c ( 3x² -6 ) - 43 = 5³

( 3x² -6 ) - 43 = 125

3x² -6=125+43

3x² -6=168

3x²=168+6

3x²=174

x²=174:3

x²=58(đề sai)

d) 3x + 2x ( 2³ . 5 - 3² . 4 ) + 5² = 4⁴

3x + 2x ( 8 . 5 - 9 . 4 ) + 25 = 256

3x + 2x.4 + 25 = 256

3x + 2x.4=256-25

3x + 2x.4=231

(3+2).x.4=231

5x.4=231

5x=231:4

5x=57,75

x=57,75:5

x=11,55

e) 720 : [ 41 - ( 2x - 5 ) ] = 2³ . 5

720 : [ 41 - ( 2x - 5 ) ] =40

41 - ( 2x - 5 )=720:40

41 - ( 2x - 5 )=18

 2x - 5=41-18

 2x - 5=23

2x=23+5

2x=28

x=28:2

x=14

câu b cách làm tương tự câu a bạn nhé!

c, |5x - 2| < hoặc bằng 13

TH1: 5x - 2 < hoặc bằng 13

        5x <= 13 +2

        5x <= 15

          x <= 15 :5

          x <= 3

TH2: 5x +2 <= 13

        5x <= 13 - 2

        5x <= 11

          x <= 11 : 5

          x <= 2,2 (loại)

Vậy x thuộc {0;1;3}

a,x={9;-4}

b,x={9}

c,x={nhìu }

x = 7 , y = 5

ta có :xy-2x+3y=13

         xy+3y-2x=13

         y(x+3)-2x=13

         y(x+3)-2x+6-6=13

         y(x+3)-2(x+3)-6=13

         (x+3)(y-2)=13+6=19

\(\Rightarrow\left(x+3\right)\left(y-2\right)\inƯ\left(19\right)\)\(=\left(-19;19;1;-1\right)\)

48∈B(2x+1)

⇒48⋮2x+1

⇒2x+1∈Ư(48)

Ư(48)={1;2;3;4;6;8;12;16;24;48}

⇒x∈{0;1}

\(2x+1\inƯ\left(48\right)\)

\(\Leftrightarrow2x+1\in\left\{1;3\right\}\)

hay \(x\in\left\{0;1\right\}\)

\(-12\left(x-5\right)+7\left(3-x\right)=15\)

\(\Rightarrow-12x+60+21-7x=15\)

\(\Rightarrow-12x-7x=15-21-60\)

\(\Rightarrow-19x=-66\)

\(\Rightarrow x=\frac{66}{19}\)

\(\left|2x-5\right|=12\)

\(\Leftrightarrow\orbr{\begin{cases}2x-5=12\\2x-5=-12\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x=17\\2x=-7\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{17}{2}\left(t/m\right)\\x=-\frac{7}{2}\left(t/m\right)\end{cases}}\)

=>-1+(3-5)+(7-9)+...+[(x-4)-(x-2)]+x=600

=>-1+(-2)+(-2)+...+(-2)+x=600                                      (\(\frac{x-1}{4}\) số hạng -2)

=>-1+(-2)\(\frac{x-1}{4}\)+x=600

=>\(\frac{-4+\left(-2\right)\left(x-1\right)+4x}{4}=600\)

=>-4+(-2)(x-1)+4(x-1)+4=600.4

=>(-2+4)(x-1)=2400+4-4

=>2(x-1)=2400

=>x-1=2400:2

=>x-1=1200

=>x=1201

x+ 7 \(⋮\)x+5

=> x+5 \(⋮\)x+5

=> ( x+7)-( x+5) \(⋮\)x+5

=> x+7 - x-5 \(⋮\)x+5

=> 2 \(⋮\)x+5

=> x+ 5 \(\in\)Ư(2)= {1; 2; -1; -2}

=>  x \(\in\){ -4; -3; -6: -7}

Vậy...

+)Ta có:x+5\(⋮\)x+5(1)

+)Theo bài ta có:x+7\(⋮\)x+5(2)

+)Từ (1) và (2)

=>(x+7)-(x+5)\(⋮\)x+5

=>x+7-x-5\(⋮\)x+5

=>2\(⋮\)x+5

=>x+5\(\in\)Ư(2)={\(\pm\)1;\(\pm\)2}

=>x\(\in\){-6;-4;-7;-3}

Vậy x​\(\in\)​ {-6;-4;-7;-3}

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