Câu 1:

\(ĐK:x\ge2\)

Áp dụng BĐT cauchy ta có:

\(\left(x+1\right)+4\ge2\sqrt{4\left(x+1\right)}=4\sqrt{x+1}\\ \Leftrightarrow2\sqrt{x+1}\le\dfrac{x+5}{2}\)

Ta có \(\left(x-2\right)+1\ge2\sqrt{x-2}\Leftrightarrow\sqrt{x-2}\le\dfrac{x-1}{2}\)

\(\Leftrightarrow P\le\dfrac{x+5}{2}+\dfrac{x-1}{2}-x+2013=x+2-x+2013=2015\)

Dấu \("="\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x-2=1\end{matrix}\right.\Leftrightarrow x=3\)

Câu 2:

\(HPT\Leftrightarrow\left\{{}\begin{matrix}10\sqrt{x}+15y^3=140\\4y^3-10\sqrt{x}=12\end{matrix}\right.\left(x\ge0\right)\\ \Leftrightarrow19y^3=152\\ \Leftrightarrow y^3=8\Leftrightarrow y=2\\ \Leftrightarrow2\sqrt{x}+24=28\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\)

Vậy \(\left(x;y\right)=\left(4;2\right)\)

Câu 3:

\(HPT\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\my+2m+y=3\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=y+2\\y=\dfrac{3-2m}{m+1}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{5}{m+1}\\x=\dfrac{3-2m}{m+1}\end{matrix}\right.\\ \Leftrightarrow xy=\dfrac{5\left(3-2m\right)}{\left(m+1\right)^2}\)

Đặt \(xy=t\)

\(\Leftrightarrow m^2t+2mt+t=15-10m\\ \Leftrightarrow m^2t+2m\left(t+5\right)+t-15=0\)

PT có nghiệm nên \(\Delta'=\left(t+5\right)^2-t\left(t-15\right)\ge0\)

\(\Leftrightarrow10t+25+15t\ge0\Leftrightarrow t\ge-1\)

Vậy \(xy_{min}=-1\Leftrightarrow\dfrac{5\left(2m-3\right)}{\left(m+1\right)^2}=1\Leftrightarrow m^2-8m+16=0\Leftrightarrow m=4\)

Câu 4: \(a^2+b^2=4a+bc+540\)

c đâu ra vậy?

Câu 5:

Thay \(x=3\Leftrightarrow P\left(2\right)+2P\left(2\right)=3^2\Leftrightarrow P\left(2\right)=3\)

Thay \(x=\sqrt{2013}\)

\(\Leftrightarrow P\left(\sqrt{2013}-1\right)+2P\left(2\right)=\left(\sqrt{2013}\right)^2=2013\\ \Leftrightarrow P\left(\sqrt{2013}-1\right)+6=2013\\ \Leftrightarrow P\left(\sqrt{2013}-1\right)=2007\)

Bài 2:

a: \(\Leftrightarrow\left\{{}\begin{matrix}2-x+y-3x-3y=5\\3x-3y+5x+5y=-2\end{matrix}\right.\)

=>-4x-2y=3 và 8x+2y=-2

=>x=1/4; y=-2

b: \(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{5}{y-1}=1\\\dfrac{1}{x-2}+\dfrac{1}{y-1}=1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-1=5\\\dfrac{1}{x-2}=1-\dfrac{1}{5}=\dfrac{4}{5}\end{matrix}\right.\)

=>y=6 và x-2=5/4

=>x=13/4; y=6

c: =>x+y=24 và 3x+y=78

=>-2x=-54 và x+y=24

=>x=27; y=-3

d: \(\Leftrightarrow\left\{{}\begin{matrix}2\sqrt{x-1}-6\sqrt{y+2}=4\\2\sqrt{x-1}+5\sqrt{y+2}=15\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-11\sqrt{y+2}=-11\\\sqrt{x-1}=2+3\cdot1=5\end{matrix}\right.\)

=>y+2=1 và x-1=25

=>x=26; y=-1

a.

ĐKXĐ: \(x;y\ge-1;xy\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+y-3=\sqrt{xy}\\x+y+2\sqrt{xy+x+y+1}=14\end{matrix}\right.\)

Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\ge0\end{matrix}\right.\) với \(u^2\ge4v\) 

\(\Rightarrow\left\{{}\begin{matrix}u-3=\sqrt{v}\\u+2\sqrt{u+v+1}=14\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=u^2-6u+9\left(u\ge3\right)\\4\left(u+v+1\right)=\left(14-u\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\4u+4\left(u^2-6u+9\right)+4=\left(14-u\right)^2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\3u^2+8u-156=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}v=\left(u-3\right)^2\\\left[{}\begin{matrix}u=6\\u=-\dfrac{26}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}u=6\\v=9\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x+y=6\\xy=9\end{matrix}\right.\) \(\Rightarrow x=y=3\)

b.

ĐKXĐ: \(x;y\ge1\)

Xét \(\sqrt{x-1}+\sqrt{y-1}=3\)

\(\Leftrightarrow x+y-2+2\sqrt{\left(x-1\right)\left(y-1\right)}=9\)

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(y-1\right)}=\dfrac{11-x-y}{2}\)

Thế vào pt đầu:

\(x+y=5+\dfrac{11-x-y}{2}\)

\(\Leftrightarrow x+y=7\Rightarrow y=7-x\)

Thế xuống pt dưới:

\(\sqrt{x-1}+\sqrt{6-x}=3\)

\(\Leftrightarrow5+2\sqrt{\left(x-1\right)\left(6-x\right)}=9\)

\(\Leftrightarrow\left(x-1\right)\left(6-x\right)=4\)

\(\Leftrightarrow...\)

a: \(\left\{{}\begin{matrix}4\sqrt{5}-y=3\sqrt{2}\\10x+\sqrt{2}\cdot y=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x+\sqrt{2}\left(4\sqrt{5}-3\sqrt{2}\right)=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\10x=-1-4\sqrt{10}+6=5-4\sqrt{10}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=4\sqrt{5}-3\sqrt{2}\\x=\dfrac{1}{2}-\dfrac{2\sqrt{10}}{5}\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}\dfrac{3}{4}x+\dfrac{2}{5}y=2,3\\x-\dfrac{3}{5}y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{9}{4}x+\dfrac{6}{5}y=6,9\\2x-\dfrac{6}{5}y=1,6\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{17}{4}x=8,5\\x-0,6y=0,8\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=8,5:\dfrac{17}{4}=8,5\cdot\dfrac{4}{17}=2\\0,6y=x-0,8=2-0,8=1,2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)

c: ĐKXĐ: y>2

\(\left\{{}\begin{matrix}\left|x-1\right|-\dfrac{3}{\sqrt{y-2}}=-1\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2\left|x-1\right|-\dfrac{6}{\sqrt{y-2}}=-2\\2\left|x-1\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{7}{\sqrt{y-2}}=-7\\2\left|1-x\right|+\dfrac{1}{\sqrt{y-2}}=5\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\sqrt{y-2}=1\\2\left|x-1\right|=5-1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y-2=1\\\left|x-1\right|=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x-1\in\left\{2;-2\right\}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=3\\x\in\left\{3;-1\right\}\end{matrix}\right.\left(nhận\right)\)

a:

ĐKXĐ: y+1>=0

=>y>=-1

 \(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}+7=0\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}2\left(x^2-2x\right)+\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}4\left(x^2-2x\right)+2\sqrt{y+1}=0\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}7\left(x^2-2x\right)=-7\\3\left(x^2-2x\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-2x=-1\\3\cdot\left(-1\right)-2\sqrt{y+1}=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x^2-2x+1=0\\2\sqrt{y+1}=-3+7=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left(x-1\right)^2=0\\\sqrt{y+1}=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-1=0\\y+1=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=3\left(nhận\right)\end{matrix}\right.\)

b: \(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\sqrt{4x^2-8x+4}+5\sqrt{y^2+4y+4}=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\2\cdot\sqrt{\left(2x-2\right)^2}+5\cdot\sqrt{\left(y+2\right)^2}=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}5\left|x-1\right|-3\left|y+2\right|=7\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}20\left|x-1\right|-12\left|y+2\right|=28\\20\left|x-1\right|+25\left|y+2\right|=65\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-37\left|y+2\right|=-37\\4\left|x-1\right|+5\left|y+2\right|=13\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\left|y+2\right|=1\\4\left|x-1\right|=13-5=8\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\left|y+2\right|=1\\\left|x-1\right|=2\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}x-1\in\left\{2;-2\right\}\\y+2\in\left\{1;-1\right\}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\in\left\{3;-1\right\}\\y\in\left\{-1;-3\right\}\end{matrix}\right.\)

c: ĐKXĐ: \(\left\{{}\begin{matrix}x< >-1\\y< >-4\end{matrix}\right.\)

\(\left\{{}\begin{matrix}\dfrac{3x}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3x+3-3}{x+1}-\dfrac{2}{y+4}=4\\\dfrac{2x+2-2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}3-\dfrac{3}{x+1}-\dfrac{2}{y+4}=4\\2-\dfrac{2}{x+1}-\dfrac{5}{y+4}=9\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{3}{x+1}+\dfrac{2}{y+4}=3-4=-1\\\dfrac{2}{x+1}+\dfrac{5}{y+4}=2-9=-7\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{6}{x+1}+\dfrac{4}{y+4}=-2\\\dfrac{6}{x+1}+\dfrac{15}{y+4}=-21\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{-11}{y+4}=19\\\dfrac{3}{x+1}+\dfrac{2}{y+4}=-1\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y+4=-\dfrac{11}{19}\\\dfrac{3}{x+1}+2:\dfrac{-11}{19}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{11}{19}-4=-\dfrac{87}{19}\\\dfrac{3}{x+1}=-1-2:\dfrac{-11}{19}=-1+2\cdot\dfrac{19}{11}=\dfrac{27}{11}\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x+1=\dfrac{11}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-\dfrac{87}{19}\\x=\dfrac{2}{9}\end{matrix}\right.\)(nhận)

d:

ĐKXĐ: x<>1 và y<>-2

\(\left\{{}\begin{matrix}\dfrac{x+1}{x-1}+\dfrac{3y}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\dfrac{x-1+2}{x-1}+\dfrac{3y+6-6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}1+\dfrac{2}{x-1}+3-\dfrac{6}{y+2}=7\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}\dfrac{2}{x-1}-\dfrac{6}{y+2}=7-4=3\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}-\dfrac{1}{y+2}=-1\\\dfrac{2}{x-1}-\dfrac{5}{y+2}=4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y+2=1\\\dfrac{2}{x-1}-5=4\end{matrix}\right.\)

=>\(\left\{{}\begin{matrix}y=-1\\\dfrac{2}{x-1}=9\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x-1=\dfrac{2}{9}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y=-1\\x=\dfrac{11}{9}\end{matrix}\right.\left(nhận\right)\)

Ta bắt đầu bằng việc giả sử một giá trị ban đầu cho x, y và z, sau đó lặp lại quá trình tính toán cho đến khi đạt được độ chính xác mong muốn.

Ví dụ, giả sử ta chọn x = 1, y = 1 và z = 1 làm giá trị ban đầu. Sau đó, ta thực hiện các bước sau:

Bước 1: Tính toán giá trị mới cho x, y và z bằng cách sử dụng các phương trình đã cho: x_new = (2y - 1) / sqrt(y) y_new = (2z - 1) / sqrt(z) z_new = (2*x - 1) / sqrt(x)

Bước 2: Kiểm tra độ chính xác của giá trị mới so với giá trị cũ. Nếu đạt được độ chính xác mong muốn, ta dừng lại. Nếu không, ta lặp lại bước 1 với giá trị mới của x, y và z.

Tiếp tục lặp lại quá trình trên cho đến khi đạt được độ chính xác mong muốn. Khi đó, ta sẽ có giá trị x, y và z tương ứng là nghiệm của hệ phương trình đã cho.

a, Áp dụng bất đẳng thức Holder cho 2 bộ số \(\left(x,y,z\right)\left(3;3;3\right)\) ta có:

\(\left(x+3\right)\left(y+3\right)\left(z+3\right)\ge\left(\sqrt[3]{xyz}+\sqrt[3]{3.3.3}\right)^3=\left(\sqrt[3]{xyz}+3\right)\)

\(\sqrt[3]{\left(x+3\right)\left(y+3\right)\left(z+3\right)}\ge3+\sqrt[3]{xyz}\)

Dấu "=" xảy ra \(\Leftrightarrow x=y=z\)

\(\Rightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}=3\sqrt{x}=\sqrt{2017}\)

\(\Rightarrow x=\frac{\sqrt{2017}}{3}\)

\(\Rightarrow\left(x,y,z\right)=\left(\frac{\sqrt{2017}}{3},\frac{\sqrt{2017}}{3},\frac{\sqrt{2017}}{3}\right)\)

P/s: Không chắc cho lắm ạ.

Vũ Minh Tuấn, Hoàng Tử Hà, đề bài khó wá, Lê Gia Bảo, Aki Tsuki, Nguyễn Việt Lâm, Lê Thị Thục Hiền,

Học 24h, @tth_new, @Akai Haruma, Nguyễn Trúc Giang, Băng Băng 2k6

Help meeee, please!

thanks nhiều

1.

ĐKXĐ: ...

Đặt \(\left\{{}\begin{matrix}\sqrt{x+1}=a\ge0\\\sqrt{y-1}=b\ge0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=a^2-1\\y=b^2+1\end{matrix}\right.\)

Hệ trở thành:

\(\left\{{}\begin{matrix}a+b=3\\a^2+b^2=5\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}b=3-a\\a^2+b^2=5\end{matrix}\right.\)

\(\Rightarrow a^2+\left(3-a\right)^2=5\)

\(\Leftrightarrow2a^2-6a+4=0\Rightarrow\left[{}\begin{matrix}a=1\Rightarrow b=2\\a=2\Rightarrow b=1\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}\sqrt{x+1}=1\\\sqrt{y-1}=2\end{matrix}\right.\\\left\{{}\begin{matrix}\sqrt{x+1}=2\\\sqrt{y-1}=1\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x=0\\y=5\end{matrix}\right.\\\left\{{}\begin{matrix}x=3\\y=2\end{matrix}\right.\end{matrix}\right.\)

2.

Pt đầu tương đương:

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

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

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

\(\Rightarrow y=2\)

Thay xuống dưới:

\(\sqrt{x^2+5}+3=3x-3\)

\(\Leftrightarrow\sqrt{x^2+5}=3x-6\) (\(x\ge2\))

\(\Leftrightarrow x^2+5=9x^2-36x+36\)

\(\Leftrightarrow8x^2-36x+31=0\Rightarrow\left[{}\begin{matrix}x=\frac{9+\sqrt{19}}{4}\\x=\frac{9-\sqrt{19}}{4}\left(l\right)\end{matrix}\right.\)

3.

ĐKXĐ: ...

Trừ vế cho vế ta được:

\(2x-2y=y-x+\sqrt{y-2}-\sqrt{x-2}\)

\(\Leftrightarrow3\left(x-y\right)+\sqrt{x-2}-\sqrt{y-2}=0\)

\(\Leftrightarrow3\left(x-y\right)+\frac{x-y}{\sqrt{x-2}+\sqrt{y-2}}=0\)

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

\(\Leftrightarrow x=y\) (ngoặc to luôn dương)

Thay vào pt đầu:

\(2x-2=x+\sqrt{x-2}\)

\(\Leftrightarrow x-2=\sqrt{x-2}\Rightarrow\left[{}\begin{matrix}x-2=0\\x-2=1\end{matrix}\right.\)

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