Bài 1: Nhân đơn thức với đa thức

dạng này chắc chắc là phải dùng AM-GM ngược dấu rồi :)

Ta có:

\(\dfrac{1+b}{1+4a^2}=1+b-\dfrac{4a^2\left(b+1\right)}{4a^2+1}\ge1+b-\dfrac{4a^2\left(b+1\right)}{4a}=1+b-a\left(b+1\right)\)

Tương tự cho 2 BĐT còn lại ta có:

\(\dfrac{1+c}{1+4b^2}\ge1+c-b\left(c+1\right);\dfrac{1+a}{1+4c^2}\ge1+a-c\left(a+1\right)\)

Cộng theo vế 3 BĐT trên ta có:

\(VT=\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+c^2}\)

\(\ge3+\left(a+b+c\right)-\left(ab+bc+ca\right)-\left(a+b+c\right)\)

\(=3-\dfrac{1}{3}\left(a+b+c\right)^2=3-\dfrac{1}{3}\cdot\dfrac{9}{4}=\dfrac{9}{4}=VP\)

Đẳng thức xảy ra khi \(a=b=c=\dfrac{1}{2}\)

\(VT=\left(\dfrac{a}{1+4c^2}+\dfrac{b}{1+4a^2}+\dfrac{c}{1+4b^2}\right)+\left(\dfrac{1}{1+4c^2}+\dfrac{1}{1+4a^2}+\dfrac{1}{1+4b^2}\right)\)

\(VT=\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)+3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)

Xét \(\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2a}{1+4c^2}\le\dfrac{4c^2a}{4c}=ca\\\dfrac{4a^2b}{1+4a^2}\le\dfrac{4a^2b}{4a}=ab\\\dfrac{4b^2c}{1+4b^2}\le\dfrac{4b^2c}{4b}=bc\end{matrix}\right.\)

\(\Rightarrow\dfrac{3}{2}-\left(\dfrac{4c^2a}{1+4c^2}+\dfrac{4a^2b}{1+4a^2}+\dfrac{4b^2c}{1+4b^2}\right)\ge\dfrac{3}{2}-\left(ab+bc+ca\right)\) (1)

Xét \(3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}1+4c^2\ge2\sqrt{4c^2}=4c\\1+4a^2\ge2\sqrt{4a^2}=4a\\1+4b^2\ge2\sqrt{4b^2}=4b\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{4c^2}{1+4c^2}\le\dfrac{4c^2}{4c}=c\\\dfrac{4a^2}{1+4a^2}\le\dfrac{4a^2}{4a}=a\\\dfrac{4b^2}{1+4b^2}\le\dfrac{4b^2}{4b}=b\end{matrix}\right.\)

\(\Rightarrow3-\left(\dfrac{4c^2}{1+4c^2}+\dfrac{4a^2}{1+4a^2}+\dfrac{4b^2}{1+4b^2}\right)\ge\dfrac{3}{2}\) (2)

Từ (1) và (2)

\(\Rightarrow VT\ge\dfrac{3}{2}-\left(ab+bc+ca\right)+\dfrac{3}{2}\)

\(\Rightarrow VT\ge3-\left(ab+bc+ca\right)\) (3)

Theo hệ quả của bất đẳng thức Cauchy

\(\Rightarrow\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{3}{4}\ge ab+bc+ca\)

\(\Rightarrow3-\dfrac{3}{4}\le3-\left(ab+bc+ca\right)\)

\(\Rightarrow\dfrac{9}{4}\le3-\left(ab+bc+ca\right)\) (4)

Từ (3) và (4)

\(\Rightarrow VT\ge\dfrac{9}{4}\)

\(\Leftrightarrow\dfrac{1+b}{1+4a^2}+\dfrac{1+c}{1+4b^2}+\dfrac{1+a}{1+4c^2}\ge\dfrac{9}{4}\) (đpcm)

Dấu " = " xảy ra khi \(a=b=c=\dfrac{1}{2}\)

\(\dfrac{x^2+x+1}{x^2+2x+1}=\dfrac{\left(x+1\right)^2-\left(x+1\right)+1}{\left(x+1\right)^2}=1-\dfrac{1}{x+1}+\dfrac{1}{\left(x+1\right)^2}\)

Đặt y=1/x+1=>\(1-y+y^2=\left(y-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}\)

Vậy GTNN của \(\dfrac{x^2+x+1}{x^2+2x+1}\) là \(\dfrac{3}{4}\) tại x=1

Ta có: \(x^{8n}+x^{4n}+1=x^{8n}+2x^{4n}+1-x^{4n}=\left(x^{4n}+1\right)^2-\left(x^{2n}\right)^2\)

\(=\left(x^{4n}+x^{2n}+1\right)\left(x^{4n}-x^{2n}+1\right)=\left(x^{4n}+2x^{2n}+1-x^{2n}\right)\left(x^{4n}-x^{2n}+1\right)=\left[\left(x^{2n}+1\right)-\left(x^n\right)^2\right]\left(x^{4n}-x^{2n}+1\right)=\left(x^{2n}+1-x^n\right)\left(x^{2n}+1+x^n\right)\left(x^{4n}-x^{2n}+1\right)\)=> \(x^{8n}+x^{4n}+1⋮x^{2n}+x^n+1\left(\forall x\right)\)

Cũng khó đấy ,để mình nghĩa chút

nhớ tìm trước khi hỏi Câu hỏi của Trần Huỳnh Cẩm Hân - Toán lớp 8 | Học trực tuyến

\(A=-x^2-4xy-5y^2-6y+1672\)

\(=-x^2-4xy-4y^2-y^2-6y-9+1681\)

\(=-\left(x^2+4xy+4y^2\right)-\left(y^2+6y+9\right)+1681\)

\(=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\)

Dễ thấy: \(\left\{{}\begin{matrix}\left(x+2y\right)^2\ge0\forall x,y\\\left(y+3\right)^2\ge0\forall y\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}-\left(x+2y\right)^2\le0\forall x,y\\-\left(y+3\right)^2\le0\forall y\end{matrix}\right.\)

\(\Rightarrow-\left(x+2y\right)^2-\left(y+3\right)^2\le0\forall x,y\)

\(\Rightarrow A=-\left(x+2y\right)^2-\left(y+3\right)^2+1681\le1681\forall x,y\)

Đẳng thức xảy ra khi \(\left\{{}\begin{matrix}-\left(x+2y\right)^2=0\\-\left(y+3\right)^2=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x+2y=0\\y+3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=-2y\\y=-3\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=6\\y=-3\end{matrix}\right.\)

a/ xet tam giac ABC VA tam giac DABco

AB chung

DAB =ABC(slt)

=>tam giac ABC DONG DANG TAM GIAC DAB(GG)

b/ap dung dinh ly pitago

bc^2=ab^2+ac^2

bc^2=15^2+20^2

bc=cang 525(tu tinh)

ta co ABC dong dang dab

=>ab/ad=bc/ab

=>ad=ab^2/bc

ad=125/cang525(tu tinh0

tuong tu

bn vẽ hình đi, mk làm cho.Mk ngại vẽ ấy mà.

c

ta có AD//BC nên theo hệ quả định lí talet:

AI/IB=SD/BC⇒AI/AI+IB=AD/AD+BC⇒AI/AB=AD/AD+BC hayAI/15=9/9+25⇒AI=9⋅15/9+25=135/34≈4(cm)

ta có:

SIAC=1/2⋅AI⋅AC=1/2⋅4⋅20=40(cm2)

SABC=1/2.AB⋅AC=1/2⋅15⋅20=150(cm2)

SBIC=SABC−SIAC=150−40=110(cm2)

a) x²-3x-x+3>0

<=> x(x-3)-(x-3)>0

<=> (x-3)(x-1)>0

\(\left[{}\begin{matrix}\left\{{}\begin{matrix}x-3>0\\x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}x-3< 0\\x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x>3\\x>1\end{matrix}\right.\\\left\{{}\begin{matrix}x< 3\\x< 1\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x>3\\x< 1\end{matrix}\right.\)

b) x³-2x²+3x-2\(\ge\) 0

<=> x³-x²-x²+x+2x-2\(\ge0\)

<=> (x-1)(x²-x+2)\(\ge0\)

Vì x²-x+2\(\ge0\)

nên x-1\(\ge0\)

x>=1

1)a²b+ab²=ab(a+b)=2ab

Ta có: (a-b)²\(\ge\)0

=>a²+b²\(\ge\)2ab

=>(a+b)²\(\ge\)4ab

=>2²\(\ge\)4ab

=>2\(\ge\)2ab

Vậy...

2)a²b³+a³b²=ab(a²b+ab²)\(\le\)1.(a²b+ab²)(từ câu 1 có 2\(\ge\)2ab)

Chứng minh tiếp tục tương tự ý 1) thì max a²b³+a³b²=2

3)2(ab³+a³b)=(a+b)(ab³+a³b)=a²b³+a³b²+2a²b²\(\le\)2+2.1²(Từ câu 2 max a²b³+a³b²=2 ; từ câu 1 thì từ câu 1 có 2\(\ge\)2ab)=4

=>ab³+a³b\(\le\)2