a,ĐKXĐ:\(x\ge2\)

\(4\sqrt{x-2}+\sqrt{9x-18}-\sqrt{\dfrac{x-2}{4}}=26\\ \Leftrightarrow4\sqrt{x-2}+3\sqrt{x-2}-\dfrac{\sqrt{x-2}}{2}=26\\ \Leftrightarrow8\sqrt{x-2}+6\sqrt{x-2}-\sqrt{x-2}=52\\ \Leftrightarrow13\sqrt{x-2}=52\\ \Leftrightarrow\sqrt{x-2}=4\\ \Leftrightarrow x-2=16\\ \Leftrightarrow x=18\left(tm\right)\)

b,ĐKXĐ:\(x\in R\)

\(3x+\sqrt{4x^2-8x+4}=1\\ \Leftrightarrow2\sqrt{x^2-2x+1}=1-3x\\ \Leftrightarrow\left|x-1\right|=\dfrac{1-3x}{2}\\ \Leftrightarrow\left[{}\begin{matrix}x-1=\dfrac{1-3x}{2}\\x-1=\dfrac{3x-1}{2}\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x-2=1-3x\\2x-2=3x-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\left(tm\right)\\x=-1\left(tm\right)\end{matrix}\right.\)

c, ĐKXĐ:\(x\ge0\)

\(\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)=7\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+1\right)-2\left(2\sqrt{x}+1\right)=7\\ \Leftrightarrow2x+\sqrt{x}-4\sqrt{x}-2=7\\ \Leftrightarrow2x-3\sqrt{x}-9=0\\ \Leftrightarrow\left(2x+3\sqrt{x}\right)-\left(6\sqrt{x}+9\right)=0\\ \Leftrightarrow\sqrt{x}\left(2\sqrt{x}+3\right)-3\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left(\sqrt{x}-3\right)\left(2\sqrt{x}+3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=3\\2\sqrt{x}=-3\left(vô.lí\right)\end{matrix}\right.\\ \Leftrightarrow x=9\left(tm\right)\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left(2x+1\right)^2=6^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(\sqrt{4x^2-4\sqrt{7}x+7}=\sqrt{7}\)

\(\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left(2x-\sqrt{7}\right)^2=\left(\sqrt{7}\right)^2\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt[]{7}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

a) \(\sqrt{4x^2+4x+1}=6\)

\(\Leftrightarrow\sqrt{\left(2x+1\right)^2}=6\)

\(\Leftrightarrow\left|2x+1\right|=6\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+1=6\\2x+1=-6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-\dfrac{7}{2}\end{matrix}\right.\)

b) \(pt\Leftrightarrow\sqrt{\left(2x-\sqrt{7}\right)^2}=\sqrt{7}\)

\(\Leftrightarrow\left|2x-\sqrt{7}\right|=\sqrt{7}\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-\sqrt{7}=\sqrt{7}\\2x-\sqrt{7}=-\sqrt{7}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{7}\\x=0\end{matrix}\right.\)

c) \(PT\Leftrightarrow\sqrt{\left(x+\sqrt{3}\right)^2}=2\sqrt{3}\)

\(\Leftrightarrow\left|x+\sqrt{3}\right|=2\sqrt{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\sqrt{3}=2\sqrt{3}\\x+\sqrt{3}=-2\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{3}\\x=-3\sqrt{3}\end{matrix}\right.\)

d) \(pt\Leftrightarrow\left|x-3\right|=9\Leftrightarrow\left[{}\begin{matrix}x-3=-9\\x-3=9\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-6\\x=12\end{matrix}\right.\)

\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)

\(\Leftrightarrow x^3+8-x^3-2x=15\)

\(\Leftrightarrow-2x=15-8=7\)

\(\Leftrightarrow x=\frac{-7}{2}\)

Vậy \(x=\frac{-7}{2}\)

a) \(\sqrt{2x-1}=\sqrt{5}\) (ĐK: \(x\ge\dfrac{1}{2}\))

\(\Leftrightarrow2x-1=5\)

\(\Leftrightarrow2x=6\)

\(\Leftrightarrow x=3\left(tm\right)\)

b) \(\sqrt{x-10}=-2\) 

⇒ Giá trị của biểu thức trong căn luôn dương nên phương trình vô nghiệm

c) \(\sqrt{\left(x-5\right)^2}=3\) 

\(\Leftrightarrow\left|x-5\right|=3\)

TH1: \(\left|x-5\right|=x-5\) với \(x-5\ge0\Leftrightarrow x\ge5\)

Pt trở thành:

\(x-5=3\) (ĐK: \(x\ge5\))

\(\Leftrightarrow x=3+5\)

\(\Leftrightarrow x=8\left(tm\right)\)

TH2: \(\left|x-5\right|=-\left(x-5\right)\) với \(x-5< 0\Leftrightarrow x< 0\)

Pt trở thành:

\(-\left(x-5\right)=3\) (ĐK: \(x< 5\))

\(\Leftrightarrow-x+5=3\)

\(\Leftrightarrow-x=-2\)

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

Vậy: \(S=\left\{2;8\right\}\)

a/ ĐKXĐ: 2x - 1 >= 0 <=> 2x > 1 <=> x>= 1/2

\(\sqrt{2x-1}=\sqrt{5}\Leftrightarrow2x-1=5\Leftrightarrow2x=6\Leftrightarrow x=3\left(tm\right)\)

b/ ĐKXĐ: x - 10 >= 0 <=> x >= 10

Biểu thức trong căn luôn nhận giá trị dương => vô nghiệm

c/ ĐKXĐ: x - 5 >=0 <=> x >= 5

\(\sqrt{x-5}=3\Leftrightarrow x-5=9\Leftrightarrow x=14\left(tm\right)\)

a) ĐKXĐ: x <= 2

pt --> 4 - 2x = 25 <=> x = -21/2 (thỏa)

??

Đề kiểu gì vậy ?

b) ĐKXĐ: x >= -1

pt <=> 8sqrt(x + 1)=16 <=> sqrt(x+1)=2 --> x + 1 = 4 <=> x = 3

10 - { [ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 } = 5

[ ( x : 3 + 17 ) : 10 + 3 : 2⁴ ] : 10 = 10 - 5 = 5

( x : 3 + 17 ) : 10 + 3 : 2⁴ = 5 x 10

( x : 3 + 17 ) : 10 + 48 = 50

( x : 3 + 17 ) : 10 = 50 - 48

( x : 3 + 17 ) : 10 = 2

x : 3 + 17 = 2 x 10

x : 3 + 17 = 20

x : 3 = 20 - 17 = 3

x = 3 x 3 = 9

a) [(2x+14) : 4 - 3] : 2 = 1

(2x+14) : 4 - 3 = 1/2

(2x+14) : 4  = 1/2 + 3

(2x+14) : 4  = 7/2

2x+14 = 7/2 . 1/4

2x = 7/8 - 1/4

2x = 5/8

x= 5/8.1/2

x= 5/16

Còn phần b nữa ak

a: ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)

b: Ta có: \(P=\left(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)

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

\(=\dfrac{-3\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\)

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

c: Thay \(x=4-2\sqrt{3}\) vào P, ta được:

\(P=\dfrac{-3}{\sqrt{3}-1+3}=\dfrac{-3}{2+\sqrt{3}}=-6+3\sqrt{3}\)

a: Để P nguyên thì \(-3⋮\sqrt{x}+3\)

\(\Leftrightarrow\sqrt{x}+3=3\)

hay x=0

\(\sqrt{-3x^3+5x+14}+\sqrt{-5x^3+6x+28}=\left(4-2x-x^2\right)\sqrt{2-x}\) (ĐKXĐ: \(x\in R,x\le2\))

\(\Leftrightarrow\sqrt{\left(2-x\right)\left(3x^2+6x+7\right)}+\sqrt{\left(2-x\right)\left(5x^2+10x+14\right)}-\left(4-2x-x^2\right)\sqrt{2-x}=0\)

\(\Leftrightarrow\sqrt{2-x}\left(\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}-4+2x+x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=2\left(tm\right)\\\sqrt{3x^2+6x+7}+\sqrt{5x^2+10x+14}=4-2x-x^2\left(1\right)\end{cases}}\)

Pt \(\left(1\right)\Leftrightarrow\sqrt{3\left(x+1\right)^2+4}+\sqrt{5\left(x+1\right)^2+9}=-\left(x+1\right)^2+5\left(2\right)\)

Ta có: \(\left(x+1\right)^2\ge0\Rightarrow\sqrt{2\left(x+1\right)^2+4}\ge\sqrt{4}=2\)

Tương tự: \(\sqrt{5\left(x+1\right)^2+9}\ge3\). Từ đó: \(VT_{\left(2\right)}\)\(\ge2+3=5\)

Mà \(VP_{\left(2\right)}=-\left(x+1\right)^2+5\le5\) nên dấu "=" xảy ra \(\Leftrightarrow\left(x+1\right)^2=0\Leftrightarrow x=-1\)(tm)

Vậy tập nghiệm của pt cho là \(S=\left\{2;-1\right\}.\)