(x²+1)²-6(x²+1)²+5

=  (x²+1)²(-6+1)+5

= -5(x²+1)²+5

= -5(x⁴+2x²+1-1)

= -5(x⁴+2x²)

= -5x⁴-10x²

\(\left(x^2+1\right)^2-6\left(x^2+1\right)^2+5\\ \Leftrightarrow\left(x^2+1\right)\left(-6+5\right)\\ \Leftrightarrow\left(x^2+1\right).-1\\ \Leftrightarrow-x^2-1\)

Đề sai à bn

đề tui đúng mà

Lời giải:

$x^3-6x^2+12x-8=0$

$\Leftrightarrow (x-2)^3=0$

$\Leftrightarrow x-2=0$

$\Leftrightarrow x=2$

\(x^3-6x^2+12x-8=0\)

`<=>x^3-3.x^3.2+3.x.2^2-2^3=0`

`<=>(x-2)^3=0`

`<=>x-2=0`

`<=>x=2`

\(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow x^3+3.x^2.3+3.x.3^2+3^3=0\)

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

`<=>x+3=0`

`<=>x=-3`

\(x^3+9x^2+27x+27=0\)

\(\Leftrightarrow x^3+3.3.x^2+3.9x+3^3=0\)

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

\(\Rightarrow x+3=0\Rightarrow x=-3\)

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

giúp vs

a⁵+b⁵=110

Cho mình xin câu trả lời chi tiết đc ko bạn???

`x-4` mũ `2=36`

TH1

`x-4=6`

`x=10`

TH2

`x-4=-6`

`x=-2`

Vậy `x=10` hoặc `x=-2`

(x - 4)² - 36 = 0

       (x - 4)² = 0 + 36

       (x - 4)² = 36

       (x - 4)² = 6² = (-6)²

TH1 : (x - 4)² = 6²

              x - 4 = 6

                   x = 6 + 4 

                   x = 10

TH2 : (x - 4)² = (-6)²

              x - 4 = -6

                   x = -6 + 4

                   x = -2

vậy x ϵ {-2;10}

Lời giải:

a. $9x^2y-3x^2y^2=3x^2y(3-y)$

b. $5x(x-2)-7(x-2)=(x-2)(5x-7)$

c. $2y(x-1)+3x-3=3y(x-1)+3(x-1)=(x-1)(3y+3)=3(x-1)(y+1)$

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

e. $x^2-4+y(x-2)=(x-2)(x+2)+y(x-2)=(x-2)(x+2+y)$

Ta có \(a^2+b^2+c^2\ge ab+bc+ca\)

<=> \(\left(a+b+c\right)^2\ge3\left(ab+bc+ca\right)\)

P = \(\dfrac{a}{bc\left(a+c\right)}+\dfrac{b}{ac\left(a+b\right)}+\dfrac{c}{ab\left(b+c\right)}\)

\(=\dfrac{a^2}{abc\left(a+c\right)}+\dfrac{b^2}{abc\left(a+b\right)}+\dfrac{c^2}{abc\left(b+c\right)}\)

\(\ge\dfrac{\left(a+b+c\right)^2}{abc\left(a+c\right)+abc\left(a+b\right)+abc\left(b+c\right)}\) (BĐT Cauchy - Schwarz) 

\(=\dfrac{\left(a+b+c\right)^2}{2abc\left(a+b+c\right)}=\dfrac{a+b+c}{2abc}=\dfrac{1}{2}\left(\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}\right)\)

\(\ge\dfrac{1}{2}.\dfrac{\left(1+1+1\right)^2}{ab+bc+ca}\) (BĐT Cauchy - Schwarz) 

\(=\dfrac{9}{2\left(ab+bc+ca\right)}\ge\dfrac{9}{\dfrac{2\left(a+b+c\right)^2}{3}}=\dfrac{27}{2\left(a+b+c\right)^2}\)