Question

1)giải phương trình

3x-15 = 2x ( x- 5)

(2x+1)^2 - ( x-1)^2= 0

 

Answer 1

 

Answer 2

3x-15= 2x( x-5)

⇔ 3x -15 = 2x² -10x

⇔ 3x -2x² +10x -15 = 0

⇔ -2x² +13x -15 = 0

⇔ -2x² +10x +3x -15 = 0

⇔ -2x(x -5) +3(x-5) = 0

⇔ (x-5).(-2x +3) = 0

TH1: x-5 = 0 ⇔ x = 5

TH2: -2x+3 = 0 ⇔ x= 3/2

Vậy S= {5; 3/2}

Answer 3

${}^{}-{}^{}=0$$\iff \left[\left(2x+1\right)+\left(x-1\right)\right]\left[\left(2x+1\right)-\left(x-1\right)\right]=0$$\iff \left(2x+1+x-1\right)\left(2x+1-x+1\right)=0$$\iff 3x\left(x+2\right)=0$$\iff 3x=0$ hoặc $x+2=0$

Answer 4

1)giải phương trình

3x-15 = 2x ( x- 5)

⇒ 2x² -10x -3x +15 =0

⇔ 2x² -13x +15 =0

⇔\(\left(x-5\right)\left(x-\dfrac{3}{2}\right)=0\)

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

Vậy \(x=5\) và \(\dfrac{3}{2}\)

(2x+1)^2 - ( x-1)^2= 0

\(\Leftrightarrow4x^2+4x+1-x^2+2x-1=0\)

⇔\(3x^2+8x=0\)

\(\Leftrightarrow x\left(3x+8\right)=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\3x+8=0\Rightarrow x=\dfrac{-8}{3}\end{matrix}\right.\)

Vay.....

 

Answer 5

a) \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3x-15=2x^2-10x\)

\(\Leftrightarrow-2x^2+3x+10x=15\)

\(\Leftrightarrow-2x^2+13x=15\)

\(\Leftrightarrow x\left(13-2x\right)=15\)

TH₁: \(x=15\)

TH₂: \(13-2x=15\)

\(\Leftrightarrow2x=13-15\)

\(\Leftrightarrow2x=-2\)

\(\Leftrightarrow x=-1\)

Vậy phương trình có nghiệm là \(x=15\) hoặc \(x=-1\)

b) \(\left(2x+1\right)^2-\left(x+1\right)^2=0\)

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

\(\Leftrightarrow x\left(3x+2\right)=0\)

TH₁: \(x=0\)

TH₂: \(3x+2=0\)

\(\Leftrightarrow3x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{3}\)

Vậy phương trình có nghiệm là \(x=0\) hoặc \(x=-\dfrac{2}{3}\)

Answer 6

\(a\))\(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

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

\(b\))\(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

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

\(\Leftrightarrow\left(x+2\right)3x=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+2=0\\3x=0\end{matrix}\right.\)

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

Answer 7

a) Ta có: \(3x-15=2x\left(x-5\right)\)

\(\Leftrightarrow3\left(x-5\right)-2x\left(x-5\right)=0\)

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

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

Vậy: \(S=\left\{5;\dfrac{3}{2}\right\}\)

b) Ta có: \(\left(2x+1\right)^2-\left(x-1\right)^2=0\)

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

\(\Leftrightarrow3x\left(x+2\right)=0\)

mà 3>0

nên x(x+2)=0

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

Vậy: S={0;-2}