a) \(1+\sqrt{3x+1}=3x\)

ĐKXĐ: \(3x+1\ge0\Leftrightarrow x\ge\frac{-1}{3}\)

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

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

\(\Leftrightarrow3x+1=9x^2-6x+1\)

\(\Leftrightarrow9x^2-9x=0\)

\(\Leftrightarrow9x\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}}\)(tm)

Vậy tập nghiệm của phương trình là \(S=\left\{0;1\right\}\)

b) \(\sqrt{\frac{5x+7}{x+3}}=4\)

ĐKXĐ: \(\hept{\begin{cases}\frac{5x+7}{x+3}>0\\x+3\ne0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x>\frac{-7}{5}\\x< -3\end{cases}}\)

\(\Rightarrow\left(\sqrt{\frac{5x+7}{x+3}}\right)^2=4^2\)

\(\Leftrightarrow\frac{5x+7}{x+3}=16\)

\(\Leftrightarrow5x+7=16\left(x+3\right)\)

\(\Leftrightarrow5x+7=16x+48\)

\(\Leftrightarrow-11x=41\)

\(\Leftrightarrow x=\frac{-41}{11}\)(tm)

Vậy phương trình có 1 nghiệm duy nhất là \(x=\frac{-41}{11}\)

1.Giải hệ phương trình:

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

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

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

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

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

g.\(\left\{{}\begin{matrix}2x^2+5y^2=129\\-3x^2+y^2=13\end{matrix}\right.\)

đặt \(P=\frac{1}{\sqrt{x^5-x^2+3xy+6}}+\frac{1}{\sqrt{y^5-y^2+3yz+6}}+\frac{1}{\sqrt{z^5-z^2+3zx+6}}\)

ta có:\(\left(x^3+2x^2+3x+3\right)\left(x-1\right)^2\ge0\)

\(\Leftrightarrow x^5-x^2\ge3x-3\)

cmtt=>\(y^5-y^2\ge3y-3;z^5-z^2\ge3z-3\)

\(\Rightarrow P\le\frac{1}{\sqrt{3x-3+3xy+6}}+\frac{1}{\sqrt{3y-3+3yz+6}}+\frac{1}{\sqrt{3z-3+3zx+6}}\)

\(=\frac{1}{\sqrt{3\left(x+xy+1\right)}}+\frac{1}{\sqrt{3\left(y+yz+1\right)}}+\frac{1}{\sqrt{3\left(z+zx+1\right)}}\)

áp dụng bunhia ta có:

\(3\left(x+xy+1\right)\ge\left(\sqrt{x}+\sqrt{xy}+1\right)^2\)

cmtt\(\Rightarrow P\le\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}\)

đặt \(\sqrt{x}=a;\sqrt{y}=b;\sqrt{z}=c\)

\(\Rightarrow\frac{1}{\sqrt{x}+\sqrt{xy}+1}+\frac{1}{\sqrt{y}+\sqrt{yz}+1}+\frac{1}{\sqrt{z}+\sqrt{zx}+1}=\frac{1}{a+ab+1}+\frac{1}{b+bc+1}+\frac{1}{c+ca+1}\)

\(=\frac{abc}{a+ab+abc}+\frac{1}{b+bc+1}+\frac{b}{bc+abc+b}=\frac{bc}{bc+b+1}+\frac{b}{bc+b+1}+\frac{1}{bc+b+1}=1\)

\(\Rightarrow P\le1\)

Cho biểu thức f(x;y) \(=\frac{2x^2+3xy^3-4x^2y-7y^3-2018}{3x-2y+\sqrt{3x^4+2y^2+3}-3x\sqrt[3]{y^2}+5}\).Gọi a,b,c là các số thực thỏa mãn:

\(\hept{\begin{cases}\sqrt{2}a+\sqrt[3]{3}b-\left(\sqrt{2}+1\right)c=\sqrt{2}\\2\sqrt{3}a-3\sqrt{2}b-\left(3-2\sqrt{7}\right)c=\sqrt{5}\\3\sqrt[3]{2}a-\left(1-3\sqrt{5}\right)b-2\sqrt{5}c=\sqrt{7}\end{cases}}\).Đặt A = f(a;b) , B = f(b;c), C = f(c;a).

Tìm min \(P=\frac{ABt^2-A^2t-C\left(A-1\right)}{Bt^2-At-C}\)

(Trích đề thi học sinh giỏi máy tính cầm tay)

1. a) cho \(1\le a,b,c\le2\). Tìm max \(P=\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\)

b) \(\left\{{}\begin{matrix}a,b,c\ge0\\a+b+c=1\end{matrix}\right.\). Cmr: \(\sqrt{\frac{3a^2+1}{3b^2+1}}+\sqrt{\frac{3b^2+1}{3c^2+1}}+\sqrt{\frac{3c^2+1}{3a^2+1}}\le\frac{7}{2}\)

2.a) \(a,b\ge0;c\ge1;a+b+c=2\). cmr: \(\left(6-a^2-b^2-c^2\right)\left(2-abc\right)\le8\)

b) \(\left\{{}\begin{matrix}a+b\le2\\a^2+b^2+ab=3\end{matrix}\right.\). Tìm max,min \(P=a^2+b^2-ab\)

Giải hpt sau:

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

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

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

d) \(\left\{{}\begin{matrix}\frac{2}{2x-y}+\frac{3}{x-2y}=\frac{1}{2}\\\frac{2}{2x-y}-\frac{1}{x-2y}=\frac{1}{18}\end{matrix}\right.\)

1. Giải các hpt sau:

a, \(\left\{{}\begin{matrix}x-y=4\\3x+4y=19\end{matrix}\right.\) b, \(\left\{{}\begin{matrix}x-\sqrt{3y}=\sqrt{3}\\\sqrt{3x}+y=7\end{matrix}\right.\)

2. Giải các hpt sau:

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

c, \(\left\{{}\begin{matrix}x+y=24\\\dfrac{x}{9}+\dfrac{y}{27}=2\dfrac{8}{9}\end{matrix}\right.\) d, \(\left\{{}\begin{matrix}\sqrt{x-1}-3\sqrt{y+2}=2\\2\sqrt{x-1}+5\sqrt{y+2=15}\end{matrix}\right.\)

3. Cho hpt \(\left\{{}\begin{matrix}\left(m+1\right)x-y=3\\mx+y=m\end{matrix}\right.\)

a, Giải hpt khi m=\(\sqrt{2}\)

b, tìm giá trị của m để hpt có nghiệm duy nhất thỏa mãn: x+y>0