tìm x biết:|x^2+|6x-2||=x^2+4
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MH
6 tháng 2 2022
\(\left|x^2+\left|6x-2\right|\right|=x^2+4\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2+\left|6x-2\right|=x^2+4\\x^2+\left|6x-2\right|=-x^2-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|6x-2\right|=4\\\left|6x-2\right|=-2x^2-4\end{matrix}\right.\)
\(\left[{}\begin{matrix}6x-2=4\\6x-2=-4\\6x-2=-2x^2-4\\6x-2=2x^2+4\end{matrix}\right.\)
Còn lại bạn tự làm tiếp ha
2 tháng 12 2023
bài 5:
1: \(\dfrac{12x^3y^2}{18xy^5}=\dfrac{12x^3y^2:6xy^2}{18xy^5:6xy^2}=\dfrac{2x^2}{3y^3}\)
2: \(\dfrac{10xy-5x^2}{2x^2-8y^2}=\dfrac{5x\cdot2y-5x\cdot x}{2\left(x^2-4y^2\right)}\)
\(=\dfrac{5x\left(2y-x\right)}{-2\left(x+2y\right)\left(2y-x\right)}=\dfrac{-5x}{2\left(x+2y\right)}\)
3: \(\dfrac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\dfrac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)
\(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
4: \(\dfrac{\left(x+1\right)\left(x^2-2x+1\right)}{\left(6x^2-6\right)\left(x^3-1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)^2}{6\left(x^2-1\right)\left(x-1\right)\left(x^2+x+1\right)}\)
\(=\dfrac{\left(x+1\right)\left(x-1\right)}{6\left(x-1\right)\left(x+1\right)\cdot\left(x^2+x+1\right)}\)
\(=\dfrac{1}{6\left(x^2+x+1\right)}\)
5: \(\dfrac{2x^2-7x+3}{1-4x^2}\)
\(=-\dfrac{2x^2-7x+3}{4x^2-1}\)
\(=-\dfrac{2x^2-6x-x+3}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{2x\left(x-3\right)-\left(x-3\right)}{\left(2x-1\right)\left(2x+1\right)}\)
\(=-\dfrac{\left(x-3\right)\left(2x-1\right)}{\left(2x-1\right)\left(2x+1\right)}=\dfrac{-x+3}{2x+1}\)
Bài 3:
1: \(9x^3-xy^2\)
\(=x\cdot9x^2-x\cdot y^2\)
\(=x\left(9x^2-y^2\right)\)
\(=x\left(3x-y\right)\left(3x+y\right)\)
2: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
3: \(x^2-3xy-6x+18y\)
\(=\left(x^2-3xy\right)-\left(6x-18y\right)\)
\(=x\left(x-3y\right)-6\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x-6\right)\)
4: \(6xy-x^2+36-9y^2\)
\(=36-\left(x^2-6xy+9y^2\right)\)
\(=36-\left(x-3y\right)^2\)
\(=\left(6-x+3y\right)\left(6+x-3y\right)\)
5: \(x^4-6x^2+5\)
\(=x^4-x^2-5x^2+5\)
\(=x^2\left(x^2-1\right)-5\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x^2-1\right)\)
\(=\left(x^2-5\right)\left(x-1\right)\left(x+1\right)\)
6: \(9x^2-6x-y^2+2y\)
\(=\left(9x^2-y^2\right)-\left(6x-2y\right)\)
\(=\left(3x-y\right)\left(3x+y\right)-2\left(3x-y\right)\)
\(=\left(3x-y\right)\left(3x+y-2\right)\)
14 tháng 11 2021
a) (x-4)(x+4)-x(x+2)=0
x2-16-x2-2x = 0
-16 - 2x = 0
2x = -16
x = -16/2
x = -8
b) 3x(x-2)-x+2=0
(3x-1)(x-2)=0
=> x ∈ {1/3 ; 2 }
c) 6x - 12x2 = 0
6x(1-2x) = 0
=> x ∈ {0; 1/2 }
d) mình thấy có vẻ hơi sai đề nên mình ko giải được, bạn thông cảm nha
19 tháng 11 2021
\(\left(x+2\right)^3-x^2\left(x-6\right)-4=0\\ \Leftrightarrow x^3-6x^2+12x-8-x^3+6x^2-4=0\\ \Leftrightarrow12x-12=0\\ \Leftrightarrow12x=12\\ \Leftrightarrow x=1\)
\(6x^2-\left(2x-3\right)\left(3x+2\right)=1\\ \Leftrightarrow6x^2-\left[3x.\left(2x-3\right)+2.\left(2x-3\right)\right]=1\\ \Leftrightarrow6x^2-\left(6x^2-9x+4x-6\right)=1\\ \Leftrightarrow6x^2-\left(6x^2-5x-6\right)=1\\ \Leftrightarrow6x^2-6x^2+5x+6=1\\ \Leftrightarrow5x=-5\\ \Leftrightarrow x=-1\)
B
1
NH
21 tháng 6 2023
a)
`4(x-2)^2 =4`
`<=>(x-2)^2 =1`
`<=>x-2=1` hoặc `x-2=-1`
`<=>x=3` hoặc `x=1`
b)
`5(x^2 -6x+9)=5`
`<=>(x-3)^2 =1`
`<=>x-3=1`hoặc `x-3=-1`
`<=>x=4` hoặc `x=2`
c)
`4x^2 +4x+1=0`
`<=>(2x+1)^2 =0`
`<=>2x+1=0`
`<=>x=-1/2`
d)
`9x^2 +6x+1=2`
`<=>(3x+1)^2 =2`
\(< =>\left[{}\begin{matrix}3x+1=\sqrt{2}\\3x+1=-\sqrt{2}\end{matrix}\right.\\ < =>\left[{}\begin{matrix}x=\dfrac{\sqrt{2}-1}{3}\\x=\dfrac{-\sqrt{2}-1}{3}\end{matrix}\right.\)
8 tháng 1 2017
\(\frac{x-1}{x^2-9x+20}+\frac{2x-2}{x^2-6x+8}+\frac{3x-3}{x^2-x-2}+\frac{4x-4}{x^2+6x+5}=0\)
\(\Leftrightarrow\frac{x-1}{\left(x-5\right)\left(x-4\right)}+\frac{2\left(x-1\right)}{\left(x-4\right)\left(x-2\right)}+\frac{3\left(x-1\right)}{\left(x-2\right)\left(x+1\right)}+\frac{4\left(x-1\right)}{\left(x+1\right)\left(x+5\right)}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{10}{x^2-25}\right)=0\)
\(\Leftrightarrow x-1=0\)
\(\Leftrightarrow x=1\)
PS: Điều kiện xác đinh bạn tự làm nhé
Uiiii....gửi nhầm chỗ :<
Sorry ha!
chờ CTV xóa baì;v