a)16.4^x=4⁸

^{\(4^2.4^x=4^{8^{ }}\)}

^{\(4^x=4^{8^{ }}:4^2\)}

\(4^x=4^{6^{ }}\)

\(x=6\)
b)(X-2)(X-5)=0

\(\Rightarrow x-2=0\rightarrow x=2\)

\(x-5=0\rightarrow x=5\)

Vậy x∈ {2;5}
c)2^x+2^x+1=96

\(2^x.1+2^{x^{ }}.2\)

\(2^x.\left(1+2\right)=96\)

\(2^x.3=96\)

\(2^x=96:3\)

\(2^x=32\)

\(2^x=2^5\)

\(x=5\)

\(Zzz\) 😪

a) x+3=12

x=12-3

x=9

b)(x-3):2=5¹⁴:5¹²

=>(x-3):2=5²

=>(x-3):2=25

=>x-3=25.2

=>x-3=50

=>x=50+3

=>x=53

c)4x+3x=30-20:10

=>x(4+3)=30-2

=>7x=28

=>x=28:7

=>x=4

d)2x-138=2³.3²

=>2x-138=8.9

=>2x-138=72

=>2x=72+138

=>2x=210

=>x=210:2

=>x=105

a) x + 3 = 12

x = 12 - 3

x = 9

b) ( x - 3 ) : 2 = 5¹⁴ : 5¹²

( x - 3 ) : 2 = 5^14-12

( x - 3 ) : 2 = 5²

( x - 3 ) : 2 = 25

x - 3 = 50

x = 53

c) 4x + 3x = 30 - 20 : 10

7x = 28

x = 4

d) 2x - 138 = 2³ x 3²

2x - 138 = 8 x 9

2x - 138 = 72

2x = 210

x = 105

a, 2x(x-5) - x ( 3 + 2x ) = 26

=> 2x^2 - 10x - 3x - 2x ^ 2 = 26 

=> - 13 x = 26 

=> x = -2

a, \(2x\left(x-5\right)-x\left(3+2x\right)=26\)

\(\Leftrightarrow2x^2-10x-3x-2x^2=26\)

\(\Leftrightarrow-13x=26\)

\(\Leftrightarrow x=-2\)

Vậy x = -2

b, \(3x^3-48x=0\)

\(\Leftrightarrow3x\left(x^2-16\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}3x=0\\x^2-16=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=4;x=-4\end{cases}}\)

Vậy x = 0 hoặc x = 4 hoặc x = -4 

a) 2x(x - 5) - x(3 + 2x) = 26

2x² - 10x - 3x - 2x² = 26

-13x = 26

=> x = 2

b) 3x³ - 48x = x(3x² - 48) = 0

=> \(\orbr{\begin{cases}x=0\\3x^2-48=0\end{cases}\Rightarrow\orbr{\begin{cases}x=0\\3x^2=48\Rightarrow x^2=16\end{cases}}}\Rightarrow\orbr{\begin{cases}x=0\\x=4;-4\end{cases}}\)

Vậy x = {0 ; 4 ; -4}

1)

xy + x - 4y = 12

x + y(x - 4) = 12

y(x - 4) = 12 - x

\(y=\dfrac{-x+12}{x-4}\)

Vì \(x,y\inℕ\) nên

\(\left(-x+12\right)⋮\left(x-4\right)\)

\(\left(-x+12\right)-\left(x-4\right)⋮\left(x-4\right)\)

\(16⋮\left(x-4\right)\)

\(\left(x-4\right)\inƯ\left(16\right)\)

\(\left(x-4\right)\in\left\{1;-1;2;-2;4;-4;8;-8;16;-16\right\}\)

\(x\in\left\{5;3;6;2;8;0;12;-4;20;-12\right\}\)

\(y\in\left\{\dfrac{-5+12}{5-4};\dfrac{-3+12}{3-4};\dfrac{-6+12}{6-4};\dfrac{-2+12}{2-4};\dfrac{-8+12}{8-4};\dfrac{-0+12}{0-4};\dfrac{-12+12}{12-4};\dfrac{4+12}{-4-4};\dfrac{-20+12}{20-4};\dfrac{12+12}{-12-4}\right\}\)

\(y\in\left\{7;-9;3;-5;1;-3;0;-2;-\dfrac{1}{2};-\dfrac{7}{5}\right\}\)

\(\left(x;y\right)\in\left\{\left(5;7\right);\left(3;-9\right);\left(6;3\right);\left(2;-5\right);\left(8;1\right);\left(0;-3\right);\left(12;0\right);\left(-4;-2\right);\left(20;-\dfrac{1}{2}\right);\left(-12;-\dfrac{7}{5}\right)\right\}\)

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(5;7\right);\left(6;3\right);\left(8;1\right);\left(12;0\right)\right\}\)

2)

(2x + 3)(y - 2) = 15

\(\left(2x+3\right)\inƯ\left(15\right)\)

\(\left(2x+3\right)\in\left\{1;-1;3;-3;5;-5;15;-15\right\}\)

Ta lập bảng

Mà \(x,y\inℕ\) nên các giá trị cần tìm là \(\left(x;y\right)\in\left\{\left(0;7\right);\left(1;5\right);\left(6;3\right)\right\}\)

các thầy cô ơi giúp em vs ạ mai em phải nộp r ạ!!!

a) x³ - 9x² + 27x - 27 = -8

<=> x³ - 3x².3 + 3x.3² - 3³ = -8

<=> (x - 3)³ = -2³

<=> x - 3 = -2

<=> x = 1 (T/m)

Vậy x = 1.

b) 64x³ + 48x² + 12x + 1 = 27

<=> (4x)³ + 3.(4x)².1 + 3.4x.1² + 1³ = 27

<=> (4x + 1)³ = 3³

<=> 4x + 1 = 3

<=> 4x = 2

<=> x = \(\frac{1}{2}\)(T/m)

Vậy x = \(\frac{1}{2}\).

c) (2x - 1)³ - 4x².(2x - 3) = 5

<=> (8x³ - 12x² + 6x - 1) - (8x³ - 12x²) = 5

<=> 8x³ - 12x² + 6x - 1 - 8x³ + 12x² = 5

<=> 6x - 1 = 5

<=> 6x = 6

<=> x = 1 (T/m)

Vậy x = 1.

\(x^3-9x^2+27x-27=-8\Leftrightarrow\left(x^3-27\right)-\left(9x^2-27x\right)=\left(x-3\right)\left(x^2+3x+9\right)-9x\left(x-3\right)=\left(x-3\right)\left(x^2-6x+9\right)=\left(x-3\right)^3=-8=\left(-2\right)^3\Rightarrow x=\left(-2\right)+3=1\)

\(64x^3+48x^2+12x+1=\left(64x^3+1\right)+\left(48x^2+12x\right)=\left(4x+1\right)\left(16x^2-4x+1\right)+12x\left(4x+1\right)=\left(4x+1\right)\left(16x^2+8x+1\right)=\left(4x+1\right)^3=27\Rightarrow4x=2\Leftrightarrow x=\frac{1}{2}\)

c) \(\left(2x-1\right)^3-4x^2.\left(2x-3\right)=5\)

\(\Leftrightarrow\left(8x^3-12x^2+6x-1\right)-\left(8x^3-12x^2\right)=5\)

\(\Leftrightarrow8x^3-12x^2+6x-1-8x^3+12x^2=5\)

\(\Leftrightarrow6x-1=5\)

\(\Leftrightarrow6x=6\)

\(\Leftrightarrow x=1\)

d) \(\left(x+4\right)^3-x^2.\left(x+12\right)=16\)

\(\Leftrightarrow\left(x^3+12x^2+48x+64\right)-\left(x^3+12x^2\right)=16\)

\(\Leftrightarrow x^3+12x^2+48x+64-x^3-12x^2=16\)

\(\Leftrightarrow48x+64=16\)

\(\Leftrightarrow48x=-48\)

\(\Leftrightarrow x=-1\)

#vì câu a,b có người làm rồi nên mình chỉ làm c,d thôi nhé ! :)

Học Tốt !!

12-x=8x-12+2

12+12-2=8x+x

22=9x

x=22/9

ma x thuoc z 

x ko co gia tri

12-x = 4(2x-3) + 2

12-x = 8x - 12 + 2

12-x-8x = -12 + 2

 12- (1-8)x = -10

        -7x     = 12 + 10

        -7x     = 22

           x     = \(\frac{22}{-7}\)

Bạn chỉ gửi 1 bài thôi chứ nhiều quá làm mỏi tay lắm

Làm bài 1 trước

\(4\cdot(-5)^2+2\cdot(-5)-20\)

\(=4\cdot25+2\cdot(-5)-20\)

\(=100+(-10)-20=100-30=70\)

\(35\cdot(14-10)-14\cdot(35-10)\)

\(=35\cdot14-35\cdot10-14\cdot35-14\cdot10\)

\(=35\cdot14-14\cdot35-35\cdot10-14\cdot10\)

\(=35\cdot10-14\cdot10=(35-14)\cdot10=210\)

\(3\cdot(-5)^2+2\cdot(-5)-20\)

Tương tự như ở câu trên

\(34\cdot(15-10)-15\cdot(34-10)\)

Tương tự như câu thứ 2

Câu cuối tự làm