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

\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)

\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)

\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)

\(a,\left(-\dfrac{2}{3}\right)^8+\left(-\dfrac{2}{3}\right)^8=2\left(-\dfrac{2}{3}\right)^8=2\cdot\dfrac{\left(-2\right)^8}{3^8}=\dfrac{2\cdot2^8}{3^8}=\dfrac{2^9}{3^8}=\dfrac{512}{6561}\)

\(\left(-\dfrac{2}{3}\right)^8+\left(-\dfrac{2}{3}\right)^8\)

\(=\left(\dfrac{2}{3}\right)^8+\left(\dfrac{2}{3}\right)^8\)

\(=2\left(\dfrac{2}{3}\right)^8\)

\(=2.\dfrac{256}{6561}\)

\(=\dfrac{512}{6561}\)

ĐKXĐ: \(x\ge-2\)

\(\Leftrightarrow x^3+3x\left(x+2\right)-4\left(x+2\right)\sqrt{x+2}=0\)

Đặt \(\sqrt{x+2}=y\ge0\) pt trở thành:

\(x^3+3xy^2-4y^3=0\)

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

\(\Leftrightarrow x=y\Leftrightarrow\sqrt{x+2}=x\) (\(x\ge0\))

\(\Leftrightarrow x^2=x+2\Leftrightarrow x=2\)

\(ĐKXĐ:x\ge-2\)

\(\Leftrightarrow x^3+3x^2+6x-4x\sqrt{x+2}-8\sqrt{x+2}=0\Leftrightarrow4x^2-4x\sqrt{x+2}+8x-8\sqrt{x+2}+x^3-x\left(x+2\right)=0\Leftrightarrow4x\left(x-\sqrt{x+2}\right)+8\left(x-\sqrt{x+2}\right)+x\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)=0\)\(\Leftrightarrow\left(x-\sqrt{x+2}\right)\left(x^2+x\sqrt{x+2}+4x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x+2}=0\left(1\right)\\x^2+x\sqrt{x+2}+4x+8=0\left(2\right)\end{matrix}\right.\) Từ (1) \(\Rightarrow x=\sqrt{x+2}\left(x\ge0\right)\Rightarrow x^2=x+2\Leftrightarrow x^2-x-2=0\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\left(TM\right)\\x=-1\left(L\right)\end{matrix}\right.\) Từ (2) \(\Rightarrow x^2+x\sqrt{x+2}+4x+8\ge\left(-2\right)^2+\left(-2\right)\sqrt{-2+2}+4\left(-2\right)+8=4>0\) \(\Rightarrow\) ko có x 

vậy...

Ta có: \(3x^2-x+2\)

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

\(=3\left(x^2-2\cdot x\cdot\dfrac{1}{6}+\dfrac{1}{36}+\dfrac{23}{36}\right)\)

\(=3\left(x-\dfrac{1}{6}\right)^2+\dfrac{23}{12}\ge\dfrac{23}{12}>0\forall x\)(đpcm)

Tham khảo:

Nguồn:Vietjack

I think that the best leisure activity for teenagers is playing sports. Firstly, they can find friends who have the same interests. Secondly, physical activities are a good way to relieve stress and reduce depression. Moreover, sports can help develop teamwork and leadership skills that may be very necessary for their future jobs. Their parents should give advice on choosing a suitable leisure activity; however, teenagers themselves will make the final decision

chụp ảnh được không

Phải đây ko

a: \(\dfrac{3x+2}{4}-\dfrac{3x+1}{3}=\dfrac{5}{6}\)

=>3(3x+2)-4(3x+1)=10

=>9x+6-12x-4=10

=>-3x+2=10

=>-3x=8

=>x=-8/3

b: \(\dfrac{x-1}{x+2}-\dfrac{x}{x-2}=\dfrac{9x-10}{4-x^2}\)

=>(x-1)(x-2)-x(x+2)=-9x+10

=>x^2-3x+2-x^2-2x=-9x+10

=>-5x+2=-9x+10

=>x=2(loại)

\(5x+\sqrt{9x^2-6x+1}\)

\(=5x+1-3x\)

=2x+1

Anh chỉ làm lời thôi nhá

Ta có 49 : 8 = 6 dư 1

Ta đổi 1 phút = 60 giây

Vậy  64 giây : 4 = 16 giây

Đ/s: 6 phút 16 giây

THi a mà gấp

49 phút 4 giây        :        8

1 phút=60 giây               6 phút 8 giây

            64 giây

             0 giây