(x^2+x)^2 +4(x^2+x)=12
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
4 x 2 x 5 = ?
Cách 1: 4 x 2 x 5 = (4 x 2) x 5 = 8 x 5 = 40
Cách 2: 4 x 2 x 5 = 4 x (2 x 5) = 4 x 10 = 40
7 x 2 x 3 = ?
Cách 1: 7 x 2 x 3 = (7 x 2) x 3 = 14 x 3 = 42
Cách 2: 7 x 2 x 3 = 7 x (2 x 3) = 7 x 6 = 42
6 x 3 x 3 = ?
Cách 1: 6 x 3 x 3 = (6 x 3) x 3 = 18 x 3 = 54
Cách 2: 6 x 3 x 3 = 6 x (3 x 3) = 6 x 9 = 54
6 x 2 x 4 = ?
Cách 1: 6 x 2 x 4 = (6 x 2) x 4 = 12 x 4 = 48
Cách 2: 6 x 2 x 4 = 6 x (2 x 4) = 6 x 8 = 48
1/(x+2)(x+3)(x+4)(x+5)-24
=(x+2)(x+5)(x+3)(x+4)
=(x+2)(x-2+7)(x+3)(x-3+7)
=[(x+2)(x-2)+7x+14][(x+3)(x-3)+7x+21]
=(x2-4+7x+14)(x2-9+7x+21)
=(x2+10+7x)(x2+12+7x)
2/(x2+x)2+4(x2+x)-12
=(x2+x)2+4(x2+x)+22-16
=(x2+x+2)2-42
=(x2+x+2+4)(x2+x+2-4)
=(x2+x+6)(x2+x-2)
3/(x2+x+1)(x2+x+2)-12
=(x2+x+1)(x2+x+-1+3)-12
=(x2+x+1)(x2+x+-1)+3(x2+x+1)-12
=(x2+x)-1+3(x2+x)+3-12
=(x2+x)(x2+x+3)-10
làm đến đây thì mk bí, bạn giúp suy nghĩ nốt nha
4/nó là nhân tử sẵn rồi mà
\(3/\)
\(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+1\right)-12\)
\(=\left(x^2+x+1\right)^2+x^2+x+1-12\)
\(=\left(x^2+x+1\right)^2+4\left(x^2+x+1\right)-3\left(x^2+x+1\right)-12\)
\(=\left(x^2+x+1\right)\left(x^2+x+1+4\right)-3\left(x^2+x+1+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
b: \(\Leftrightarrow\dfrac{-3x^2+36x+12}{3\left(x+4\right)\left(x-1\right)}=\dfrac{36\left(x-1\right)}{3\left(x+4\right)\left(x-1\right)}+\dfrac{12\left(x+4\right)}{3\left(x-1\right)\left(x+4\right)}\)
\(\Leftrightarrow-3x^2+36x+12=36x-36+12x+48\)
\(\Leftrightarrow-3x^2+36x+12-48x-12=0\)
\(\Leftrightarrow3x\left(x+4\right)=0\)
=>x=0(nhận) hoặc x=-4(loại)
a,3/4 . (-5/12)+3/4.(-7/12)
` 3/4 . [ - ( 5/12 + 7/12 ) ] `
`3/4 . (-1) = -3/4 `
`2/3 . x - 0,5 = 3/4 `
` x - 0,5 = 3/4 - 2/3 `
` x-0,5 = 1/12 `
` x = 1/12 + 0,5 `
` x= 7/12 `
Bài 2:
a: =>2/3x=3/4+1/2=3/4+2/4=5/4
=>x=5/4:2/3=5/4*3/2=15/8
b:=>-2x+4=3x-12
=>-5x=-16
=>x=16/5
\(\left(1+2\right),y^2-13y+12=y^2-12y-y-12=y\left(y-12\right)+\left(y-12\right)=\left(y+1\right)\left(y-12\right)\)
\(3,x^2-x-30=x^2-6x+5x-30=x\left(x-6\right)+5\left(x-6\right)=\left(x+5\right)\left(x-6\right)\)
\(4,y^2+y-42=y^2-6y+7y-42=y\left(y-6\right)+7\left(y-6\right)=\left(y+7\right)\left(y-6\right)\)
\(5,x^2+3x-10=x^2-2x+5x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
\(6,x^2-8x+15=x^2-5x-3x+15=x\left(x-5\right)-3\left(x-5\right)=\left(x-3\right)\left(x-5\right)\)
1:
\(\Leftrightarrow\left(x^2+5x+6\right)\left(x^2+5x+4\right)=24\)
\(\Leftrightarrow\left(x^2+5x\right)^2+10\left(x^2+5x\right)=0\)
\(\Leftrightarrow x^2+5x=0\)
=>x=0 hoặc x=-5
3: \(\Leftrightarrow\left(x^2+x+6\right)\left(x^2+x-2\right)=0\)
=>(x+2)(x-1)=0
=>x=-2 hoặc x=1
1) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left[\left(x+2\right)\left(x+5\right)\right]\left[\left(x+3\right)\left(x+4\right)\right]-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\)
Đặt \(x^2+7x+11=a\), ta có:
\(=\left(a+1\right)\left(a-1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)\)
\(=\left(x^2+7x+6\right)\left(x^2+7x+16\right)\)
\(=\left(x^2+6x+x+6\right)\left(x^2+7x+16\right)\)
\(=\left[x\left(x+6\right)+\left(x+6\right)\right]\left(x^2+7x+16\right)\)
\(=\left(x+6\right)\left(x+1\right)\left(x^2+7x+16\right)\)
2) \(\left(x^2+x\right)^2+4\left(x^2+x\right)-12\)
\(=\left(x^2+x\right)^2+2\left(x^2+x\right).2+4-4-12\)
\(=\left(x^2+x+2\right)^2-16\)
\(=\left(x^2+x+2\right)^2-4^2\)
\(=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+6\right)\)
3) \(\left(x^2+x+1\right)\left(x^2+x+2\right)-12\)
Đặt \(x^2+x+1=a\), ta được
\(=a\left(a+1\right)-12\)
\(=a^2+a-12\)
\(=a^2+2.a.\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{4}-12\)
\(=\left(a+\dfrac{1}{2}\right)^2-\dfrac{49}{4}\)
\(=\left(a+\dfrac{1}{2}\right)^2-\left(\dfrac{7}{2}\right)^2\)
\(=\left(a+\dfrac{1}{2}-\dfrac{7}{2}\right)\left(a+\dfrac{1}{2}+\dfrac{7}{2}\right)\)
\(=\left(a-3\right)\left(a+4\right)\)
\(=\left(x^2+x+1-3\right)\left(x^2+x+1+4\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x+5\right)\)
4) \(\left(a^2-4\right)\left(a^2+6a+5\right)\)
\(=\left(a-2\right)\left(a+2\right)\left(a^2+5a+a+5\right)\)
\(=\left(a-2\right)\left(a+2\right)\left[a\left(a+5\right)+\left(a+5\right)\right]\)
\(=\left(a-2\right)\left(a+2\right)\left(a+5\right)\left(a+1\right)\)
Câu 2: Y^2+4y+4=16=>(y+2)^2=16 =>y=2;hoặc y=6
pt mới 1: X^2+x+1/4=2+1/4=9/4=>x=2 hoặc x=-5/2
pt mới 2: X^+x+1/4=1/4-5/2=>vô nghiệm
@ngonhuminh Xin lỗi nhưng tui không hiểu. Làm gì có Y, là X mà