Bài 4: Phương trình tích

\(2x^2-3x-5=0 \\ \Leftrightarrow2x^2+2x-5x-5=0\\ \Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\\ \Leftrightarrow\left(2x-5\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}2x-5=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}2x=5\\x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=-1\end{matrix}\right.\\ Vậy.S=\left\{\dfrac{5}{2};-1\right\}\)

\(2x^2-3x-5=0\)

\(\Leftrightarrow2x^2+2x-5x-5=0\)

\(\Leftrightarrow2x\left(x+1\right)-5\left(x+1\right)=0\)

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

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

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

Vậy \(x=\dfrac{5}{2};x=-1\) là các nghiệm của phương trình.

#\(Toru\)

b: =>(x+9)(x-3)(x+21)=0

=>x+9=0; x-3=0; x+21=0

=>x=-9; x=3;x=-21

g: =>(x+2)(x+3)=0

=>x=-2 hoặc x=-3

d: =>x^2+4x+4=0

=>(x+2)^2=0

=>x+2=0

=>x=-2

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

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

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

Trong các pt sau, pt tích là

A.(x-2)^2(x+2)=2

B.0=(x-2)^2(x+2)

C.(x-2)^2(x+2)=2(x+2)

D. (x-2)^2(x+2)+(x+2)

B. 0 = (x - 2)²(x + 2)

C. (x - 2)²(x + 2) = 2(x + 2)

\(2x^2+8x-10=0\)

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

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

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

D

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

\(\Leftrightarrow2x^2-2x-x^2-2x+4=0\)

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

\(\Leftrightarrow\left(x-2\right)^2=0\)

\(\Leftrightarrow x-2=0\)

\(\Leftrightarrow x=2\)

Vậy \(S=\left\{2\right\}\)

\(\left(x^2-2\right)-\left(x-2\right)\left(3x-7\right)=0\\ \Leftrightarrow x^2-2-\left(3x^2-6x-7x+14\right)=0\\ \Leftrightarrow x^2-2-3x^2+13x-14=0\\ \Leftrightarrow-2x^2+13x-16=0\\\Leftrightarrow \left[{}\begin{matrix}x=\dfrac{13+\sqrt{41}}{4}\\x=\dfrac{13-\sqrt{41}}{4}\end{matrix}\right.\)

xem lại đề ạ

=>19x=7

=>x=7/19

\(19x-7=0\)

\(\Leftrightarrow19x=7\)

\(\Leftrightarrow x=\dfrac{7}{19}\)