\(\hept{\begin{cases}x-2\sqrt{y+1}=3\\x^3-4x^2\sqrt{y+1}-9-8y=-52-4xy\end{cases}\left(ĐK\hept{\begin{cases}x\in R\\y\ge-1\end{cases}}\right)}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\\left(x^3-4x^2\sqrt{y+1}+4xy+4x\right)-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\x\left(x-2\sqrt{y+1}\right)^2-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\x\left(3+2\sqrt{y+1}-2\sqrt{y+1}\right)^2-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\9x-13x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\-4x-8y+52=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\-x-2y+13=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\-2\sqrt{y+1}-2y+10=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=3+2\sqrt{y+1}\\\sqrt{y+1}=y-5\left(1\right)\end{cases}}\)

\(\left(1\right)\Leftrightarrow y+1=y^2-10y+25\left(y\ge1\right)\)

\(\Leftrightarrow y^2-11y+24=0\)

\(\Delta=11^2-4.24=25>0\)

\(\Rightarrow\)pt (1) có 2 nghiệm pb \(\orbr{\begin{cases}y=\frac{11+5}{2}=8\left(tm\right)\\y=\frac{11-5}{2}=3\left(tm\right)\end{cases}}\)

+) \(y=8\Rightarrow x=9\left(tm\right)\)

+) \(y=3\Rightarrow x=7\left(tm\right)\)

Vậy hệ pt có no (x,y) =( 9,8) ; ( 7,3) 

\(\hept{\begin{cases}x^2+y^2+\frac{2xy}{x+y}=1\left(1\right)\\\sqrt{x+y}=x^2-y\left(2\right)\end{cases}}\)\(ĐK:x+y>0\)

Từ \(\left(1\right)\Rightarrow x^2+y^2+2xy-1-2xy+\frac{2xy}{x+y}=0\)

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

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

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

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

\(\Leftrightarrow\left(x+y-1\right)\left(x^2+y^2+x+y\right)=0\)( vì x+y >0 )

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

TH1: Xét pt:  \(x^2+y^2+x+y=0\)

Ta thấy \(\hept{\begin{cases}x^2\ge0;\forall x\\y^2\ge0;\forall y\\x+y>0\end{cases}\Rightarrow}x^2+y^2+x+y>0\)

\(\Rightarrow\)pt vô nghiệm

TH2: x=1-y thay vào (2) ta được :

\(\left(1-y\right)^2-y=1\)

\(\Leftrightarrow y^2-3y=0\)

\(\Leftrightarrow y\left(y-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}y=0\Rightarrow x=1\\y=3\Rightarrow x=-2\end{cases}}\)

Vậy hệ pt có no (x,y) = (-2;3) ; (1;0)

\(\hept{\begin{cases}\left(x+y\right)\sqrt{x-y+2}=x+3y+2\\\left(x-y\right)\sqrt{x-y+2}=\left(x+y+1\right)\sqrt{x+y-2}\end{cases}\left(1\right)ĐK:\hept{\begin{cases}x-y\ge-2\\x+y\ge2\end{cases}}}\)

Đặt \(\hept{\begin{cases}a=x+y\left(a\ge2\right)\\b=\sqrt{x-y+2}\left(b\ge0\right)\end{cases}\Rightarrow}\hept{\begin{cases}a=x+y\\b^2-2=x-y\end{cases}}\)

Hệ (1) trở thành:\(\hept{\begin{cases}ab=2a-b^2+4\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}2a-b^2-ab+4=0\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}a\left(2-b\right)-\left(b-2\right)\left(b+2\right)=0\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\left(2-b\right)\left(a+b+2\right)=0\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\) hoặc \(\hept{\begin{cases}a+b+2=0\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

TH1: \(\hept{\begin{cases}b=2\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}b=2\\\left(a+1\right)\sqrt{a-2}=4\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\\left(a+1\right)^2\left(a-2\right)=16\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\a^3-3a-18=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\\left(a^3+3a^2+6a\right)-\left(3a^2+9a+18\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\\left(a-3\right)\left(a^2+3a+6\right)=0\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=2\\a=3\end{cases}}\)( vì \(a^2+3a+6>0\))

\(\Rightarrow\hept{\begin{cases}x+y=3\\\sqrt{x-y+2}=2\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}x=\frac{5}{2}\\y=\frac{1}{2}\end{cases}}\)

TH2:\(\hept{\begin{cases}a+b+2=0\\\left(b^2-2\right)b=\left(a+1\right)\sqrt{a-2}\end{cases}}\)

Vì \(a\ge2;b\ge0\)

\(\Rightarrow a+b+2>0\)

\(\Rightarrow\)hệ pt vô nghiệm

Vậy hệ pt có no (x,y) = \(\left(\frac{5}{2};\frac{1}{2}\right)\)

ĐKXĐ : x ≥ 0 ; x ≠ 4

\(\frac{1}{\sqrt{x}-2}\ge1\)<=> \(\frac{1}{\sqrt{x}-2}-1\ge0\)

<=> \(\frac{1}{\sqrt{x}-2}-\frac{\sqrt{x}-2}{\sqrt{x}-2}\ge0\)

<=> \(\frac{-\sqrt{x}+3}{\sqrt{x}-2}\ge0\)

TH1. \(\hept{\begin{cases}-\sqrt{x}+3\ge0\\\sqrt{x}-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\le9\\x>4\end{cases}}\Rightarrow4< x\le9\)

TH2. \(\hept{\begin{cases}-\sqrt{x}+3\le0\\\sqrt{x}-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x\ge9\\x< 4\end{cases}}\left(loai\right)\)

Vậy với 4 < x ≤ 9 thì B ≥ 1

M=x+yxy.1z≥2√xyxy.1z=2z√xy≥2z(x+y2)=4z(x+y)M=x+yxy.1z≥2xyxy.1z=2zxy≥2z(x+y2)=4z(x+y)

=4z(1−z)=414−(z−12)2≥16=4z(1−z)=414−(z−12)2≥16

Min M= 16 khi  z=1/2 và  x=y =1/4

a) Gọi \(A\in Ox;B\in Oy\Rightarrow\Delta OAB\)vuông tại O

Đường thẳng (d) giao Ox tại điểm \(A\left(x;0\right)\)-> thay y=0 vào hàm số ta được: 0=(m+2)x+3 -> (m+2)x=-3 -> \(x=\frac{-3}{m+2}\)

-> Điểm \(A\left(\frac{-3}{m+2};0\right)\)-> \(OA=|\frac{-3}{m+2}|\)(OA>0)

Đường thẳng (d) giao Oy tại điểm \(B\left(0;y\right)\)-> thay x=0 vào hàm số ta được: y=(m+2).0+3=3

-> Điểm \(B\left(0;3\right)\)-> \(OB=3\)

Có: \(S_{\Delta OAB}=\frac{3}{4}=\frac{1}{2}OA\cdot OB=\frac{1}{2}\cdot3\cdot\frac{|-3|}{|m+2|}=\frac{3\cdot3}{2|m+2|}=\frac{9}{2|m+2|}\)

\(\Rightarrow6|m+2|=36\Leftrightarrow|m+2|=6\Leftrightarrow\orbr{\begin{cases}m+2=6\\m+2=-6\end{cases}}\Leftrightarrow\orbr{\begin{cases}m=4\\m=-8\end{cases}}\)(TM)

Vậy...

b) ĐK: OA>0

\(\Delta OAB\)vuông tại O -> \(AB=\sqrt{OA^2+OB^2}=\sqrt{3^2+\left(\frac{-3}{m+2}\right)^2}=\sqrt{9+\frac{9}{\left(m+2\right)^2}}\)

Kẻ \(OH\perp d\)tại H -> OH là khoảng cách từ đường thẳng từ O đến d

Áp dụng htl trong \(\Delta OAB\)vuông tại O, đường cao OH -> \(OA.OB=OH.AB\)

\(\rightarrow3\cdot\frac{|-3|}{|m+2|}=\frac{3\sqrt{2}}{2}.\sqrt{9+\frac{9}{\left(m+2\right)^2}}\)

\(\Leftrightarrow\left(3\cdot\frac{|-3|}{|m+2|}\right)^2=\left(\frac{3\sqrt{2}}{2}\right)^2\left(9+\frac{9}{\left(m+2\right)^2}\right)\)

\(\Leftrightarrow\frac{81}{\left(m+2\right)^2}=\frac{9\cdot9}{2}+\frac{9\cdot9}{2\left(m+2\right)^2}\Leftrightarrow\frac{81}{\left(m+2\right)^2}=\frac{81}{2}+\frac{81}{2\left(m+2\right)^2}\)

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

\(\Rightarrow1-\left(m+2\right)^2=0\Rightarrow\left(m+2\right)^2=1\Leftrightarrow\orbr{\begin{cases}m+2=1\\m+2=-1\end{cases}}\)     

\(\Leftrightarrow\orbr{\begin{cases}m=-1\\m=-3\end{cases}}\)(TM)

Vậy...

Hì cậu kiểm tra xem tớ có sai dấu hay sai bước chỗ nào với nhé vì tớ hay cẩu thả lắm:'33