\(\left(x^2+1\right)^2-\left(2x+100\right)^2=0\)

\(\Leftrightarrow\left(x^2+1-2x-100\right)\left(x^2+1-2x+100\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2-2x-99=0\\x^2-2x+101=0\left(loại\right)\end{cases}}\)

\(\Leftrightarrow\left(x-1\right)^2=100\)

\(\Leftrightarrow\orbr{\begin{cases}x=11\\x=-9\end{cases}}\) ( thỏa mãn )

a ) \(4\left(x+5\right)-3\left|2x-1\right|=0\)

\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)

\(\Leftrightarrow\left|2x-1\right|=\frac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=\frac{4}{3}\left(x+5\right)\\2x-1=-\frac{4}{3}\left(x+5\right)\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-1=\frac{4}{3}x+\frac{20}{3}\\2x-1=-\frac{4}{3}x-\frac{20}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}\frac{2}{3}x=-\frac{23}{3}\\\frac{2}{3}x=-\frac{17}{3}\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{23}{2}\left(l\right)\\x=-\frac{17}{10}\left(n\right)\end{cases}}\)

Vậy \(x=-\frac{17}{10}\)

b ) \(\frac{2-x}{2007}-1=\frac{1-x}{2008}-\frac{x}{2009}\)

\(\Leftrightarrow\frac{2-x}{2007}+1=\left(\frac{1-x}{2008}+1\right)+\left(1-\frac{x}{2009}\right)\)

\(\Leftrightarrow\frac{2009-x}{2007}=\frac{2009-x}{2008}=\frac{2009-x}{2009}\)

\(\Leftrightarrow\left(2009-x\right)\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\right)=0\)

\(\Leftrightarrow2009-x=0\left(\frac{1}{2007}-\frac{1}{2008}-\frac{1}{2009}\ne0\right)\)

\(\Leftrightarrow x=2019\)

Vậy phương trình có nghiệm \(x=2019\)

c ) \(x^4+4x^2-5=0\)

\(\Leftrightarrow x^4-x^2+5x^2-5=0\)

\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+5=0\left(l\right)\\x=1\end{cases}}\)

            \(x=-1\)

Vậy \(x=1\) hoặc \(x=-1\)

Chúc bạn học tốt !!!

\(x^4+4x^2-5=0\)

\(\Leftrightarrow x^4-x^2+5x^2-5=0\)

\(\Leftrightarrow x^2\left(x^2-1\right)+5\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x^2+5\right)\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x^2+5=0\left(l\right)\\x=1\\x=-1\end{matrix}\right.\)

\(4\left(x+5\right)-3\left|2x-1\right|=0\)

\(\Leftrightarrow3\left|2x-1\right|=4\left(x+5\right)\)

\(\Leftrightarrow\left|2x-1\right|=\dfrac{4}{3}\left(x+5\right)\left(ĐK:x\ge-5\right)\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}\left(x+5\right)\\2x-1=-\dfrac{4}{3}\left(x+5\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-1=\dfrac{4}{3}x+\dfrac{20}{3}\\2x-1=-\dfrac{4}{3}x-\dfrac{20}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{2}{3}x=-\dfrac{23}{3}\\\dfrac{2}{3}x=-\dfrac{17}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{23}{2}\left(l\right)\\x=-\dfrac{17}{10}\left(n\right)\end{matrix}\right.\)

Vậy: \(x=-\dfrac{17}{10}\)

\(\dfrac{2-x}{2007}-1=\dfrac{1-x}{2008}-\dfrac{x}{2009}\\ \Leftrightarrow\dfrac{2-x}{2007}+1=\left(\dfrac{1-x}{2008}+1\right)+\left(1-\dfrac{x}{2009}\right)\\ \Leftrightarrow\dfrac{2009-x}{2007}=\dfrac{2009-x}{2008}+\dfrac{2009-x}{2009}\\ \Leftrightarrow\left(2009-x\right)\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\right)=0\\ \Leftrightarrow2009-x=0\left(\dfrac{1}{2007}-\dfrac{1}{2008}-\dfrac{1}{2009}\ne0\right)\\ \Leftrightarrow x=2019\)

Vậy phương trình có nghiệm \(x=2019\)

Lời giải:

a) 

PT \(\Leftrightarrow \frac{(x+2)^3}{8}-\frac{x^3+8}{2}=0\)

\(\Leftrightarrow (x+2)^3-4(x^3+8)=0\)

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

\(\Leftrightarrow (x+2)[(x+2)^2-4(x^2-2x+4)]=0\)

\(\Leftrightarrow (x+2)(-3x^2+12x-12)=0\)

\(\Leftrightarrow (x+2)(x^2-4x+4)=0\Leftrightarrow (x+2)(x-2)^2=0\Rightarrow x=\pm 2\)

b) Bạn kiểm tra lại xem có sai đề không?

1) =0

3)=(3⁷⁹⁸⁸-1)/2

3) Q=(3+1)(3^2+1)(3^4+1)....(3^3994+1)

=(3-1)(3+1)(3^2+1)(3^4+1)...(3^3994+1)

=(3^2-1)(3^2+1)(3^4+1)...(3^3994+1)

=(3^4-1)(3^4+1)...(3^3994+1)

=.........

=(3^3994-1)(3^3994+1)

=3^7988-1

quá khó vì em mới học lớp 6

a, \(\left(x-1\right).\left(x+2\right)\)\(>0\Rightarrow\orbr{\begin{cases}x-1< 0;x+2< 0\left(loai\right)\Rightarrow x< 1\\x-1>0;x+2>0\Rightarrow x>1;x>-2\end{cases}}\)

=> -2 < x < 1

Câu b và câu d làm tương tự nha bạn(Câu b thì xét khác dấu) 

a) a=  2 và 1

b)    =      7

c=     5600 và 7899

d  5 và 6 

Bài 1:

a) Ta có: 4x-20=0

\(\Leftrightarrow4\left(x-5\right)=0\)

mà \(4\ne0\)

nên x-5=0

hay x=5

Vậy: x=5

b) Ta có: 3-2x=3(x+1)-x-2

\(\Leftrightarrow3-2x=3x+3-x-2\)

\(\Leftrightarrow3-2x=2x+1\)

\(\Leftrightarrow3-2x-2x-1=0\)

\(\Leftrightarrow2-4x=0\)

\(\Leftrightarrow2\left(1-2x\right)=0\)

mà \(2\ne0\)

nên 1-2x=0

\(\Leftrightarrow2x=1\)

hay \(x=\frac{1}{2}\)

Vậy: Tập nghiệm \(S=\left\{\frac{1}{2}\right\}\)

c) Ta có: \(\frac{x+2}{2008}+\frac{x+3}{2007}+\frac{x+4}{2006}+\frac{x+2028}{6}=0\)

\(\Leftrightarrow\frac{x+2}{2008}+1+\frac{x+3}{2007}+1+\frac{x+4}{2006}+1+\frac{x+2028}{6}-3=0\)

\(\Leftrightarrow\frac{x+2010}{2008}+\frac{x+2010}{2007}+\frac{x+2010}{2006}+\frac{x+2010}{6}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\right)=0\)

mà \(\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}+\frac{1}{6}\ne0\)

nên x+2010=0

hay x=-2010

Vậy: Tập nghiệm S={-2010}

d) Ta có: 2x(x+3)+5(x+3)=0

\(\Leftrightarrow\left(x+3\right)\left(2x+5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\2x+5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\2x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{-5}{2}\end{matrix}\right.\)

Vậy: Tập nghiệm \(S=\left\{-3;\frac{-5}{2}\right\}\)

Bài 2:

ĐKXĐ: \(x\notin\left\{-1;1\right\}\)

Bài 6:

ĐKXĐ: \(x\notin\left\{1;2;-1;-2\right\}\)

Ta có: \(\frac{1}{x-1}+\frac{1}{x-2}=\frac{1}{x+2}+\frac{1}{x+1}\)

\(\Leftrightarrow\frac{x-2}{\left(x-1\right)\left(x-2\right)}+\frac{x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1}{\left(x+1\right)\left(x+2\right)}+\frac{x+2}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{x-2+x-1}{\left(x-1\right)\left(x-2\right)}=\frac{x+1+x+2}{\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow\frac{2x-3}{\left(x-1\right)\left(x-2\right)}-\frac{2x+3}{\left(x+1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{\left(2x-3\right)\left(x^2+3x+2\right)}{\left(x^2-1\right)\left(x^2-4\right)}-\frac{\left(2x+3\right)\left(x^2-3x+2\right)}{\left(x^2-1\right)\left(x^2-4\right)}=0\)

\(\Leftrightarrow2x^3+3x^2-5x-6-\left(2x^3-3x^2-5x+6\right)=0\)

\(\Leftrightarrow2x^3+3x^2-5x-6-2x^3+3x^2+5x-6=0\)

\(\Leftrightarrow6x^2-12=0\)

\(\Leftrightarrow6x^2=12\)

\(\Leftrightarrow x^2=2\)

hay \(x=\pm\sqrt{2}\)

Vậy: Tập nghiệm \(S=\left\{\sqrt{2};-\sqrt{2}\right\}\)