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Bài 1 :
\(A=\left(x-1\right)\left(x-2\right)\left(x+7\right)\left(x+8\right)+8\)
\(A=\left[\left(x-1\right)\left(x+7\right)\right]\left[\left(x-2\right)\left(x+8\right)\right]+8\)
\(A=\left(x^2+6x-7\right)\left(x^2+6x-16\right)+8\)
Đặt \(a=x^2+6x-7\)
\(A=a\left(a-9\right)+8\)
\(A=a^2-9a+8\)
\(A=a^2-8a-a+8\)
\(A=a\left(a-8\right)-\left(a-8\right)\)
\(A=\left(a-8\right)\left(a-1\right)\)
Thay a vào là xong bạn :)
3a) x2 (x-1) - 4x2 + 8x - 4
= x2(x-1) - ( 2x - 2)2
= (x\(\sqrt{x-1}\))2 -( 2x - 2)2
= (x\(\sqrt{x-1}\)- 2x+2) ( x\(\sqrt{x-1}\)+ 2x - 2)
3b) = x3 +33 + (x+3) (x-9)
= (x + 3)( x2 - 3x + 9) + (x+3)(x-9)
= (x+3)(x2 -2x) = (x + 3)(x - 2)x
a) (2x + 1)2 - 4(x + 2)2 = 99
=> 4x2 + 4x + 1 - 4(x2 + 4x + 4) = 99
=> 4x2 + 4x + 1 - 4x2 - 16x - 16 = 99
=> -12x = 114
=> x = -9,5
b) (x - 3)2 - (x - 4)(x + 8) = 1
=> x2 - 6x + 9 - (x2 + 4x - 32) = 1
=> x2 - 6x + 9 - x2 - 4x + 32 = 1
=> -10x = -40
=> x = 4
c) 3(x + 2)2 + (2x - 1)2 - 7(x - 3)(x + 3) = 36
=> 3(x2 + 4x + 4) + 4x2 - 4x + 1 - 7(x2 - 9) = 36
=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
=> 8x = -40
=> x = -5
a) ( 2x + 1 ) - 4( x + 2 )2 = 99
<=> 4x2 + 4x + 1 - 4( x2 + 4x + 4 ) = 99
<=> 4x2 + 4x + 1 - 4x2 - 16x - 16 = 99
<=> -12x - 15 = 99
<=> -12x = 114
<=> x = -114/12 = -19/2
b) ( x + 3 )2 - ( x - 4 )( x + 8 ) = 1
<=> x2 + 6x + 9 - ( x2 + 4x - 32 ) = 1
<=> x2 + 6x + 9 - x2 - 4x + 32 = 1
<=> 2x + 41 = 1
<=> 2x = -40
<=> x = -20
c) 3( x + 2 )2 + ( 2x - 1 )2 - 7( x + 3 )( x - 3 ) = 36
<=> 3( x2 + 4x + 4 ) + 4x2 - 4x + 1 - 7( x2 - 9 ) = 36
<=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
<=> 8x + 76 = 36
<=> 8x = -40
<=> x = -5
b: \(\Leftrightarrow2\left(x^2-2x+1\right)-3x^2+5x-1=0\)
\(\Leftrightarrow2x^2-4x+2-3x^2+5x-1=0\)
\(\Leftrightarrow-x^2+x+1=0\)
\(\Leftrightarrow x^2-x-1=0\)
\(\text{Δ}=\left(-1\right)^2-4\cdot1\cdot\left(-1\right)=5\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{1-\sqrt{5}}{2}\\x_2=\dfrac{1+\sqrt{5}}{2}\end{matrix}\right.\)
c: \(\Leftrightarrow x^2+6x+9-1-\left(x^2+8x-4x-32\right)=0\)
\(\Leftrightarrow x^2+6x+8-x^2-4x+32=0\)
=>2x+40=0
hay x=-20
d: \(\Leftrightarrow3x^2+12x+12+4x^2-4x+1-7\left(x^2-9\right)=36\)
\(\Leftrightarrow7x^2+8x+13-7x^2+63=36\)
=>8x+76=36
hay x=-5
a) x^3-2x^2+x
= x(x^2-2x+1)
x(x-1)^2
b) x^2-2x-15
= (x^2-2x+1)-16
= (x-1)^2-4^2
= ( x-5)(x+3)
d) 5(x-y)-y(x-y)
=(x-y)(5-y)
e) 27x^2(y-1)-9x^3(1-y)
= 27x^2(y-1)+9x^3(y-1)
= (y-1)(27x^2+9x^3)
= 9x^2(y-1)(3+x)
f) 36-12x+x^2
= x^2- 6.2x+36
= ( x - 6)^2
g) 125^3+27y^3
125^3+ ( 9y)^3
= 125^3+ 3.125^2.9y+3.125.(9y)2+ (9y)3
i) xy+xz+3y+3z
= x( x+y) + 3(y+z)
= (y+z)(x+3)
a/ \(x^3-2x^2+x\)
= \(x\left(x^2-2x+1\right)\)
= \(x\left(x-1\right)^2\)
b/ \(x^2-2x-15\)
= \(x^2-2x+1-16\)
= \(\left(x-1\right)^2-4^2\)
= \(\left(x-1-4\right)\left(x-1+4\right)\)
= \(\left(x-5\right)\left(x+3\right)\)
c/ \(x^2-2x-y^2+1\)
= \(\left(x-1\right)^2-y^2\)
= \(\left(x-y-1\right)\left(x+y-1\right)\)
d/ \(5\left(x-y\right)-y\left(x-y\right)\)
= \(\left(x-y\right)\left(5-y\right)\)
e/ \(27x^2\left(y-1\right)-9x^3\left(1-y\right)\)
= \(27x^2\left(y-1\right)+9x^3\left(y-1\right)\)
= \(\left(y-1\right)\left(27x^2+9x^3\right)\)
f/ \(36-12x+x^2\)
= \(x^2-12x+6^2\)
= \(\left(x-6\right)^2\)
Bài 2
a) 4x(x-3)-3x+9
=4x(x-3)-3(x-3)
= (x-3)(4x-3)
b) x3+2x2-2x-4
=(x3+2x2)-(2x+4)
=x2(x+2)-2(x+2)
=(x+2)(x2-2)
c) 4x2-4y+4y-1
=4x2-1
=(2x-1)(2x+1)
d) x5-x
=x(x4-1)
=x(x2-1)(x2+1)
a) 4x(x-3)-3x+9
= 4x(x-3) - 3(x-3)
= (x-3)(4x-3)
b)x3 + 2x2 - 2x - 4
= x2(x + 2) - 2(x + 2)
= (x+2)(x2-2)
c) 4x2 - 4y +4y -1
= [(2x)2-12] + (-4y+4y)
= (2x+1)(2x-1)
d) x5-x
= x(x4 - 1)
a) \(\left(-4\right)^{\left(x+3\right)}=\left(-8\right)^{-2}\Rightarrow\left(-2\right)^{\left(2x+4\right)}=\left(-2\right)^{-6}\Rightarrow2x+4=-6\Rightarrow x=-5\)