a: \(\Leftrightarrow\dfrac{2x-3}{x-1}=4\)

=>4x-4=2x-3

=>2x=1

hay x=1/2

b: \(\Leftrightarrow\sqrt{\dfrac{2x-3}{x-1}}=2\)

=>(2x-3)=4x-4

=>4x-4=2x-3

=>2x=1

hay x=1/2(nhận)

c: \(\Leftrightarrow\sqrt{2x+3}\left(\sqrt{2x-3}-2\right)=0\)

=>2x+3=0 hoặc 2x-3=4

=>x=-3/2 hoặc x=7/2

e: \(\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

=>căn (x-5)=2

=>x-5=4

hay x=9

a: Ta có: \(\sqrt{4x+20}-3\sqrt{x+5}+\dfrac{4}{3}\sqrt{9x+45}=6\)

\(\Leftrightarrow2\sqrt{x+5}-3\sqrt{x+5}+4\sqrt{x+5}=6\)

\(\Leftrightarrow3\sqrt{x+5}=6\)

\(\Leftrightarrow x+5=4\)

hay x=-1

b: Ta có: \(\dfrac{1}{2}\sqrt{x-1}-\dfrac{3}{2}\sqrt{9x-9}+24\sqrt{\dfrac{x-1}{64}}=-17\)

\(\Leftrightarrow\dfrac{1}{2}\sqrt{x-1}-\dfrac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow\sqrt{x-1}=17\)

\(\Leftrightarrow x-1=289\)

hay x=290

\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)

\(ĐK:x\ge\dfrac{3}{2}\)

\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)

\(\Leftrightarrow4x^2-9=4x+12\)

\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)

\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)

\(ĐK:x\ge5\)

\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)

\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)

\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)

ĐK:x>=1

\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)

\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)

\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)

\(ĐK:x\ge3\)

\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)

\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)

\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)

\(\Leftrightarrow\sqrt{x-3}=0\)    (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))

\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)

Câu 1:

ĐK: \(x\geq \frac{-3}{2}\)

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

\(\Rightarrow 2x+3=(3-\sqrt{5})^2=14-6\sqrt{5}\)

\(\Rightarrow x=\frac{11-6\sqrt{5}}{2}\)

Câu 2: ĐK: \(x\geq 0\)

\(\sqrt{5+\sqrt{7x}}=2+\sqrt{7}\)

\(\Rightarrow 5+\sqrt{7x}=(2+\sqrt{7})^2=11+4\sqrt{7}\)

\(\Rightarrow \sqrt{7x}=6+4\sqrt{7}\)

\(\Rightarrow 7x=(6+4\sqrt{7})^2\Rightarrow x=\frac{(6+4\sqrt{7})^2}{7}\)

Câu 3: ĐK: \(x\geq 0\)

\((\sqrt{x}-2)(5-\sqrt{x})=4-x\)

\(\Leftrightarrow 5\sqrt{x}-x-10+2\sqrt{x}=4-x\)

\(\Leftrightarrow 7\sqrt{x}=14\Rightarrow \sqrt{x}=2\Rightarrow x=4\)

Câu 4: ĐK: \(x\ge 1\)

Sửa đề \(\frac{1}{2}\sqrt{x-1}-\frac{3}{2}\sqrt{9x-9}+24\sqrt{\frac{x-1}{64}}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{3}{2}\sqrt{9}.\sqrt{x-1}+24\sqrt{\frac{1}{64}}\sqrt{x-1}=-17\)

\(\Leftrightarrow \frac{\sqrt{x-1}}{2}-\frac{9\sqrt{x-1}}{2}+3\sqrt{x-1}=-17\)

\(\Leftrightarrow \sqrt{x-1}(\frac{1}{2}-\frac{9}{2}+3)=-17\)

\(\Leftrightarrow -\sqrt{x-1}=-17\Rightarrow \sqrt{x-1}=17\Rightarrow x=17^2+1=290\)

a. ĐKXĐ: $x\geq 1$

PT $\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{3}{2}.\sqrt{9}.\sqrt{x-1}+24.\sqrt{\frac{1}{64}}.\sqrt{x-1}=-17$

$\Leftrightarrow \frac{1}{2}\sqrt{x-1}-\frac{9}{2}\sqrt{x-1}+3\sqrt{x-1}=-17$

$\Leftrightarrow -\sqrt{x-1}=-17$

$\Leftrightarrow \sqrt{x-1}=17$

$\Leftrightarrow x-1=289$

$\Leftrightarrow x=290$

b. ĐKXĐ: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{9}.\sqrt{2x-1}-0,5\sqrt{2x-1}+\frac{1}{2}.\sqrt{25}.\sqrt{2x-1}+\sqrt{49}.\sqrt{2x-1}=24$

$\Leftrightarrow 3\sqrt{2x-1}-0,5\sqrt{2x-1}+2,5\sqrt{2x-1}+7\sqrt{2x-1}=24$
$\Leftrightarrow 12\sqrt{2x-1}=24$

$\Leftrihgtarrow \sqrt{2x-1}=2$

$\Leftrightarrow x=2,5$ (tm)

c. ĐKXĐ: $x\geq 2$

PT $\Leftrightarrow \sqrt{36}.\sqrt{x-2}-15\sqrt{\frac{1}{25}}\sqrt{x-2}=4(5+\sqrt{x-2})$

$\Leftrightarrow 6\sqrt{x-2}-3\sqrt{x-2}=20+4\sqrt{x-2}$

$\Leftrightarrow \sqrt{x-2}=-20< 0$ (vô lý)

Vậy pt vô nghiệm

d. ĐKXĐ: $x>\frac{-2}{3}$

PT $\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{1}{2}\sqrt{9}.\sqrt{\frac{1}{3x+2}}+\sqrt{16}.\sqrt{\frac{1}{3x+2}}-5\sqrt{\frac{1}{4}}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}-\frac{3}{2}\sqrt{\frac{1}{3x+2}}+4\sqrt{\frac{1}{3x+2}}-\frac{5}{2}\sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \sqrt{\frac{1}{3x+2}}=1$

$\Leftrightarrow \frac{1}{3x+2}=1$

$\Leftrightarrow 3x+2=1$

$\Leftrightarrow x=-\frac{1}{3}$

a) điều kiện xác định \(x-2\ge0vàx^2-4x+3\ge0\)

\(pt\Leftrightarrow x^2-4x+3=x-2\Leftrightarrow x^2-5x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5+\sqrt{5}}{2}\\x=\dfrac{5-\sqrt{5}}{2}\left(L\right)\end{matrix}\right.\) bạn giải nó bằng cách giải den ta nha .

vậy \(x=\dfrac{5+\sqrt{5}}{2}\)

b) điều kiện xác định : \(x\ge1\)

đặc \(\sqrt{x-1}=t\left(t\ge0\right)\)

\(pt\Leftrightarrow2\left(\dfrac{t}{2}-3\right)=\dfrac{2.2t}{3}-\dfrac{1}{3}\) giải phương trình này rồi thế ngược lại là xong

c) điều kiện xác định : \(x\ge\dfrac{7}{9}\)

\(pt\Leftrightarrow9x-7=7x+5\Leftrightarrow x=6\) vậy \(x=6\)

d) câu cuối chờ nhát h mk chưa nghỉ ra

d) Ta có pt \(4+\sqrt{2x+6-6\sqrt{2x-3}}=\sqrt{2x-2+2\sqrt{2x-3}}=0\)

\(\Leftrightarrow4+\sqrt{2x-3-6\sqrt{2x-3}+9}=\sqrt{2x-3-2\sqrt{2x-3}+1}\Leftrightarrow4+\left|\sqrt{2x-3}-3\right|=\left|\sqrt{2x-3}-1\right|\)

Đặt \(\sqrt{2x-3}=a\left(a\ge0\right),pt\Leftrightarrow4+\left|a-3\right|=\left|a-1\right|\)

xét \(a\ge3,pt\Leftrightarrow4+a-3=a-1\Leftrightarrow0a=1\left(VN\right)\)

xét \(a\le1.pt\Leftrightarrow4+3-a=1-a\Leftrightarrow0a=6\left(VN\right)\)

xét \(3>x>1,pt\Leftrightarrow4+3-a=a-1\Leftrightarrow a=1\)(k thỏa mãn )

=> pt vô nghiệm !

1,\(K=\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{x}}\)

\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}\right)\)\(=\dfrac{1}{\sqrt{2}}\left(\sqrt{\left(\sqrt{5}-1\right)^2}+\sqrt{\left(\sqrt{5}+1\right)^2}\right)\)

\(=\dfrac{1}{\sqrt{2}}\left(\left|\sqrt{5}-1\right|+\sqrt{5}+1\right)\)\(=\dfrac{1}{\sqrt{2}}\left|\sqrt{5}-1+\sqrt{5}+1\right|=\dfrac{1}{\sqrt{2}}.2\sqrt{5}\)\(=\sqrt{10}\)

2, \(\sqrt{x-3}-2\sqrt{x^2-3x}=0\left(đk:x\ge3\right)\)

\(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\1-2\sqrt{x}=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=\left(\dfrac{1}{2}\right)^2=\dfrac{1}{4}\left(ktm\right)\end{matrix}\right.\)

Vậy pt có nghiệm x=3

3, \(\dfrac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\left(đk:x>-\dfrac{5}{7}\right)\)

\(\Leftrightarrow9x-7=7x+5\)

\(\Leftrightarrow x=6\left(tm\right)\)

4, \(x-5\sqrt{x}+4=0\)(đk: \(x\ge0\))

\(\Leftrightarrow\left(\sqrt{x}-1\right)\left(\sqrt{x}-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=1\\\sqrt{x}=4\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=16\end{matrix}\right.\) (tm)

Vậy...

1) Bạn tự làm

2) ĐK: \(x\ge3\)

PT \(\Leftrightarrow\sqrt{x-3}\left(1-2\sqrt{x}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\2\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy ...

3) ĐK: \(x>-\dfrac{5}{7}\)

PT \(\Rightarrow9x-7=7x+5\) \(\Leftrightarrow x=6\)

  Vậy ...

4) ĐK: \(x\ge0\)

PT \(\Leftrightarrow x-4\sqrt{x}-\sqrt{x}+4=0\)

      \(\Leftrightarrow\left(\sqrt{x}-4\right)\left(\sqrt{x}-1\right)=0\)

      \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=4\\\sqrt{x}=1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=16\\x=1\end{matrix}\right.\)

  Vậy ...

a)

ĐKXĐ: \(x> \frac{-5}{7}\)

Ta có: \(\frac{9x-7}{\sqrt{7x+5}}=\sqrt{7x+5}\)

\(\Rightarrow 9x-7=\sqrt{7x+5}.\sqrt{7x+5}=7x+5\)

\(\Rightarrow 2x=12\Rightarrow x=6\) (hoàn toàn thỏa mãn)

Vậy......

b) ĐKXĐ: \(x\geq 5\)

\(\sqrt{4x-20}+3\sqrt{\frac{x-5}{9}}-\frac{1}{3}\sqrt{9x-45}=4\)

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

\(\Leftrightarrow 2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)

\(\Leftrightarrow 2\sqrt{x-5}=4\Rightarrow \sqrt{x-5}=2\Rightarrow x-5=2^2=4\Rightarrow x=9\)

(hoàn toàn thỏa mãn)

Vậy..........

c) ĐK: \(x\in \mathbb{R}\)

Đặt \(\sqrt{6x^2-12x+7}=a(a\geq 0)\Rightarrow 6x^2-12x+7=a^2\)

\(\Rightarrow 6(x^2-2x)=a^2-7\Rightarrow x^2-2x=\frac{a^2-7}{6}\)

Khi đó:

\(2x-x^2+\sqrt{6x^2-12x+7}=0\)

\(\Leftrightarrow \frac{7-a^2}{6}+a=0\)

\(\Leftrightarrow 7-a^2+6a=0\)

\(\Leftrightarrow -a(a+1)+7(a+1)=0\Leftrightarrow (a+1)(7-a)=0\)

\(\Rightarrow \left[\begin{matrix} a=-1\\ a=7\end{matrix}\right.\) \(\Rightarrow a=7\) vì \(a\geq 0\)

\(\Rightarrow 6x^2-12x+7=a^2=49\)

\(\Rightarrow 6x^2-12x-42=0\Leftrightarrow x^2-2x-7=0\)

\(\Leftrightarrow (x-1)^2=8\Rightarrow x=1\pm 2\sqrt{2}\)

(đều thỏa mãn)

Vậy..........

d)

ĐKXĐ: \(x^2+5x+2\ge 0\)

\((x+1)(x+4)-3\sqrt{x^2+5x+2}=6\)

\(\Leftrightarrow (x^2+5x+4)-3\sqrt{x^2+5x+2}=6\)

Đặt \(\sqrt{x^2+5x+2}=a(a\geq 0)\Rightarrow x^2+5x+2=a^2\)

PT trở thành:

\(a^2+2-3a=6\)

\(\Leftrightarrow a^2-3a-4=0\Leftrightarrow (a-4)(a+1)=0\)

\(\Rightarrow a=4\) vì \(a\geq 0\)

\(\Rightarrow x^2+5x+2=a^2=16\)

\(\Rightarrow x^2+5x-14=0\Leftrightarrow (x-2)(x+7)=0\)

\(\Rightarrow \left[\begin{matrix} x=2\\ x=-7\end{matrix}\right.\) (đều thỏa mãn)

Vậy................