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}

${}^{}-{}^{}=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 $x+2=0$

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

\(\Leftrightarrow3x-x^2+2x=-x^2-2x-1\)

\(\Leftrightarrow-x^2+x^2+3x+2x+2x+1=0\)

\(\Leftrightarrow7x+1=0\)

\(\Leftrightarrow x=-\dfrac{1}{7}\)

Vậy \(S=\left\{-\dfrac{1}{7}\right\}\)

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

\(\Leftrightarrow3x-x^2+2x=-\left(x^2+2x+1\right)\)

\(\Leftrightarrow5x-x^2=-x^2-2x-1\)

\(\Leftrightarrow-x^2+x^2+5x+2x=-1\)

\(\Leftrightarrow7x=-1\)

\(\Leftrightarrow x=\left(-1\right)\div7\)

\(\Leftrightarrow x=-\dfrac{1}{7}\)

Ko bt đúng or sai :>

3x -x(x-2)= -(x+1)^2

<=>3x -x^2 +2x= -x^2-2x -1

<=> -x^2 +x^2 +5x +2x=-1

<=>7x= -1

<=>x= -1/7

Giải HPT:

\(\left\{{}\begin{matrix}3x-6y=1959\\x+7y=2019\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}3x-6y=1959\\3x+21y=6057\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}27y=4098\\x+7y=2019\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y\approx152\\x=955\end{matrix}\right.\)

Mik chỉ làm gần bằng đc thôi vì y là số thập phân.

1) \(A=\dfrac{\sqrt{2+\sqrt{3}}}{\sqrt{2}}=\dfrac{\sqrt{4+2\sqrt{3}}}{2}=\dfrac{\sqrt{\left(\sqrt{3}+1\right)^2}}{2}=\dfrac{\sqrt{3}+1}{2}\)

2) \(\left\{{}\begin{matrix}3x-6y=1959\\x+7y=2019\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}3x-6y=1959\\3x+21y=6057\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+7y=2019\\27x=4098\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8609}{9}\\y=\dfrac{1366}{9}\end{matrix}\right.\)

Lời giải:

b/

\(\frac{3x+5}{2x^2-5x+3}\geq 0\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} 3x+5\geq 0\\ 2x^2-5x+3>0\end{matrix}\right.\\ \left\{\begin{matrix} 3x+5\leq 0\\ 2x^2-5x+3<0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow \left[\begin{matrix} \left\{\begin{matrix} x\geq \frac{-5}{3}\\ x>\frac{3}{2}(\text{hoặc}) x< 1\end{matrix}\right.\\ \left\{\begin{matrix} x\leq \frac{-5}{3}\\ 1< x< \frac{3}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Leftrightarrow \left[\begin{matrix} x>\frac{3}{2}\\ \frac{-5}{3}\leq x< 1\end{matrix}\right.\ \)

c/

$2x^3+x+3>0$

$\Leftrightarrow 2x^2(x+1)-2x(x+1)+3(x+1)>0$

$\Leftrightarrow (x+1)(2x^2-2x+3)>0$

$\Leftrightarrow (x+1)[x^2+(x-1)^2+2]>0$

$\Leftrightarrow x+1>0$

$\Leftrightarrow x>-1$

2:

a: =>2x^2-4x-2=x^2-x-2

=>x^2-3x=0

=>x=0(loại) hoặc x=3

b: =>(x+1)(x+4)<0

=>-4<x<-1

d: =>x^2-2x-7=-x^2+6x-4

=>2x^2-8x-3=0

=>\(x=\dfrac{4\pm\sqrt{22}}{2}\)

x-1 + x-3 =1 <=> 2x -4=1 tu giai not