Question

Tìm x,biết:

a/\(x+5x^2=0\Leftrightarrow......\)
b/\(x+1=\left(x+1\right)^2\Leftrightarrow..........\)
c/\(x^3+x=0\Leftrightarrow.......\)
d/\(5x\left(x-2\right)-\left(2-x\right)=0\)
e/\(x\left(2x-1\right)+\frac{1}{3}-\frac{2}{3}x=0\Leftrightarrow........\)
g/\(x\left(x-4\right)+\left(x-4\right)^2=0\Leftrightarrow.....\)
h/\(x^2-3x=0\Leftrightarrow.....\)
i/\(4x\left(x+1\right)=8\left(x+1\right)\Leftrightarrow.....\)

Answer 1

Tìm x,biết:

a/ $x+5$$x^2=0$

$\iff x\; (\; 1\; +\; 5x\; )\; =\; 0$

\(\Leftrightarrow\) x = 0 hoặc 1 + 5x = 0

1) x = 0

2) 1+ 5x = 0 \(\Leftrightarrow\) x = \(\frac{-1}{5}\)

Vậy: S = \(\left\{0;\frac{-1}{5}\right\}\)

b/$x+1={\left(x+1\right)}^2$

\(\Leftrightarrow\) (x+1) - (x+1)² = 0

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

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

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

\(\Leftrightarrow\) x= -1 ; 0

Vậy: S=\(\left\{-1;0\right\}\)

c/ $x^3+x=0$

\(\Leftrightarrow\) x(x² + 1) = 0

\(\Leftrightarrow\) x = 0 hoặc x² + 1 = 0

Ta có : x² + 1 \(\ge\) 0 vs mọi x

Vậy: S = \(\left\{0\right\}\)

d/$5x\left(x-2\right)-\left(2-x\right)=$

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

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

\(\Leftrightarrow\) x - 2 = 0 hoặc 5x+ 1 = 0

\(\Leftrightarrow\) x = 2 hoặc x = \(\frac{-1}{5}\)

Vậy: S = \(\left\{\frac{-1}{5};2\right\}\)

g/ $x\left(x-4\right)+{\left(x-4\right)}^2=0$

$\iff \; (x\; -\; 4)(\; x+x-4)\; =\; 0$

$\backslash (\backslash Leftrightarrow\backslash )\; x\; -\; 4\; =\; 0\; hoặc\; 2x-4=0$

$\backslash (\backslash Leftrightarrow\backslash )$ x = 4 hoặc x = 2

Vậy: S= \(\left\{2;4\right\}\)

h/ $x^2-3x=0$

$\iff x\; (x-3)\; =\; 0$

\(\Leftrightarrow\) x = 0 hoặc x = 3

Vậy: S = \(\left\{0;3\right\}\)

Vậy: S= \(\left\{0;3\right\}\)
i/$4x\left(x+1\right)=8\left(x+1\right)$

$\iff$4x(x+1)-8(x+1) = 0

\(\Leftrightarrow\) 4(x+1) (x - 2) = 0

\(\Leftrightarrow\) x+1 = 0 hoặc x - 2 = 0

\(\Leftrightarrow\) x= -1 hoặc x = 2

Vậy: S=\(\left\{-1;2\right\}\)