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1, a, ( x2 - 1 )*( x2 + 2x)
= x4 + 2x3 - x2 - 2x
b, ( 2x - 1 )*( 3x + 2 )*( 3 - x )
= ( 6x2 + 4x - 3x - 2 )*( 3 - x )
= ( 6x2 + x - 2 )*( 3 - x )
= 18x2 - 6x3 + 3x - x2 - 6 + 2x
= -6x3 - 17x2 + 5x - 6
2, b, B = \(8x^3+48x^2+96x+64\)
= \(\left(2x+4\right)^3\)
Thay x = 8 vào B ta có:
B = \(\left(2\cdot8+4\right)^3\)
= ( 16 + 4 )3
= 203
= 8000
mình chỉ làm đc câu b thôi câu a bạn xem lại đề đi hình như câu a sai đề rồi
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\(a,17x^3y^2-34x^2y^2+51x^2y^3\)
\(=17x^2y^2\left(x-2+3y\right)\)
\(b,16x^2\left(x^2-y\right)-10y\left(y-x^2\right)\)
\(=16x^2\left(x^2-y\right)+10y\left(x^2-y\right)\)
\(=\left(x^2-y\right)\left(16x^2+10y\right)\)
\(=2\left(x^2-y\right)\left(8x^2+5y\right)\)
\(c,x^2+2xy+y^2-xz-yz\)
\(=\left(x^2+2xy+y^2\right)-\left(xz+yz\right)\)
\(=\left(x+y\right)^2-z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y-z\right)\)
\(d,64xy-96x^2y+48x^3y-8x^4y\)
\(=8xy\left(8-12x+6x^2-x^3\right)\)
\(=-8xy\left(x^3-6x^2+12x-8\right)\)
\(=-8xy\left(x-2\right)^3\)
a) 17 x3y2-34x2y2+51x2y3
= 17x2y2 ( x - 2 + 3y )
b) 16x2(x2-y)-10y(y-x2)
= 16x2( x2 - y ) + 10y ( x2 - y )
= (x2 - y ) (16x2 + 10y )
c)x2+2xy+y2-xz-yz
= ( x2 + 2xy + y2 ) - ( xz + yz )
= ( x + y )2 - z ( x + y )
= ( x + y ) ( x + y - z )
d)64xy-96x2y+48x3y-8x4y ( Bài này mình không chắc =)) )
= 8xy ( 8 - 12x + 6x2 - x3 )
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a) Ta có: \(x^3+12x^2+48x+64\)
\(=x^3+3\cdot x^2\cdot4+3\cdot x\cdot4^2+4^3\)
\(=\left(x+4\right)^3\)
b) Ta có: \(x^3-12x^2+48x-64\)
\(=x^3-3\cdot x^2\cdot4+3\cdot x\cdot4^2-4^3\)
\(=\left(x-4\right)^3\)
c) Ta có: \(8x^3+12x^2y+6xy^2+y^3\)
\(=\left(2x\right)^3+3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2+y^3\)
\(=\left(2x+y\right)^3\)
d)Sửa đề: \(x^3-3x^2+3x-1\)
Ta có: \(x^3-3x^2+3x-1\)
\(=x^3-3\cdot x^2\cdot1+3\cdot x\cdot1^2-1^3\)
\(=\left(x-1\right)^3\)
e) Ta có: \(8-12x+6x^2-x^3\)
\(=2^3-3\cdot2^2\cdot x+3\cdot2\cdot x^2-x^3\)
\(=\left(2-x\right)^3\)
f) Ta có: \(-27y^3+9y^2-y+\frac{1}{27}\)
\(=\left(\frac{1}{3}\right)^3+3\cdot\left(\frac{1}{3}\right)^2\cdot\left(-3y\right)+3\cdot\frac{1}{3}\cdot\left(-3y\right)^{^2}+\left(-3y\right)^3\)
\(=\left(\frac{1}{3}-3y\right)^3\)
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1a) 8xy(8-12x+6x*x-x*x*x)
chú thích x*x là x bình phương
x*x*x là x lập phương
2. a) 3x (x-5)- (x-1)(2+3x)=30
3x*x-15x-2x-3x*x+2+3x=30
14x=28
x=2
b) (x+2)(x-3)-(x-2)(x+5)=0
x*x-3x+2x-6-x*x-5x+2x+10=0
2x=-4
x=-2
còn mấy bài còn lại mình không biết
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a) Ta có: \(3x^2-6xy+3y^2\)
\(=3\left(x^2-2xy+y^2\right)\)
\(=3\left(x-y\right)^2\)
b) Ta có: \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
c) Ta có: \(64xy-96x^2y+48x^3y-8x^4y\)
\(=8xy\left(8-12x+6x^2-x^3\right)\)
\(=8xy\left(2-x\right)^3\)
d) Ta có: \(54x^3+16y^3\)
\(=2\left(27x^3+8y^3\right)\)
\(=2\left(3x+2y\right)\left(9x^2-6xy+4y^2\right)\)
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Cô hướng dẫn câu tìm x:
\(\left(x^2-4x\right)^2-8\left(x^2-4x\right)+15=0\)
Đặt \(x^2-4x=t\), pt trở thành \(t^2-8t+15=0\Leftrightarrow\left(t-3\right)\left(t-5\right)=0\Leftrightarrow\orbr{\begin{cases}t=3\\t=5\end{cases}}\)
Với t = 3, ta có phương trình \(x^2-4x=3\Leftrightarrow x^2-4x-3=0\Leftrightarrow\orbr{\begin{cases}x=\sqrt{7}+2\\x=-\sqrt{7}+2\end{cases}}\)
Với t = 5, ta có \(x^2-4x=5\Leftrightarrow x^2-4x-5=0\Leftrightarrow\orbr{\begin{cases}x=5\\x=-1\end{cases}}\)
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bài 1 điền vào chỗ trống
a) x2 + 4x + 4
= (x + 2)2
b) x2 - 8x + 16
= (x - 4)2
c) x3 +12x2 + 48x + 64
= (x + 4)3
d) x3 - 6x + 12x - 8
= (x - 2)3
e) x2 + 2x + 1
= (x + 1)2
f) x2 - 1
= (x - 1)(x + 1)
g) x2 - 4x + 4
= (x - 2)2
h) x2 - 4
= (x - 2)(x + 2)
i) x2 + 6x + 9
= (x + 3)2
j) 4x2 - 9
= (2x - 3)(2x + 3)
k) 16x2 - 8x + 1
= (4x - 1)2
l) 9x2 + 6x + 1
= (3x + 1)2
m) 36x2 + 36x + 9
= (6x + 3)2
n) x3 + 27
= (x + 3)(x2 - 3x + 9)
o) 17x3 + 27 (Đề sai)
Giải:
\(8x^3-48x^2+96x-64=0\)
\(\Leftrightarrow\left(2x\right)^3-3.\left(2x\right)^2.4+3.2x.4^2-4^3=0\)
\(\Leftrightarrow\left(2x-4\right)^3=0\)
\(\Leftrightarrow2x-4=0\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\).
Chúc bạn học tốt!
cả câu này nx ạ: 49x2-25(x+1)2=0
4x2-25-(2x-5)(2x+7)=0