ĐK : x ≥ 0 ; x khác 1

\(\frac{\sqrt{x}-4}{\sqrt{x}-1}\ge-2\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-1}+2\ge0\)

\(\Leftrightarrow\frac{\sqrt{x}-4}{\sqrt{x}-1}+\frac{2\sqrt{x}-2}{\sqrt{x}-1}\ge0\Leftrightarrow\frac{3\sqrt{x}-6}{\sqrt{x}-1}\ge0\)

Xét hai trường hợp :

1.\(\hept{\begin{cases}3\sqrt{x}-6\ge0\\\sqrt{x}-1>0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}\ge2\\\sqrt{x}>1\end{cases}}\Leftrightarrow x\ge4\)

2.\(\hept{\begin{cases}3\sqrt{x}-6\le0\\\sqrt{x}-1< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}\sqrt{x}\le2\\\sqrt{x}< 1\end{cases}}\Leftrightarrow x< 1\)

Kết hợp với ĐK =>Với \(\orbr{\begin{cases}x\ge4\\0\le x< 1\end{cases}}\)thì tmđb

34,64101615

\(\frac{2}{3}\sqrt{9x}-3\sqrt{4x}=\sqrt{25x}-16\)

\(\Leftrightarrow2\sqrt{x}-6\sqrt{x}=5\sqrt{x}-16\)

\(\Leftrightarrow9\sqrt{x}=16\Leftrightarrow\sqrt{x}=\frac{16}{9}\Leftrightarrow x=\frac{256}{81}\)

\(a,\sqrt{\frac{x-2}{25}}+2\sqrt{4x-8}=2\sqrt{x-2}+11\)

\(ĐKXĐ:x\ge2\)

\(\frac{1}{5}\sqrt{x-2}+4\sqrt{x-2}-2\sqrt{x-2}=11\)

\(\frac{11}{5}\sqrt{x-2}=11\)

\(\sqrt{x-2}=5\)

\(x-2=25\)

\(x=27\left(TM\right)\)

\(b,\sqrt{x^2-2x+1}=3x-2\)

\(ĐKXĐ:x\ge\frac{3}{2}\)

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

\(\left|x-1\right|=3x-2\)

\(x-1=3x-2\)

\(x=\frac{1}{2}\left(KTM\right)\)vậy pt vô nghiệm

b, đk  : x >= 2/3

|x - 1| = 3x - 2

=> x - 1 = 3x - 2 hoặc x - 1 = 2 - 3x

=> 2x = 1 hoặc 4x = 3

=> x = 1/2 (ktm) hoặc x = 3/4 (tm)

(1)=x^3-y^3=7
<=>(x-y)(x^2+y^2+xy)=7
<=>(X-y)^3+3xy(x-y)=7
thay(2)vào
=>(x-y)^3+3.2=7
=>x-y=1
thay vào (2)=>=xy=2
=>y^2+y-2=0
___y=1 &-2
=>x=2&-1

(1)=x^3-y^3=7

<=>(x-y)(x^2+y^2+xy)=7

<=>(X-y)^3+3xy(x-y)=7

thay(2)vào

=>(x-y)^3+3.2=7

=>x-y=1

thay vào (2)=>=xy=2

=>y^2+y-2=0

y=1 &-2

=>x=2&-1