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\(a,36-4x^2+20xy-25y^2\\ =36-\left(4x^2-20xy+25y^2\right)\\ =6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]\\ =6^2-\left(2x-5y\right)^2\\ =\left[6-\left(2x-5y\right)\right]\left[6+\left(2x-5y\right)\right]\\ =\left(6-2x+5y\right).\left(6+2x-5y\right)\)
a/
\(=6^2-\left[\left(2x\right)^2-2.2x.5y+\left(5y\right)^2\right]=\)
\(6^2-\left(2x-5y\right)^2=\left[6-\left(2x-5y\right)\right].\left[6+\left(2x-5y\right)\right]\)
Lời giải:
$M=(x^{10}-24x^9)-(x^9-24x^8)+(x^8-24x^7)-(x^7-24x^6)+(x^6-24x^5)-(x^5-24x^4)+(x^4-24x^3)-(x^3-24x^2)+(x^2-24x)-(x-24)+1$
$=x^9(x-24)-x^8(x-24)+x^7(x-24)-.....+x(x-24)-(x-24)+1$
$=(x-24)(x^9-x^8+x^7-...+x-1)+1$
$=0.(x^9-x^8+....+x-1)+1=1$
\(M=x^{10}-25x^9+25x^8-25x^7+...-25x^3+25x^2-25x+25\)
Ta thấy : \(x=24\Rightarrow x+1=25\)
\(\Rightarrow M=x^{10}-\left(x+1\right)x^9+\left(x+1\right)x^8-\left(x+1\right)x^7+...-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+\left(x+1\right)\)
\(M=x^{10}-x^{10}-x^9+x^9+x^8-x^8-x^7+...-x^4-x^3+x^3+x^2-x^2-x+x+1\)
\(\Rightarrow M=1\)
Vậy \(M=1\left(tạix=24\right)\)
M=x
10
−25x
9
+25x
8
−25x
7
+...−25x
3
+25x
2
−25x+25
Ta thấy :
x
=
24
⇒
x
+
1
=
25
x=24⇒x+1=25
⇒
M
=
x
10
−
(
x
+
1
)
x
9
+
(
x
+
1
)
x
8
−
(
x
+
1
)
x
7
+
.
.
.
−
(
x
+
1
)
x
3
+
(
x
+
1
)
x
2
−
(
x
+
1
)
x
+
(
x
+
1
)
⇒M=x
10
−(x+1)x
9
+(x+1)x
8
−(x+1)x
7
+...−(x+1)x
3
+(x+1)x
2
−(x+1)x+(x+1)
M
=
x
10
−
x
10
−
x
9
+
x
9
+
x
8
−
x
8
−
x
7
+
.
.
.
−
x
4
−
x
3
+
x
3
+
x
2
−
x
2
−
x
+
x
+
1
M=x
10
−x
10
−x
9
+x
9
+x
8
−x
8
−x
7
+...−x
4
−x
3
+x
3
+x
2
−x
2
−x+x+1
⇒
M
=
1
⇒M=1
Vậy
M
=
1
(
t
ạ
i
x
=
24
)
M=1(tạix=24)