Lời giải:

\(\frac{4}{9}\times \frac{3}{7}\times \frac{7}{4}=\frac{1}{3}\)

\(\frac{6}{5}\times \frac{4}{5}\times \frac{25}{16}=\frac{3}{2}\)

\(\frac{7}{8}\times \frac{16}{9}\times \frac{3}{14}=\frac{1}{3}\)

a) x – 9 = -14

x = -14 + 9

x = -5

b) 2( x + 7 ) = -16

2( x + 7 ) = 2 . ( -8 )

x + 7 = -8

x = -8 – 7 = -15

c) | x – 9 | = 7

x – 9 = 7 hoặc x – 9 = -7

x = 7 + 9 hoặc x = -7 + 9

x = 16 hoặc x = 2

d) ( x – 5 )( x + 7 ) = 0

x – 5 = 0 hoặc x + 7 = 0

x = 5 hoặc x = -7

a: Đ

b: S

c: S

d: 88/66=44/33=4/3

a.\(\left(-12\right)x-14=-2\)

\(\left(-12\right)x=-2+14\)

\(\left(-12\right)x=12\)

\(x=12:\left(-12\right)\)

\(x=-1\)

\(b,\left(-8\right)x=\left(-5\right)\left(-7\right)-3\)

\(\left(-8\right)x=35-3\)

\(\left(-8\right)x=32\)

\(x=32:\left(-8\right)\)

\(x=-4\)

\(c,\left(-9\right)x+3=\left(-2\right)\left(-7\right)+16\)

\(\left(-9\right)x+3=14+16\)

\(\left(-9\right)x+3=30\)

\(\left(-9\right)x=30-3\)

\(\left(-9\right)x=27\)

\(x=27:\left(-9\right)\)

\(x=-3\)

A,x=6/7

B, x=61/60

a)=(3/8+10/16)+(7/12+10/24)

=1+1=2

c)=(4/6+14/6)+(7/13+19/13)+(17/9+1/9)

=3+2+2=7

\(\text{6) 27,693 + 8,54 = 36.233}\)

\(\text{7) (-81,73) - 17,249= -98.979}\)

\(\text{8) 9,371 . 8,65 = 81.05915}\)

\(\text{9) (-14,29) . 73,6 = -1051.744}\)

\(\text{10) 24,108 : 6,15= 3.92}\)

\(\text{11) 324,82 + 312,25= 637.07}\)

\(\text{13) 8,446 : 4,12= 2.05}\)

\(\text{14) (-882) : 3,6 = -245}\)

\(\text{15) 19,427 . 1,8 = 34.9686}\)

\(6) 27,693 + 8,54 = 36,233\\ 7) (-81,73)-17,249 = -98,979\\ 8) 9,371.8,65 = 81,05915\\ 9) (-14,29).73,6 = -1051,744\\ 10) 24,108:6,15 = 3,92\\ 11) 324,82+ 312,25 = 637,07\\ 13) 8,446 : 4,12 = 2,05\\ 14) (-882):3,6 = -245\\15) 19,427.1,8 = 34,9686\)

Câu 1:

$150\vdots x; 84\vdots x, 30\vdots x$
$\Rightarrow x=ƯC(150,84,30)$

$\Rightarrow ƯCLN(150,84,30)\vdots x$

$\Rightarrow 6\vdots x$. Kết hợp với điều kiện $x<16$ suy ra:
$x\in \left\{1; -1; 2; -2; 3;-3;6;-6\right\}$

Câu 2:

$x\vdots 15; x\vdots 14; x\vdots 20$

$\Rightarrow x=BC(15,14,20)$

$\Rightarrow x\vdots BCNN(15,14,20)$

$\Rightarrow x\vdots 420$

$\Rightarrow x=420k$ với $k$ là số tự nhiên.

Mà $400\leq x\leq 1200$

$\Rightarrow 400\leq 420k\leq 1200$

$\Rightarrow 0< k< 3$

Mà $k$ là stn nên $k=1$ hoặc $k=2$

$\Rightarrow x=420.1=420$ hoặc $x=420.2=840$

Câu 3:

$9\vdots x+1$

$\Rightarrow x+1\in Ư(9)$

$\Rightarrow x+1\in \left\{1; -1; 3; -3; 9;-9\right\}$

$\Rightarrow x\in \left\{0; -2; 2; -4; 8; -10\right\}$