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

b) \(\Rightarrow\left(x-3\right)\left(5x-1\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\)

c) \(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\)

d) \(\Rightarrow\left(x-7\right)\left(3x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

\(a,\Leftrightarrow\left(x-2\right)\left(x+1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ b,\Leftrightarrow\left(x-3\right)\left(5x-1\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=\dfrac{1}{5}\end{matrix}\right.\\ c,\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{5}{3}\\x=\dfrac{4}{3}\end{matrix}\right.\\ d,\Leftrightarrow\left(x-7\right)\left(3x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=7\\x=\dfrac{2}{3}\end{matrix}\right.\)

a) X : \(\frac{7}{9}=\frac{32}{99}:\left(2+\frac{4}{9}\right)\) => X : \(\frac{7}{9}=\frac{32}{99}:\frac{22}{9}\)=> X : \(\frac{7}{9}=\frac{64}{81}\) => X = \(\frac{64}{81}.\frac{7}{9}=\frac{64}{63}\)

b) \(\frac{17}{99}:\left(2+\frac{3}{9}\right)=X:\frac{3}{9}\)=> \(\frac{17}{99}:\frac{7}{3}=X:\frac{1}{3}\)=> \(\frac{17}{231}=X:\frac{1}{3}\)=> X = \(\frac{17}{231}.\frac{1}{3}=\frac{17}{693}\)

Vậy...

a, \(x=\frac{64}{63}\)

b, \(\frac{17}{693}\)
 

a)X:O,(7)=O,(32):2,(4)

X:O,(7)=32/99:0,(4)+2

X:0,(7)=32/99:4/9+2

X:0,(7)=128/891+2

X=1910/891.7/9

X=1,667290186

b)0,(17)/2,(3)=X/0,(3)

=>X=17/9.3/9:7/3

X=17/63

Lời giải:
a.

$0< x< \frac{1}{4}+\frac{4}{5}$

$\Rightarrow 0< x< \frac{21}{20}$ hay $0< x< 1,05$

$\Rightarrow x=1$

b.

$\frac{4}{7}+\frac{3}{7}< x< \frac{5}{3}+\frac{2}{3}$
$\Rightarrow 1< x< \frac{7}{3}$
$\Rightarrow x=2$

cha loi di

a) x: 7/9 = 32/99 : 22/9

<=>x * 9/7= 32/99 * 9/22

<=>x* 9/7 = 16/121

<=>x=16/121 : 9/7

<=>x=112/1089

b) 17/99 : 7/3= x: 1/3

<=> 17/99 * 3/7 = x*3

<=> 17/231 = 3x

<=>x= 17/231 : 3

<=>x=17/693

b) \(x^3-x^2-x+1=0\Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\)

\(\Leftrightarrow x-1=0\)   hoặc   \(x+1=0\)

\(\Leftrightarrow x=1\)         hoặc    \(x=-1\)

c) \(x^2-6x+8=0\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)

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

a) \(x^3+x^2+x+1=0\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

(do \(x^2+1\ge1>0\))

a: Ta có: \(x^3+x^2+x+1=0\)

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

\(\Leftrightarrow x+1=0\)

hay x=-1

b: Ta có: \(x^3-x^2-x+1=0\)

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

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

c: Ta có: \(x^2-6x+8=0\)

\(\Leftrightarrow\left(x-2\right)\left(x-4\right)=0\)

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

_a)x=x

_b)x=x^1

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

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

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

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

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

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

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

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

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

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

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

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

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

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

\(\Leftrightarrow x+2=0\)

\(\Leftrightarrow x=-2\)

a. (x - 2²) - 1 = 0

<=> x - 4 - 1 = 0

<=> x = 5

b. 4 - (x - 2)² = 0

<=> 2² - (x - 2)² = 0

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

<=> x(4 - x) = 0

<=> \(\left[{}\begin{matrix}x=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)

d. (3x - 2)² - (2x + 3)² = 5(x + 4)(x - 4)

<=> (3x - 2 - 2x - 3)(3x - 2 + 2x + 3) = 5(x² - 16)

<=> (x - 5)(5x + 1) = 5x² - 80

<=> 5x² + x - 25x - 5 = 5x² - 80

<=> 5x² - 5x² + x - 25x = -80 + 5

<=> -24x = -75

<=> x = \(\dfrac{25}{8}\)

a)\(\left(x-2^2\right)-1=0\Rightarrow x-4-1=0\Rightarrow x=5\)