( 3x-1) ( x²+ 9) = (3x-1) (7x-10)

⇒( 3x-1) ( x²+ 9) - (3x-1) (7x-10) = 0

⇒( 3x-1) (( x²+ 9)-(7x-10)) = 0

⇒( 3x-1)(x²+9-7x+10)=0

⇒( 3x-1)(x²-7x+19)=0

⇒\(\left[{}\begin{matrix}3x-1=0\\x^2-7x+19=0\end{matrix}\right.\)

3x-1=0

⇒x=\(\dfrac{1}{3}\)

x²-7x+19=0

⇒ \(x^2-\dfrac{7}{2}x-\dfrac{7}{2}x+\left(\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}=0\)

vì \(\left(x-\dfrac{7}{2}\right)^2\ge0\); \(\dfrac{27}{4}>0\)

⇒ \(\left(x-\dfrac{7}{2}\right)^2+\dfrac{27}{4}>0\)

⇒ x vô nghiệm

Vậy x= \(\dfrac{1}{3}\)

\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Ta có: \(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+9-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+19\right)=0\)

\(\Leftrightarrow3x-1=0\)

hay \(x=\dfrac{1}{3}\)

khó quá

xin loi to h cau 3 lan het luot roi\

de mai nha

Bn giải câu của mih trc đc k ?

Tìm x: 4x^2(x-5)-(5-x)^2 =0 Mình cảm ơn

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

<=> \(\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

<=> \(\left(3x-1\right)\left(x^2-7x+12\right)=0\)

<=> \(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

<=> \(\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

<=> \(\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

<=>  3x -1 = 0 hoặc x - 3 = 0 hoặc x - 4 = 0

<=> x = 1/3 hoặc x = 3 hoặc x = 4 

Vậy S = { 1/3 ; 3; 4 }

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)[x\left(x-4\right)-3\left(x-4\right)]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\)

Tương đương với 1 trong 3 biểu thức trên bằng 0.

Giải ra 3 nghiệm là \(x=\frac{1}{3};x=4;x=3\)

3x + 1 chia hêt cho 2x - 1

2.(3x + 1) chia hết cho 2x - 1

6x + 2 chia hết cho 2x - 1

6x - 3 + 3 + 3 chia hết cho 2x - 1

3.(2x - 1) + 6 chia hết cho 2x - 1

=> 6 chia hết cho 2x - 1

=> 2x - 1 thuộc Ư(6) = {1 ; -1 ; 2 ; -2 ; 3 ; -3}

Ta có bảng sau :

b) xy + 3x = 0

x.( y + 3) = 0

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

Vậy nếu x = 0 hoặc y = -3

Thì xy + 3y = 0

\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)

Vậy tập nghiệm của ohuowng trình là \(S=\left\{\dfrac{1}{3};3;4\right\}\)

a: 7x+35=0

=>7x=-35

=>x=-5

b: \(\dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\)

=>8-x-8(x-7)=1

=>8-x-8x+56=1

=>-9x+64=1

=>-9x=-63

hay x=7(loại)

a, \(7x=-35\Leftrightarrow x=-5\)

b, đk : x khác 7 

\(8-x-8x+56=1\Leftrightarrow-9x=-63\Leftrightarrow x=7\left(ktm\right)\)

vậy pt vô nghiệm 

2, thiếu đề 

1.

\(a,7x+35=0\\ \Rightarrow7x=-35\\ \Rightarrow x=-5\\ b,ĐKXĐ:x\ne7\\ \dfrac{8-x}{x-7}-8=\dfrac{1}{x-7}\\ \Leftrightarrow\dfrac{8-x}{x-7}-\dfrac{8\left(x-7\right)}{x-7}-\dfrac{1}{x-7}=0\\ \Leftrightarrow\dfrac{8-x-8x+56-1}{x-7}=0\\ \Rightarrow-9x+63=0\\ \Leftrightarrow-9x=-63\\ \Leftrightarrow x=7\left(ktm\right)\)

2.đề thiếu

\(a.\left(3x+2\right)\left(x^2-1\right)=\left(9x^2-4\right)\left(x+1\right)\)

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

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

\(\Leftrightarrow2x=1\)

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

c: =>x-3=0

hay x=3

d: \(\Leftrightarrow\left(3x-1\right)\cdot\left(x^2+2-7x+10\right)=0\)

\(\Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)

hay \(x\in\left\{\dfrac{1}{3};3;4\right\}\)