x^2+6x-8
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13 tháng 11 2021
\(x\left(x-2\right)+x-2=0\\ \Leftrightarrow x\left(x-2\right)+\left(x-2\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
\(x^2-2x+1=9\\ \Leftrightarrow\left(x-1\right)^2=9\\ \Leftrightarrow\left[{}\begin{matrix}x-1=-3\\x-1=3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x=4\end{matrix}\right.\)
\(7x^2=2x\\ \Leftrightarrow7x^2-2x=0\\ \Leftrightarrow x\left(7x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\7x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{2}{7}\end{matrix}\right.\)
\(x^2-6x=8\\ \Leftrightarrow x^2-6x-8=0\\ \left(x^2-6x+9\right)-17=0\\ \Leftrightarrow\left(x-3\right)^2-\sqrt{17^2}=0\\ \Leftrightarrow\left(x-3-\sqrt{17}\right)\left(x-3+\sqrt{17}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x-3-\sqrt{17}=0\\x-3+\sqrt{17}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3+\sqrt{17}\\x=3-\sqrt{17}\end{matrix}\right.\)
3 tháng 10 2021
\(a,\left(-\dfrac{2}{3}\right)^8+\left(-\dfrac{2}{3}\right)^8=2\left(-\dfrac{2}{3}\right)^8=2\cdot\dfrac{\left(-2\right)^8}{3^8}=\dfrac{2\cdot2^8}{3^8}=\dfrac{2^9}{3^8}=\dfrac{512}{6561}\)
NV
Nguyễn Việt Lâm
Giáo viên
26 tháng 2 2021
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:
\(x^3+3xy^2-4y^3=0\)
\(\Leftrightarrow\left(x-y\right)\left(x^2+xy+4y^2\right)=0\)
\(\Leftrightarrow x=y\Leftrightarrow\sqrt{x+2}=x\) (\(x\ge0\))
\(\Leftrightarrow x^2=x+2\Leftrightarrow x=2\)
26 tháng 2 2021
\(ĐKXĐ:x\ge-2\)
\(\Leftrightarrow x^3+3x^2+6x-4x\sqrt{x+2}-8\sqrt{x+2}=0\Leftrightarrow4x^2-4x\sqrt{x+2}+8x-8\sqrt{x+2}+x^3-x\left(x+2\right)=0\Leftrightarrow4x\left(x-\sqrt{x+2}\right)+8\left(x-\sqrt{x+2}\right)+x\left(x-\sqrt{x+2}\right)\left(x+\sqrt{x+2}\right)=0\)\(\Leftrightarrow\left(x-\sqrt{x+2}\right)\left(x^2+x\sqrt{x+2}+4x+8\right)=0\Leftrightarrow\left[{}\begin{matrix}x-\sqrt{x+2}=0\left(1\right)\\x^2+x\sqrt{x+2}+4x+8=0\left(2\right)\end{matrix}\right.\) 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...
LA
1
15 tháng 7 2021
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)
NA
3
20 tháng 6 2021
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
30 tháng 7 2023
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)
SD
3
24 tháng 3 2022
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
\(=x^2+6x+9-17=\left(x+3\right)^2-17=\left(x+3-\sqrt{17}\right)\left(x+3+\sqrt{17}\right)\)