ĐKXĐ bạn tự tìm nhé.

a, \(x^2-7x=6\sqrt{x+5}-30\)

\(\Leftrightarrow x^2-7x+30-6\sqrt{x+5}=0\)

\(\Leftrightarrow\left(x+5-6\sqrt{x+5}+9\right)+\left(x^2-8x+16\right)=0\)

\(\Leftrightarrow\left(\sqrt{x+5}-3\right)^2+\left(x-4\right)^2=0\)

\(\Leftrightarrow\hept{\begin{cases}\sqrt{x+5}-3=0\\x-4=0\end{cases}}\Leftrightarrow x=4\)

b, \(\Leftrightarrow\sqrt{x^2-3x+2}-\sqrt{x-2}=\sqrt{x^2+2x-3}-\sqrt{x+3}\left(x\ge2\right)\)

\(\Leftrightarrow\sqrt{x-2}\left(\sqrt{x-1}-1\right)=\sqrt{x+3}\left(\sqrt{x-1}-1\right)\)

\(\Leftrightarrow\left(\sqrt{x-2}-\sqrt{x+3}\right)\left(\sqrt{x-1}-1\right)=0\)

\(\Leftrightarrow\sqrt{x-1}-1=0\Leftrightarrow x=2\)

c, Đặt :  \(\hept{\begin{cases}\sqrt{2x^2-9x+4}=a\\\sqrt{2x-1}=b\end{cases}\left(a;b\ge0\right)\Rightarrow\hept{\begin{cases}2x^2-9x+4=a^2\\2x-1=b^2\end{cases}}}\)

Pt đã cho trở thành: \(a+3b=\sqrt{2x^2+21x-11}\Rightarrow a^2+6ab+9b^2=2x^2+21x-11\)

Lại có: \(2x^2-9x+4+15\left(2x-1\right)=2x^2+21x-11\)

\(\Rightarrow a^2+15b^2=2x^2+21x-11\)

Do đó: \(a^2+6ab+9b^2=a^2+15b^2\Leftrightarrow6ab-6b^2=0\)

\(\Leftrightarrow6b\left(a-b\right)=0\Leftrightarrow\orbr{\begin{cases}b=0\\a=b\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=0\\\sqrt{2x^2-9x+4}=\sqrt{2x-1}\end{cases}}\)

Bạn tự giải tiếp.

d, \(x^2+2015x-2014=2\sqrt{2017x-2016}\)

\(\Leftrightarrow x^2+2017x-2016-2x+2=2\sqrt{2017x-2016}\)

\(\Leftrightarrow\left(x^2-2x+1\right)+\left(2017x-2016-2\sqrt{2017x-2016}+1\right)=0\)

\(\Leftrightarrow\left(x-1\right)^2+\left(\sqrt{2017x-2016}-1\right)^2=0\)

\(\Leftrightarrow x=1\)

Chúc bạn học tốt.

Ta có:

\(\frac{2}{\sqrt{a}}+\frac{2}{\sqrt{b}}+\frac{2}{\sqrt{c}}=\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}\right)+\left(\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)+\left(\frac{1}{\sqrt{c}}+\frac{1}{\sqrt{a}}\right)\)

\(\ge\frac{\left(1+1\right)^2}{\sqrt{a}+\sqrt{b}}+\frac{\left(1+1\right)^2}{\sqrt{b}+\sqrt{c}}+\frac{\left(1+1\right)^2}{\sqrt{c}+\sqrt{a}}\)

\(=\frac{4}{\sqrt{a}+\sqrt{b}}+\frac{4}{\sqrt{b}+\sqrt{c}}+\frac{4}{\sqrt{c}+\sqrt{a}}\)

=> \(2\left(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\right)\)\(\ge4\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)

=> \(\frac{1}{\sqrt{a}}+\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\)\(\ge2\left(\frac{1}{\sqrt{a}+\sqrt{b}}+\frac{1}{\sqrt{b}+\sqrt{c}}+\frac{1}{\sqrt{c}+\sqrt{a}}\right)\)

"=" xảy ra <=> a =b =c.

Nhân cả 2 vế với xyz bất đẳng thức sẽ thành yz+ xz+xy+yz\(\sqrt{1+x^2}\)+xz\(\sqrt{1+y^2}+xy\sqrt{1+z^2}\le x^2y^2z^2\)

Ta có yz\(\sqrt{1+x^2}=\sqrt{yz}.\sqrt{yz+x^2yz}=\sqrt{yz}.\sqrt{yz+x\left(x+y+z\right)}=\)\(\sqrt{yz}.\sqrt{\left(x+y\right)\left(x+z\right)}\)\(\le\)\(yz+\frac{\left(x+y\right)\left(x+z\right)}{4}\)(2ab\(\le a^2+b^2\))

làm tương tự ta được xz\(\sqrt{1+x^2}\le xz+\frac{\left(x+y\right)\left(y+z\right)}{4};xy\sqrt{1+z^2}\le xy+\frac{\left(y+z\right)\left(z+x\right)}{4}.\)

vế trái \(\le\) 2(xy+yz+zx) + \(\frac{\left(x+y\right)\left(x+z\right)+\left(y+x\right)\left(y+z\right)+\left(z+x\right)\left(z+y\right)}{4}\)\(\le2.\frac{1}{3}.\left(x+y+z\right)^2+\frac{\frac{1}{3}\left(x+y+y+z+z+x\right)^2}{4}=\left(x+y+z\right)^2=x^2y^2z^2.\)

[ (a-b)² +(b-c)² +(c-a)² \(\ge0\)<=>\(ab+bc+ca\le\frac{1}{3}\left(a+b+c\right)^2\) áp dụng vào trên)

dấu '=' xảy ra khi x=y=z \(\sqrt{3}\)

\(DK:x\in\left[\frac{1}{2};4\right]\)

PT

\(\Leftrightarrow\left(\sqrt{x^2+x+2}-2\right)+\left(\sqrt{2x-1}-1\right)+\left(x-1\right)=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x+2\right)}{\sqrt{x^2+x+2}+2}+\frac{2\left(x-1\right)}{\sqrt{2x-1}+1}+\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(\frac{x+2}{\sqrt{x^2+x+2}}+\frac{2}{\sqrt{2x-1}+1}+1\right)=0\)

Vì \(\frac{x+2}{\sqrt{x^2+x+2}}+\frac{2}{\sqrt{2x-1}+1}+1>0\)

\(\Rightarrow x=1\left(n\right)\)

Vay nghiem cua PT la \(x=1\)

F=x³+y³+2xy=(x+y)³-3xy(x+y)+2xy

=(x+y)³-xy(3x+3y-2)

=2007³-xy[3.2007-2]

làm tiếp đi 

chú ý \(xy\le\frac{\left(x+y\right)^2}{4}\)(bđt AM-GM)

Đầu tiên tìm GTLN, GTNN của xy.

Không mất tính tổng quát giả sử:

\(x\ge y+1\)

\(\Leftrightarrow x-y-1\ge0\)

\(\Leftrightarrow x-y-1+xy\ge xy\)

\(\Leftrightarrow\left(x-1\right)\left(y+1\right)\ge xy\)

Từ đây ta suy được:

\(2006.1< 2005.2< 2004.3< ...< 1003.1004\)

Vậy \(min_{xy}=2006.1;max_{xy}=1003.1004\)

Ta lại có:

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

Thế vô là xong

khai triển và rút gọn ta được:

\(4a^3+4b^3+4c^3+24abc\ge\left(a+b+c\right)^3.\)<=> \(a^3+b^3+c^3+8abc\ge\left(a+b\right)\left(b+c\right)\left(c+a\right)\)<=> a(a-b)(a-c) + b(b-a)(b-c) +c(c-a)(c-b) +3abc\(\ge0\)

giả sử \(a\ge b\ge c\)

c(c-a)(c-b)\(\ge0\)

a(a-b)(a-c) + b(b-a)(b-c) = (a-b)(a² - b² + bc-ac) = (a-b)²(a+b-c) \(\ge0\)

3abc\(\ge0\)

cộng vế theo vế ta được bdt cần chứng minh

dâu '=' khi \(\hept{\begin{cases}c\left(c-a\right)\left(c-b\right)=0\\\left(a-b\right)^2\left(a+b-c\right)=0\\3abc=0\end{cases}}\)=> a=b; c=0

điều kiện x\(x^3\ge0< =>x\ge0\)

pt <=> x³ +\(\sqrt{x^3}=\left(x+4\right)^2+\left(x+4\right)\)

đặt \(\sqrt{x^3}=a\left(a\ge0\right)=>a^2=x^3\)ta có hệ \(\hept{\begin{cases}a^2+a=\left(x+4\right)^2+\left(x+4\right)\\a^2=x^3\end{cases}< =>\hept{\begin{cases}\left(a-x-4\right)\left(a+x+4+1\right)=0\\a^2=x^3\end{cases}< =>}}\)

\(\hept{\begin{cases}a-x-4=0\\a^2=x^3\end{cases}\left(a+x+5>0\right)}\)vì a; x\(\ge0\) <=> \(\hept{\begin{cases}a=x+4\\\left(x+4\right)^2=x^3\end{cases}< =>x^3-x^2-8x-16=0}\)<=> (x-4)(x² +3x +4) = 0 <=> x=4 (vì x² +3x+4>0 với x\(\ge0\))

vậy x=4 là nghiệm duy nhất

cam on anh

goi giao MF voi ABla H , giao ME voi AC la K, MD voi BC la I

Do tam giac ABC noi tiep (O) ma M thuoc (o) nen ABMC noi tiep

xet tam giac MDF co \(\hept{\begin{cases}H.la.trung.diem.MF\\I.la.trung.diem.DM\end{cases}\Rightarrow HI//DF}\) (1)

tuong tu cung co \(IK//ED\) va  \(HK//EF\) ( do tinh chat duong trung binh)          (2)

Xet tu giac HBIM co \(\widehat{BHM}+\widehat{BIM}=90+90=180^o\)

=> HBIM la tu giac noi tiep => \(\widehat{HIB}=\widehat{BMH}\)  (cung chan \(\widebat{BH}\) )   (4)

tuong tu cung chung minh duoc tu giac MIKC la tu giac noi tiep => \(\widehat{KIC}=\widehat{KMC}\left(cung.chan.\widebat{KC}\right)\)(3)

Lai co \(\widehat{HBM}=\widehat{MAH}+\widehat{AMB}\) (tinh chat goc ngoai)

va \(\widehat{MCK}=\widehat{MCB}+\widehat{ACB}\) 

ma ABMC noi tiep suy ra \(\hept{\begin{cases}\widehat{AMB}=\widehat{ACB}\\\widehat{MAB}=\widehat{MCB}\end{cases}}\)

=> \(\widehat{MHB}=\widehat{MCK}\)

xet tam giac MHB va tam giac MKC co

\(\widehat{H}=\widehat{K}=90\)

\(\widehat{MHB}=\widehat{MCK}\) (cmt)

=> \(\widehat{HMB}=\widehat{KMC}\) (5)

tu (3),(4),(5)  =>\(\widehat{HIB}=\widehat{KIC}\)

=> H,I,K thang hang (6)

tu (1),(2),(6)

suy ra F,D,E thang hang ( tien de Oclit)

chuc ban hoc tot

Cần gấp !!