\)

Theo Cauchy ta có : \(\frac{b}{4a}+\frac{c}{4b}+\frac{a}{4c}\ge3\sqrt[3]{\frac{b}{4a}\cdot\frac{c}{4b}+\frac{a}{4c}}=\frac{3}{4}\)

=> \(A+\frac{3}{4}+\frac{3}{4}\ge3\)=> \(A\ge\frac{3}{2}\)(3)

Từ (1), (2) và (3) => \(P\ge\frac{3}{16}\)

Đẳng thức xảy ra <=> a = b = c

Vậy MinP = 3/16 <=> a = b = c 

Ta có:

\(P=\left(\frac{a}{a+b}\right)^4+\left(\frac{b}{b+c}\right)^4+\left(\frac{c}{c+a}\right)^4=\left(\frac{1}{1+\frac{b}{a}}\right)^4+\left(\frac{1}{1+\frac{c}{b}}\right)^4+\left(\frac{1}{1+\frac{a}{c}}\right)^4\)

Đặt \(\left(\frac{b}{a},\frac{c}{b},\frac{a}{c}\right)=\left(x,y,z\right)\left(x,y,z>0\right)\) \(\Rightarrow xyz=1\)

Khi đó: \(P=\frac{1}{\left(1+x\right)^4}+\frac{1}{\left(1+y\right)^4}+\frac{1}{\left(1+z\right)^4}\)

\(\ge3\left[\frac{1}{\left(1+x\right)^2}+\frac{1}{\left(1+y\right)^2}+\frac{1}{\left(1+z\right)^2}\right]^2\)

Ta có: \(\left(1+xy\right)\left(1+\frac{x}{y}\right)\ge\left(1+x\right)^2\Leftrightarrow\left(1+x\right)^2\le\frac{\left(1+xy\right)\left(x+y\right)}{y}\)( Bunyakovsky)

\(\Leftrightarrow\frac{1}{\left(1+x\right)^2}\ge\frac{y}{\left(1+xy\right)\left(x+y\right)}\) ; tương tự: \(\frac{1}{\left(1+y\right)^2}\ge\frac{x}{\left(1+xy\right)\left(x+y\right)}\)

Áp dụng BĐT Cauchy: \(\frac{1}{\left(1+z\right)^2}+\frac{1}{4}\ge2\sqrt{\frac{1}{\left(1+z\right)^2}\cdot\frac{1}{4}}=\frac{1}{1+z}\)

\(\Rightarrow\frac{1}{\left(1+z\right)^2}\ge\frac{1}{1+z}-\frac{1}{4}\)

Khi đó: \(P\ge\frac{1}{3}\left[\frac{x}{\left(1+xy\right)\left(x+y\right)}+\frac{y}{\left(1+xy\right)\left(x+y\right)}+\frac{1}{1+z}-\frac{1}{4}\right]^2\)

\(=\frac{1}{3}\left(\frac{1}{1+xy}+\frac{1}{1+z}-\frac{1}{4}\right)^2=\frac{1}{3}\left(\frac{xyz}{xyz+xy}+\frac{1}{1+z}-\frac{1}{4}\right)^2\)

\(=\frac{1}{3}\left(\frac{z}{1+z}+\frac{1}{1+z}-\frac{1}{4}\right)^2=\frac{1}{3}\left(1-\frac{1}{4}\right)^2=\frac{3}{16}\)

Dấu "=" xảy ra khi: a = b = c

Vậy Min(P) = 3/16 khi a = b = c

Bổ đề: \(\left(x+y\right)^2\le2\left(x^2+y^2\right)\)

1. \(\left(\sqrt{a}+\sqrt{b}\right)^2\le2\left(a+b\right)=4\Rightarrow\sqrt{a}+\sqrt{b}\le2\)

Đẳng thức xảy ra khi \(a=b=1\)

2. \(\left(\sqrt{a-1}+\sqrt{b-1}\right)^2\le2\left(a-1+b-1\right)=2\left(a+b-2\right)=2\left(4-2\right)=4\)

\(\Rightarrow\sqrt{a-1}+\sqrt{b-1}\le2\)

Đẳng thức xảy ra khi \(a=b=2\)

3,4,5 tương tự.

phiền bạn giải hộ luôn mấy bài còn lại nha, mình ngu lắm, sorry vì phiền bn

Ta có: \(P=\frac{2x^2+7x+23}{x^2+2x+10}\Leftrightarrow P\left(x^2+2x+10\right)=2x^2+7x+23\)

\(\Leftrightarrow Px^2+2Px+10P-2x^2-7x-23=0\)

\(\Leftrightarrow\left(P-2\right)x^2+\left(2P-7\right)x+\left(10P-23\right)=0\)

\(\Delta=\left(2P-7\right)^2-4\left(P-2\right)\left(10P-23\right)\ge0\)

\(\Leftrightarrow4P^2-28P+49-4\left(10P^2-43P+46\right)\ge0\)

\(\Leftrightarrow4P^2-28P+49-40P^2+173P-184\ge0\)

\(\Leftrightarrow-36P^2+145P-135\ge0\)

\(\Rightarrow36P^2-145P+135\ge0\)

\(\Leftrightarrow P^2-\frac{145}{36}P+\frac{27}{29}\ge0\)

\(\Leftrightarrow\left(P^2-2\cdot\frac{145}{72}+\frac{21025}{5184}\right)-\frac{469757}{150336}\ge0\)

\(\Leftrightarrow\left(P-\frac{145}{72}\right)^2\ge\frac{469757}{150336}\)

\(\Rightarrow-\sqrt{\frac{469757}{150336}}\le P-\frac{145}{72}\le\sqrt{\frac{469757}{150336}}\)

\(\Leftrightarrow\frac{145}{72}-\sqrt{\frac{469757}{150336}}\le P\le\frac{145}{72}+\sqrt{\frac{469757}{150336}}\)

Vậy \(Min_P=\frac{145}{72}-\sqrt{\frac{469757}{150336}}\) và \(Max_P=\frac{145}{72}+\sqrt{\frac{469757}{150336}}\)

Bất đẳng thức cần chứng minh có thể được viết lại thành: \(32\left(a^5+b^5\right)\ge2\left(a+b\right)^5\)

* Xét biểu thức: \(32\left(a^5+b^5\right)-2\left(a+b\right)^5=10\left(a-b\right)^2\left(a+b\right)\left(3a^2+3b^2+2ab\right)\ge0\forall a,b\inℝ\)do \(3a^2+3b^2+2ab=2a^2+2b^2+\left(a+b\right)^2\ge0\forall a,b\inℝ\)

Đẳng thức xảy ra khi a = b = 1