Bài 1: 

b: \(=\dfrac{x+3-4-x}{x-2}=\dfrac{-1}{x-2}\)

Bài 2: 

a: \(=\dfrac{x+1}{2\left(x+3\right)}+\dfrac{2x+3}{x\left(x+3\right)}\)

\(=\dfrac{x^2+x+4x+6}{2x\left(x+3\right)}=\dfrac{x^2+5x+6}{2x\left(x+3\right)}=\dfrac{x+2}{2x}\)

d: \(=\dfrac{3}{2x^2y}+\dfrac{5}{xy^2}+\dfrac{x}{y^3}\)

\(=\dfrac{3y^2+10xy+2x^3}{2x^2y^3}\)

e: \(=\dfrac{x^2+2xy+x^2-2xy-4xy}{\left(x+2y\right)\left(x-2y\right)}=\dfrac{2x^2-4xy}{\left(x+2y\right)\cdot\left(x-2y\right)}=\dfrac{2x}{x+2y}\)

a) 1,2.2,5 = 3;

125 : 0,25 = 500

b)

\(1,2.2,5 = \dfrac{6}{5}.\dfrac{5}{2} = \dfrac{{30}}{{10}} = 3\)

\(125:0,25 = 125:\dfrac{1}{4} = 125.4 = 500\)

a)  3 6 5 .2 = 3 + 6 5 .2 = 3.2 + 6 5 .2 = 6 + 12 5 = 42 5

b)  5 4 7 : 2 = 5 + 4 7 : 2 = 5 2 + 2 7 = 39 14

\(a,\left(x-1\right)\left(2x-1\right)\)

\(=2x^2-x-2x+1\)

\(=2x^2-3x+1\)

\(b,\left(9x^4+12x^3-15x^2-3x\right):3x\)

\(=3x^3+4x^2-5x-1\)

a) \(\left(2\sqrt{2}-3\right)^2\)

\(=\left(2\sqrt{2}\right)^2-2\cdot2\sqrt{2}\cdot3+3^2\)

\(=4\cdot2-12\sqrt{2}+9\)

\(=17-12\sqrt{2}\)

b) \(\sqrt{\left(\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right)^2}\)

\(=\left|\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\right|\)

\(=\dfrac{1}{\sqrt{2}}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{2}}{2}-\dfrac{1}{2}\)

\(=\dfrac{\sqrt{2}-1}{2}\)

c) \(\sqrt{\left(0,1-\sqrt{0,1}\right)^2}\)

\(=\left|0,1-\sqrt{0,1}\right|\)

\(=0,1-\sqrt{0,1}\)

Lời giải:

$10+2.4^2=10+2.16=10+32=42$

Đáp án A

$50-4.3^2=50-4.9=50-36=14$

Đáp án A.

a) Ta có 8 x 15 y 3 . 4 y 2 x 2 = 8 x . 4 y 2 15 y 3 . x 2 = 32 15 xy  

b) Ta có  9 a 2 a + 3 . a 2 − 9 6 a 3 = 9 a 2 . ( a − 3 ) ( a + 3 ) ( a + 3 ) 6 a 3 = 3 ( a − 3 ) 2 a

a) 12,3 + 5,67 = 17,97                           

12,3 - 5,67 = 6,63

b) ( -12,3) + (-5,67) = -(12,3 + 5,67) = -17,97                  

5,67 - 12,3 = -(12,3 - 5,67)= - 6,63

Bài 3: 

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\dfrac{a}{6}=\dfrac{b}{3}=\dfrac{c}{4}=\dfrac{c-b}{4-3}=10\)

Do đó: a=60; b=30; c=40