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Bài 1:a) \(\left(x-3\right)^3=x^3-9x^2+27x-27\)
b)\(\left(\frac{1}{5}x-1\right)\left(\frac{1}{5}x+1\right)=\frac{1}{25}x^2-1\)
Bài 3:
a)x(x-6) + 10x - 60 =0
\(\Rightarrow x^2-6x+10x-60=0\)
\(\Rightarrow x^2+4x+60=0\)
\(\Rightarrow\left(x+2\right)^2+54=0\)
Vì \(\left(x+2\right)^2+54\ge54\forall x\)
\(\Rightarrow\)không có giá trị nào của x thỏa mãn.
a) -25x6 - y8 + 10x3y4 = -25x6 + 10x3y4 - y8
= - ( 25x6 - 10x3y4 + y8 )
= - [ ( 5x3 )2 - 2 . 5x3y4 + ( y4 )2 ]
= - ( 5x3 - y4 )2
b) \(\dfrac{1}{4}\)x2 - 5xy + 25y2 = (\(\dfrac{1}{2}\)x)2 - 2 . \(\dfrac{1}{2}\) x . 5y + ( 5y )2
= (\(\dfrac{1}{2}\) x - 5y )2
c) ( x - 5 )2 - 16 = ( x - 5 )2 - 42
= ( x - 5 - 4 ) . ( x - 5 + 4 )
= ( x - 9 ) . ( x - 1 )
d) 25 - ( 3 - x )2 = 52 - ( 3 - x )2
= ( 5 - 3 + x ) . ( 5 + 3 - x )
= ( x + 2 ) . ( 8 - x )
\(5x^2y^3-25x^2y^2+10x^2y^4=5x^2y^2\left(y-5+2y^2\right)\)
\(12a^4-24a^2b^2-6ab=6a\left(2a^3-4ab^2-3b\right)\)
mk chỉnh đề
\(-25x^6-y^8+10x^3y^4=-\left(5x^3-y^4\right)^2\)
\(25-\left(3-x\right)^2=\left(5-3+x\right)\left(5+3-x\right)=\left(2+x\right)\left(8-x\right)\)
a) \(x^3-5x^2+8x-4=\left(x^3-x^2\right)-4\left(x^2-x\right)+4\left(x-1\right)=x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-4x+4\right)=\left(x-1\right)\left(x-2\right)^2\)
b) \(A=5x\left(2x-3\right)+4\left(2x-3\right)+7\) chia hết cho 2x-3 => 7 chia hết cho 2x -3
=> 2x -3 thuộc U(7) ={-7;-1;1;7}
+2x-3 =-7 => x =-2
+2x-3 =-1 => x =1
+2x-3 =1 => x =2
+2x -3 =7 => x =5
Bài 1
A,7x − 6x 2 − 2 = −(6x 2 − 7x + 2)
= −(6x 2 − 3x − 4x + 2)
= −[3x(2x − 1) − 2(2x − 1)] = −(3x − 2)(2x −1)
b,\(2x^2+3x-5\)
=\(2x^2-2x+5x-5\)=\(2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\)
a/ x4 +5x3 +10x-4
=(x4- 4)+(5x3 + 10x)
=(x2+2) (x2-2) + 5x(x2 +2 )
=(x2+2)(x2 -2 +5x)
b/x5 - x4 +x3 -x2 +x-1
=x4(x-1)+x3(x-1)+(x-1)
=(x-1)(x4+x3+1)
a) \(\left(x+y\right)^2-\left(x+y\right)\)
\(=\left(x+y\right).\left(x+y-x-y\right)\)
\(=x+y.\)
c) \(x^3-x\)
\(=x.\left(x^2-1\right).\)
d) \(x^2-36\)
\(=x^2-6^2\)
\(=\left(x-6\right).\left(x+6\right)\)
e) \(x^2-3\)
\(=x^2-\left(\sqrt{3}\right)^2\)
\(=\left(x-\sqrt{3}\right).\left(x+\sqrt{3}\right)\)
h) \(x^3-27.b^3\)
\(=\left(x^3-27\right).b^3\)
\(=\left(x^3-3^3\right).b^3\)
\(=\left(x-3\right).\left(x^2+3x+3^2\right).b^3\)
\(=\left(x-3\right).\left(x^2+3x+9\right).b^3\)
Chúc bạn học tốt!
Mấy bài kia phá tung tóe rồi rút gọn hết sức xong thay x vào, làm câu c thôi nhé:
c) \(C=x^{14}-10x^{13}+10x^{12}-10x^{11}+...+10x^2-10x+10\)
riêng câu này ta thay x = 9 vào luôn, vậy ta có:
\(C=9^{14}-10\cdot9^{13}+10\cdot9^{12}-10\cdot9^{11}+...+10\cdot9^2-10\cdot9+10\)
\(=9^{14}-\left(9+1\right)\cdot9^{13}+\left(9+1\right)\cdot9^{12}-\left(9+1\right)\cdot9^{11}+...+\left(9+1\right)\cdot9^2-\left(9+1\right)\cdot9+10\)
\(=9^{14}-9^{14}-9^{13}+9^{13}+9^{12}-9^{12}-9^{11}+...+9^3+9^2-9^2-9+10\)
\(=-9+10\)
\(=1\)
a,A=x3+11x2+30x
A=x2(x+5)+6x2+30x
A=x2(x+5)+6x(x+5)
A=(x2+6x)(x+5)=x(x+5)(x+6)
e,( x+1)(x+3)(x+5)(x+7)+15
=(x2+8x+7)(x2+8x+15)+15
=(x2+8x+11-4)(x2+8x+11+4)+15
=(x2+8x+11)-1=(x2+8x+10)(x2+8x+12)
\(5x^3.A=25x^6-30x^5+10x^3\)
=> \(A=\dfrac{25x^6-30x^5+10x^3}{5x^3}\)
=> \(A=\dfrac{25x^6}{5x^3}-\dfrac{30x^5}{5x^3}+\dfrac{10x^3}{5x^3}\)
=> \(A=5x^3-6x^2+2\)