Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
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 3:
b: \(\Leftrightarrow x^2\left(x+1\right)^2=0\)
hay \(x\in\left\{0;-1\right\}\)
c: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=0\)
=>x-1=0
hay x=1
d: \(\Leftrightarrow6x^2-3x-4x+2=0\)
\(\Leftrightarrow\left(2x-1\right)\left(3x-2\right)=0\)
hay \(x\in\left\{\dfrac{1}{2};\dfrac{2}{3}\right\}\)
a. \(8x\left(x-2007\right)-2x+4034=0\)
\(\Rightarrow\left(x-2017\right)\left(4x-1\right)\)
\(\Rightarrow\left[{}\begin{matrix}x-2017=0\\4x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=2017\\4x=1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\)
Vậy x=2017 hoặc x=1/4
b.\(\dfrac{x}{2}+\dfrac{x^2}{8}=0\)
\(\Rightarrow\dfrac{x}{2}\left(1+\dfrac{x}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}\dfrac{x}{2}=0\\1+\dfrac{x}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{x}{4}=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy x=0 hoặc x=-4
c.\(4-x=2\left(x-4\right)^2\)
\(\Rightarrow\left(4-x\right)-2\left(x-4\right)^2=0\)
\(\Rightarrow\left(4-x\right)\left(2x-7\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4-x=0\\2x-7=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{7}{2}\end{matrix}\right.\)
Vậy x=4 hoặc x=7/2
d.\(\left(x^2+1\right)\left(x-2\right)+2x=4\)
\(\Rightarrow\left(x-2\right)\left(x^2+3\right)=0\)
Nxet: (x2+3)>0 với mọi x
=> x-2=0 <=>x=2
Vậy x=2
a, 8\(x\).(\(x-2007\)) - 2\(x\) + 4034 = 0
4\(x\)(\(x\) - 2007) - \(x\) + 2017 = 0
4\(x^2\) - 8028\(x\) - \(x\) + 2017 = 0
4\(x^2\) - 8029\(x\) + 2017 = 0
4(\(x^2\) - 2. \(\dfrac{8029}{8}\) \(x\) +( \(\dfrac{8029}{8}\))2) - (\(\dfrac{8029}{4}\))2 + 2017 = 0
4.(\(x\) + \(\dfrac{8029}{8}\))2 = (\(\dfrac{8029}{4}\))2 - 2017
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
a) \(\left(2x-3\right)^2-\left(2x+5\right)^2=10\)
\(\Leftrightarrow4x^2-12x+9-4x^2-20x-25-10=0\)
\(\Leftrightarrow-32x-26=0\)
\(\Leftrightarrow-32x=26\)
\(\Rightarrow x=-\frac{13}{16}\)
b) \(4\left(x+1\right)^2+\left(2x-1\right)^2+8\left(x-1\right)\left(x+1\right)=11\)
\(\Leftrightarrow4x^2+8x+4+4x^2-4x+1+8x^2-8=0\)
\(\Leftrightarrow16x^2+4x-3=0\)
\(\Leftrightarrow4\left(4x^2+x+\frac{1}{16}\right)-\frac{13}{4}=0\)
\(\Leftrightarrow\left[2\left(2x+\frac{1}{4}\right)\right]^2-\left(\frac{\sqrt{13}}{2}\right)^2=0\)
\(\Leftrightarrow\left(4x+\frac{1}{2}-\frac{\sqrt{13}}{2}\right)\left(4x+\frac{1}{2}+\frac{\sqrt{13}}{2}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}4x+\frac{1-\sqrt{13}}{2}=0\\4x+\frac{1+\sqrt{13}}{2}=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{13}-1}{8}\\x=\frac{-1-\sqrt{13}}{8}\end{cases}}\)
c) \(\left(x+5\right)^2=45+x^2\)
\(\Leftrightarrow x^2+10x+25-x^2-45=0\)
\(\Leftrightarrow10x-20=0\)
\(\Leftrightarrow10x=20\)
\(\Rightarrow x=2\)
d) \(\left(2x-3\right)^2-\left(2x-1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2+4x-1+3=0\)
\(\Leftrightarrow-8x+11=0\)
\(\Leftrightarrow-8x=-11\)
\(\Rightarrow x=\frac{11}{8}\)
e) \(\left(x-1\right)^2-\left(5x-3\right)^2=0\)
\(\Leftrightarrow\left(x-1-5x+3\right)\left(x-1+5x-3\right)=0\)
\(\Leftrightarrow\left(-4x+2\right)\left(6x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-4x+2=0\\6x-4=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{2}{3}\end{cases}}\)
`đk:x ne 0,-2`
`a)D=(x/(x+2)+(8x+8)/(x^2+2x)-(x+2)/x):((x^2-x-3)/(x^2+2x)+1/x)`
`=((x^2+8x+8-x^2-4x-4)/(x(x+2))):((x^2-x-3+x+2)/(x(x+2)))`
`=(4x+4)/(x(x+2)):(x^2-1)/(x(x+2))`
`=(4x+4)/(x^2-1)(x ne +-1)`
`=4/(x-1)`
`b)x(x-2)-(x-2)=0`
`<=>(x-2)(x-1)=0`
Vì `x ne 1=>x-1 ne 0`
`=>x-2=0<=>x=2`
`=>D=4/(2-1)=4`
`c)D<0`
Mà `4>0`
`=>x-1<0`
`=>x<1`
Kết hợp đkxđ:
`=>x<1,x ne 0,x ne -2`
`d)D=2`
`<=>4/(x-1)=2`
`<=>2/(x-1)=1`
`<=>x-1=2`
`<=>x=3(tm)`
\(a,\Leftrightarrow9x^2=-36\Leftrightarrow x\in\varnothing\\ b,\Leftrightarrow3\left(x+4\right)-x\left(x+4\right)=0\\ \Leftrightarrow\left(3-x\right)\left(x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-4\end{matrix}\right.\\ c,\Leftrightarrow2x^2-x-2x^2+3x+2=0\\ \Leftrightarrow2x=-2\Leftrightarrow x=-1\\ d,\Leftrightarrow\left(2x-3-2x\right)\left(2x-3+2x\right)=0\\ \Leftrightarrow-3\left(4x-3\right)=0\\ \Leftrightarrow x=\dfrac{3}{4}\\ e,\Leftrightarrow\dfrac{1}{3}x\left(x-9\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=9\end{matrix}\right.\\ f,\Leftrightarrow x^2\left(x-1\right)-\left(x-1\right)=0\\ \Leftrightarrow\left(x^2-1\right)\left(x-1\right)=0\\ \Leftrightarrow\left(x-1\right)^2\left(x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
a)Đặt A= \(x^2+2x+11=\left(x+1\right)^2+10\)
vì \(\left(x+1\right)^2\ge0;\forall x\)
\(\Rightarrow\left(x+1\right)^2+11\ge11;\forall x\)
Hay \(A\ge11>0;\forall x\)
phần b và c mình sẽ giải ra hằng đẳng thức lập luận tương tự phần a
b)\(4x^2+8x+5\)
\(\left(2x\right)^2+2.2x.2+2^2+1\)
\(=\left(2x+2\right)^2+1\)
c) \(x^2+x+2=x^2+2.x.\frac{1}{2}+\frac{1}{4}-\frac{1}{4}+2\)
\(=\left(x+\frac{1}{2}\right)^2+\frac{7}{4}\)