a) \(\frac{3}{4}\sqrt{x}-\sqrt{9x}+5=\frac{1}{4}\sqrt{9x}\)

ĐK : x ≥ 0

⇔ \(\frac{3}{4}\sqrt{x}-\sqrt{3^2x}-\frac{1}{4}\sqrt{3^2x}=-5\)

⇔ \(\frac{3}{4}\sqrt{x}-3\sqrt{x}-\frac{1}{4}\cdot3\sqrt{x}=-5\)

⇔ \(-\frac{9}{4}\sqrt{x}-\frac{3}{4}\sqrt{x}=-5\)

⇔ \(-3\sqrt{x}=-5\)

⇔ \(\sqrt{x}=15\)

⇔ \(x=225\)( tm )

b) \(\sqrt{3-x}-\sqrt{27-9x}+1,25\sqrt{48-16x}=6\)

ĐK : x ≤ 3

⇔ \(\sqrt{3-x}-\sqrt{3^2\left(3-x\right)}+\frac{5}{4}\sqrt{4^2\left(3-x\right)}=6\)

⇔ \(\sqrt{3-x}-3\sqrt{3-x}+\frac{5}{4}\cdot4\sqrt{3-x}=6\)

⇔ \(-2\sqrt{3-x}+5\sqrt{3-x}=6\)

⇔ \(3\sqrt{3-x}=6\)

⇔ \(\sqrt{3-x}=2\)

⇔ \(3-x=4\)

⇔ \(x=-1\)( tm )

c) \(\sqrt{9x^2+12x+4}=4\)

⇔ \(\sqrt{\left(3x+2\right)^2}=4\)

⇔ \(\left|3x+2\right|=4\)

⇔ \(\orbr{\begin{cases}3x+2=4\\3x+2=-4\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{3}\\x=-2\end{cases}}\)

d) \(\frac{1}{3}\sqrt{x-1}+2\sqrt{4x-4}-12\sqrt{\frac{x-1}{25}}=\frac{29}{15}\)

ĐK : x ≥ 1

⇔  \(\frac{1}{3}\sqrt{x-1}+2\sqrt{2^2\left(x-1\right)}-12\sqrt{\left(\frac{1}{5}\right)^2\cdot\left(x-1\right)}=\frac{29}{15}\)

⇔  \(\frac{1}{3}\sqrt{x-1}+2\cdot2\sqrt{x-1}-12\cdot\frac{1}{5}\sqrt{x-1}=\frac{29}{15}\)

⇔  \(\frac{1}{3}\sqrt{x-1}+4\sqrt{x-1}-\frac{12}{5}\sqrt{x-1}=\frac{29}{15}\)

⇔ \(\frac{29}{15}\sqrt{x-1}=\frac{29}{15}\)

⇔ \(\sqrt{x-1}=1\)

⇔ \(x-1=1\)

⇔ \(x=2\)( tm )

a) \(\sqrt{\left(2x-1\right)^2}=3\)

⇔ \(\left|2x-1\right|=3\)

⇔ \(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

⇔ \(\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

b) \(3\sqrt{x}-2\sqrt{9x}+\sqrt{16x}=5\)

ĐKXĐ : \(x\ge0\)

⇔ \(3\sqrt{x}-2\sqrt{3^2x}+\sqrt{4^2x}=5\)

⇔ \(3\sqrt{x}-2\cdot3\sqrt{x}+4\sqrt{x}=5\)

⇔ \(7\sqrt{x}-6\sqrt{x}=5\)

⇔ \(\sqrt{x}=5\)

⇔ \(x=25\)( tm )

c) \(\sqrt{4x+20}-3\sqrt{5+x}+\frac{3}{4}\sqrt{9x+45}=6\)

ĐKXĐ : \(x\ge-5\)

⇔ \(\sqrt{2^2\left(x+5\right)}-3\sqrt{x+5}+\frac{3}{4}\sqrt{3^2\left(x+5\right)}=6\)

⇔ \(2\sqrt{x+5}-3\sqrt{x+5}+\frac{3}{4}\cdot3\sqrt{x+5}=6\)

⇔ \(-\sqrt{x+5}+\frac{9}{4}\sqrt{x+5}=6\)

⇔ \(\frac{5}{4}\sqrt{x+5}=6\)

⇔ \(\sqrt{x+5}=\frac{24}{5}\)

⇔ \(x+5=\frac{576}{25}\)

⇔ \(x=\frac{451}{25}\left(tm\right)\)

ĐK: \(x\ge0\)\(4\sqrt{x}-2\sqrt{9x}+16\sqrt{x}=5\)  5  (=) \(\sqrt{x}\left(4-2\sqrt{9}+16\right)=5\) (=) \(\sqrt{x}.14=5\)(=) x=\(\frac{25}{196}\)

ĐK: \(x\ge-5\)PT(=) \(\sqrt{5+x}\left(\sqrt{4}-3+\frac{4}{3}.3\right)=6\) (=) \(\sqrt{5+x}.3=6\) (=)\(\sqrt{5+x}=2\)(=) X = -1 (nhận)

1.

ĐKXĐ: \(5\leq x\leq 1\) (vô lý) nên PT sai ngay từ đầu.

2.

ĐKXĐ: \(x\geq -1\)

PT \(\Leftrightarrow \sqrt{9}.\sqrt{x+1}+\sqrt{4}.\sqrt{x+1}=\sqrt{x+4}\)

\(\Leftrightarrow 3\sqrt{x+1}+2\sqrt{x+1}=\sqrt{x+4}\)

\(\Leftrightarrow 5\sqrt{x+1}=\sqrt{x+4}\)

\(\Rightarrow 25(x+1)=x+4\) (bình phương 2 vế)

\(\Leftrightarrow x=\frac{-7}{8}\) (thỏa mãn)

Vậy..........

1.a) \(\sqrt{x^2-4}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\sqrt{x-2}.\sqrt{x+2}-\sqrt{x-2}=0\)

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

\(\Leftrightarrow\orbr{\begin{cases}\sqrt{x-2}=0\\\sqrt{x+2}-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\\sqrt{x+2}=1\end{cases}}\)

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

\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=-1\end{cases}}\)

Vậy x=2 hoặc x=-1

1 + 1=

Ai có nhu cầu tình dục cao thì liên hẹ vs e nha, e làm cho, 20k thôi, e cần tiền chữa bệnh cho mẹ

d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

ĐK:\(x\ge-3\)

\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)

\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)

\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)

d)\(2x^2+4x=\sqrt{\frac{x+3}{2}}\)

ĐK:\(x\ge-3\)

\(\Leftrightarrow4x^4+16x^3+16x^2=\frac{x+3}{2}\)

\(\Leftrightarrow\frac{8x^4+32x^3+32x^2-x-3}{2}=0\)

\(\Leftrightarrow8x^4+32x^3+32x^2-x-3=0\)

\(\Leftrightarrow\left(2x^2+3x-1\right)\left(4x^2+10x+3\right)=0\)