(3x+4)²-(3x-1).(3x+1)=49​

<=> 9x²+24x+16-(9x²-1)=49

<=>9x²+24x+16-9x²+1=49

<=>24x+17=49

<=>24x     =32

<=>x        =4/3

Vậy ...

​(x+2).(x^2-2x+4)-x.(x+3).(x-3)

=x³+8-x(x²-9)

=x³+8-x³+9x

=9x+8

(3x+4)²-(3x-1).(3x+1)=49​

<=> 9x²+24x+16-(9x²-1)=49

<=>9x²+24x+16-9x²+1=49

<=>24x+17=49

<=>24x     =32

<=>x        =4/3

Vậy ...

​(x+2).(x^2-2x+4)-x.(x+3).(x-3)

=x³+8-x(x²-9)

=x³+8-x³+9x

=9x+8

lớp 8 đến bây giờ đã học hằng đẳng thức chưa ạ ? Để mình đưa bạn hướng giải câu đầu ! 

câu sau : x^2 -9^2 = 0<=> (x-9)^2=0 <=>x-9=0 <=> x=9

X^2-81=0

=>X^2-9^2=0

=>(X-9)^2=0

=>X-9=0

=>X=9

(3x+4)\(^2\) - (3x-1)(3x+1)=49

=>\(9\text{x}^2+24x+16-9\text{x}^2+1\)\(=49\)

=>\(24\text{x}+17=49\)

=> 24x = 32

=> x = \(\dfrac{4}{3}\)

b) \(\left(3\text{x}-1\right)^2-\left(3\text{x}-2\right)^2=0 \)

\(=>9\text{x}^2-6\text{x}+1-9\text{x}^2+12\text{x}-4=0\)

\(=>6\text{x}-3=0\)

=> 6x = 3

=> x = \(\dfrac{1}{2}\)

c) \(\left(2\text{x}+1\right)^2-\left(x-1\right)^2=0\)

\(=>4\text{x}^2+4\text{x}+1-x^2+2\text{x}-1=0\)

=> \(3\text{x}^2+6\text{x}=0\)

=> \(3\text{x}\left(x+2\right)=0\)

=> 3x=0 hoặc x+2 = 0

+) 3x = 0 => x =0

+) x+2 = 0 => x = -2

a)

Theo bài ra ta có :

\(\left(x+7\right)\left(3x-1\right)-x^2+49=0\)

\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(x^2-49\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(3x-1\right)-\left(\left(x-7\right)\left(x+7\right)\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(3x-1-x+7\right)=0\)

\(\Leftrightarrow\left(x+7\right)\left(2x+6\right)=0\)

\(\Leftrightarrow\left[\begin{matrix}x+7=0\\2x+6=0\end{matrix}\right.\)

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

Vậy \(S=\left\{-3;-7\right\}\)

Chúc bạn học tốt =))

a/

<=>(x+7)(3x-1)-(x^2-7^2)=0

<=>(x+7)(3x-1)-(x-7)(x+7)=0

<=>(x+7)(3x-1-x+7)=0

<=>(x+7)(2x+6)=0

<=>x+7=0 hoặc 2x+6=0

<=>x=-7 2x=-6

<=> x=-3

=>S (-7;-3)

Cho hỏi cái đề câu b là đây hả:

\(\frac{x^2-3x+1}{x^2}-\frac{3}{x}=-4\)

ĐKXĐ: \(x\ne0\)

\(\Leftrightarrow\)\(\frac{x^2-3x+1}{x^2}-\frac{3x}{x^2}=\frac{-4x^2}{x^2}\)

\(\Leftrightarrow\)\(x^2-3x+1-3x=-4x^2\)

\(\Leftrightarrow x^2+4x^2-6x+1=0\)

\(\Leftrightarrow5x^2-6x+1=0\)

\(\Leftrightarrow5x^2-5x-x+1=0\)

\(\Leftrightarrow5x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow\left(5x-1\right)\left(x-1\right)=0\left[\begin{matrix}5x-1=0\Rightarrow x=0,2\\x-1=0\Rightarrow x=1\end{matrix}\right.\)

b: \(\Leftrightarrow\dfrac{7x+10}{x+1}\left(x^2-x-2-2x^2+3x+5\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(-x^2+2x+3\right)=0\)

\(\Leftrightarrow\left(7x+10\right)\left(x^2-2x-3\right)=0\)

=>(7x+10)(x-3)=0

hay \(x\in\left\{-\dfrac{10}{7};3\right\}\)

d: \(\Leftrightarrow\dfrac{13}{2x^2+7x-6x-21}+\dfrac{1}{2x+7}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow\dfrac{13}{\left(2x+7\right)\left(x-3\right)}+\dfrac{1}{\left(2x+7\right)}-\dfrac{6}{\left(x-3\right)\left(x+3\right)}=0\)

\(\Leftrightarrow26x+91+x^2-9-12x-14=0\)

\(\Leftrightarrow x^2+14x+68=0\)

hay \(x\in\varnothing\)

Ta có : 

\(\left(x+7\right)\left(3x-1\right)=49-x^2\)

\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1\right)=7^2-x^2\)

\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1\right)=\left(x+7\right)\left(7-x\right)\)

\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1\right)-\left(x+7\right)\left(7-x\right)=0\)

\(\Leftrightarrow\)\(\left(x+7\right)\left(3x-1-7+x\right)=0\)

\(\Leftrightarrow\)\(\left(x+7\right)\left(4x-8\right)=0\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x+7=0\\4x-8=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\4x=8\end{cases}}}\)

\(\Leftrightarrow\)\(\orbr{\begin{cases}x=-7\\x=\frac{8}{4}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-7\\x=2\end{cases}}}\)

Vậy \(x=2\) hoặc \(x=-7\)

Chúc bạn học tốt ~ 

Ta có: (x+7)(3x-1)=49-x^2

\(\Rightarrow\left(x+7\right)\left(3x-1\right)-49+x^2=0\)

\(\Rightarrow4x^2+20-56=0\)

\(\Rightarrow\left(2x\right)^2+2.2x.5+5^2-81=0\)

\(\Rightarrow\left(2x+5\right)^2=81\)

\(\Rightarrow2x+5=9\)hoặc \(2x+5=-9\)

\(\Rightarrow x=2\)hoặc \(x=-7\)

( 3x + 4 )² - ( 3x - 1 )( 3x + 1 ) = 49

<=> 9x² + 24x + 16 - ( 9x² - 1 ) - 49 = 0

<=> 9x² + 24x - 33 - 9x² + 1 = 0

<=> 24x - 32 = 0 <=> x = 4/3

\(\left(3x+4\right)^2-\left(3x-1\right)\left(3x+1\right)=49\)

\(\Leftrightarrow9x^2+24x+16-9x^2+1-49=0\)

\(\Leftrightarrow24x=32\)

\(\Leftrightarrow x=\frac{3}{2}\)

#H

 tìm x 

[3x+ 4] mũ 2-[3x-1][3x+1]=49

x=1,(3)

a) (3x + 4)² - (3x - 1).(3x + 1) = 49

=> (3x + 4).3x + (3x + 4).4 - (9x² - 1) = 49

=> 9x² + 12x + 12x + 16 - 9x² + 1 = 49

=> 24x + 17 = 49

=> 24x = 49 - 17

=> 24x = 32

=> \(x=\frac{32}{24}=\frac{4}{3}\)

Vậy \(x=\frac{4}{3}\)

b) (2x + 1)² - (x - 1)² = 0

=> (2x + 1 - x + 1).(2x + 1 + x - 1) = 0

=> (x + 2).3x = 0

=> (x + 2).x = 0

\(\Rightarrow\left[\begin{array}{nghiempt}x+2=0\\x=0\end{array}\right.\) \(\Rightarrow\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)

Vậy \(\left[\begin{array}{nghiempt}x=-2\\x=0\end{array}\right.\)

a)

➜9x²+24x+16-9x²+1=49

➜24x +17=49

⇒24x=49

⇒x=4/3

b)⇒4x²+4x+1-x²+2x-1=0

⇒3x²+6x=0

⇒3x(x+3)=0

⇒3x=0 hoặc x+3 =0

⇒x=0 hoặc x=-3