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

b: \(=2\sqrt{2}+2\sqrt{2}+5\sqrt{2}=9\sqrt{2}\)

c: \(=4\sqrt{3}+3\sqrt{3}-3\sqrt{5}+2\sqrt{5}=7\sqrt{3}-\sqrt{5}\)

d: \(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}=-\sqrt{3}\)

e: \(=\left(\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)

=7-2*căn 21+2*căn 21

=7

f: \(=\left(2\sqrt{11}-3\sqrt{2}\right)\cdot\sqrt{11}+3\sqrt{22}\)

=22-3*căn 22+3*căn 22

=22

a) \(3\sqrt{5}+\sqrt{20}-2\sqrt{5}\)

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

\(=3\sqrt{5}\)

b) \(2\sqrt{2}+\sqrt{8}+\sqrt{50}\)

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

\(=9\sqrt{5}\)

c) \(4\sqrt{3}+\sqrt{27}-\sqrt{45}+2\sqrt{5}\)

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

\(=7\sqrt{3}-\sqrt{5}\)

d) \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)

\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)

\(=-\sqrt{3}\)

e) \(\left(\sqrt{28}-\sqrt{12}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(2\sqrt{7}-2\sqrt{3}-\sqrt{7}\right)\sqrt{7}+2\sqrt{21}\)

\(=\left(\sqrt{7}-2\sqrt{3}\right)\sqrt{7}+2\sqrt{21}\)

\(=7-2\sqrt{21}+2\sqrt{21}\)

\(=7\)

f) \(\left(\sqrt{99}-\sqrt{18}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(3\sqrt{11}-3\sqrt{2}-\sqrt{11}\right)\sqrt{11}+3\sqrt{22}\)

\(=\left(2\sqrt{11}-3\sqrt{2}\right)\sqrt{11}+3\sqrt{22}\)

\(=22-3\sqrt{22}+3\sqrt{22}\)

\(=22\)

g) \(3\sqrt{45}-5\sqrt{125x}+7\sqrt{20x}+28\)

\(=9\sqrt{5}-25\sqrt{5x}+14\sqrt{5x}+28\)

\(=9\sqrt{5}-11\sqrt{5x}+28\)

1) \(\left(5x-4\right)\left(4x+6\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}5x-4=0\\4x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=4\\4x=6\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{4}{5}\\x=\dfrac{3}{2}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{4}{5};\dfrac{3}{2}\right\}\)

2) \(\left(4x-10\right)\left(24+5x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-10=0\\24+5x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}4x=10\\5x=-24\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=\dfrac{-24}{5}\end{matrix}\right.\)

Vậy phương trình có tập nghiệm S = \(\left\{\dfrac{5}{2};\dfrac{-24}{5}\right\}\)

3) \(\left(x-3\right)\left(2x+1\right)=0\)

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

Vậy phương trình có tập nghiệm S = \(\left\{3;\dfrac{-1}{2}\right\}\)

1)6x-8=3x+1

6x-3x=1+8

3x=9

x=3

Vậy x=3

2: 12-10x=25-30x

=>20x=13

=>x=13/20

3: \(3\left(2x+3\right)-2\left(4x-5\right)=10x+21\)

=>6x+9-8x+10=10x+21

=>10x+21=-2x+19

=>12x=-2

=>x=-1/6

4: \(\Leftrightarrow25x-15-6x+12=11-5x\)

=>19x-3=11-5x

=>24x=14

=>x=7/12

5: \(\Leftrightarrow8-12x-5+10x=4-6x\)

=>4-6x=-2x+3

=>-4x=-1

=>x=1/4

6: \(\Leftrightarrow32x-24-6+9x=13-40x\)

=>41x-30=13-40x

=>81x=43

=>x=43/81

7: \(\Leftrightarrow10x-5+20x=5x-11\)

=>30x-5=5x-11

=>25x=-6

=>x=-6/25

a; -2\(x\) - 3.(\(x-17\)) = 34 - 2.( - \(x\) + 25)

    - 2\(x\) - 3\(x\) + 51 = 34 + 2\(x\) - 50

       2\(x\) + 2\(x\) + 3\(x\) = - 34 + 50 + 51

            7\(x\)            = 67

               \(x\)           =  67 : 7

                \(x\)         =  \(\dfrac{67}{7}\)

Vậy \(x\) = \(\dfrac{67}{7}\) 

b; 17\(x\) + 3.(- 16\(x\) - 37) = 2\(x\) + 43 - 4\(x\)

    17\(x\) - 48\(x\)  - 111 =  2\(x\) - 4\(x\) + 43

    - 31\(x\) - 2\(x\) + 4\(x\) = 111 + 43

        - \(x\) x (31 + 2 - 4) = 154

        - \(x\) x (33 - 4) =  154

         - \(x\) x 29 = 154

         - \(x\)         = 154 : (-29)

           \(x\)        = - \(\dfrac{154}{29}\)

          Vậy \(x=-\dfrac{154}{29}\) 

c; {-3\(x\) + 2.[45 - \(x\) - 3(3\(x\) + 7) - 2\(x\)] + 4\(x\) } = 55

    {-3\(x\) + 2.[45 - \(x\) - 9\(x\) - 21 - 2\(x\)] + 4\(x\)} = 55

    {- 3\(x\) + 2.[(45 - 21) - (\(x+9x\)+2\(x\))] + 4\(x\) } = 55

     { (- 3\(x\) + 4\(x\)) + 2.[24 - 12\(x\)] } = 55

                   \(x\) + 48 - 24\(x\) = 55

                   \(x-24x\)         = 55 - 48

                     - 23\(x\)           = 7

                            \(x\)           = 7 : - 23

                             \(x=-\dfrac{7}{23}\)

Vậy \(x=-\dfrac{7}{23}\)