Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) \(A=\left(x^3+x^2\right)-\left(x+1\right)=x\left(x+1\right)-\left(x+1\right)=\left(x-1\right)\left(x+1\right)\)
b) \(B=\left(x^3-3x^2\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)=\left(x^2-4\right)\left(x-3\right)=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
Bạn tải ứng dụng PhotoMath về nha. Ứng dụng này sẽ giải toán số chi tiết
a) \(x^3-4x^2-12x+27\)
\(=\left(x^3+27\right)-\left(4x^2+12x\right)\)
\(=\left(x+3\right)\left(x^2-3x+9\right)-4x\left(x+3\right)\)
\(=\left(x+3\right)\left(x^2-7x+9\right)\)
b) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x+2\right)\left(x-2\right)\left(x-3\right)\)
a) \(9x^2+6xy+y^2=\left(3x+y\right)^2\)
b) \(6x-9-x^2=-\left(x-3\right)^2\)
\(x^8+x^7+1\)
\(=x^8+x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-xx+1\)
\(=\left(x^8-x^6+x^5-x^3+x^2\right)\)
\(+\left(x^7-x^5+x^4-x^2+x\right)\)
\(+\left(x^6-x^4+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^4+x^3-x+1\right)\)
2:
a: =>x-1=0 hoặc 3x+1=0
=>x=1 hoặc x=-1/3
b: =>x-5=0 hoặc 7-x=0
=>x=5 hoặc x=7
c: =>\(\left[{}\begin{matrix}x-1=0\\x+5=0\\3x-8=0\end{matrix}\right.\Leftrightarrow x\in\left\{1;-5;\dfrac{8}{3}\right\}\)
d: =>x=0 hoặc x^2-1=0
=>\(x\in\left\{0;1;-1\right\}\)
Bài 4.
a) 3xy2 - 45x2y = 3xy( y - 15x )
b) 25y2 - 4x2 + 4x - 1
= 25y2 - ( 4x2 - 4x + 1 )
= ( 5y )2 - ( 2x - 1 )2
= ( 5y - 2x + 1 )( 5y + 2x - 1 )
c) x2 - 5x + xy - 5y
= x( x - 5 ) + y( x - 5 )
= ( x - 5 )( x + y )
d) x2 - 8x - 33
= x2 + 3x - 11x - 33
= x( x + 3 ) - 11( x + 3 )
= ( x + 3 )( x - 11 )
Bài 5.
a) A = ( x - 2 )3 - x2( x - 4 ) + 8
= x3 - 6x2 + 12x - 8 - x3 + 4x2 + 8
= -2x2 + 12x
B = ( x2 - 6x + 9 ) : ( x - 3 ) - x( x + 7 ) - 9
= ( x - 3 )2 : ( x - 3 ) - x2 - 7x - 9
= x - 3 - x2 - 7x - 9
= -x2 - 6x - 12
b) Với x = -1 thì A = -2.(-1)2 + 12.(-1) = -2 - 12 = -14
Giải như sau.
(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y
⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn !
\(\left(x+6\right)\left(2x+1\right)=0\)
<=> \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)
<=> \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)
Vậy....
hk tốt
^^
a) 4(x+2) - 7(2x - 1) + 9(3x - 4)=30
⇔4x+8 - 14x + 7 + 27x - 36 = 30
⇔ 17x = 51
⇔ x = 3
b) 2(5x - 8) - 3(4x - 5) = 4(3x - 4) + 11
⇔ 10x - 16 - 12x + 15 = 12x - 16 + 11
⇔ -14x = -4
⇔ x= \(\frac{2}{7}\)
c) 5x(1 - 2x) - 3x(x + 18) = 0
⇔ 5x - 10x\(^2\) - 3x\(^2\) -54x =0
⇔ -13x\(^2\) -49 x = 0
⇔ -x ( 13x + 49 ) =0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\13x+49=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{-49}{13}\end{matrix}\right.\)
d) 5x - 3{4x - 2[4x - 3(5x - 2)]} = 182
⇔ 5x - 3[ 4x - 2( 4x - 15x + 6 ) ]= 182
⇔5x - 3 ( 4x - 8x + 30x - 12 ) = 182
⇔ 5x - 3 ( 26x - 12 ) = 182
⇔ 5x - 78x + 36 = 182
⇔ - 73x = 146
⇔ x = -2
a) Ta có: \(a\left(m-n\right)+m-n\)
\(=a\left(m-n\right)+\left(m-n\right)\)
\(=\left(m-n\right)\left(a+1\right)\)
b) Ta có: \(mx+my+5x+5y\)
\(=m\left(x+y\right)+5\left(x+y\right)\)
\(=\left(x+y\right)\left(m+5\right)\)
c) Ta có: \(ma+mb-a-b\)
\(=m\left(a+b\right)-\left(a+b\right)\)
\(=\left(a+b\right)\left(m-1\right)\)
d) Ta có: \(1-xa-x+a\)
\(=\left(a+1\right)-x\left(a+1\right)\)
\(=\left(a+1\right)\left(1-x\right)\)
e) Ta có: \(\left(a-b\right)^2-\left(b-a\right)\left(a+b\right)\)
\(=\left(a-b\right)^2+\left(a-b\right)\left(a+b\right)\)
\(=\left(a-b\right)\left(a-b+a+b\right)\)
\(=2a\left(a-b\right)\)
f) Ta có: \(a\left(a-b\right)\left(a+b\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab\right)-\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(=\left(a+b\right)\left(a^2-ab-a^2+ab-b^2\right)\)
\(=b^2\cdot\left(a+b\right)\)
g) Ta có: \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)
\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]\)
\(=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)
\(=\left(x+7\right)\left(-8x^2+21x+9\right)\)
\(=\left(x+7\right)\left(-8x^2+24x-3x+9\right)\)
\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]\)
\(=\left(x+7\right)\left(x-3\right)\left(-8x-3\right)\)
h) Ta có: \(\left(x+5\right)^2-3\left(x+5\right)\)
\(=\left(x+5\right)\left(x+5-3\right)\)
\(=\left(x+5\right)\left(x+2\right)\)
i) Ta có: \(2x\left(x-3\right)-3\left(x-3\right)^2\)
\(=\left(x-3\right)\left[2x-3\left(x-3\right)\right]\)
\(=\left(x-3\right)\left(2x-3x+9\right)\)
\(=\left(x-3\right)\left(9-x\right)\)
j) Ta có: \(x\left(x-7\right)+\left(7-x\right)^2\)
\(=x\left(x-7\right)+\left(x-7\right)^2\)
\(=\left(x-7\right)\left(x+x-7\right)\)
\(=\left(x-7\right)\left(2x-7\right)\)
k) Ta có: \(3x\left(x-9\right)^2-\left(9-x\right)^3\)
\(=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\cdot\left(3x+x-9\right)\)
\(=\left(x-9\right)^2\cdot\left(4x-9\right)\)