20) -5-(x + 3) = 2 - 5x ⇔ -5 - x - 3 = 2 -5x ⇔ 4x = 10 ⇔ x = \(\frac{5}{2}\)

Vậy...

1) 16 - 8x = 0 ⇔ 8(2 - x) = 0⇔ 2 - x = 0 ⇔ x = 2

Vậy phương trình có nghiệm là x = 2

a) 6x(5x + 3) + 3x(1 – 10x) = 7  

⇒ 30x²+18x+3x-30x²=7

⇒21x=7

⇒x=\(\dfrac{7}{21}\)

⇒x= \(\dfrac{1}{3}\)

b) (3x – 3)(5 – 21x) + (7x + 4)(9x – 5) = 44

⇒15x-63x²-15+63x + 63x²-35x+36x-20=44

⇒79x-35=44

⇒79x=44+35

⇒79x=79

⇒x=1

d) 5x(12x + 7) – 3x(20x – 5) = - 100

⇒60x²+35x-60x²+15=-100

⇒35x+15=-100

⇒35x=-100-15

⇒35x=-115

⇒x=\(\dfrac{-115}{35}\)

⇒x=\(\dfrac{-23}{7}\)

Mấy cái này chuyển vế đổi dấu là xong í mà :3

1,

16-8x=0

=>16=8x

=>x=16/8=2

2, 

7x+14=0

=>7x=-14

=>x=-2

3,

5-2x=0

=>5=2x

=>x=5/2

Mk làm 3 cau làm mẫu thôi

Lúc đăng đừng đăng như v :>

chi ra khỏi ngt nản

từ câu 1 đến câu 8 cs thể làm rất dễ,bn tham khảo bài của bn muwaa r làm những câu cn lại

1, 16 - 8x = 0

<=>-8x = 16

<=> x = -2

Vậy_

2, 7x + 14 = 0

<=> 7x = -14

<=> x = -2

3, 5 - 2x = 0

<=> - 2x = -5

<=> x =\(\frac{5}{2}\)

Vậy_

4, 3x - 5 = 7

<=> 3x = 7 + 5

<=> 3x = 12

<=> x = 4

Vậy...

5, 8 - 3x = 6

<=> - 3x = 6 - 8

<=> -3x = - 2

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

Vậy......

a)  3x – 15 = 25 – 5x 

=> 3x + 5x = 25 + 15

=> 8x = 40

=> x = 5

 b) 3x - 17 = 2x – 7     

=> 3x - 2x = -7 + 17

=> x = 10

 c) 2x – 17 =  – (3x – 18)

=> 2x - 17 = -3x + 18

=> 2x + 3x = 18 + 17

=> 5x = 35

=> x = 7

d) 3x – 14 = 2(x – 9) + 1

=> 3x - 14 = 2x - 18 + 1

=> 3x - 2x = -18 + 1 + 14

=> x = -3

f) (x – 5)² = 9          

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

a) Ta có: \(3x-15=25-5x\)

\(\Leftrightarrow3x-15-25+5x=0\)

\(\Leftrightarrow8x-40=0\)

\(\Leftrightarrow8x=40\)

hay x=5

Vậy: x=5

b) Ta có: \(3x-17=2x-7\)

\(\Leftrightarrow3x-17-2x+7=0\)

\(\Leftrightarrow x-10=0\)

hay x=10

Vậy: x=10

c) Ta có: \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=-3x+18\)

\(\Leftrightarrow2x-17+3x-18=0\)

\(\Leftrightarrow5x-35=0\)

\(\Leftrightarrow5x=35\)

hay x=7

Vậy: x=7

d) Ta có: \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14-2x+18-1=0\)

\(\Leftrightarrow x+3=0\)

\(\Leftrightarrow x=-3\)

Vậy: x=-3

f) Ta có: \(\left(x-5\right)^2=9\)

\(\Leftrightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy: \(x\in\left\{2;8\right\}\)

Bài 2. Tìm x, biết :                                

a) \(3x-15=25-5x\)

\(\Leftrightarrow8x=40\)

\(\Leftrightarrow x=5\)

Vậy x = 5

b) \(3x-17=2x-7\)

\(\Leftrightarrow x=10\)

Vậy x = 10

c) \(2x-17=-\left(3x-18\right)\)

\(\Leftrightarrow2x-17=18-3x\)

\(\Leftrightarrow5x=35\)

\(\Leftrightarrow x=7\)

Vậy x = 7

d) \(3x-14=2\left(x-9\right)+1\)

\(\Leftrightarrow3x-14=2x-18+1\)

\(\Leftrightarrow3x-14=2x-17\)

\(\Leftrightarrow x=-3\)

Vậy x = -3

e) \(\left(x-5\right)^2=9\)

\(\Rightarrow\left[{}\begin{matrix}x-5=3\\x-5=-3\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=8\\x=2\end{matrix}\right.\)

Vậy x = {8; 2}

a: \(4x^3+12=120\)

=>\(4x^3=108\)

=>\(x^3=27=3^3\)

=>x=3

b: \(\left(x-4\right)^2=64\)

=>\(\left[{}\begin{matrix}x-4=8\\x-4=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-4\end{matrix}\right.\)

c: (x+1)^3-2=5^2

=>\(\left(x+1\right)^3=25+2=27\)

=>x+1=3

=>x=2

d: 136-(x+5)^2=100

=>(x+5)^2=36

=>\(\left[{}\begin{matrix}x+5=6\\x+5=-6\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=-11\end{matrix}\right.\)

e: \(4^x=16\)

=>\(4^x=4^2\)

=>x=2

f: \(7^x\cdot3-147=0\)

=>\(3\cdot7^x=147\)

=>\(7^x=49\)

=>x=2

g: \(2^{x+3}-15=17\)

=>\(2^{x+3}=32\)

=>x+3=5

=>x=2

h: \(5^{2x-4}\cdot4=10^2\)

=>\(5^{2x-4}=\dfrac{100}{4}=25\)

=>2x-4=2

=>2x=6

=>x=3

i: (32-4x)(7-x)=0

=>(4x-32)(x-7)=0

=>4(x-8)*(x-7)=0

=>(x-8)(x-7)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-7=0\end{matrix}\right.\)

=>\(\left[{}\begin{matrix}x=8\\x=7\end{matrix}\right.\)

k: (8-x)(10-2x)=0

=>(x-8)(x-5)=0

=>\(\left[{}\begin{matrix}x-8=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=5\end{matrix}\right.\)

m: \(3^x+3^{x+1}=108\)

=>\(3^x+3^x\cdot3=108\)

=>\(4\cdot3^x=108\)

=>\(3^x=27\)

=>x=3

n: \(5^{x+2}+5^{x+1}=750\)

=>\(5^x\cdot25+5^x\cdot5=750\)

=>\(5^x\cdot30=750\)

=>\(5^x=25\)

=>x=2

a) \(\Rightarrow72-20x-36x+84=30x-240-6x-84\)

\(\Rightarrow80x=480\Rightarrow x=6\)

b) \(\Rightarrow15x+25-8x+12=5x+6x+36+1\)

\(\Rightarrow4x=0\Rightarrow x=0\)

c) \(\Rightarrow10x-16-12x+15=12x-16+11\)

\(\Rightarrow14x=4\Rightarrow x=\dfrac{2}{7}\)

\(\text{a) 5(2x-3)-4(5x-7)=19-2(x+11)}\)

\(10x-15-20x+28=19-2x-22\)

\(10x-20x+2x=19-22-28+15\)

\(-8x=-16\)

\(\Rightarrow x=2\)

\(\text{b) 4(x+3)-7x+17=8(5x-1)+166}\)

\(4x+12-7x+17=40x-8+166\)

\(4x-7x-40x=-8+166-17-12\)

\(-43x=129\)

\(x=-3\)

\(\text{c) 17-14(x+1)=13-4(x+1)-5(x-3)}\)

\(17-14x+14=13-4x-4-5x+15\)

\(-14x+4x+5x=13-4+15-14-17\)

\(-5x=-7\)

\(x=\frac{7}{5}\)

\(\text{d) 5x+3,5+(3x-4)=7x-3(x-0,5)}\)

\(5x+3,5+3x-4=7x-3x+1,5\)

\(5x+3x-7x+3x=1,5-3,5\)

\(x=-2\)

\(\text{e) 7(4x+3)-4(x-1)=15(x+0,75)+7}\)

\(28x+21-4x+4=15x+11,25+7\)

\(28x-4x-15x=11,25+7-4-21\)

\(9x=\frac{-27}{4}\)

\(x=\frac{-3}{4}\)

\(\text{f) 3x+2,42+o,8x=3,38-0,2x}\)

\(3x+0,8x+0,2x=3,38-2,42\)

\(4x=\frac{24}{25}\)

\(x=\frac{6}{25}\)

chúc bạn học tốt !!

Nhìu quá bạn ơi

nhiều vậy ai mà làm dc bạn ơi

làm hai câu ba câu cũng được

Bạn gõ ct lại dưới dạng latex (kí hiệu này gọi là latex Σ ) nhé

minh moi hoc lop 5 thoi

1) 3x - 6=  5x + 2

5x - 3x = -6 - 2

2x = -8

x = -4

2) 15 - x = 4x - 5

4x + x = 15 + 5

5x = 20

x = 4

Tương tự như trên 

1) 3x - 6 = 5x + 2

  3x - 5x = 2 + 6

        -2x = 8

           x = 8 : (-2)

           x = -4

2) 15 - x = 4x - 5

    -x - 4x = -5 - 15

         -5x = -20

            x = -20 : (-5)

            x = 4