Ta có  3 y − 5 + 2 x − 3 = 0 7 x − 4 + 3 x + y − 1 − 14 = 0 ⇔ 3 y − 15 + 2 x − 6 = 0 7 x − 28 + 3 x + 37 − 3 − 14 = 0 ⇔ 2 x + 3 y = 21 10 x + 3 y = 45

⇔ 3 y = 21 − 2 x 10 x + 21 − 2 x = 45 ⇔ 3 y = 21 − 2 x 8 x = 24 ⇔ x = 3 3 y = 15 ⇔ x = 3 y = 5

Vậy hệ phương trình có nghiệm duy nhất (x; y) = (3; 5)

⇒ x 2   +   y 2   =   32   +   52   =   34

Đáp án: B

a) \(x:13=41\) .

\(\Rightarrow x=41.13=533\)

b) \(1428:x=14\)

\(\Rightarrow x=1428:14=102\)

c) \(4x:17=0\) 

\(\Rightarrow4x=0\)

\(\Rightarrow x=0\)

d) \(7x-8=713\)

\(7x=721\)

\(\Rightarrow x=103\) 

e) \(8\left(x-3\right)=0\)

\(\Rightarrow x-3=0\)

\(\Rightarrow x=3\)

g) \(0:x=0\)

\(\Rightarrow x\in R\)

a)x:13=41

x=533

b)1428:x=14

x=102

c)4x:17=0

x=0

8(x-3)=0

x-3=0/8

x-3=0

x=3+0

x=3

a. x - (17 - 8) = 5 + (10 - 3x)

x - 9 = 5 + 10 - 3x

x - 9 = 15 - 3x

x + 3x = 15 + 9

4x = 24

x = 24 : 4

x = 6

b. 25 - (30 + x) = x - (27 - 8)

25 - 30 - x = x - 19

-5 - x = x - 19

-5 + 19 = x + x

14 = 2x

14 : 2 = x

7 = x

c. |x + 5| = 7

\(\Rightarrow\left\{\begin{matrix}x+5=7\\-\left(x+5\right)=7\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=2\\-x-5=7\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=2\\x=-12\end{matrix}\right.\)

d. x(x - 1) = 0

\(\Rightarrow\left\{\begin{matrix}x=0\\x-1=0\end{matrix}\right.\Rightarrow\left\{\begin{matrix}x=0\\x=1\end{matrix}\right.\)

a) x-(17-8)=5+(10-3x)

x-9=5+10-3x

x-9=15-3x

x+3x=15+9

4x=24

x=6

4.x:17= 0

4.x=0:17

4.x=0

x=0:4

x=0

7.x-8=713

7.x=713+8

7.x=721

x=721:7=103

x-3=0:8

x.3=0

x=0:3

x=0

a: \(\Leftrightarrow\dfrac{1}{x+2}-\dfrac{1}{x+5}+\dfrac{1}{x+5}-\dfrac{1}{x+10}+\dfrac{1}{x+10}-\dfrac{1}{x+17}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

=>\(\dfrac{x+17-x-2}{\left(x+2\right)\left(x+17\right)}=\dfrac{x}{\left(x+2\right)\left(x+17\right)}\)

=>x=15

b: \(\Leftrightarrow-\dfrac{1}{x-1}+\dfrac{1}{x-3}-\dfrac{1}{x-3}+\dfrac{1}{x-8}-\dfrac{1}{x-8}+\dfrac{1}{x-20}-\dfrac{1}{x-20}=\dfrac{-3}{4}\)

=>1/x-1=3/4

=>x-1=4/3

=>x=7/3

4x :17=0  

4x=0x17

4x=0

x=0:4

x=0

7x -8=713

7x=713+8

7x=721

x=721:7

x=103

8(x-3)=0

(x-3)=0:8

(x-3)=0

x=0+3

x=3

0:x=0

x=0:0

x=0;1;2;3;4;5;6;7...

\(4x:17=0\)                        \(7x-8=713\)                          \(8\left(x-3\right)=0\)

\(4x=0\)                               \(7x=721\)                                     \(x-3=0\)

\(x=0\)                                   \(x=103\)                                       \(x=3\)

\(0:x=0\)

\(x=0\)