\(\Leftrightarrow2\left(x+1\right)^3=56\Leftrightarrow\left(x+1\right)^3=28\Leftrightarrow\)

=5*[55+13+32]

=5*100                     [* là dấu nhân]

=500

5x55+13x5+32

=5x(55+13+32)

=5x100

=500

5 x 55 + 13 x 5 + 5 x 32 = 5 x ( 55 + 13 + 32 )

                                       = 5 x 100

                                       = 500

Chúc học tốt!

a)x^m+4+x^m+3-x-1

=(x^m+4-x)+(x^m+3-1)

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

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

Với x=-2 ta có:... bn tự thay

b)x⁶-x⁴+2x³+2x²=x⁶-2x⁵+2x⁴+2x⁵-4x⁴+4x³+x⁴-2x³+2x²

=x⁴(x²-2x+2)+2x³(x²-2x+2)+x²(x²-2x+2)

=(x⁴+2x³+x²)(x²-2x+2)

=[x²(x²+2x+1)](x²-2x+2)

=x²(x+1)²(x²-2x+2)

Với x=-2 bn tự thay nhé h mk bận

\(A=3x-x^2\)

\(=-\left(x^2-2.x.\frac{3}{2}+\left(\frac{3}{2}\right)^2-\frac{9}{4}\right)\)

\(=-\left(\left(x-\frac{3}{2}\right)^2-\frac{9}{4}\right)\)

\(=\frac{9}{4}-\left(x-\frac{3}{2}\right)^2\ge\frac{9}{4}\)

Min A = \(\frac{9}{4}\)khi \(x-\frac{3}{2}=0=>x=\frac{3}{2}\)

\(B=25+2x-x^2\)

\(=-\left(x^2-2x+1-26\right)\)

\(=-\left(\left(x-1\right)^2-26\right)\)

\(=26-\left(x-1\right)^2\ge26\)

Min A = 26 khi \(x-1=0=>x=1\)

\(C=x^2-5x+19\)

\(=x^2-2.x.\frac{5}{2}+\left(\frac{5}{2}\right)^2+\frac{51}{4}\)

\(=\left(x+\frac{5}{2}\right)^2+\frac{51}{4}\ge\frac{51}{4}\)

Min C = \(\frac{51}{4}\)khi \(x+\frac{5}{2}=0=>x=\frac{-5}{2}\)

@@@ nha các bạn . Thanks

cảm ơn bạn nhiều lắm

1/x+x:12-30=61

  x+x:12      =61+30

 x               =84

\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\\ \dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\\ \Rightarrow x=-5\)

\(\dfrac{-1}{4}< \dfrac{x}{24}< \dfrac{-1}{6}\)
\(\Rightarrow\dfrac{-6}{24}< \dfrac{x}{24}< \dfrac{-4}{24}\)
\(\Rightarrow-6< x< -4\)
\(\Rightarrow x=-5\)

`(-1)/4 < x/24 < (-1)/6`

`(-6)/24 < x/24 < (-4)/24`

`->x=-5`

( 11 x 9 - 100 +1 ) x ( 1 x 2 x ... x10 ) = (99 + 1 - 100) x 1 x 2 x .... x 10

                                                             =  0 x 1 x 2 x ....... x 10

                                                             = 0

Ta có 

(11x9 -100+1)x (1x2x3x..x100)

= 0x ( 1x2x3x..x100)

=0

Minh lon la nhan 10 moi dung

Đề ko rõ ràng \(\sqrt{x^2}+x+\dfrac{1}{4}\) hay \(\sqrt{x^2+x+\dfrac{1}{4}}\)??

m??