A)\(\dfrac{7}{20}-\left(\dfrac{5}{8}-\dfrac{2}{5}\right)\)

\(=\dfrac{7}{20}-\left(\dfrac{25}{40}-\dfrac{16}{40}\right)\)

\(=\dfrac{7}{20}-\dfrac{9}{40}\)

\(=\dfrac{14}{40}-\dfrac{9}{40}=\dfrac{5}{40}=\dfrac{1}{8}\)

B) \(\dfrac{5}{6}+\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\)

\(=\dfrac{5}{6}+\left(\dfrac{20}{36}-\dfrac{9}{36}\right)\)

\(=\dfrac{5}{6}+\dfrac{11}{36}\).

\(=\dfrac{30}{36}+\dfrac{11}{36}=\dfrac{41}{36}\)

C) \(\dfrac{9}{10}-\left(\dfrac{2}{5}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)

\(=\dfrac{9}{10}-\left(\dfrac{4}{10}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)

\(=\dfrac{9}{10}-\dfrac{1}{10}+\dfrac{7}{20}\)

\(=\dfrac{18}{20}-\dfrac{2}{20}+\dfrac{7}{20}=\dfrac{23}{20}\)

a: =7/20-5/8+2/5

=14/40-25/40+16/40

=5/40=1/8

b: =5/6+5/9-1/4

=30/36+20/36-9/36

=41/36

c: =9/10-2/5+3/10+7/20

=12/10-2/5+7/20

=7/20+6/5-2/5

=7/20+4/5

=7/20+16/20

=23/20

A) \(\frac{4}{5}+\frac{5}{4}-2=\frac{16}{20}+\frac{25}{20}-\frac{40}{20}\)

\(=\frac{1}{20}\)

B) \(3-\frac{2}{3}-\frac{3}{2}=\frac{18}{6}-\frac{4}{6}-\frac{9}{6}\)

\(=\frac{5}{6}\)

C) \(\frac{7}{8}-\left(\frac{1}{4}+\frac{2}{5}\right)=\frac{7}{8}-\frac{13}{20}\)

\(=\frac{35}{40}-\frac{26}{40}=\frac{9}{40}\)

D) \(\frac{9}{10}-\left(\frac{2}{5}+\frac{3}{10}\right)+\frac{7}{10}=\frac{9}{10}-\frac{7}{10}+\frac{7}{10}\)

\(=\frac{9}{10}\)

Bài 1: 

a: x+1/2=5/6

nên x=5/6-1/2=1/3

b: x+1/4=3/4

nên x=3/4-1/4=2/4=1/2

c: x+3/10=1/2

nên x=1/2-3/10=5/10-3/10=1/5

d: x+1/4=3/8

nên x=3/8-1/4=3/8-2/8=1/8

TOT LAM

Là sao!!!!!???????!!!!!???

bài 1 : 

B=15-3x-3y

a) x+y-5=0 

=>x+y=-5

B=15-3x-3y <=> B=15-3(x+y)

Thay x+y=-5 vào biểu thức  B ta được :

B=15-3(-5)

B=15+15

B=30

Vậy giá trị của biểu thức B=15-3x-3y tại x+y+5=0 là 30

b)Theo đề bài ; ta có :

B=15-3x-3.2=10

15-3x-6=10

15-3x=16

3x=-1

\(x=\frac{-1}{3}\)

Bài 2:

a)3x²-7=5

3x²=12

x²=4

x=\(\pm2\)

b)3x-2x²=0

=> 3x=2x²

=>\(\frac{3x}{x^2}=2\)

=>\(\frac{x}{x^2}=\frac{2}{3}\)

=>\(\frac{1}{x}=\frac{2}{3}\)

=>\(3=2x\)

=>\(\frac{3}{2}=x\)

c) 8x² + 10x + 3 = 0

=>\(8x^2-2x+12x-3=0\)

\(\Rightarrow\left(2x+3\right)\left(4x-1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x+3=0\\4x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}2x=-3\\4x=1\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{2}\\x=\frac{1}{4}\end{cases}}}\)

vậy \(x\in\left\{-\frac{3}{2};\frac{1}{4}\right\}\)

Bài 5 đề  sai  vì  |1| không thể =2

a, a x 6 = 3 x 6 = 18

b, a + b = 4 + 2 = 6

c, b + a = 2 + 4 = 6

d, a - b = 8 - 5 = 3

e, m x n = 5 x 9 = 45

a. 18

b. 6

c. 6

d. 3

e. 45

Sửa đề:

A=/x+5/+10

Ta có: /x+5/>= 0 với mọi x>=0

=> A=/x+5/+10 >= 10

=> Amin=10. Dấu "=" xảy ra <=> x+5=0<=> x=-5

Vậy...

\(\text{a) }A=\left|x+5\right|+10\)

\(\text{Vì }\left|x+5\right|\ge0\forall x\)

\(\Rightarrow A=\left|x+5\right|+10\ge10\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left|x+5\right|=0\)

\(\Rightarrow x=-5\)

\(\text{Vậy Min}_A=10\Leftrightarrow x=-5\)

\(\text{b) }\left|3-x\right|+5\)

\(\text{Vì }\left|3-x\right|\ge0\forall x\)

\(\Rightarrow\left|3-x\right|+5\ge5\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left|3-x\right|=0\)

\(\Rightarrow x=3\)

\(\text{Vậy Min}_B=5\Leftrightarrow x=3\)

\(\text{d) }D=\left(x+2\right)^2+15\)

\(\text{Vì ( x + 2 )}^2\ge0\forall x\)

\(\Rightarrow\left(x+2\right)^2+15\ge15\)

\(\text{Dấu ''='' xảy ra khi :}\)

\(\left(x+2\right)^2=0\)

\(\Rightarrow x+2=0\)

\(\Rightarrow x=-2\)

\(C=\left|x+1\right|+\left|y+4\right|\)

Ta có +) \(\left|x+1\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

+)\(\left|y+4\right|\ge0\).Dấu "=" xảy ra \(\Leftrightarrow y+4=0\Leftrightarrow y=-4\)

\(\Rightarrow B=\left|x+1\right|+\left|y+2\right|\ge0\)

\(\Rightarrow minB=0\Leftrightarrow\hept{\begin{cases}x=-1\\y=-4\end{cases}}\)

P/s: ko chắc nha