bạn viết rõ đề câu a;b nhé 

c, \(2x\left(x-5\right)-\left(x-5\right)=0\Leftrightarrow\left(2x-1\right)\left(x-5\right)=0\Leftrightarrow x=\dfrac{1}{2};x=5\)

d, \(\left(x+3\right)\left(x+3-5+x\right)=0\Leftrightarrow\left(x+3\right)\left(2x-2\right)=0\Leftrightarrow x=-3;x=1\)

e, \(\left(x+2\right)\left(3-4x\right)=\left(x+2\right)^2\Leftrightarrow\left(x+2\right)\left(3-4x-x-2\right)=0\)

\(\Leftrightarrow\left(x+2\right)\left(-5x+1\right)=0\Leftrightarrow x=-2;x=\dfrac{1}{5}\)

giúp mik với

a) (2^x)⁵ : 4³ = 8¹⁵ => 2^5x = 8¹⁵.4³ = (2³)¹⁵.(2²)³ = 2⁴⁵.2⁶ = 2⁵¹ => 5x = 51 => x = 10,2 

b) (3²)^x .9³ = 243⁹ => 3^2x = 243⁹ : 9³ = (3⁵)⁹ : (3²)³ = 3⁴⁵ : 3⁶ = 3³⁹ => 2x = 39 => x = 19,5

c) (1/125)³.5^x = 25⁵ => 5^x = 25⁵ : (1/125)³ = (5²)⁵ : (1/5³)³ = 5¹⁰ : (5^-3)³ = 5¹⁰ : 5^-9 = 5¹⁹ => x = 19

d) 1/81 : 3^x = 1/729 => 3^x = 1/81 : 1/729 = 1/3⁴.729 = 3^-4.3⁶ = 3² => x = 2

e) (5x - 2)⁴ = 16⁸ = (16²)⁴ = 256⁴

=> 5x - 2 = -256 ; 256 => 5x = -254 ; 258 => x = -50,8 ; 51,6

P/S : Thay x = 10,2 vào câu a , x = 19,5 vào câu b sẽ thấy điều hư cấu : 2^10,2 và 9^19,5.Ko thể tính được giá trị của 2 lũy thừa này.

692157

roois vãi

-Đăng tách câu hỏi bạn nhé.

rối thế bn

a, \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x\right)^2-\left(3x+3\right)^2=0\Leftrightarrow\left(4x-3x-3\right)\left(4x+2x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(6x+3\right)=0\Leftrightarrow x=-\frac{1}{2};x=3\)

b, \(\left(5x-4\right)^2-49x^2=0\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(-2x-4\right)\left(12x-4\right)=0\Leftrightarrow x=-2;x=\frac{1}{3}\)

c, \(5x^3-20x=0\Leftrightarrow5x\left(x^2-4\right)=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\Leftrightarrow x=0;x=\pm2\)

1: Ta có: \(16x^2-9\left(x+1\right)^2=0\)

\(\Leftrightarrow\left(4x-3x-3\right)\left(4x+3x+3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(7x+3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{7}{3}\end{matrix}\right.\)

2: Ta có: \(\left(5x-4\right)^2-49x^2=0\)

\(\Leftrightarrow\left(5x-4-7x\right)\left(5x-4+7x\right)=0\)

\(\Leftrightarrow\left(2x+4\right)\left(12x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-2\\x=\dfrac{1}{3}\end{matrix}\right.\)

3: Ta có: \(5x^3-20x=0\)

\(\Leftrightarrow5x\left(x-2\right)\left(x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)

\(a,=5x\left(x-1\right)+y\left(x-1\right)=\left(5x+y\right)\left(x-1\right)\\ b,=\left(x-y\right)^2-9=\left(x-y-3\right)\left(x-y+3\right)\)

a)5x²-5x+xy-y=(5x²-5x)+(xy-y)=5x(x-1)+y(x-1)=(x-1)(5x+y)

b)x²-2xy+y²-9=(x²-2xy+y²)-9=(x-y)²-3²=(x-y-3)(x-y+3)

Câu 2,3,4 nx thôi ạ. Câu 1 có bạn giúp r ạ 

1)\(\sqrt{4x^2+12x+9}=2-x\)

\(\Leftrightarrow\sqrt{\left(2x+3\right)^2}=2-x\)

\(\Leftrightarrow\left|2x+3\right|=2-x\)

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=2-x\\2x+3=x-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=-1\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=-5\end{matrix}\right.\)

\(\)

2)\(\sqrt{x^4+2x^2+1}=x^2+5x+4\)       ĐK:\(x\ge-1\)

\(\Leftrightarrow\sqrt{\left(x^2+1\right)^2}=x^2+5x+4\)

\(\Leftrightarrow\left|x^2+1\right|=x^2+5x+4\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+1=x^2+5x+4\\x^2+1=-x^2-5x-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=-3\\2x^2+5x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{3}{5}\\2\left(x+\dfrac{5}{4}\right)^2+\dfrac{15}{8}=0\left(voli\right)\end{matrix}\right.\)