(3x-1)(x^2+2)=(3x-1)(7x-10)
< giải các phương trình > giup minh nhe cac ban
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( 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= \(\dfrac{1}{3}\)
\(\left(3x-1\right)\left(x^2+9\right)=\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2+9\right)-\left(3x-1\right)\left(7x-10\right)\\ \Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\\ \Leftrightarrow\left(3x-1\right)\left[x\left(x-4\right)-3\left(x-4\right)\right]=0\\ \Leftrightarrow\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}3x-1=0\\x-3=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\x=3\\x=4\end{matrix}\right.\)
Áp dụng đẳng thức: \(\frac{a}{b}=\frac{c}{d}=ad=bc\) để tìm x
\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
<=> \(\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
<=> \(\left(3x-1\right)\left(x^2-7x+12\right)=0\)
<=> \(\left(3x-1\right)\left(x^2-3x-4x+12\right)=0\)
<=> \(\left(3x-1\right)\left[x\left(x-3\right)-4\left(x-3\right)\right]=0\)
<=> \(\left(3x-1\right)\left(x-3\right)\left(x-4\right)=0\)
<=> 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 }
1. \(\Leftrightarrow\left(2x-1\right)\left(3x+1\right)< 0\)
\(\Rightarrow-\frac{1}{3}< x< \frac{1}{2}\)
2. \(\Leftrightarrow\left(x-2\right)\left(3-2x\right)>0\)
\(\Rightarrow\frac{3}{2}< x< 2\)
3. \(\Leftrightarrow\left(5x-3\right)^2>0\)
\(\Rightarrow x\ne\frac{3}{5}\)
4. \(\Leftrightarrow-3\left(x-\frac{1}{6}\right)-\frac{59}{12}< 0\)
\(\Rightarrow x\in R\)
5. \(\Leftrightarrow2\left(x-1\right)^2+5\ge0\)
\(\Rightarrow x\in R\)
6. \(\Leftrightarrow\left(x+2\right)\left(8x+7\right)\le0\)
\(\Rightarrow-2\le x\le-\frac{7}{8}\)
7.
\(\Leftrightarrow\left(x-1\right)^2+2>0\)
\(\Rightarrow x\in R\)
8. \(\Leftrightarrow\left(3x-2\right)\left(2x+1\right)\ge0\)
\(\Rightarrow\left[{}\begin{matrix}x\le-\frac{1}{2}\\x\ge\frac{2}{3}\end{matrix}\right.\)
9. \(\Leftrightarrow\frac{1}{3}\left(x+3\right)\left(x+6\right)< 0\)
\(\Rightarrow-6< x< -3\)
10. \(\Leftrightarrow x^2-6x+9>0\)
\(\Leftrightarrow\left(x-3\right)^2>0\)
\(\Rightarrow x\ne3\)
\(\left(3x-1\right)\left(x^2+2\right)=\left(3x-1\right)\left(7x-10\right)\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2\right)-\left(3x-1\right)\left(7x-10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2+2-7x+10\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-7x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x^2-4x-3x+12\right)=0\)
\(\Leftrightarrow\left(3x-1\right)[x\left(x-4\right)-3\left(x-4\right)]=0\)
\(\Leftrightarrow\left(3x-1\right)\left(x-4\right)\left(x-3\right)=0\)
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 :
2x - 1 | 1 | -1 | 2 | -2 | 3 | -3 |
x | 1 | 0 | 3/2 | -1/2 | 2 | -1 |
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