Câu 1: 

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

Câu 2:

A = (1 + 2 + 3 + 4)⁶² 

B = 1³ + 2³ + 3³ + 4³

B = (1 + 2 + 3 + 4)³ < (1 + 2 + 3 + 4)⁶²

Vậy A < B

Phương trình có 2 nghiệm dương pb khi:

\(\left\{{}\begin{matrix}\Delta'=\left(m+1\right)^2-\left(2m-5\right)>0\\x_1+x_2=2\left(m+1\right)>0\\x_1x_2=2m-5>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m^2+6>0\\m>-1\\m>\dfrac{5}{2}\end{matrix}\right.\)

\(\Rightarrow m>\dfrac{5}{2}\)

Khi đó:

\(\left|\dfrac{1}{\sqrt{x_1}}-\dfrac{1}{\sqrt{x_2}}\right|=\sqrt{6}\Rightarrow\left|\dfrac{\sqrt{x_1}-\sqrt{x_2}}{\sqrt{x_1x_2}}\right|=\sqrt{6}\)

\(\Rightarrow\dfrac{\left(\sqrt{x_1}-\sqrt{x_2}\right)^2}{x_1x_2}=6\Rightarrow x_1+x_2-2\sqrt{x_1x_2}=6x_1x_2\)

\(\Rightarrow2\left(m+1\right)-2\sqrt{2m-5}=6\left(2m-5\right)\)

\(\Leftrightarrow5\left(2m-5\right)+2\sqrt{2m-5}-7=0\)

Đặt \(\sqrt{2m-5}=t>0\Rightarrow5t^2+2t-7=0\Rightarrow\left[{}\begin{matrix}t=1\\t=-\dfrac{7}{5}\left(loại\right)\end{matrix}\right.\)

\(\Rightarrow\sqrt{2m-5}=1\Rightarrow2m-5=1\)

\(\Rightarrow m=3\) (thỏa mãn)

Câu 1: (2x-3)-(x-5)=(x+2)-(x-1)

2x -3 -x+5 = x+2 -x +1

2x -x -x +x = 2+1 +3 -5

x= 1

Câu 2:     2(x-1)-5(x+2)=10

2x -2 -5x -10 =10

2x -5x = 10 +2 +10

(2-5) x = 22

-3x= 22

x= 22/-3

Câu 1: ( 2x - 3 ) - ( x - 5 ) = ( x + 2 ) - ( x - 1 )
=> ( 2x - x ) - ( 3 - 5 ) = ( x - x ) + ( 2 + 1 )
=> x + 2 = 3
=> x = 1

Thử lại: ( 2 - 3 ) - ( 1 - 5 ) = ( 1 + 2 ) - ( 1 - 1 )
=> -1 + 4 = 3 - 0
=> 3 = 3    ( thoả mãn )

Câu 2: 2 ( x - 1 ) - 5 ( x + 2 ) = 10
=> ( 2x - 2 ) - ( 5x + 10 ) = 10
=> ( 2x - 5x ) - ( 2 + 10 ) = 10
=> -3x - 12 = 10
=> -3x = 22
=> x = ^-22/₃

Thử lại: 2 ( ^-22/₃ - 1 ) - 5 ( ^-22/₃ + 2 ) = 10
=> 2 * ^-25/₃ - 5 * ^-16/₃ = 10
=> ^-50/₃ - ^-80/₃ = 10
=> ^{(-50) - (-80)}/₃ = 10
=> 30 / 3 = 10    ( thoả mãn )