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:\(x\ge0\)

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

      \(\dfrac{3}{5}\) x 0,25\(x\) = - \(\dfrac{1}{2}\)

            0,25\(x\) = - \(\dfrac{1}{2}\) : \(\dfrac{3}{5}\)

             0,25\(x\) =  - \(\dfrac{1}{2}\) x \(\dfrac{5}{3}\)

             0,25\(x\) = - \(\dfrac{5}{6}\)

                   \(x\) = - \(\dfrac{5}{6}\) : 0,25

                   \(x\) = - \(\dfrac{5}{6}\) x 4

                   \(x\) = - \(\dfrac{10}{3}\)

Vậy \(x\) = - \(\dfrac{10}{3}\)

2626 : (\(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\)) = 26

            \(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\) = 2626 : 26

             \(\dfrac{1}{2}\)\(x\) + \(\dfrac{5}{2}\)\(x\) = 101

        \(x\) x ( \(\dfrac{1}{2}\) + \(\dfrac{5}{2}\)) = 101

          \(x\) x 3 = 101

          \(x\)       = 101 : 3

          \(x\)       = \(\dfrac{101}{3}\)

Vậy \(x\) = \(\dfrac{101}{3}\) 

\(5x^2+3x-26=0\)

\(\Leftrightarrow5x^2+13x-10x-26=0\)

\(\Leftrightarrow x\left(5x+13\right)-2\left(5x+13\right)=0\)

\(\Leftrightarrow\left(x-2\right)\left(5x+13\right)=0\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x-2=0\\5x+13=0\end{array}\right.\)

\(\Leftrightarrow\left[\begin{array}{nghiempt}x=2\\x=-\frac{13}{5}\end{array}\right.\)

thanks

\(\Leftrightarrow9x+3=-20x+16\)

=>29x=13

hay x=13/29

\(\Leftrightarrow\dfrac{3x-1}{4}-\dfrac{3x-6}{8}-\dfrac{5-3x}{2}>1\)

\(\Leftrightarrow\dfrac{\left(3x-1\right).2-\left(3x-6\right)-\left(5-3x\right).4}{8}>1\)

\(\Leftrightarrow\dfrac{6x-2-3x+6-20+12x}{8}>1\)

<=> 15x - 16 > 8

<=> 15x > 24

<=> x > 8/5

Ta có: \(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)

\(\Leftrightarrow2\left(3x-1\right)-3\left(x-2\right)-8>4\left(5-3x\right)\)

\(\Leftrightarrow6x-2-3x+6-8>20-12x\)

\(\Leftrightarrow3x-4-20+12x>0\)

\(\Leftrightarrow15x>24\)

hay \(x>\dfrac{8}{5}\)

\(\dfrac{3x-1}{4}-\dfrac{3\left(x-2\right)}{8}-1>\dfrac{5-3x}{2}\)

MTC : 8

\(\Rightarrow\dfrac{2\left(3x-1\right)}{8}-\dfrac{3\left(x-2\right)}{8}-\dfrac{8}{8}>\dfrac{4\left(5-3x\right)}{8}\)

Suy ra : 2(3x - 1) - 3(x - 2) - 8 > 4(5 - 3x)

          \(\Leftrightarrow\) 6x - 2 - 3x + 6 - 8 > 20 - 12x

          \(\Leftrightarrow\) 6x - 3x + 12x > 20 + 2 - 6 + 8

          \(\Leftrightarrow\) 15x > 24

          \(\Leftrightarrow\) x > \(\dfrac{24}{15}=\dfrac{8}{5}\)

 Vay x >\(\dfrac{8}{5}\)

Chuc ban hoc tot

Cảm ơn bạn

a) (x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210

Các phần tử của phép tính trên là: (26 - 4) : 2 + 1 = 12(phần tử) ⇒ 12x

Tổng của phép tính trên là: (26 + 4) x 12 : 2 = 180

Tính: 

(x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210

12x + (4 + 6 + 8 + ... + 26) = 210

 12x + 180 = 210

12x = 210 - 180

12x = 30

x = 30 : 12

x = 2,5

a) (x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210

Các phần tử của phép tính trên là: (26 - 4) : 2 + 1 = 12(phần tử) ⇒ 12x

Tổng của phép tính trên là: (26 + 4) x 12 : 2 = 180

Tính: 

(x + 4) + (x + 6) + (x + 8) + ... + (x + 26) = 210

12x + (4 + 6 + 8 + ... + 26) = 210

 12x + 180 = 210

12x = 210 - 180

12x = 30

x = 30 : 12

x = 2,5

b) x + 280 : 25 - 7,2 = 15

x + 11,2 - 7,2 = 15

x + 11,2 = 15 + 7,2

x + 11,2 = 22,2

x = 22,2 - 11,2

x = 11

ĐKXĐ: \(-1\le x\le\dfrac{5}{2}\)

\(\Leftrightarrow\sqrt{3x+3}-3+1-\sqrt{5-2x}=x^3-3x^2-10x+24\)

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

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

Xét (1), ta có:

\(\dfrac{3}{\sqrt{3x+3}+3}+\dfrac{2}{1+\sqrt{5-2x}}>0\)

\(-1\le x\le\dfrac{5}{2}\Rightarrow\left\{{}\begin{matrix}x+3>0\\x-4< 0\end{matrix}\right.\) \(\Rightarrow\left(x+3\right)\left(x-4\right)< 0\)

\(\Rightarrow\left(1\right)\) vô nghiệm hay pt có nghiệm duy nhất \(x=2\)

Em cảm ơn