Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 14:Tìm x
a,\(x-3=\left(3-x\right)^2\)
\(\Rightarrow\left(x-3\right)-\left(3-x\right)^2=0\)
\(\Rightarrow\left(x-3\right)+\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(1+x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
b,\(\left(2x-5\right)-\left(5+2x\right)^2=0\)
\(\Rightarrow\left(2x-5\right)+\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)\left(1+2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=5\\2x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
Bài 8:
b. 1+8x6y3 = 13+23(x2)3y3 = 13+(2x2y)3
= (1+2x2y)(1-2x2y+4x4y2)
e. 27x3+\(\dfrac{y^3}{8}\)\(=\left(3x\right)^3+\left(\dfrac{y}{2}\right)^3\)
= (3x+\(\dfrac{y}{2}\))(9x2-\(\dfrac{3xy}{2}\)+\(\dfrac{y^2}{4}\))
Bài 9:
c. 1- 9x +27x2 -27x3 = 13-3.12.3x+3.(3x)2-(3x)3
= (1-3x)3
d. x3+\(\dfrac{3}{2}x^2\)+\(\dfrac{3}{4}x+\dfrac{1}{8}\) = x3+\(3x^2.\dfrac{1}{2}\)+\(3x.\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3\)
= (x+\(\dfrac{1}{2}\))3
f. x2 - 2xy +y2 -4m2 +4m.n - n2 = (x2 - 2xy +y2)-((2m)2 -2.2m.n + n2)
= (x-y)2-(2m-n)2 = (x-y-2m+n)(x-y+2m-n)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
a) 5x - 15y = 5(x - 3y)
b) \(\dfrac{3}{5}\)x2 + 5x4 - x2 - y
= \(\dfrac{3}{5}\)x2 + 5x2.x2 - x2 - y
= x2(\(\dfrac{3}{5}\) + 5x2 -1) - y
c) 14x2y2 - 21xy2 + 28x2y
= 7xy.xy - 7xy.3y + 7xy.4x
= 7xy(xy - 3y + 4x)
= 7xy[(xy - 3y) + 4x]
= 7xy[y(x - 3) +4x]
d) \(\dfrac{2}{7}x\)(3y - 1) - \(\dfrac{2}{7}y\)(3y - 1)
= (3y - 1).(\(\dfrac{2}{7}x\) - \(\dfrac{2}{7}y\) )
= (3y - 1).[\(\dfrac{2}{7}\)(x - y)]
e) x3 - 3x2 + 3x - 1
= x2.x - 3x.x + 3.x - 1
= x(x2-3x+3) - 1
g) 27x3 + \(\dfrac{1}{8}\)
= (3x)3 + \(\left(\dfrac{1}{2}\right)^3\)
= (3x + \(\dfrac{1}{2}\)).(9x2 - \(\dfrac{3}{2}\)x + \(\dfrac{1}{4}\))
h) (x+y)3 - (x-y)3
= 2(3x2y) + 2y3
f) (x+y)2 - 4x2
= -3x2 + y(2x + y)
a, bằng cách tìm nhân tử chung
1,\(x^2-3x\)
=x.(\(\left(x-3\right)\)
2,\(15x^2-6x\)
=3x.(5x-2)
3,\(4x\left(x-y\right)\)\(+2y\left(x-y\right)\)
=(x-y).(4x+2y)
=2(x-y).(x+y)
=2(\(x^2-y^2\left(\right)\)
b, dùng hằng đẳng thức
1,\(64x^2-25y^2\)
=\(\left(8x\right)^2-\left(5y\right)^2\)
=(8x-5y)(8x+5y)
2,\(9x^2-30x-25\)
=\(\left(3x-5\right)^2\)
3,
\(\dfrac{1}{4}x^2+2x+4\)
=\(\left(\dfrac{1}{2}x+2\right)^2\)
4,\(25a^2-2a+\dfrac{1}{25}\)
=(\(\left(5a-\dfrac{1}{5}\right)^2\)
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 )
a) (x+2) \(\left(x^2-2x+4\right)\)
b) (3 - 2y) \(\left(9+6y+4y^2\right)\)
d) (4x - y) \(\left(16x^2+4xy+y^2\right)\)
bạn giải chi tiết hộ mình với nha.
mk sắp phải nộp bài r. huhuhuhu
a) Có lẽ câu này sai đề rồi. Dấu chia phải là dấu nhân chứ?
Mình sửa đề luôn nhé.
\(15\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)-25\dfrac{1}{4}.\left(-\dfrac{7}{5}\right)\)
= \(-\dfrac{7}{5}.\left(15\dfrac{1}{4}-25\dfrac{1}{4}\right)\)
= \(-\dfrac{7}{5}.\left(\dfrac{61}{4}-\dfrac{101}{4}\right)\)
= \(-\dfrac{7}{5}.\left(-10\right)\)
= \(14\)
b) \(38-2y=\dfrac{4}{3}\)
\(2y=38-\dfrac{4}{3}=\dfrac{110}{3}\Rightarrow y=\dfrac{55}{3}\)
a) x4 - x3y - x + y
= x3(x - y) - (x - y)
=(x - y)(x3 - 1)
=(x - y)(x - 1)(x2 + x + 1)
b) x3 - 4x2 - 8x + 8
= (x3 + 8) - (4x2 + 8x)
= (x + 2)(x2 - 2x + 4) - 4x(x + 2)
= (x + 2)(x2 - 2x + 4 - 4x)
= (x + 2)(x2 - 6x + 4)
a, = (x^4-x^3y)-(x-y)
=x^3.(x-y)-(x-y) = (x-y).(x^3-1) = (x-y).(x-1).(x^2+x+1)
b, = (x^3+2x^2)-(6x^2+12x)+(4x+8)
= x^2.(x+2)-6x.(x+2)+4.(x+2) = (x+2).(x^2-6x+4)
k mk nha
Lời giải:
a.
$=(x^2)^2+(\frac{1}{2}y^4)^2+2.x^2.\frac{1}{2}y^4-x^2y^4$
$=(x^2+\frac{1}{2}y^4)^2-(xy^2)^2$
$=(x^2+\frac{1}{2}y^4-xy^2)(x^2+\frac{1}{2}y^4+xy^2)$
b.
$=(\frac{1}{2}x^2)^2+(y^4)^2+2.\frac{1}{2}x^2.y^4-x^2y^4$
$=(\frac{1}{2}x^2+y^4)^2-(xy^2)^2$
$=(\frac{1}{2}x^2+y^4-xy^2)(\frac{1}{2}x^2+y^4+xy^2)$
c.
$=(8x^2)^2+(y^2)^2+2.8x^2.y^2-16x^2y^2$
$=(8x^2+y^2)^2-(4xy)^2=(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
d.
$=\frac{64x^4+y^4}{64}=\frac{1}{64}(8x^2+y^2-4xy)(8x^2+y^2+4xy)$
c: \(64x^4+y^4\)
\(=64x^4+16x^2y^2+y^4-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2\)
\(=\left(8x^2+y^2-4xy\right)\left(8x^2+y^2+4xy\right)\)