23/4+2/3=77/12

\(=\dfrac{23}{4}+\dfrac{2}{3}=\dfrac{77}{12}\)

a) (2/5 x 25/29) + (3/5 x 25/29)
= (50/145) + (75/145)
= 125/145

b) (5/2 x 3/7) - (3/14 : 6/7)
= 15/14 - (3/14 x 7/6)
= 15/14 - 1/2
= (30/28) - (14/28)
= 16/28
= 4/7

c) (15/4 : 5/12) - (6/5 : 11/15)
= (15/4 x 12/5) - (6/5 x 15/11)
= 180/20 - 90/55
= 9 - 18/11
= (99/11) - (18/11)
= 81/11
= 7 4/11

a) (2/3) + (20/21 x 3/2 x 7/5)
= 2/3 + (60/210)
= 2/3 + 2/7
= (14/21) + (6/21)
= 20/21

b) (5/17 x 21/32 x 47/24 x 0)
= 0

c) (11/3 x 26/7) - (26/7 x 8/3)
= (286/21) - (208/21)
= 78/21
= 3 9/21
= 3 3/7

a) (25/8) : x = 5/16
=> (25/8) x (16/5) = x
=> 4 = x

b) x + (7/15) = 6/15
=> x = (6/15) - (7/15)
=> x = -1/15

c) x : (28/49) = 7/12
=> x x (49/28) = 7/12
=> x = (7/12) x (28/49)
=> x = 1/2

a) 6 x x = (5/8) : (3/4)
=> 6x = (5/8) x (4/3)
=> 6x = 20/24
=> 6x = 5/6
=> x = (5/6) / 6
=> x = 5/36

câu,b,không,đủ,thông,tin,nhan,bạn.

\(13-\sqrt{\left(\sqrt{2}-3\right)^2}\)
=13-/\(\sqrt{2}\)-3/
=13-(3-\(\sqrt{2}\))
=13-3+\(\sqrt{2}\)
=10\(\sqrt{2}\)

\(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}\)
\(=\sqrt{2-2\sqrt{2}+1}-\sqrt{4-2.2\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=\sqrt{2}-1-2+\sqrt{2}\)
\(=2\sqrt{2}-3\)

1) \(13-\sqrt{\left(\sqrt{2}-3\right)^2}=13-\left|\sqrt{2}-3\right|=13+\sqrt{2}-3=10+\sqrt{2}\)

2) \(\sqrt{3-2\sqrt{2}}-\sqrt{6-4\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}-\sqrt{\left(2-\sqrt{2}\right)^2}=\left|\sqrt{2}-1\right|-\left|2-\sqrt{2}\right|=\sqrt{2}-1-2+\sqrt{2}=2\sqrt{2}-3\)

1) 

2) 

x=7

=>x+1=8

=> A= x^15 - 8x^14 + 8x^13 - 8x^12 +....- 8x^2 + 8x - 5 

=x¹⁵-(x+1)x¹⁴+(x+1)x¹³-(x+1)x¹²+...-(x+1)x²+(x+1)x-5

=x¹⁵-x¹⁵-x¹⁴+x¹⁴+x¹³-x¹³-x¹²+...-x³-x²+x²+x-5

=x-5

=>A=7-5=2

Vậy A=2 khi x=7

\(B=x^{15}-8x^{14}+8x^{13}-8x^{12}+...+8x-5\)

\(=x^{15}-\left(x+1\right)x^{14}+\left(x+1\right)x^{13}-\left(x+1\right)x^{12}+...+\left(x+1\right)x-x+2\)

\(=x^{15}-x^{15}-x^{14}+x^{14}+x^{13}-x^{13}-x^{12}+...+x^2+x-x+2\)

\(=2\)

\(C=\left(1-2-3-4\right)+...+\left(197-198-199-200\right)\)

=-8x25=-200

\(D=-\left(11+13+...+99\right)+\left(10+12+...+98\right)\)

=(-1)+(-1)+...+(-1)

=-1x45=-45

Bài 1:

\(\left(\frac{9}{4}:\frac{6}{5}\right)-1=\frac{45}{24}-1=\frac{15}{8}-1=\frac{7}{8}\)

\(2+\left(\frac{1}{4}\times\frac{2}{5}\right)=2+\frac{2}{10}=2+\frac{1}{5}=\frac{11}{5}\)

Bài 2: 

\(\frac{3}{5}+\frac{2}{3}=\frac{9}{15}+\frac{10}{15}=\frac{19}{15}\)

\(1-\frac{9}{11}=\frac{1}{1}-\frac{9}{11}=\frac{11}{11}-\frac{9}{11}=\frac{2}{11}\)

\(\frac{7}{9}\times\frac{3}{14}=\frac{21}{126}=\frac{3}{18}\)

\(\frac{15}{7}:\frac{5}{21}=\frac{315}{35}=9\)

1 Tính giá trị biểu thức

( 9/4 : 6/5 ) - 1 = 45/24 - 1 = 7/8

2 + ( 1/4 × 2/5 ) = 2 + 1/10 = 21/10

2 Tính

3/5 + 2/3 = 

1 - 9/11 = 2/11

7/9 × 3/14 = 1/6

15/7 : 5/21 = 9

(9/4 : 6/5) - 1 = (9/4 . 5/6) - 1

= 15/8 - 1 = 7/8

2 + (1/4 x 2/5)  = 2 + 1/10

= 21/10

3/5 + 2/3 = 19/15

1 - 9/11 = 2/11

7/9 x 3/14 = 1/6

15/7 : 5/21 = 9

Ta có: A\(=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{15}=\dfrac{2}{45}\)

\(A=\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{15}\)

\(=\dfrac{1}{9}-\dfrac{1}{15}\)

\(=\dfrac{2}{45}\)

-Chúc bạn học tốt-

A = \(\dfrac{1}{9}.\dfrac{1}{10}+\dfrac{1}{10}.\dfrac{1}{11}+\dfrac{1}{11}.\dfrac{1}{12}+\dfrac{1}{12}.\dfrac{1}{13}+\dfrac{1}{13}.\dfrac{1}{14}+\dfrac{1}{14}.\dfrac{1}{15}\)

= \(\dfrac{1}{9.10}+\dfrac{1}{10.11}+\dfrac{1}{11.12}+\dfrac{1}{12.13}+\dfrac{1}{13.14}+\dfrac{1}{14.15}\)

= \(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{11}+...+\dfrac{1}{14}-\dfrac{1}{15}\)

= \(\dfrac{1}{9}-\dfrac{1}{15}\)

= \(\dfrac{2}{45}\)

2, 

a) \(315-\left(135-x\right)=215\)

\(\Rightarrow135-x=315-215\)

\(\Rightarrow135-x=100\)

\(\Rightarrow x=135-100\)

\(\Rightarrow x=35\)

b) \(x-320:32=25\cdot16\)

\(\Rightarrow x-10=5^2\cdot4^2\)

\(\Rightarrow x-10=20^2\)

\(\Rightarrow x-10=400\)

\(\Rightarrow x=410\)

c) \(3\cdot x-2018:2=23\)

\(=3\cdot x-1009=23\)

\(\Rightarrow3\cdot x=1032\)

\(\Rightarrow x=1032:3\)

\(\Rightarrow x=344\)

d) \(280-9\cdot x-x=80\)

\(\Rightarrow280-x\cdot\left(9+1\right)=80\)

\(\Rightarrow280-10\cdot x=80\)

\(\Rightarrow10\cdot x=280-80\)

\(\Rightarrow10\cdot x=200\)

\(\Rightarrow x=20\)

e) \(38\cdot x-12\cdot x-x\cdot16=40\)

\(\Rightarrow x\cdot\left(38-12-16\right)=40\)

\(\Rightarrow x\cdot10=40\)

\(\Rightarrow x=40:10\)

\(\Rightarrow x=4\)

bài 1 cậu ko giải giúp mình dc à