a) Ta có: \(x^2+3x-10=0\)

\(\Leftrightarrow x^2+5x-2x-10=0\)

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

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

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

Vậy: S={-5;2}

b) Ta có: \(3x^2-7x+1=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{1}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{1}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}-\dfrac{37}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=\dfrac{37}{36}\)

\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{7}{6}=\dfrac{\sqrt{37}}{6}\\x-\dfrac{7}{6}=-\dfrac{\sqrt{37}}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{37}+7}{6}\\x=\dfrac{-\sqrt{37}+7}{6}\end{matrix}\right.\)

Vậy: \(S=\left\{\dfrac{\sqrt{37}+7}{6};\dfrac{-\sqrt{37}+7}{6}\right\}\)

c) Ta có: \(3x^2-7x+8=0\)

\(\Leftrightarrow3\left(x^2-\dfrac{7}{3}x+\dfrac{8}{3}\right)=0\)

mà 3>0

nên \(x^2-\dfrac{7}{3}x+\dfrac{8}{3}=0\)

\(\Leftrightarrow x^2-2\cdot x\cdot\dfrac{7}{6}+\dfrac{49}{36}+\dfrac{47}{36}=0\)

\(\Leftrightarrow\left(x-\dfrac{7}{6}\right)^2=-\dfrac{47}{36}\)(vô lý)

Vậy: \(x\in\varnothing\)

ko bt

<=>(3x-1)(x²+2-7x+10)=0

<=>(3x-1)(x²-7x+12)=0

<=>(3x-1)(x-4)(x-3)=0

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

<=> 3x=1 x=4 x=3

<=> x=1/3

S=(1/3;4;3)

Bài này trong bài 25 luyện tap

tìm x nha

c, \(3x^2-7x+10=0\)

\(\Leftrightarrow3x^2+3x-10x+10=0\)

\(\Leftrightarrow3x\left(x+1\right)-10\left(x+1\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)

d, \(2x\left(x-10\right)-x+10=0\)

\(\Leftrightarrow2x\left(x-10\right)-\left(x-10\right)=0\)

\(\Leftrightarrow\left(x-10\right)\left(2x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=10\\x=\dfrac{1}{2}\end{matrix}\right.\)

(3x-1).(x²+2)=(3x-1).(7x-10)

<=>(3x-1)(x²+2)-(3x-1)(7x-10)=0

<=>(3x-1)(x²+2-7x+10)=0

<=>(3x-1)(x²-7x+12)=0

<=>(3x-1)(x-4)(x-3)=0

Ta có:

+3x-1=0 => x=1/3

+x-4=0 => x=4

+x-3=0 => x=3

k nha..!

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

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

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

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

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

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

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

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

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

4(3x-2)-3(x-4)=7x-10

⇔12x-8-3x+12=7x-10

⇔2x=-14

⇔x=-7

các câu còn lại tương tự nha bạn, không khó đâu, cố gắng thì bạn sẽ làm được, chúc bạn học giỏi

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

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

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

\(\Leftrightarrow\left(3x-1\right)\left(x-2\right)\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1=0\\x-2=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=2\\x=5\end{matrix}\right.\)

Vậy ...............

Mộc Lung Hoa bài của Thư đúng rồi đó nhé, mình làm thiếu

(3x-1)(x²+2)-(3x-1)(7x-10)=0

(3x-1)(x²+2-7x+10)=0

(3x-1)(x²-7x+12)=0

(3x-1)(x²-3x-4x+12)=0

(3x-1)x(x-3)-4(x-3)=0

(3x-1)(x-3)(x-4)=0

->x=1/3 ; x=3 ;x=4