\(a,3\left|2x-4\right|-5=7\\ 3\left|2x-4\right|=12\\ \left|2x-4\right|=4\\ \left|2x-4\right|=\left(\pm2\right)^2\\ \Rightarrow\left[{}\begin{matrix}2x-4=2\\2x-4=-2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=6\\2x=2\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

Vậy x={3;1}

\(a,3\left|2x-4\right|-5=7\\ \Rightarrow3\left|2x-4\right|=12\\ \Rightarrow\left|2x-4\right|=4\\ \Rightarrow\left[{}\begin{matrix}2x-4=4\\2x-4=-4\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}2x=8\\2x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=4\\x=0\end{matrix}\right.\)

Vậy x={4;0}

So sánh

\(A=\dfrac{1999^{1999}+1}{1999^{1998}+1}\) ; \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}\)

Ta có: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>1\) ( vì tử > mẫu )

Do đó: \(B=\dfrac{1999^{2000}+1}{1999^{1999}+1}>\dfrac{1999^{2000}+1+1998}{1999^{1999}+1+1998}=\dfrac{1999^{2000}+1999}{1999^{1999}+1999}=\dfrac{1999.\left(1999^{1999}+1\right)}{1999.\left(1999^{1998}+1\right)}=\dfrac{1999^{1999}+1}{1999^{1998}+1}=A\)

Vậy B > A

Chúc bạn học tốt

x - 56/5 = -3x + 4

x + 3x = 4 + 56/5

4x = 76/5

x = 76/5 : 4

x = 19/5

`@` `\text {Ans}`

`\downarrow`

\(x-\dfrac{56}{5}=-3x+4\)

`\Rightarrow `\(x-\dfrac{56}{5}+3x-4=0\)

`\Rightarrow `\(\left(x+3x\right)+\left(-\dfrac{56}{5}-4\right)=0\)

`\Rightarrow `\(4x-\dfrac{76}{5}=0\)

`\Rightarrow `\(4x=\dfrac{76}{5}\)

`\Rightarrow `\(x=\dfrac{76}{5}\div4\)

`\Rightarrow `\(x=\dfrac{19}{5}\)

Vậy, `x=`\(\dfrac{19}{5}\)

\(\dfrac{1}{27}\cdot3^x=81\)

\(\Rightarrow3^x=81:\dfrac{1}{27}\)

\(\Rightarrow3^x=2187\)

\(\Rightarrow3^x=3^7\)

\(\Rightarrow x=7\)

Xét tam giác ADB và tam giác AEB có:

+góc DAB=góc DEB(=90^o)

+BD chung

+góc DBA=góc EBD(BD là tia pgiac)

=>tam giác ADB=tam giác EDB(ch-gn)

=>BA=BE(2 cạnh tương ứng)

b)Từ 2 tam giác ta chứng minh trên ta có:

DA=DE(2 cạnh t/ứng)

Mà BA=BE(Cmt)

=>BD là đường trung trực của AE

C A B D E

a) \(\left(2x-3\right)\left(2x+3\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)

b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)

\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)

\(\Rightarrow x\left(6x-2-15-6x\right)\)

\(\Rightarrow-16x=0\)

\(\Rightarrow x=0\)

d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)

\(\Rightarrow9x^2-4-4x+4=0\)

\(\Rightarrow9x^2-4x=0\)

\(\Rightarrow x\left(9x-4\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)

\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)

\(B=1\cdot2\cdot3+2\cdot3\cdot4+3\cdot4\cdot5+...+98\cdot99\cdot100\)

\(\Rightarrow4B=4\cdot\left(1\cdot2\cdot3+2\cdot3\cdot4+...+98\cdot99\cdot100\right)\)

\(\Rightarrow4B=1\cdot2\cdot3\cdot\left(4-0\right)+2\cdot3\cdot4\cdot\left(5-1\right)+3\cdot4\cdot5\cdot\left(6-2\right)+...+98\cdot99\cdot100\cdot\left(101-97\right)\)

\(\Rightarrow4B=1\cdot2\cdot3\cdot4+2\cdot3\cdot4\cdot5-1\cdot2\cdot3\cdot4-....+98\cdot99\cdot100\cdot101-97\cdot98\cdot99\cdot100\)

\(\Rightarrow4B=98\cdot99\cdot100\cdot101\)

\(\Rightarrow B=\dfrac{98\cdot99\cdot100\cdot101}{4}\)

\(\Rightarrow B=25\cdot98\cdot99\cdot101\)

B=1x2x3+2x3x4+...+98x99x100

=>4B=1x2x3x(4-0)+2x3x4x(5-1)+...+98x99x100x(101-97)

4B=1x2x3x4+2x3x4x5-1x2x3x4+...+98x99x100x101-97x98x99x100

4B=98x99x100x101

=>B=\(\dfrac{98\cdot99\cdot100\cdot101}{4}\)=24497550.

\(\dfrac{1}{2}-\dfrac{1}{2}\left(3-2x\right)=0\)

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

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

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

\(\Rightarrow2x-2=0\Rightarrow2x=2\Rightarrow x=1\)

x = 1 nha Nhật

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2008}-1\right)\left(\dfrac{1}{2009}-1\right)\\ =-\dfrac{1}{2}.\left(-\dfrac{2}{3}\right)...\left(-\dfrac{2007}{2008}\right)\left(-\dfrac{2008}{2009}\right)\\ =\dfrac{1}{2}.\dfrac{2}{3}...\dfrac{2007}{2008}.\dfrac{2008}{2009}\\ =\dfrac{1.2...2007.2008}{2.3...2008.2009}=\dfrac{1}{2009}\)

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)...\left(\dfrac{1}{2008}-1\right)\left(\dfrac{1}{2009}-1\right)\)

`=`\(\left(\dfrac{1}{2}-\dfrac{2}{2}\right)\left(\dfrac{1}{3}-\dfrac{3}{3}\right)...\left(\dfrac{1}{2008}-\dfrac{2008}{2008}\right)\left(\dfrac{1}{2009}-\dfrac{2009}{2009}\right)\)

`=`\(-\dfrac{1}{2}\cdot\left(-\dfrac{2}{3}\right)\cdot...\cdot\left(-\dfrac{2007}{2008}\right)\cdot\left(-\dfrac{2008}{2009}\right)\)

`=`\(-\dfrac{1}{2009}\)

\(x=-7\)