a, \(\left(2x+\dfrac{1}{2}\right)^2-\left(1-2x\right)^2=0\)

\(\Leftrightarrow\left[\left(2x+\dfrac{1}{2}\right)-\left(1-2x\right)\right]\cdot\left[\left(2x+\dfrac{1}{2}\right)+\left(1-2x\right)\right]=0\)

\(\Leftrightarrow\left(2x+\dfrac{1}{2}-1+2x\right)\cdot\left(2x+\dfrac{1}{2}+1-2x\right)=0\)

\(\Leftrightarrow\left[\left(4x+-\dfrac{1}{2}\right)\right]-\dfrac{3}{2}=0\)

\(\Leftrightarrow4x+\left(-\dfrac{1}{2}\right)=0\)

\(\Leftrightarrow4x=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{8}\)

b, \(x.\left(x-2\right)+2-x=0\)

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

\(\Leftrightarrow x.\left(x-2\right)-\left(x-2\right)=0\)

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

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

Vậy x = 2 hoặc 1

a, Ta có : \(4n^2.\left(n+2\right)+4n.\left(n+2\right)\)

\(=\left(n+2\right).\left(4n^2+4n\right)\)

\(=4n.\left(n+2\right).\left(n+1\right)\)

\(=4n.\left(n+1\right).\left(n+2\right)⋮4\)

\(n.\left(n+1\right).\left(n+2\right)\) là tích của ba số liên tiếp

\(\Rightarrow n.\left(n+1\right).\left(n+2\right)⋮2\) và \(3\)

mà \(n.\left(n+1\right).\left(n+2\right)⋮\left(2.3\right)\)

Vậy \(4n^2.\left(n+2\right)+4n.\left(n+2\right)⋮24\left(đpcm\right)\)

b,

+ Thực hiện phép tính :

Ta có : \(\dfrac{6n^2+n-1}{3n+2}=2n-1+\dfrac{1}{3n+2}\)

Để \(\left(6n+n-1\right)⋮\left(3n+2\right)\) thì \(\dfrac{1}{3n+2}\in Z\)

\(\Rightarrow3n+2\inƯ\left(1\right)\)

\(\Rightarrow3n+2\in\left\{\pm1\right\}\)

Ta có bảng sau :

Vậy n = -1

Hỗn, láo, mất dạy!

Vớ va vớ vẩn, toàn đăng thứ linh tinh

a, \(9x^2+6x-8\)

\(=\left(9x^2+6x+1\right)-9\)

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

\(=\left(3x+1-3\right).\left(3x+1+3\right)\)

\(=\left(3x-2\right).\left(3x+4\right)\)

b, \(x^7+x^2+1\)

\(=x^7+x^2+x+1-x\)

\(=\left(x^7-x\right)+\left(x^2+x+1\right)\)

\(=x.\left(x^6-1\right)+\left(x^2+x+1\right)\)

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

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

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

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

\(=\left(x^2+x+1\right).\left(x^5+x^2-x^4-x+1\right)\)

\(=\left(x^2+x+1\right).\left(x^5-x^4+x^2-x+1\right)\)

c, \(x^5+x^4+1\)

\(=x^5+x^4+x^2+x+1-x^2-x\)

\(=\left(x^5-x^2\right)+\left(x^4-x\right)+\left(x^2+x+1\right)\)

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

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

\(=\left(x^2+x+1\right).\left[x^2.\left(x-1\right)+x.\left(x-1\right)+1\right]\)

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

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

\(\sqrt{x+1}-\sqrt{x-2}=\sqrt{x+3}\)

\(\Leftrightarrow\sqrt{x+3}+\sqrt{x+2}=\sqrt{x+1}\)

ĐKXĐ : \(\left\{{}\begin{matrix}x+1\ge0\\x+3\ge0\\x-2\ge0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-1\\x\ge-3\\x\ge2\end{matrix}\right.\)\(\Leftrightarrow x\ge2\)

\(pt\Leftrightarrow\left(\sqrt{x+3}+\sqrt{x-2}\right)^2=x+1\)

\(\Leftrightarrow x+3+2.\sqrt{\left(x+3\right).\left(x-2\right)}+x-2=x+1\)

\(\Leftrightarrow2.\sqrt{\left(x+3\right).\left(x-2\right)}=-x\)

\(\Leftrightarrow\sqrt{\left(x+3\right).\left(x-2\right)}=\dfrac{-1}{2}x\)

\(\Leftrightarrow\left(x+3\right).\left(x-2\right)=\dfrac{1}{4}x^2\)

\(\Leftrightarrow4.\left(x+3\right).\left(x-2\right)=x^2\)

\(\Leftrightarrow\left(4x+12\right).\left(x-2\right)-x^2=0\)

\(\Leftrightarrow4x^2-8x+12x-24-x^2=0\)

\(\Leftrightarrow3x^2+4x-24=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-2+\sqrt{76}}{3}\left(\text{thỏa mãn}\right)\\x=\dfrac{-2-\sqrt{76}}{3}\left(\text{loại }\right)\end{matrix}\right.\)

Vậy ......

Thừa số là:

305,24:(4+3+6)=23,48

Tích đúng là:

436*23,48=10237,28

**** nhé Trọng Thái 

kẻ AH vuông góc BC

có B = 60⁰nên BH=\(\frac{1}{2}AB=8cm\)

ta có AH²=AC²-CH²=AB²-HB²

=>16²-8²=14²-CH²

=>192=196-CH

=>CH=2cm

BC=CH+HB=2+8=10cm

tick nha

\(P=\dfrac{4xy}{y^2-x^2}\div\left(\dfrac{1}{\left(y-x\right).\left(y+x\right)}+\dfrac{1}{\left(x+y\right)^2}\right)\)

\(P=\dfrac{4xy}{y^2-x^2}\div\dfrac{x+y+y-x}{\left(y-x\right).\left(x+y\right)^2}\)

\(P=\dfrac{4xy}{y^2-x^2}\div\dfrac{2y}{\left(y-x\right).\left(x+y\right)^2}\)

\(P=\dfrac{4xy}{\left(y-x\right).\left(y+x\right)}\cdot\dfrac{\left(y-x\right).\left(x+y\right)^2}{2y}\)

\(P=\dfrac{2x.\left(x+y\right)}{1}=2x.\left(x+y\right)\)

Xong rồi đó

\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)

\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)

\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)

\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)

\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)

\(\Rightarrow x-2009=0\Rightarrow x=2009\)

\(tan\cdot\left(x+\dfrac{\pi}{4}\right)+cot\cdot\left(2x-\dfrac{\pi}{3}\right)=0\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=-cot\cdot\left(2x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=cot\cdot\left(-2x+\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{2}+2x-\dfrac{\pi}{3}\right)\)

\(\Leftrightarrow tan\cdot\left(x+\dfrac{\pi}{4}\right)=tan\cdot\left(\dfrac{\pi}{6}+2x\right)\)

\(\Leftrightarrow x+\dfrac{\pi}{4}=\dfrac{\pi}{6}+2x+k\pi\)

\(\Leftrightarrow-x=\dfrac{-\pi}{12}+k\pi\)

\(\Leftrightarrow x=\dfrac{\pi}{12}-k\pi\left(k\in Z\right)\)