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

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

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

\(\Leftrightarrow7x+1=0\)

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

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

3x-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^3-6x^2+x^2-2x+x-2>0

=>(x-2)(3x^2+x+1)>0

=>x-2>0

=>x>2

 \(3x^3-5x^2-x-2>0\)

\(\Leftrightarrow3x^3-6x^2+x^2-2x+x-2>0\)

\(\Leftrightarrow3x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)>0\)

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

Mặt khác: \(3x^2+x+1=2x^2+\left(x^2+x+1\right)\)

Ta lại có: \(x^2+x+1=x^2+2x\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{3}{4}=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\)

\(\Rightarrow3x^2+x+1>0\)

\(\Rightarrow x-2>0\)

\(\Leftrightarrow x>2\)

Vậy bpt có nghiệm là \(x>2\)

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{8609}{9}\\y=\dfrac{1366}{9}\end{matrix}\right.\)

minh biet

ta có : 

\(\left|x+1\right|+\left|x-1\right|=1+\left|\left(x-1\right)\left(x+1\right)\right|\)

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

\(\Leftrightarrow\left(\left|x-1\right|-1\right)\left(\left|x+1\right|-1\right)=0\Leftrightarrow\orbr{\begin{cases}\left|x-1\right|=1\\\left|x+1\right|=1\end{cases}}\)

\(\Leftrightarrow x\in\left\{-2,0,2\right\}\)

lx

lỗi r bn

lx

a: =>(x^2+x)^2-2(x^2+x)+(x^2+x)-2=0

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

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

=>x=-2 hoặc x=1

b: ĐKXĐ: x<>4; x<>1

PT =>\(\dfrac{x+3+3x-12}{x-4}=\dfrac{6}{1-x}\)

=>(4x-9)(1-x)=6(x-4)

=>4x-4x^2-9+9x=6x-24

=>-4x^2+13x-9-6x+24=0

=>-4x^2+7x+15=0

=>x=3(nhận) hoặc x=-5/4(nhận)