a

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

b

x^3 chứ: )

\(x^3+27+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow x^3+3^3+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\\ \Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\\ \Leftrightarrow\left(x+3\right).x.\left(x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)

a: 78x(x-97)-x+97=0

=>(x-97)(78x-1)=0

=>\(\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

b: \(\dfrac{2}{3}x\left(x^2-4\right)=0\)

=>\(x\left(x-2\right)\left(x+2\right)=0\)

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

c: \(\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

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

=>4x+8=0

=>x+2=0

=>x=-2

\(a,78x\left(x-97\right)-x+97=0\)

\(\Leftrightarrow78x\left(x-97\right)-\left(x-97\right)=0\)

\(\Leftrightarrow\left(x-97\right)\left(78x-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-97=0\\78x-1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=97\\x=\dfrac{1}{78}\end{matrix}\right.\)

\(b,\dfrac{2}{3}x\left(x^2-4\right)=0\)

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

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

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

\(c,\left(x+2\right)^2-\left(x-2\right)\left(x+2\right)=0\)

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

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

\(\Leftrightarrow\left(x+2\right)\cdot4=0\)

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

Vi (x-2).(x+4)=0

=>x-2=0    hoac x+4=0

        x=2               x=-4

Vay x=2 hoac x=-4

( x - 2 ) . ( x + 4 )=0

Th1:x-2=0

=>x=2

Th2:x+4=0

=>x=-4

vậy x=2 hoặc -4

\(3\left(2x-3\right)\left(3x+2\right)-2\left(x+4\right)\left(4x-3\right)+9x\left(4-x\right)=0\)

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

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

\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=3\\x=2\end{cases}}\)

xin lỗi toán lớp 8 thì mk chịu

x²–4=0 

=> x²=4

=>\(\orbr{\begin{cases}x=2\\x=-2\end{cases}}\)

=>\(^{x^2-2^2=0}\)

(x+2)(x-2)=0

\(\orbr{\begin{cases}x+2=0\\x-2=0\end{cases}}\)=>\(\orbr{\begin{cases}x=-2\\x=2\end{cases}}\)

a, 1 - 2x < 7

=> -2x < 6

=> x < -3

=> x thuộc {-4; -5; -6; ...}

b, \(\left(x-1\right)\left(x-2\right)>0\)

th1 :

\(\hept{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Rightarrow\hept{\begin{cases}x< 1\\x< 2\end{cases}\Rightarrow}x< 1\Rightarrow x\in\left\{0;-1;-2;...\right\}}\)

th2 :

\(\hept{\begin{cases}x-1>0\\x-2>0\end{cases}\Rightarrow\hept{\begin{cases}x>1\\x>2\end{cases}\Rightarrow}x>2\Rightarrow x\in\left\{3;4;5;...\right\}}\)

vậy_

c tương tự b

\(a.1-2x< 7\Leftrightarrow2x< 7+1=8\Leftrightarrow x< 8:2\Leftrightarrow x< 4\)

Vậy x < 4

\(b.\left(x-1\right)\left(x-2\right)>0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1>0;x-2>0\\x-1< 0;x-2< 0\end{cases}}\)

\(TH1\Leftrightarrow\orbr{\begin{cases}x-1>0\\x-2>0\end{cases}\Leftrightarrow\orbr{\begin{cases}x>0+1=1\\x>0+2=2\end{cases}\Rightarrow x>2}}\)

\(TH2\Leftrightarrow\orbr{\begin{cases}x-1< 0\\x-2< 0\end{cases}\Leftrightarrow\orbr{\begin{cases}x< 0+1=1\\x< 0+2=2\end{cases}\Rightarrow}}x< 2\)

Vậy \(x\ne2\)

A= x²+x-2-x+4

  =x²+2

Vì x² >=0 => x²+2>0

 Vậy pj]ơng trình vô nghiệm.

A= (x-1).(x+2)-(x-4)=0

^{x2+2X-X-2-X+4=0}

x2-2=0

x=\(\sqrt{2}\)

đặt A=(x+1)+(X+2)+(x+3)+....+(x+99)

=> A= x+1+x+2+x+3+....+x+100

=x+x+x+x+...+x+(1+2+3+4+..+99)( có 99x)

=> 99x+4950=0

=> 99x=-4950

=> x=-50

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

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

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

\(\Leftrightarrow\orbr{\begin{cases}x-\frac{1}{3}=\frac{1}{2}\\x-\frac{1}{3}=\frac{-1}{2}\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{5}{6}\\x=\frac{-1}{6}\end{cases}}}\)

Vậy x= 5/6 hoặc -1/6

b) - Nếu x=0 thì \(5^y=2^0+624=1+624=625=5^4\Rightarrow y=4\left(y\in N\right)\)

- Nếu x \(\ne\) 0 thì vế trái là số chẵn , vế phải là số lẻ \(\forall x;y\inℕ\) ( vô lí)

Vậy x=0, y=4

thank you bạn.