\(\sqrt{36x-72}-15\sqrt{\dfrac{x-2}{25}}=20+4\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}-3\sqrt{x-2}-4\sqrt{x-2}=20\)

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

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

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

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

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

$\Leftrightarrow 3x+2=1$

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

Đặt \(\sqrt{x-2}=\:a\)(a >= 0)

Ta có 6a - 3a = 4(5 + a)

<=> a = - 20 (không thỏa điều kiện)

Vậy phương trình vô nghiệm

bạn giải rõ hơn chút nữa được không? Mình cám ơn nhiều

\(\sqrt{36-72x}\)- \(15\sqrt{\frac{x-2}{25}}\)= \(4\left(5+\sqrt{X+2}\right)\)

<=> \(\sqrt{36\left(x-2\right)}\)- \(15×\frac{\sqrt{x-2}}{5}\)= \(4\left(5+\sqrt{X+2}\right)\)

<=> \(6\sqrt{x-2}\)- \(3\sqrt{x-2}\) = \(4\left(5+\sqrt{X+2}\right)\)

<=> 6a - 3a = 20 + 4a 

<=> a = -20

a: Ta có: \(\sqrt{\left(x-3\right)^2}=3-x\)

\(\Leftrightarrow\left|x-3\right|=3-x\)

\(\Leftrightarrow x-3\le0\)

hay \(x\le3\)

b: Ta có: \(\sqrt{4x^2-20x+25}+2x=5\)

\(\Leftrightarrow\left|2x-5\right|=5-2x\)

\(\Leftrightarrow2x-5\le0\)

hay \(x\le\dfrac{5}{2}\)

Lời giải:

a) ĐK: $x\geq 2$

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

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

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

Vậy pt vô nghiệm.

b) ĐK: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{2x-2\sqrt{2x-1}}=2$

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

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

$\Leftrightarrow |\sqrt{2x-1}-1|=2$

$\Leftrightarrow \sqrt{2x-1}-1=\pm 2$

$\Leftrightarrow \sqrt{2x-1}=3$ (chọn) hoặc $\sqrt{2x-1}=-1$

$\Rightarrow x=5$ (thỏa mãn)

3.

PT \(\left\{\begin{matrix} x+2\geq 0\\ 3x^2=(x+2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ 2x^2-4x-4=0\end{matrix}\right.\Rightarrow x=1\pm \sqrt{3}\)

Lời giải:

a) ĐK: $x\geq 2$

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

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

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

Vậy pt vô nghiệm.

b) ĐK: $x\geq \frac{1}{2}$

PT $\Leftrightarrow \sqrt{2x-2\sqrt{2x-1}}=2$

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

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

$\Leftrightarrow |\sqrt{2x-1}-1|=2$

$\Leftrightarrow \sqrt{2x-1}-1=\pm 2$

$\Leftrightarrow \sqrt{2x-1}=3$ (chọn) hoặc $\sqrt{2x-1}=-1$

$\Rightarrow x=5$ (thỏa mãn)

3.

PT \(\left\{\begin{matrix} x+2\geq 0\\ 3x^2=(x+2)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq -2\\ 2x^2-4x-4=0\end{matrix}\right.\Rightarrow x=1\pm \sqrt{3}\)

a,bạn viết thiếu đầu bài

b,<=>3x-2=4

<=>3x=6

<=>x=2

vậy...........................

c,=>\(5\left(2\sqrt{x}-19\right)=4-\sqrt{x}\)ĐKXĐ x>=0 x khác 16

<=>\(10\sqrt{x}-95-4+\sqrt{x}=0\)

<=>\(11\sqrt{x}-99=0\)

<=>\(11\sqrt{x}=99\)

<=>\(\sqrt{x}=9< =>x=81\)

vậy.............

k mk nha

#quynh tong ngoc ơi, câu a đề bài là vậy rồi nhé >< Mình viết đúng đấy bạn ạ

nếu             câu a                       bạn                      viết               đúng 

thì                             phương                     trình                            ấy vô nghiệm 

vì 4-8=-2 thì căn ko có nghĩa => phương trình vô nghiệm

mk làm đúng đó ko sai đâu      ^ ^

\(1,ĐKx\ge5\)

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

\(\Rightarrow\sqrt{x-5}\left(\sqrt{x+5}+2\right)-3\left(\sqrt{x+5}+2\right)=0\)

\(\Rightarrow\left(\sqrt{x+5}+2\right)\left(\sqrt{x-5}-3\right)=0\)

\(\left[{}\begin{matrix}\sqrt{x+5}=-2loại\\\sqrt{x-5}=3\end{matrix}\right.\)\(\Rightarrow x-5=9\Rightarrow x=14\)(TMĐK)

2a,ĐK \(x\ge0;x\ne9\)

,\(B=\dfrac{7\left(3-\sqrt{x}\right)-12}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}\)

\(M=\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(3-\sqrt{x}\right)}=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{9-7\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{x-6\sqrt{x}+9}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)

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