ta có : (x²+y²)^2\(\ge\)4x²y²

\(\Leftrightarrow\)25\(\ge\)4x²y²

\(\Leftrightarrow\)x²y²\(\le\)\(\frac{25}{4}\)

ta lại có: S=(x²)³+(y²)³=(x²+y²)³−3(x²+y²)x²y²  = 5³ - 3.5.x²y² = 125 - 15.x²y²\(\le\)125

vậy...

đó nha bn

a,Xét tg DHB và tg DCA có: ^HDB=^CDA=90 độ, ^DBH=^DAC ( cùng phụ với hai góc bằng nhau BHD=^AHE)

Do đó: tg HDB đồng dạng tg DCA (g.g)

Suy ra: HD/DC=BD/DA-> bd*dc=dh*da

b, HD/HA=SBHC/SABC

HE/BE=SAHC/SABC

HF/CF=SHAB/SABC

HD/HA+HE/BE+HF/CF=SBHC/SABC+SAHC/SABC+SAHB/SABC=1

a, \(2+\frac{2\left(x+3\right)}{6}\le2-\frac{x-3}{5}\)

\(\Leftrightarrow\frac{60+10\left(x+3\right)}{30}\le\frac{60-6\left(x-3\right)}{30}\Leftrightarrow60+10x+30\le60-6x+18\)

\(\Leftrightarrow90+10x\le78-6x\Leftrightarrow12\le-16x\Leftrightarrow x\ge-\frac{12}{16}=-\frac{3}{4}\)

b, \(\frac{3-5x}{-4}\le0\Leftrightarrow3-5x\le0\left(-4\le0\right)\Leftrightarrow x\ge\frac{3}{5}\)

c, \(\frac{2x+1}{2}+3\ge\frac{3-5x}{3}-\frac{4x+1}{4}\)

\(\Leftrightarrow\frac{12x+6-36}{12}\ge\frac{12-20x-12x-3}{12}\Leftrightarrow12x-30\ge-32x+9\)

\(\Leftrightarrow44x\le39\Leftrightarrow x\le\frac{39}{44}\)

d, MTC là 18 quy đồng lên nhé 

=) vào ngay quả bảng phá dấu GTTĐ, cay thế :< 

a, \(3x+\frac{2x}{3}-3=\frac{5}{2}x-2\Leftrightarrow\frac{18x+4x-18}{6}=\frac{15x-12}{6}\)

\(\Rightarrow22x-18=15x-12\Leftrightarrow7x=6\Leftrightarrow x=\frac{6}{7}\)

Vậy pt có nghiệm x = 6/7 

b, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}+\frac{x+1}{3}=\frac{x+7}{12}\)

\(\Leftrightarrow\frac{9\left(2x+1\right)-2\left(5x+3\right)+4\left(x+1\right)}{12}=\frac{x+7}{12}\)

\(\Rightarrow18x+9-10x-6+4x+4=x+7\)

\(\Leftrightarrow12x+7=x+7\Leftrightarrow11x=0\Leftrightarrow x=0\)

Vậy pt có nghiệm là x = 0 

c, \(\frac{3x}{x-3}-\frac{x-3}{x+3}=2\)ĐK : \(x\ne\pm3\)

\(\Leftrightarrow\frac{3x\left(x+3\right)-\left(x-3\right)^2}{\left(x-3\right)\left(x+3\right)}=\frac{2\left(x-3\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}\)

\(\Rightarrow3x^2+9x-x^2+6x-9=2\left(x^2-9\right)\)

\(\Leftrightarrow2x^2+15x-9=2x^2-18\Leftrightarrow15x+9=0\Leftrightarrow x=-\frac{9}{15}=-\frac{3}{5}\)

Vậy pt có nghiệm là x = -3/5 

d, Sửa đề :  \(\frac{x+10}{2003}+\frac{x+6}{2007}+\frac{x+2}{2011}+3=0\)

\(\Leftrightarrow\frac{x+10}{2003}+1+\frac{x+6}{2007}+1+\frac{x+2}{2011}+1=0\)

\(\Leftrightarrow\frac{x+2013}{2003}+\frac{x+2013}{2007}+\frac{x+2013}{2011}=0\)

\(\Leftrightarrow\left(x+2013\right)\left(\frac{1}{2003}+\frac{1}{2007}+\frac{1}{2011}\ne0\right)=0\Leftrightarrow x=-2013\)

Vậy pt có nghiệm là x = -2013 

e, \(4\left(x+5\right)-3\left|2x-1\right|=10\)

\(\Leftrightarrow4x+20-3\left|2x-1\right|=10\Leftrightarrow-3\left|2x-1\right|=-10-4x\)

\(\Leftrightarrow\left|2x-1\right|=\frac{10+4x}{3}\)

ĐK : \(\frac{10+4x}{3}\ge0\Leftrightarrow10+4x\ge0\Leftrightarrow x\ge-\frac{10}{4}=-\frac{5}{2}\)

TH1 : \(2x-1=\frac{10+4x}{3}\Rightarrow6x-3=10+4x\Leftrightarrow2x=13\Leftrightarrow x=\frac{13}{2}\)( tm )

TH2 : \(2x-1=\frac{-10-4x}{3}\Rightarrow6x-3=-10-4x\Leftrightarrow10x=-7\Leftrightarrow x=-\frac{7}{10}\)( tm )

f, để mình xem lại đã, quên cách phá GTTĐ rồi :v :> 

Tự làm đi e

\(\left(x+\frac{1}{9}\right)\left(2x-5\right)< 0\)

TH1 : \(\hept{\begin{cases}x+\frac{1}{9}< 0\\2x-5>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -\frac{1}{9}\\x>\frac{5}{2}\end{cases}}}\)vô lí 

TH2 : \(\hept{\begin{cases}x+\frac{1}{9}>0\\2x-5< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-\frac{1}{9}\\x< \frac{5}{2}\end{cases}\Leftrightarrow}-\frac{1}{9}< x< \frac{5}{2}}\)

Vậy tập nghiệm của bft là S = { x | -1/9 < x < 5/2 } 

Ta có A = -x² + 2x + 9 = -x² + 2x - 1 + 10 = -(x² - 2x + 1) + 10 = -(x - 1)² + 10 \(\le\)10

Dấu "=" xảy ra <=> x - 1 = 0 

=> x = 1

Vậy Max A = 10 <=> x = 1

Ta có: \(A=-x^2+2x+9\)

\(A=\left(-x^2+2x-1\right)+10\)

\(A=-\left(x^2-2x+1\right)+10\)

\(A=-\left(x-1\right)^2+10\le10\left(\forall x\right)\)

Dấu "=" xảy ra khi: x = 1