\(P+2x^2=2+2x+2x^2+2x^2=4x^2+2x+2\)

\(=\left(4x^2+2x+\dfrac{1}{4}\right)+\dfrac{7}{4}=\left(2x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\)

Vì: \(\left(2x+\dfrac{1}{2}\right)^2\ge0\Rightarrow\left(2x+\dfrac{1}{2}\right)^2+\dfrac{7}{4}\ge\dfrac{7}{4}\)

dấu ''='' xảy ra khi \(2x+\dfrac{1}{2}=0\Rightarrow x=-\dfrac{1}{4}\)

Vậy \(Min_{P+2x^2}=\dfrac{7}{4}\) khi \(x=-\dfrac{1}{4}\)

Đặt \(\dfrac{x}{2}=\dfrac{2y}{3}=\dfrac{3z}{4}=k\). Khi đó ta có:

\(x=2k;2y=3k\Rightarrow y=\dfrac{3k}{2};3z=4k\Rightarrow z=\dfrac{4k}{3}\)

\(\Rightarrow xyz=108\Leftrightarrow2k\cdot\dfrac{3k}{2}\cdot\dfrac{4k}{3}=108\)

\(\Rightarrow\dfrac{24k^3}{6}=108\Rightarrow k^3=27\Rightarrow k=3\)

\(\Rightarrow\left\{{}\begin{matrix}x=2\cdot3=6\\y=\dfrac{3\cdot3}{2}=\dfrac{9}{2}\\z=\dfrac{4\cdot3}{3}=4\end{matrix}\right.\)

Vậy....

\(\left(x+5\right)⋮\left(x+1\right)\Rightarrow\dfrac{x+5}{x+1}\in Z\)

Ta co: \(\dfrac{x+5}{x+1}=\dfrac{x+1+4}{x+1}=\dfrac{x+1}{x+1}+\dfrac{4}{x+1}=1+\dfrac{4}{x+1}\)

\(\left(1+\dfrac{4}{x+1}\right)\in Z\Rightarrow\left(x+1\right)\in U\left(4\right)\)

=> \(x+1=\left\{\pm1;\pm2;\pm4\right\}\)

\(\Rightarrow x=\left\{-5;-3;-2;0;1;3\right\}\)

\(x^4-32x^2-16x+255=x^4+5x^3-7x^2-51x-5x^3-25x^2+35x+255\)

\(=\left(x^4+5x^3-7x^2-51x\right)-\left(5x^3+25x^2-35x-255\right)\)

\(=x\left(x^3+5x^2-7x-51\right)-5\left(x^3+5x^2-7x-51\right)\)

\(=\left(x^3+5x^2-7x-51\right)\left(x-5\right)\)

\(=\left[\left(x^3+8x^2+17x\right)-\left(3x^2-24x-51\right)\right]\left(x-5\right)\)

\(=\left[x\left(x^2+8x+17\right)-3\left(x^2+8x+17\right)\right]\left(x-5\right)\)

\(=\left(x^2+8x+17\right)\left(x-3\right)\left(x-5\right)\)

a) \(x^2-x-6=0\)

\(\Leftrightarrow x^2-3x+2x-6=0\)

\(\Leftrightarrow\left(x^2-3x\right)+\left(2x-6\right)=0\)

\(\Leftrightarrow x\left(x-3\right)+2\left(x-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Vậy....

b) \(x\left(x-2\right)\left(x-1\right)\left(x+1\right)=24\)

\(\Leftrightarrow\left(x^2-2x\right)\left(x^2-1\right)-24=0\)

\(\Leftrightarrow x^4-2x^3-x^2+2x-24=0\)

\(\Leftrightarrow\left(x-3\right)\left(x+2\right)\left(x^2-x+4\right)=0\)

\(\left[{}\begin{matrix}x-3=0\\x+2=0\\x^2-x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)

Có: \(x^2-x+4=\left(x^2-x+\dfrac{1}{4}\right)+\dfrac{15}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{15}{4}\ge\dfrac{15}{4}>0\)

-> vô nghiệm

Vậy pt có 2 nghiệm....

a) \(M=\left(x-3\right)^3-\left(x+1\right)^3+12x\left(x-1\right)\)

\(=x^3-9x^2+27x-27-x^3-3x^2-3x-1+12x^2-12x\)

\(=12x-28\)

b) Khi \(x=-\dfrac{2}{3}\Rightarrow M=12\cdot\left(-\dfrac{2}{3}\right)-28=-8-28=-36\)

c) Có: \(12x-28=-16\Rightarrow12x=12\Rightarrow x=1\)

Vậy x = 1 thì M = -16

Áp dụng bđt Cauchy-Schwarz có:

\(F^2=\left(\sqrt{a+b}+\sqrt{a+c}+\sqrt{b+c}\right)^2\)

\(=2+2\sqrt{\left(a+b\right)\left(a+c\right)}+2\sqrt{\left(a+c\right)\left(b+c\right)}+2\sqrt{\left(a+b\right)\left(b+c\right)}\le2+2a+b+c+a+b+2c+a+c+2b\)

\(=2+4a+4b+4c=2+4\left(a+b+c\right)=6\)

\(\Rightarrow F\le\sqrt{6}\)

''='' xảy ra khi a = b = c = \(\dfrac{1}{3}\)

Vậy \(F_{max}=\sqrt{6}\)

Câu 1:

\(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\le0\)

Vì \(\left\{{}\begin{matrix}\left|3x-5\right|\ge0\forall x\\\left(2y+5\right)^{208}\ge0\forall y\\\left(4z-3\right)^{20}\ge0\forall z\end{matrix}\right.\)

=> \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4x-3\right)^{20}\ge0\)

mà theo đề thì: \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}\le0\)

=> \(\left|3x-5\right|+\left(2y+5\right)^{208}+\left(4z-3\right)^{20}=0\)

\(\Rightarrow\left\{{}\begin{matrix}\left|3x-5\right|=0\\\left(2y+5\right)^{208}=0\\\left(4z-3\right)^{20}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}3x-5=0\\2y+5=0\\4z-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=\dfrac{5}{3}\\y=-\dfrac{5}{2}\\z=\dfrac{3}{4}\end{matrix}\right.\)

Vậy.....

P/s: mấy câu kia dễ tự làm, câu 6 có đầy trên gu gồ nhé, tự tìm

Giải:

a) tứ giác ADHE có: g A = g ADH = g HEA =90 độ

=> ADHE là hcn

=> AD = HE (đpcm)

b) Vì I là trung điểm của HB

=> DI là đường trung tuyến ứng vs cạnh huyền của tg DBH

=> DI = IB => \(\Delta IDB\) cân tại I

Cmtt ta có: \(\Delta KEH\) cân tại K

Ta có: \(\widehat{DBI}=\widehat{EHK}\) (đồng vị)

=> \(\widehat{DBI}=\widehat{BDI}=\widehat{EHK}=\widehat{HEK}\)

=> \(\widehat{BID}=\widehat{EKH}\) ( = \(180^o-2\widehat{DBI}\) hoặc \(=180^o-2\widehat{EHK}\))

mà 2 góc này đồng vị

=> DI // EK --> ddpcm

đăng nhầm mục?

Câu 1: a) Có: \(\sqrt{x}\ge0\Rightarrow\dfrac{1}{2}+\sqrt{x}\ge\dfrac{1}{2}\)

''='' xảy ra khi x = 0

Vậy \(P_{min}=\dfrac{1}{2}\) khi x = 0

b) Có: \(-2\sqrt{x-1}\le0\Rightarrow7-2\sqrt{x-1}\le7\)

''='' xảy ra khi x = 1

Vậy \(Q_{max}=7\) khi x = 1

Câu 2: \(M\in Z\) khi \(\left\{{}\begin{matrix}\sqrt{x-1}\in Z\\\sqrt{x-1}⋮2\end{matrix}\right.\)

mà \(x< 50\) => Để \(\left\{{}\begin{matrix}\sqrt{x-1}\in Z\\\sqrt{x-1}⋮2\end{matrix}\right.\) thì \(x-1=\left\{4;16;36\right\}\)

\(\Rightarrow x=\left\{5;17;37\right\}\)

Vậy....