a) 731 – 680 + 19 = 51 + 19

                              = 70

b) 63 x 2 : 7 = 126 : 7

                    = 18

c) 14 x 6 – 29 = 84 – 29

                       = 55

d) 348 + 84 : 6 = 348 + 14

                        = 362

Lời giải:
$|x-2|\geq 0$ với mọi $x\in\mathbb{R}$ (tính chất trị tuyệt đối)

$\Rightarrow A=|x-2|+5\geq 5$

Vậy $A_{\min}=5$ khi $x-2=0\Leftrightarrow x=2$

1.a,=(54+45+1).113

=100.113

=11300

b,=(3/7+8/14)+(4/9+10/18)

=1+1

=2

2.a,=13/10+1/3

=49/30

b,=12/9.(1/12+1/6)

=12/9.1/4

=1/3

c,=3/4.3/2

=9/8

d,=3/2-1/3

=7/6

1:tính bằng cách thuận tiện nhất:

a)54 x 113 + 45 x 113 + 113

= 54 x 113 + 45 x 113 + 113x1

=113 x(54+45+1)

= 113x100

=1300

 b)3/7 + 4/9 + 8/14 + 10/18

=(3/7+8/14)+(4/9+10/18)

=    1           + 1

=2

\(a)2^3\cdot15-[115-(12-5)^2]\\=8\cdot15-(115-7^2)\\=120-(115-49)\\=120-66\\=54\\---\\b)5\cdot[(85-35:7):8+90]-60\\=5\cdot[(85-5):8+90]-60\\=5\cdot(80:8+90)-60\\=5\cdot(10+90)-60\\=5\cdot100-60\\=500-60\\=440\)

\(---\)

\(c,\left\{\left[261-\left(36-31\right)^3\cdot2\right]-9\right\}\cdot1001\)

\(=\left[\left(261-5^3\cdot2\right)-9\right]\cdot1001\)

\(=\left[\left(261-125\cdot2\right)-9\right]\cdot1001\)

\(=\left[\left(261-250\right)-9\right]\cdot1001\)

\(=\left(11-9\right)\cdot1001\)

\(=2\cdot1001\)

\(=2002\)

\(---\)

\(d,3\cdot10^2-\left[1200-\left(4^2-2\cdot3\right)^3\right]\)

\(=3\cdot100-\left[1200-\left(16-6\right)^3\right]\)

\(=300-\left(1200-10^3\right)\)

\(=300-\left(1200-1000\right)\)

\(=300-200\)

\(=100\)

#\(Toru\)

a) 2³.15 -[115-(12-5)² ]

= 2³.15 -[115-36]

= 8.15 -79

= 120-79

=41

b)5.[(85 - 35 : 7) :8 + 90 ] - 50

=5 .[80:8+90]-50

=5.100-50

=500-50

=450

c){[261 - ( 36-31)³.2 ]-9}.1001

={[261 - 125.2 ]-9}.1001

={[261 -250 ]-9}.1001

={11-9}.1001

=2.1001

=2002

d)3.10² - [1200 - ( 4² - 2.3)^3]

=3.100-[1200 -(16-6)³ ]

=300-[1200-1000]

=300-200

=100

Bài 1: 

\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

\(\Rightarrow-x=1-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

\(x+4=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}-4\)

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

\(x-\dfrac{1}{5}=2\)

\(\Rightarrow x=2+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

Giúp vs ạ

\(A=-3\left(3+1\right)+5\left(1-1\right)+8\left(-1+1-2\right)\)

\(A=-28\)

A=9x-3x²-5x²-5x+8x²-8x-16

   =-4x-16

   =-4(x-4)

Thay x=-1 vào A, ta được:

A(-1)=-4(-1-4)=(-4).(-5)=20

Vậy A(-1)=20

a/\(240\times20-\left(846+202\times2,5\right)\)
\(=4800-\left(846+505\right)\)
\(=4800-1351\)
\(=3449\)
b/\(2000+\left(32,4:3-2,8\right)\)
\(=2000+\left(10,8-2,8\right)\)
\(=2000+8\)
\(=2008\)
c/\(320\times1,25-\left(933+302\times3,5\right)\)
\(=400-\left(933+1057\right)\)
\(=400-1990\)
\(=-1590\)

nhanh chưa

a) 21/20 : 4/5

= 21/16

b) x = 8/7+3/9

    x = 31/21

\(a,\Leftrightarrow\left[{}\begin{matrix}3x+2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{2}{3}\left(l\right)\\x=-2\left(l\right)\end{matrix}\right.\Leftrightarrow x\in\varnothing\Leftrightarrow A\in\varnothing\\ b,\text{ý bạn là rút gọn A hả?}\\ A=\dfrac{x-2+2x+3x+6}{\left(x-2\right)\left(x+2\right)}=\dfrac{6x+4}{\left(x-2\right)\left(x+2\right)}\)