bai EZ​ qua

mà bài dài quá sao làm hết dc

1) 5 + (-4) = 1

2) (-8) + 2 = -6

3) 8 + (-2) = 6

4) 11 + (-3) = 8

5) (-11) + 2 = -9

6) (-7) + 3 = -4

7) (-5) + 5 = 0

8) 11 + (-12) = -1

9) (-18) + 20 = 2

10) (15) + (-12) = 3

11) (-17) + 17 = 0

12) 16 + (-2) = 14

13) (30) + (-14) = 16

14) (-19) + 20 = 1

15) (-18) + 15 = -3

16) (10) + (-6) = 4

17) (-28) + 14 = -14

18) 15 + (-30) = -15

19) (15) + (-4) = 11

20) (-21) + 11 = -10

21) 8 + (-22) = -14

22) (-15) + 4 = -11

23) (-3) + 2 = -1

24) 17 + (-14) = 3

25) 17 + (-14) = 3

thế này đọc đc mới đỉnh

a.=3/20=0,15

b.-1/4=-0,25

kb nha

LÀM CHI TIẾT HỘ MÌNH VỚI

a, 2 : 5 + (-1) : 6 - 3 : 4 - (-2 : 3)

= 0,4 + 0,16 + 0,75 - 0,6

= 0,71

b, 7 : 10 - ( -3 : 4 ) + (-5 : 6) - 1 : 5 + (-2) : 3

= 0,7 - (-0,75) + 0,8 - 0,2 + -0,6

= 1,45

\(\Rightarrow\)a, 2 : 5 + (-1) : 6 -  3 : 4 - (-2) : 3 \(\approx0,71\)

        b, 7 : 10 - (-3 : 4) + (-5 : 6) - 1 : 5 + (-2) : 3 \(\approx1,45\)

Bạn cần viết đề bằng công thức toán để được hỗ trợ tốt hơn (biểu tượng $\sum$ góc trái khung soạn thảo)

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

Lời giải:
a.

$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$

b.

$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$

c.

$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$

$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$

d.

$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$

$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$

e.

$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$

$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$

\(B=\left(\dfrac{4}{1-\sqrt{5}}+\dfrac{1}{2+\sqrt{5}}-\dfrac{4}{3-\sqrt{5}}\right)\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{\left(1-\sqrt{5}\right)\left(1+\sqrt{5}\right)}+\dfrac{2-\sqrt{5}}{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}-\dfrac{4\left(3+\sqrt{5}\right)}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[\dfrac{4\left(1+\sqrt{5}\right)}{1-5}+\dfrac{2-\sqrt{5}}{4-5}-\dfrac{4\left(3+\sqrt{5}\right)}{9-5}\right]\left(\sqrt{5}-6\right)\)

\(B=\left[-\dfrac{4\left(1+\sqrt{5}\right)}{4}-\dfrac{2-\sqrt{5}}{1}-\dfrac{4\left(3+\sqrt{5}\right)}{4}\right]\left(\sqrt{5}-6\right)\)

\(B=\left(-1-\sqrt{5}-2+\sqrt{5}-3-\sqrt{5}\right)\left(\sqrt{5}-6\right)\)

\(B=\left(-\sqrt{5}-6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(\sqrt{5}+6\right)\left(\sqrt{5}-6\right)\)

\(B=-\left(5-36\right)\)

\(B=-\left(-31\right)\)

\(B=31\)

_____________________________

\(\sqrt{48}-\dfrac{\sqrt{21}-\sqrt{15}}{\sqrt{7}-\sqrt{5}}+\dfrac{2}{\sqrt{3}+1}\)

\(=4\sqrt{3}-\dfrac{\sqrt{3}\left(\sqrt{7}-\sqrt{5}\right)}{\sqrt{7}-\sqrt{5}}+\dfrac{2\left(\sqrt{3}-1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

\(=4\sqrt{3}-\sqrt{3}-\dfrac{2\left(\sqrt{3}-1\right)}{2}\)

\(=3\sqrt{3}-\sqrt{3}+1\)

\(=2\sqrt{3}+1\)

     A = 1/99 - 1/99.98 - 1/98.97 - ............... - 1/3.2 - 1/2.1

\(A=\frac{1}{99}-\left(\frac{1}{99.98}+\frac{1}{98.97}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

đặt \(B=\frac{1}{99.98}+\frac{1}{97.87}+...+\frac{1}{3.2}+\frac{1}{2.1}\)

\(B=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}\)

\(B=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}\)

\(B=1-\frac{1}{99}\)

\(B=\frac{98}{99}\)

\(\Rightarrow A=\frac{1}{99}-\frac{98}{99}=\frac{-97}{99}\)