K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Dưới đây là một vài câu hỏi có thể liên quan tới câu hỏi mà bạn gửi lên. Có thể trong đó có câu trả lời mà bạn cần!
23 tháng 11 2021
câu 1 B
câu 2 D
câu 3 ko bt
câu 4 x=-1/2; x = -(căn bậc hai(3)*i-1)/4;x = (căn bậc hai(3)*i+1)/4;
câu 5 x=-5/3, x=0, x=1
23 tháng 11 2021
Câu 1: x2 + 2 xy + y2 bằng:
A. x2 + y2 B.(x + y)2 C. y2 – x2 D. x2 – y2
Câu 2: (4x + 2)(4x – 2) bằng:
A. 4x2 + 4 B. 4x2 – 4 C. 16x2 + 4 D. 16x2 – 4
Câu 3: 25a2 + 9b2 - 30ab bằng:
A.(5a-9b)2 B.(5a – 3b)2 C.(5a+3b)2 D.(5a)2 – (3b)2
Câu 4: 8x3 +1 bằng
A.(2x+1).(4x2-2x+1) B. (2x-1).(4x2+2x+1) C.(2x+1)3 D.(2x)3-13
Câu 5:Thực hiện phép nhân x(3x2 + 2x - 5) ta được:
A.3x3 - 2x2 – 5x B. 3x3 + 2x2 – 5x C. 3x3 - 2x2 +5x D. 3x3 + 2x2 + 5x
10 tháng 10 2021
\(6x^3-9x^2=3x^2\left(2x-3\right)\\ 25x^2-0,09=\left(5x-0,3\right)\left(5x+0,3\right)\\ x^2-x-y^2-y=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)\\ \left(x^2+4\right)^2-16x^2=\left(x^2-4x+4\right)\left(x^2+4x+4\right)=\left(x-2\right)^2\left(x+2\right)^2\)
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
a) \(4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
b) \(x^2-3y^2\)
\(=x^2-\left(y\sqrt{3}\right)^2\)
\(=\left(x-y\sqrt{3}\right)\left(x+y\sqrt{3}\right)\)
c) \(9x^2-\dfrac{1}{4}\)
\(=\left(3x\right)^2-\left(\dfrac{1}{2}\right)^2\)
\(=\left(3x-\dfrac{1}{2}\right)\left(3x+\dfrac{1}{2}\right)\)
d) \(\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
e) \(9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3+x-y\right)\left(3-x+y\right)\)
f) \(\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2-4x+4\right)\left(x^2+4x+4\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
T
26 tháng 9 2023
\(a,4x^2-1\)
\(=\left(2x\right)^2-1^2\)
\(=\left(2x-1\right)\left(2x+1\right)\)
\(b,25x^2-0,09\)
\(=\left(5x\right)^2-\left(0,3\right)^2\)
\(=\left(5x-0,3\right)\left(5x+0,3\right)\)
\(d,\left(x-y\right)^2-4\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
\(e,9-\left(x-y\right)^2\)
\(=3^2-\left(x-y\right)^2\)
\(=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]\)
\(=\left(3-x+y\right)\left(3+x-y\right)\)
\(=\left(-x+y+3\right)\left(x-y+3\right)\)
\(f,\left(x^2+4\right)^2-16x^2\)
\(=\left(x^2+4\right)^2-\left(4x\right)^2\)
\(=\left(x^2+4-4x\right)\left(x^2+4+4x\right)\)
\(=\left(x^2-2\cdot x\cdot2+2^2\right)\left(x^2+2\cdot x\cdot2+2^2\right)\)
\(=\left(x-2\right)^2\left(x+2\right)^2\)
#\(Toru\)
23 tháng 6 2023
này mình có vài câu không làm được, xin lỗi bạn nha
\(b,16x^2-8x+1=\left(4x-1\right)^2\\ c,4x^2+12xy+9y^2=\left(2x+3y\right)^2\\ e,=x^2+2x+1+y^2+2y+1+2\left(x+1\right)\left(y+1\right)\\ =\left(x+1\right)^2+2\left(x+1\right)\left(y+1\right)+\left(y+1\right)^2\\ =\left[\left(x+1\right)+\left(y+1\right)\right]^2=\left(x+y+2\right)^2\\ g,=x^2-2x\left(y+2\right)+\left(x+2\right)^2=\left[x-\left(y+2\right)\right]^2=\left(x-y-2\right)^2\\ h,=\left[x+\left(y+1\right)\right]^2=\left(x+y+1\right)^2\)
XO
19 tháng 6 2021
a) (x + y)2 - 2(x + y) + 1
= (x + y)2 - 2.1.(x + y) + 1
= (x + y - 1)2
b) x3 + 1 - x2 - x
= (x3 - x2) - (x - 1)
= x2(x - 1) - (x - 1)
= (x2 - 1)(x - 1) = (x - 1)(x + 1)(x - 1) = (x - 1)2(x + 1)
c) 27x3 - 0,001
= \(\left(3x\right)^3-\frac{1}{1000}=\left(3x\right)^3-\left(\frac{1}{10}\right)^3=\left(3x-\frac{1}{10}\right)\left(9x^2+\frac{3}{10}x+\frac{1}{100}\right)\)
d) 125x3 - 1 =(5x)3 - 1 = (5x - 1)(25x2 + 5x + 1)
e) (x2 + 4)2 - 16x2
= (x2 + 4)2 - (4x)2
= (x2 - 4x + 4)(x2 + 4x + 4)
= (x - 2)2(x + 2)2
= [(x - 2)(x + 2)]2
NC
19 tháng 6 2021
a.\(\left(x+y\right)^2-2\left(x+y\right)+1\)
\(=\left(x+y\right)^2-2.\left(x+y\right).1+1^2\)
\(=\left[\left(x+y\right)-1\right]^2\)
\(=\left(x+y-1\right)^2\)
b.\(x^3+1-x^2-x\)
\(=\left(x^3-x^2\right)+\left(1-x\right)\)
\(=x^2\left(x-1\right)-\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-1\right)\)
\(=\left(x-1\right)^2\left(x+1\right)\)
H
8 tháng 8 2023
`4-x=2(x-4)^2`
`<=>4-x=2(x^2-8x+16)`
`<=> 4-x=2x^2 - 16x+32`
`<=> 4-x-2x^2+16x-32=0`
`<=> -2x^2 +15x-28=0`
`<=> -(2x^2-15x+28)=0`
`<=>-(2x^2-7x-8x+28)=0`
`<=> - [x(2x-7) - 4(2x-7)]=0`
`<=> -(2x-7)(x-4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}-2x+7=0\\x-4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-2x=-7\\x=4\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
__
`(x^2 +1) (x-2)+2x=4`
`<=> x^3 -2x^2 +x-2+2x-4=0`
`<=> x^3 -2x^2 +3x-6=0`
`<=> (x^3+3x)-(2x^2+6)=0`
`<=> x(x^2 +3) -2(x^2+3)=0`
`<=>(x^2+3)(x-2)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^2+3=0\\x-2=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)
__
`x^4 -16x^2=0`
`<=> x^2 (x^2 -16)=0`
`<=>x^2(x-4)(x+4)=0`
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
H9
HT.Phong (9A5)
CTVHS
8 tháng 8 2023
\(4-x=2\left(x-4\right)^2\)
\(\Leftrightarrow4-x=2\left(x^2-8x+16\right)\)
\(\Leftrightarrow4-x=2x^2-16x+32\)
\(\Leftrightarrow2x^2-15x+28=0\)
\(\Leftrightarrow2x^2-7x-8x+28=0\)
\(\Leftrightarrow x\left(2x-7\right)-4\left(2x-7\right)=0\)
\(\Leftrightarrow\left(2x-7\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-7\\x=4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{2}\\x=4\end{matrix}\right.\)
___________
\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Leftrightarrow x^3-2x^2+x-2+2x=4\)
\(\Leftrightarrow x^3-2x^2+3x-2-4=0\)
\(\Leftrightarrow x^3-2x^2+3x-6=0\)
\(\Leftrightarrow x^2\left(x-2\right)+3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x^2+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=-3\left(\text{vô lý}\right)\\x=2\left(tm\right)\end{matrix}\right.\)
\(\Leftrightarrow x=2\)
________________
\(x^4-16x^2=0\)
\(\Leftrightarrow\left(x^2\right)^2-\left(4x\right)^2=0\)
\(\Leftrightarrow\left(x^2-4x\right)\left(x^2+4x\right)=0\)
\(\Leftrightarrow x\left(x-4\right)x\left(x+4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
22 tháng 12 2022
a: \(=5x\left(xy^2+3x+6y^2\right)\)
b: \(=\left(x-2\right)\left(x+3\right)-\left(x-2\right)\left(x+2\right)=\left(x-2\right)\left(x+3-x-2\right)=\left(x-2\right)\)
c: \(=\left(x-3\right)\left(x-4\right)\)
d: \(=x\left(x^2-2xy+y^2-9\right)\)
=x(x-y-3)(x-y+3)
e: \(=\left(x+y\right)^2-25=\left(x+y+5\right)\left(x+y-5\right)\)
f: \(=\left(x-4\right)\left(x+3\right)\)
AH
Akai Haruma
Giáo viên
5 tháng 8 2021
a.
$x^2+20x+100=(x+10)^2$
b.
$16x^2+24xy+9y^2=(4x+3y)^2$
c.
$y^2-14y+49=(y-7)^2$
d.
$9x^2-42xy+49y^2=(3x-7y)^2$
e.
$4x^2-9y^2=(2x-3y)(2x+3y)$
AH
Akai Haruma
Giáo viên
5 tháng 8 2021
f.
$16-x^2=(4-x)(4+x)$
g.
$49x^2-1=(7x-1)(7x+1)$
h.
$16x^2-25=(4x-5)(4x+5)$
i.
$8x^3+24x^2y+54xy^2+27y^3=(2x+3y)^3$
k.
$x^3-6x^2y+12xy^2-8y^3=(x-2y)^3$
l.
$(2a+b)(4a^2-2ab+b^2)=(2a)^3+b^3=8a^3+b^3$
m.
$(3x-4y)(9x^2+12xy+16y^2)=(3x)^3-(4y)^3=27x^3-64y^3$
\(\left(x-y\right)^2-4=\left(x-y-2\right)\left(x-y+2\right)\)
\(9-\left(x-y\right)^2=\left(3-x+y\right)\left(3+x-y\right)\)
\(\left(x^2+4\right)^2-16x^2=\left(x^2-4x+4\right)\left(x^2+4x+4\right)=\left(x-2\right)^2\left(x+2\right)^2\)
\((X-y)^2-4=(x-y-2)(x-y+2)\)\((X^2+4)^2-16x^2=(x^2+4)^2-(4x)^2=(x^2+4-4x)(x^2+4+4x)\)
\(9-(x-y)^2=(3-x+y)(3+x-y)\)