a/ Ta có : \(49.x^2-4=0\)

\(\Rightarrow49x^2=4\)

\(\Rightarrow x^2=\frac{4}{49}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{2}{7}\end{cases}}\)

b/ \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)

\(\left(x+3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)=11\)

\(\Rightarrow\left(x^2+2.3.x+3^2\right)-\left(x^2-2^2\right)=11\)

\(\Rightarrow x^2+6x+9-x^2+4=11\)

\(\Rightarrow6x+13=11\)

\(\Rightarrow6x=11-13\)

\(\Rightarrow x=\frac{-2}{6}=\frac{-1}{3}\)

c/ \(\left(2x+1\right)^2-\left(x-3\right)^2-3\left(x+5\right)\left(x-5\right)=5\)

\(\Rightarrow\left(2x+1\right)\left(2x+1\right)-\left(x-3\right)\left(x-3\right)-3\left[\left(x+5\right)\left(x-5\right)\right]=5\)

\(\Rightarrow\left(4x^2+2.2x+1\right)-\left(x^2-2.3x+9\right)-3\left(x^2-25\right)\)\(=5\)

\(\Rightarrow\left(4x^2+4x+1\right)-\left(x^2-6x+9\right)-\left(3x^2-75\right)=5\)

\(\Rightarrow4x^2+4x+1-x^2+6x-9-3x^2+75=5\)

\(\Rightarrow\left(4x^2-x^2-3x^2\right)+\left(4x+6x\right)+\left(1-9+75\right)=5\)

\(\Rightarrow10x+67=5\)

\(\Rightarrow10x=5-67=-62\)

\(\Rightarrow x=\frac{-62}{10}=\frac{-31}{5}\)

d/ \(\left(3x+1\right)\left(3x-1\right)=8\)

\(\Rightarrow9x^2-1=8\)

\(\Rightarrow9x^2=8+1=9\)

\(\Rightarrow x^2=\frac{9}{9}=1\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)

Ai đó bấm hộ mình cái nút đúng đi!

Ta có : 49x² - 4 = 0

=> 49x² = 4

=> x² = 196

=> x² = 14² ; (-14)²

=> x = 14 ; -14

a) 49 . x² - 4 = 0

49 . x² = 4

7 ² . x² = 2²

(7 . x)² = 2²

7 . x = 2

x = 2 : 7

x = \(\frac{2}{7}\)

a) \(x^2-36=0\)

\(\Leftrightarrow x^2=36\)

\(\Leftrightarrow x=\pm\sqrt{36}=\pm6\)

b) \(\left(3x-5\right)^2-\left(x+6\right)^2=0\)

\(\Leftrightarrow\left(3x-5-x-6\right)\left(3x-5+x+6\right)=0\)

\(\Leftrightarrow\left(2x-11\right)\left(4x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=\frac{-1}{4}\end{cases}}\)

d) \(4x^3-36x=0\)

\(\Leftrightarrow4x\left(x^2-9\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4x=0\\x^2-9=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=\pm3\end{cases}}}\)

Vậy...

#)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

de qua

x.(2.x-1)+1/3-2/3.x=0

\(a,x^3+3x^2=4x+12\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+3=0\\x^2-4=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)

\(b,49x^2=\left(3x+2\right)^2\)

\(7x=3x+2\)

\(\Rightarrow7x-3x=2\)

\(\Rightarrow4x=2\)

\(\Rightarrow x=\frac{1}{2}\)

các câu còn lại tương tự nha

\(a,x^3+3x^2=4x+12\)

\(x^3+3x^2-4x-12=0\)

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

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

\(\Rightarrow\orbr{\begin{cases}x+3=0\\\left(x+2\right)\left(x-2\right)=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm2\end{cases}}\)

\(b,49x^2=\left(3x+2\right)^2\)

\(\Rightarrow\left(7x\right)^2=\left(3x+2\right)^2\)

\(\Rightarrow7x=3x+2\)

\(\Rightarrow7x-3x=2\)

\(\Rightarrow4x=2\)

\(\Rightarrow x=\frac{1}{2}\)

\(c,3x^2\left(x-5\right)+12\left(5-x\right)=0\)

\(3x^2\left(x-5\right)-12\left(x-5\right)=0\)

\(\left(x-5\right)\left(3x^2-12\right)=0\)

\(\Rightarrow3.\left(x-5\right)\left(x^2-4\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=\pm2\end{cases}}}\)

\(d,x^2\left(x-5\right)+45-9x=0\)

\(x^2\left(x-5\right)+9\left(5-x\right)=0\)

\(x^2\left(x-5\right)-9\left(x-5\right)=0\)

\(\left(x-5\right)\left(x^2-9\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x-5=0\\x^2-9=0\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=\pm3\end{cases}}\)

b: \(3x^2-2x-1=0\)

=>\(3x^2-3x+x-1=0\)

=>\(\left(x-1\right)\left(3x+1\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{3}\end{matrix}\right.\)

a: Bạn ghi lại đề đi bạn

a: \(x^3-4x^2-x+4=0\)

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

=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)

=>\(\left(x-4\right)\left(x^2-1\right)=0\)

=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)

=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)

\(\left(3x+2\right)-\left(x-1\right)=49x+1\)
\(\Leftrightarrow3x+1-x+1-49x-1=0\)
\(\Leftrightarrow-47x-1=0\)
\(\Leftrightarrow-47x=1\)
\(\Leftrightarrow x=-\dfrac{1}{47}\)

Ta có : (3x+2) - (x-1) = 49x + 1

\(\Leftrightarrow\) 3x + 2 - x +1 = 49x + 1

\(\Leftrightarrow\) 2x + 3 = 49x +1

\(\Leftrightarrow\) 49x + 1 - 2x -3 = 0

\(\Leftrightarrow\) 47x - 2 = 0

\(\Leftrightarrow\) x = \(\dfrac{2}{47}\)

Chuẩn 100%

Làm 1 câu thuii nha mik nhát quá!! nhưng các bài còn lại tương tự nha!!

a. \(\left(x+1\right)\left(3-x\right)=0\)

\(\Rightarrow\orbr{\begin{cases}x+1=0\\3-x=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

Vậy..

hok tốt!!

\(\left(x+1\right)\left(3-x\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\3-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-1\\x=3\end{cases}}}\)

vậy x=-1 hoặc x=3

\(\left(x-2\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\2x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\2x=1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=\frac{1}{2}\end{cases}}}\)

vậy x=2 hoặc x=1/2

câu c tương tự

\(\left(x^2+1\right)\left(81-x^2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\81-x^2=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x^2=-1\\x^2=81\end{cases}}}\Leftrightarrow\orbr{\begin{cases}x\in\varnothing\\x=\pm9\end{cases}}\)

vậy x=9 hoặc x=-9

Những cách dành cho trâu bò xD

\(a,\left(x+1\right)\left(3-x\right)=0\)

\(< =>3x-x^2+3-x=0\)

\(< =>2x-x^2+3=0\)

\(< =>-x^2+2x+3=0\)

Ta có : \(\Delta=2^2-4.\left(-1\right).3=4+12=16\)

Vì \(\Delta>0\)nên pt có 2 nghiệm phân biệt 

\(x_1=\frac{-2+\sqrt{16}}{-2}=\frac{2}{-2}=-1\)

\(x_2=\frac{-2-\sqrt{16}}{-2}=\frac{-6}{-2}=3\)

Vậy ...