\(\pi< x< \dfrac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sinx< 0\\cosx< 0\end{matrix}\right.\)

\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\dfrac{4}{5}\)

\(\Rightarrow tanx=\dfrac{sinx}{cosx}=\dfrac{4}{3}\) ; \(cotx=\dfrac{1}{tanx}=\dfrac{3}{4}\)

\(P=\dfrac{4}{3}+\dfrac{3}{4}=\dfrac{25}{12}\)

Áp dụng BĐT \(\left|a\right|+\left|b\right|\ge\left|a+b\right|\):

\(A=\left|x-3\right|+\left|x-1\right|+\left|x+1\right|+\left|x+3\right|\)

\(=\left|3-x\right|+\left|x+3\right|+\left|1-x\right|+\left|x+1\right|\)

\(\ge\left|3-x+x+3\right|+\left|1-x+x+1\right|=8\)

\(minA=8\Leftrightarrow\left\{{}\begin{matrix}\left(3-x\right)\left(x+3\right)\ge0\\\left(1-x\right)\left(x+1\right)\ge0\end{matrix}\right.\Leftrightarrow-1\le x\le1\)

Điều kiện: \(\left\{ \begin{array}{l} x > - 2\\ y > 1\\ x + y > 0 \end{array} \right.\)

Hệ phương trình tương đương: \(\left\{ \begin{array}{l} \sqrt {\dfrac{{x + y}}{{x + 2}}} + \sqrt {\dfrac{{x + y}}{{y - 1}}} = 2\\ {\left( {\dfrac{{x + 2}}{{x + y}}} \right)^2} + \left( {\dfrac{{y - 1}}{{x + y}}} \right)^2 = 2 \end{array} \right.\). Đặt \(\left\{ \begin{array}{l} a = \sqrt {\dfrac{{x + y}}{{x + 2}}} \\ b = \sqrt {\dfrac{{x + y}}{{y - 1}}} \end{array} \right.\) (với \(a,b > 0\))

Ta có hệ phương trình: \(\left\{ \begin{array}{l} a + b = 2\\ \dfrac{1}{{{a^4}}} + \dfrac{1}{{{b^4}}} = 2 \end{array} \right.\left( * \right)\)

Áp dụng BĐT AM - GM, ta có:

\(\begin{array}{l} 2 = a + b \geqslant 2\sqrt {ab} \Rightarrow ab \leqslant 1\\ 2 = \dfrac{1}{{{a^4}}} + \dfrac{1}{{{b^4}}} \geqslant 2\sqrt {\dfrac{1}{{{a^4}}}.\dfrac{1}{{{b^4}}}} \Rightarrow ab \geqslant 1 \end{array}\)

Thế nên \(\left( * \right) \Leftrightarrow a = b = 1\)

Ta lại có hệ phương trình: \(\left\{ \begin{array}{l} \dfrac{{x + y}}{{x + 2}} = 1\\ \dfrac{{x + y}}{{y - 1}} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = - 1\\ y = 2 \end{array} \right.\)

Vậy hệ phương trình có nghiệm là \((-1;2)\)

Đk: \(\left\{{}\begin{matrix}x>-2\\y>1\\x+y>0\end{matrix}\right.\)

hpt\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\2\left(x+y\right)^2=\left(x+2\right)^2+\left(y-1\right)^2\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{x+y}{x+2}}+\sqrt{\dfrac{x+y}{y-1}}=2\\\left(\dfrac{x+2}{x+y}\right)^2+\left(\dfrac{y-1}{x+y}\right)^2=2\end{matrix}\right.\)

Đặt \(a=\sqrt{\dfrac{x+y}{x+2}},b=\sqrt{\dfrac{x+y}{y-1}}\left(a,b>0\right)\)

Ta có hệ: \(\left\{{}\begin{matrix}a+b=2\\\dfrac{1}{a^4}+\dfrac{1}{b^4}=2\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4+b^4=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left[\left(a+b\right)^2-2ab\right]^2-2a^2b^2=2a^4b^4\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(4-2ab\right)^2-2a^2b^2=2a^4b^4\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^4b^4=a^2b^2-8ab+8\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\a^2b^2\left(a^2b^2-1\right)+8\left(ab-1\right)=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\\left(ab-1\right)\left[a^2b^2\left(ab+1\right)+8\right]=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}a+b=2\\ab-1\end{matrix}\right.\left(a,b>0\right)\Leftrightarrow\left\{{}\begin{matrix}a=1\\b=1\end{matrix}\right.\)

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

f(8 ) = 8 + 3 = 11

ĐK: \(x\ne\dfrac{1}{2};x\ne-\dfrac{1}{3}\)

\(\dfrac{x+2}{3x+1}\ge\dfrac{x-2}{2x-1}\)

\(\Leftrightarrow\dfrac{\left(x+2\right)\left(2x-1\right)-\left(x-2\right)\left(3x+1\right)}{\left(3x+1\right)\left(2x-1\right)}\ge0\)

\(\Leftrightarrow\dfrac{2x^2+3x-2-3x^2+5x+2}{6x^2-x-1}\ge0\)

\(\Leftrightarrow\dfrac{-x^2+8x}{6x^2-x-1}\ge0\)

\(\Leftrightarrow\left\{{}\begin{matrix}-x^2+8x\ge0\\6x^2-x-1>0\end{matrix}\right.\left(1\right)\) hoặc \(\left\{{}\begin{matrix}-x^2+8x\le0\\6x^2-x-1< 0\end{matrix}\right.\left(2\right)\)

\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}0\le x\le8\\\left[{}\begin{matrix}x>\dfrac{1}{2}\\x< -\dfrac{1}{3}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x\le8\)

\(\left(2\right)\Leftrightarrow\left\{{}\begin{matrix}\left[{}\begin{matrix}x\le0\\x\ge8\end{matrix}\right.\\-\dfrac{1}{3}< x< \dfrac{1}{2}\end{matrix}\right.\Leftrightarrow-\dfrac{1}{3}< x\le0\)

Vậy ...

\(-4\le\dfrac{x^2-2x-7}{x^2+1}\le1\)

\(\Leftrightarrow-4\left(x^2+1\right)\le x^2-2x-7\le x^2+1\)

\(\Leftrightarrow-4x^2-4\le x^2-2x-7\le x^2+1\)

\(\Leftrightarrow\left[{}\begin{matrix}-4x^2-4-x^2+2x+7\le0\\x^2-2x-7-x^2-1\le0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-5x^2+2x+3\le0\\-2x-8\le0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left[{}\begin{matrix}x\le-\dfrac{3}{5}\\x\ge1\end{matrix}\right.\\x\ge-4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}-4\le x\le-\dfrac{3}{5}\\x\ge1\end{matrix}\right.\)

\(-4\le\dfrac{x^2-2x-7}{x^2+1}\)

​\(\Leftrightarrow-4\left(x^2+1\right)\le x^2-2x-7\)​

\(\Leftrightarrow5x^2-2x-3\ge0\)

\(\Leftrightarrow\left(x-1\right)\left(5x+3\right)\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ge1\\x\le-\dfrac{3}{5}\end{matrix}\right.\)

\(\dfrac{x^2-2x-7}{x^2+1}\le1\)

\(\Leftrightarrow x^2-2x-7\le x^2+1\)

\(\Leftrightarrow x\ge-4\)

Vậy \(S=[1;+\infty)\cup\left[-4;-\dfrac{3}{5}\right]\)

Đề kiểu j vậy, ghi cho đầy đủ rồi người ta mới làm cho chứ, sáng giờ toàn gặp kiểu này

Gui đầy đủ mà này nó v á bn

\(\left\{{}\begin{matrix}2x^3+x^2y=3\left(1\right)\\2y^3+xy^2=3\end{matrix}\right.\)

Trừ vế theo vế hai phương trình ta được:

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

\(\Leftrightarrow2\left(x-y\right)\left(x^2+xy+y^2\right)+xy\left(x-y\right)=0\)

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

\(\Leftrightarrow\left(x-y\right)\left[2\left(x+\dfrac{9}{16}y\right)^2+\dfrac{7}{8}y^2\right]=0\left(2\right)\)

Do \(2\left(x+\dfrac{9}{16}y\right)^2+\dfrac{7}{8}y^2\ge0\), đẳng thức xảy ra khi \(x=y=0\)

Thay vào phương trình ta thấy \(x=y=0\) không phải là nghiệm 

\(\Rightarrow2\left(x+\dfrac{9}{16}y\right)^2+\dfrac{7}{8}y^2>0\)

Khi đó \(\left(2\right)\Leftrightarrow x=y\)

\(\left(1\right)\Leftrightarrow2x^3+x^3=3\Leftrightarrow x=y=1\)

\(\Rightarrow x_0^3+y_0^3=2\)

tập làm quen gõ công thức toán học đi bạn? :D 

Ta có: \(x-1=0\Rightarrow x=1\),\(x+3=0 \Rightarrow x = - 3\)

BXD:

Vậy \(T=(-\infty;-3]\cup[1;+\infty)\)

- Đặt \(f\left(x\right)=\left(x-1\right)\left(x+3\right)\)

- Cho \(f\left(x\right)=0\Rightarrow\left\{{}\begin{matrix}x=1\\x=-3\end{matrix}\right.\)

- Lập bảng xét dấu : 

x___________-3_________________1______________

x-1____-_____|________-_________0______+___________

x+3___-______0_______+_________|_____+____________

f(x)___+______0_______-__________0_____+____________

- Từ bảng xét dấu :- Để f(x) \(\ge0\)

Vậy phương trình có tập nghiệm \((-\infty;-3]\cup[1;+\infty)\)