#)Giải :

Câu 1 :

5x(1 - 2x ) - 3x ( x+18) = 0 

<=> 5x - 10x^2 - 3x^2 - 54x = 0 

<=> -13x^2 - 49x = 0 

<=> x= 0 hoặc x = - 49/13

Vậy x có hai giá trị là 0 và - 49/13

#)Giải :

Câu 2 :

( 12x - 5 )( 4x - 1 ) + ( 3x - 7 )( 1 - 16x ) = 81

<=> 48x² - 32x + 5 - 48x + 115x - 7 = 81

<=> 83x - 2 = 81

<=> x = 1

Vậy x = 1

1)5x(1 - 2x ) - 3x ( x+18) = 0 
<=> 5x - 10x^2 - 3x^2 - 54x = 0 
<=> -13x^2 - 49x = 0 
<=> x= 0 hoặc x = -49/13

2) <=> 48x^2 - 12x - 20x + 5 + 3x - 48x^2 - 7 + 112x = 81 

<=> -32x + 115x = 81 + 2 
<=> 83x = 83 
<=> x = 1

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: \(5x^3-20x=0\)

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

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

Bài 2: Tính giá trị của biểu thức sau:

\(16x^2-y^2=\left(4x+y\right)\left(4x-y\right)\)

Thay \(\hept{\begin{cases}x=87\\y=13\end{cases}}\)

\(\Rightarrow\left(4.87+13\right)\left(4.87-13\right)=361.335=120935\)

Bài 4: Tìm x

a) \(9x^2+x=0\)

\(\Rightarrow x\left(9x+1\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x=0\\9x+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\frac{-1}{9}\end{cases}}\)

b) \(27x^3+x=0\)

\(\Rightarrow x\left(27x^2+1=0\right)\)

\(\Rightarrow\orbr{\begin{cases}x=0\\27x^2+1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\27x^2=\left(-1\right)\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x^2=\frac{-1}{27}\end{cases}}\)

Ta có: \(\frac{-1}{27}\) loại vì \(x^2\ge0\forall x\)

Vậy \(x=0\)

d: ta có: \(x^2-4x+4=9\left(x-2\right)\)

\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)

a: =>2^4x<2^28

=>4x<28

=>x<7

b: =>5^3x+3<5

=>3x+3<1

=>3x<-2

=>x<-2/3

a) \(16^x< 128^4\)

= (24)^x < (27)⁴

= 2^4x < 2²⁸

= 4x < 28

= x < 7 

Vậy \(x=\left\{0;1;2;3;4;5;6;\right\}\)

\(#Tuyết\)

a.   2x+\(\dfrac{4}{5}\)=0 hoặc 3x-\(\dfrac{1}{2}\)=0

2x=- 4/5 hoặc 3x=1/2

x=-2/5 hoặc x=\(\dfrac{1}{6}\)

b. x-\(\dfrac{2}{5}\)=0 hoặc x+\(\dfrac{4}{7}\)=0

x=2/5 hoặc x=-\(\dfrac{4}{7}\)

d. x(1+5/8-12/16)=1

\(\dfrac{7}{8}\)x=1=> x=8/7