P=x3+y3+2xy=(x+y)3−3xy(x+y)+2xy=2013−601xy" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">P=x3+y3+2xy=(x+y)3−3xy(x+y)+2xy=2013−601xy

S=xy=x(201−x)" role="presentation" style="border:0px; direction:ltr; display:inline-table; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">S=xy=x(201−x)

1≤x≤200" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">1≤x≤200

S=200−(x−1)(x−200)≥0⇒Smin=200" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">S=200−(x−1)(x−200)≥0⇒Smin=200

x≤y⇒x≤100" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">x≤y⇒x≤100

S=100.101−(x−100)(x−101)≤100.101⇒Smax=100.101" role="presentation" style="border:0px; direction:ltr; display:inline-block; float:none; font-size:16.38px; line-height:0; margin:0px; max-height:none; max-width:none; min-height:0px; min-width:0px; padding:1px 0px; position:relative; white-space:nowrap; word-spacing:normal; word-wrap:normal" class="MathJax_CHTML mjx-chtml">S=100.101−(x−100)(x−101)≤100.101⇒Smax=100.101

\(x^6+2x^3+1=0\)

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

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

\(\Leftrightarrow x^3=\left(-1\right)^3\)

\(\Leftrightarrow x=-1\)

___________

\(x\left(x-5\right)=4x-20\)

\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)

\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=5\end{matrix}\right.\)

_____________

\(x^4-2x^2=8-4x^2\)

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

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

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

\(\Leftrightarrow x^2=2\)

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

_______________

\(\left(x^3-x^2\right)-4x^2+8x-4\)

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

\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)

Hô nhiều nhưng cố lên các bn ơi các bn sẽ làm được mà

Bạn ghi rõ dấu nhân với x ra , mk chẳng phân biệt đc

(X:10)+37=60<=>X:10=23=>X=230

25xX-15xX=72<=>10xX=72=>X=7,2

138-(Xx5)=38<=>Xx5=100=>X=20

(Xx9):52=18<=>Xx9:52=18=>x:52=2=>x=104

52xX+48xX=100=>100xX=100=>X=1

623xX-123xX=1000=>500xX=100=>X=2

Xx16+84xX=700=>100xX=700=>X=7

236xX-Xx36=2000=>200xX=2000=>X=10

216:X+34:X=10=>250:X=10=>X=25

2125:X-125:X=100=>2000:X=100=>X=20

Ta có:\(1001=1000+1=x+1\)

\(x^8-1001x^7+1001x^6+...+1001x^2-1001x+250\\ =x^8-\left(x+1\right)x^7+\left(x+1\right)x^6+...+\left(x+1\right)x^2-\left(x+1\right)x\\ =x^8-x^8-x^7+x^7+x^6+...+x^3+x^2-x^2-x+250\\ =-x+250=-1000+250\\ =-750\)

Vì số tận cùng = 0 nhân với số nào cũng tận cùng = 0

==> 1 x 2 x 3 x ... x 1000 =  . . . 0

Hay tích đó có tận cùng là 0

- Chúc bạn học tốt -

Tích \(1\cdot2\cdot3...999\cdot1000\)có số 1000 tận cùng là 0 nên nó tận cùng bằng 0

vì là 1000 có tận cùng là không mà khii nhân vẫn có tận cùng là0

ta có: 20=21-1=x-1

B=x⁶-20x⁵-20x⁴-20x³-20x²-20x+3 

= x⁶-(x-1)x⁵-(x-1)x⁴-(x-1)x³-(x-1)x²-(x-1)x+3 

=x⁶-x⁶+x⁵-x⁵+x⁴-x⁴+x³-x³+x²-x²+x+3

=x+3

=21+3

=24

a (X:10)+37=60

X:10=23

X=2,3

b 25xX-15xX=72

X(x25-15)=72

Xx10=72

X=7,2

c 138-(Xx5)=38

Xx5=100

X=20

d (Xx9):52=18

Xx9=936

X=104

e 52xX+48xX=100

Xx(52+48)=100

Xx100=100

X=1

f 623xX-123xX=1000

Xx(623-123)=1000

Xx500=1000

X=2

g Xx16+84xX=700

Xx(16+84)=700

Xx100=700

X=7

h 236xX-Xx36=2000

Xx(236-26)=2000

Xx200=2000

X=10

i 216:X+34:X=10

(216+34):X=10

250:X=10

X=25

f 2125:X-125:X=100

(2125-125):X=100

2000:X=100

X=20

x=21

=>x-1=20

B=x^6-x^5(x-1)-x^4(x-1)-...-x(x-1)+3

=x^6-x^6+x^5-x^5+x^5-...-x^2+x+3

=x+3

=21+3=24