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/$4$

a) y(x²-y²)(x²+y²)-y(x⁴-y⁴)=y[(x²)²-(y²)²] - y(x⁴-y⁴)=y(x⁴-y⁴)-y(x⁴-y⁴)=0

vậy giá trị biểu thức không phụ thuộc vào biến (đpcm)

b) \(\left(\frac{1}{3}+2x\right)\left(4x^2-\frac{2}{3}x+\frac{1}{9}\right)-\left(8x^3-\frac{1}{27}\right)\)

\(=\left[\left(2x\right)^3+\left(\frac{1}{3}\right)^3\right]-\left(8x^3-\frac{1}{27}\right)=8x^3+\frac{1}{27}-8x^3+\frac{1}{27}=\frac{1}{54}\)

vậy giá trị biểu thức không phụ thuộc vào biến (đpcm)

c) (x - 1)^3 - (x - 1)(x^2 + x + 1) - 3(1 - x)x

= (x - 1)(x^2 + x + 1) - (x - 1)(x^2 + x + 1) - 3x(1 - x)

= x^3 - 3x^2 + 3x - 1 - x^3 + 1 - 3x + 3x^2

= 0 (đpcm)

a) sửa đề: \(\frac{x^2}{\left(x-y\right)\left(x-z\right)}+\frac{y^2}{\left(y-x\right)\left(y-z\right)}+\frac{z^2}{\left(z-x\right)\left(z-y\right)}\)

=\(\frac{-x^2\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-y^2\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}+\frac{-z^2\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(-\frac{\left\lbrace x^2\left(y-z\right)+y^2\left(z-x\right)+z^2\left(x-y\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

xét tử số:

Tử=\(x^2y-x^2z+y^2z-y^2x+z^2x-z^2y\)

=\(x^2\left(y-z\right)-x\left(y^2-z^2\right)+yz\left(y-z\right)\)

=\(x^2\left(y-z\right)-x\left(y-z\right)\left(y+z\right)+yz\left(y-z\right)\)

=\(\left(y-z\right)\left\lbrace x^2-x\left(y+z\right)+yz\right\rbrace\)

=\(\left(y-z\right)\left\lbrace x\left(x-y\right)-z\left(x-y\right)\right\rbrace\)

=\(\left(y-z\right)\left(x-y\right)\left(x-z\right)\)

=\(-\left(x-y\right)\left(y-z\right)\left(z-x\right)\)

thay lại vào biểu thức cũ:

\(\Rightarrow-\frac{\left\lbrace-\left(x-y\right)\left(y-z\right)\left(z-x\right)\right\rbrace}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=\frac{\left(x-y\right)\left(y-z\right)\left(z-x\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)

=\(1\)

b) \(\frac{1}{\left(a-b\right)\left(b-c\right)}+\frac{1}{\left(b-c\right)\left(c-a\right)}+\frac{1}{\left(c-a\right)\left(a-b\right)}\)

=\(\frac{\left(c-a\right)+\left(a-b\right)+\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(c-a\right)}=0\)

a) \(\frac{5-x}{4x^2-8x}\) + \(\frac{7}{8x}\) = \(\frac{x-1}{2x\left(x-2\right)}\) +\(\frac{1}{8x-16}\)                               ĐKXĐ : x #0, x#2, x#-2

<=> \(\frac{5-x}{4x\left(x-2\right)}\) + \(\frac{7}{8x}=\frac{x-1}{2x\left(x-2\right)}\) + \(\frac{1}{8\left(x-2\right)}\)

<=> \(\frac{2\left(5-x\right)}{8x\left(x-2\right)}+\frac{7\left(x-2\right)}{8x\left(x-2\right)}=\frac{4\left(x-1\right)}{8x\left(x-2\right)}+\frac{x}{8x\left(x-2\right)}\)

=> 10 - 2x + 7x - 14 = 4x - 4 + x

<=>-2x + 7x - 4x + x  = -4 - 10 + 14

<=>x=-14

\(a,\left(2x^2+1\right)+4x>2x\left(x-2\right)\)

\(\Leftrightarrow2x^2+1+4x>2x^2-4x\)

\(\Leftrightarrow4x+4x>-1\)

\(\Leftrightarrow8x>-1\)

\(\Leftrightarrow x>-\frac{1}{8}\)

\(b,\left(4x+3\right)\left(x-1\right)< 6x^2-x+1\)

\(\Leftrightarrow4x^2-4x+3x-3< 6x^2-x+1\)

\(\Leftrightarrow4x^2-x-3< 6x^2-x+1\)

\(\Leftrightarrow4x^2-6x^2< 1+3\)

\(\Leftrightarrow-2x^2< 4\)

\(\Leftrightarrow x^2>2\)

\(\Leftrightarrow x>\pm\sqrt{2}\)