Chắc sai ;-;

\(x\ne\pm y\)

\(\hept{\begin{cases}\frac{80}{x+y}+\frac{48}{x-y}=7\\\frac{100}{x+y}-\frac{32}{x-y}=3\end{cases}}\)\(\Rightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{x-y}\\\frac{\frac{5}{4}.80}{x+y}-\frac{32}{x-y}=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{x-y}\\\frac{5}{4}\left(7-\frac{48}{x-y}\right)-\frac{32}{x-y}=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{x-y}\\\frac{35}{4}-\frac{60}{x-y}-\frac{32}{x-y}=3\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{x-y}\\\frac{35}{4}-\frac{92}{x-y}=3\end{cases}}\)\(\Leftrightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{x-y}\\\frac{92}{x-y}=\frac{23}{4}\end{cases}\Leftrightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{x-y}\\x-y=16\end{cases}}}\)

\(\Leftrightarrow\hept{\begin{cases}\frac{80}{x+y}=7-\frac{48}{16}\\x-y=16\end{cases}\Leftrightarrow\hept{\begin{cases}x+y=20\\x-y=16\end{cases}\Leftrightarrow}\hept{\begin{cases}x=18\\y=2\end{cases}}}\)

ĐK: \(x\ne\pm y\)

Đặt \(\frac{1}{x+y}=a;\frac{1}{x-y}=b\)

\(\Rightarrow\hept{\begin{cases}80a+48b=7\\100a-32b=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}100a+60b=\frac{35}{4}\\100a-32b=3\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}92b=\frac{23}{4}\\80a+48b=7\end{cases}}\)

\(\Leftrightarrow\hept{\begin{cases}b=\frac{1}{16}\\a=\frac{1}{20}\end{cases}}\)

Trở lại cách đặt :

\(\Rightarrow\hept{\begin{cases}\frac{1}{x+y}=\frac{1}{20}\\\frac{1}{x-y}=\frac{1}{16}\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x+y=20\\x-y=16\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}2x=36\\x+y=20\end{cases}}\)

\(\Rightarrow\hept{\begin{cases}x=18\\y=2\end{cases}}\)(TMĐK)

Vậy hệ phương trình có nghiệm duy nhất (18;2)

Đk: x \(\ge\)0; x \(\ne\)9; y \(\ne\)1/2

Đặt \(\frac{4}{\sqrt{x}-3}=a\); \(\frac{1}{\left|2y-1\right|}=b\)(b \(\ge\)0)

Khi đó, ta có: \(\hept{\begin{cases}2a+b=5\\a+b=3\end{cases}}\) <=> \(\hept{\begin{cases}a=2\\b=3-b=1\end{cases}}\)

<=> \(\hept{\begin{cases}\frac{4}{\sqrt{x}-3}=2\\\frac{1}{\left|2y-1\right|}=1\end{cases}}\) <=> \(\hept{\begin{cases}2\sqrt{x}-6=4\\\left|2y-1\right|=1\end{cases}}\)

<=>> \(\hept{\begin{cases}\sqrt{x}=5\\\left|2y-1\right|=1\end{cases}}\) <=> x = 25 (tm)) và \(\orbr{\begin{cases}2y-1=1\\2y-1=-1\end{cases}}\) <=> x = 25 và \(\orbr{\begin{cases}y=1\\y=0\end{cases}}\left(tm\right)\)

Vậy (a;b) = {(25; 0); (25; 1)}

\(\frac{3\sqrt{x}+6-6}{\sqrt{x}+2}=3-\frac{6}{\sqrt{x}+2}\)

để biểu thức MIN thì \(\frac{6}{\sqrt{x}+2}\)là MAX

mà \(\sqrt{x}+2\ge2\)

để \(\frac{6}{\sqrt{x}+2}\)MAX

\(< =>\sqrt{x}+2=2< =>x=0\)

\(MIN:\frac{3\sqrt{0}}{\sqrt{0}+2}=0\)với x=0

sai đề mọi người ơi 

\(2\sqrt{x}^2-5\sqrt{x}+2=0\)

\(2\sqrt{x}^2-4\sqrt{x}-\sqrt{x}+2=0\)

\(\sqrt{x}\left(2\sqrt{x}-1\right)-2\left(2\sqrt{x}-1\right)=0\)

\(\left(2\sqrt{x}-1\right)\left(\sqrt{x}-2\right)=0\)

\(\orbr{\begin{cases}2\sqrt{x}-1=0\\\sqrt{x}-2=0\end{cases}\orbr{\begin{cases}x=\frac{1}{4}\left(TM\right)\\x=4\left(TM\right)\end{cases}}}\)

a) Đk: x \(\le\)1/4

Ta có: \(\sqrt{1-4x}+2x=3\)

<=> \(\sqrt{1-4x}=3-2x\)(đk: x \(\le\)1/4)

<=> \(1-4x=\left(3-2x\right)^2\)

<=> \(4x^2-12x+9=1-4x\)

<=> \(4x^2-8x+8=0\)

<=> \(x^2-2x+2=0\)

<=> \(\left(x-1\right)^2+1=0\) => pt vn

b) Đk: x \(\ge\)5

Ta có: \(\sqrt{2x+1}-\sqrt{x-5}=\sqrt{3x-2}\) (đk: \(x\ge5\))

<=> \(2x+1+x-5-2\sqrt{\left(2x+1\right)\left(x-5\right)}=3x-2\)

<=> \(2\sqrt{\left(2x+1\right)\left(x-5\right)}=-2\) => pt vn

c) Đk: x \(\ge\)-6/5

Ta có: \(\sqrt{5x+6}+\sqrt{9x+5}=\sqrt{6x+7}+\sqrt{2x+8}\)

<=> \(\sqrt{5x+6}-\sqrt{6x+7}=\sqrt{2x+8}-\sqrt{9x+5}\)(đk: \(-\frac{6}{5}\le x\le-1\)

<==> \(5x+6+6x+7-2\sqrt{\left(5x+6\right)\left(6x+7\right)}=2x+8+9x+5-2\sqrt{\left(2x+8\right)\left(9x+5\right)}\)

<=> \(\sqrt{30x^2+71x+42}=\sqrt{18x^2+82x+40}\)

<=> \(30x^2+71x+42=18x^2+82x+40\)

<=> \(12x^2-11x+2=0\)

\(\Delta=\left(-11\right)^2-4.2.12=25>0\) => pt có 2 nghiệm pb

x1 = 2/3 (ktm) ; x2 = 1/4 (ktm)

=> pt vn

d) \(\sqrt[3]{x^2+1}=\sqrt[3]{3x^2-x-1}-\sqrt[3]{2x^2-x-2}\)

<=> \(x^2+1=x^2+1-3\sqrt[3]{\left(3x^2-x-1\right)\left(2x^2-x-2\right)}.\left(\sqrt[3]{3x^2-x-1}-\sqrt[3]{2x^2-x-2}\right)\)

<=> \(\sqrt[3]{\left(3x^2-x-1\right)\left(2x^2-x-2\right)}.\sqrt[3]{x^2+1}=0\)

<=> \(\left(3x^2-x-1\right)\left(2x^2-x-2\right)\left(x^2+1\right)=0\)

<==> \(\orbr{\begin{cases}3x^2-x-1=0\\2x^2-x-2=0\end{cases}}\)

(còn lại tự giải)

Câu c em thử lại thấy x1=2/3 vẫn đúng ạ ?

ĐK: \(-1\le x\le1\).

Đặt \(\sqrt{1-x}=a\ge0,\sqrt{x+1}=b\ge0\).

Phương trình đã cho tương đương với: 

\(2a-b+3ab=2a^2+b^2\)

\(\Leftrightarrow2a-b=\left(2a-b\right)\left(a-b\right)\)

\(\Leftrightarrow\left(2a-b\right)\left(a-b-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2a=b\\a=b+1\end{cases}}\)

TH1: \(2a=b\)

\(2\sqrt{1-x}=\sqrt{x+1}\)

\(\Leftrightarrow4\left(1-x\right)=x+1\)

\(\Leftrightarrow x=\frac{3}{5}\)(thỏa mãn) 

TH2: \(a=b+1\)

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

\(\Leftrightarrow1-x=x+1+1+2\sqrt{x+1}\)

\(\Leftrightarrow-2x-1=2\sqrt{x+1}\)

\(\Rightarrow4x^2+4x+1=4x+4\)

\(\Leftrightarrow x=\frac{\pm\sqrt{3}}{2}\)

Thử lại chỉ có \(x=-\frac{\sqrt{3}}{2}\)thỏa mãn.