3/4x-1/4=2(x-3)+1/4x
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
HP
2 tháng 2 2021
1.
\(x^4-6x^2-12x-8=0\)
\(\Leftrightarrow x^4-2x^2+1-4x^2-12x-9=0\)
\(\Leftrightarrow\left(x^2-1\right)^2=\left(2x+3\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-1=2x+3\\x^2-1=-2x-3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2-2x-4=0\\x^2+2x+2=0\end{matrix}\right.\)
\(\Leftrightarrow x=1\pm\sqrt{5}\)
HP
2 tháng 2 2021
3.
ĐK: \(x\ge-9\)
\(x^4-x^3-8x^2+9x-9+\left(x^2-x+1\right)\sqrt{x+9}=0\)
\(\Leftrightarrow\left(x^2-x+1\right)\left(\sqrt{x+9}+x^2-9\right)=0\)
\(\Leftrightarrow\sqrt{x+9}+x^2-9=0\left(1\right)\)
Đặt \(\sqrt{x+9}=t\left(t\ge0\right)\Rightarrow9=t^2-x\)
\(\left(1\right)\Leftrightarrow t+x^2+x-t^2=0\)
\(\Leftrightarrow\left(x+t\right)\left(x-t+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-t\\x=t-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\sqrt{x+9}\\x=\sqrt{x+9}-1\end{matrix}\right.\)
\(\Leftrightarrow...\)
VP
3
KK
6 tháng 9 2021
1. \(\sqrt{x^2-4x+3}=x-2\)
<=> x2 - 4x + 3 = (x - 2)2
<=> x2 - 4x + 3 = x2 - 4x + 4
<=> x2 - x2 - 4x + 4x = 1
<=> 0 = 1 (Vô lí)
vậy PT có nghiệm là S = \(\varnothing\)
KK
6 tháng 9 2021
2. \(\sqrt{4x^2-4x+1}=x-1\)
<=> \(\sqrt{\left(2x-1\right)^2}=x-1\)
<=> 2x - 1 = x - 1
<=> 2x - x = -1 + 1
<=> x = 0
1 tháng 8 2023
1: =>x^2-x=3-x
=>x^2=3
=>x=căn 3 hoặc x=-căn 3
2: =>x^2-4x+3=x^2-4x+4 và x>=2
=>3=4(vô lý)
3: =>2|x-1|=6
=>|x-1|=3
=>x-1=3 hoặc x-1=-3
=>x=-2 hoặc x=4
4: =>|2x-3|=|x-2|
=>2x-3=x-2 hoặc 2x-3=-x+2
=>x=1 hoặc x=5/3
5: =>\(\sqrt{x+2}\left(\sqrt{x-2}+\sqrt{x+2}\right)=0\)
=>x+2=0
=>x=-2
12 tháng 8 2021
\(a/\)
\(4x-4y+x^2-2xy+y^2\)
\(=\left(4x-4y\right)+\left(x^2-2xy+y^2\right)\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
\(b/\)
\(x^4-4x^3-8x^2+8x\)
\(=\left(x^4+8x\right)-\left(4x^3+8x^2\right)\)
\(=x\left(x^3+8\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4\right)-4x^2\left(x+2\right)\)
\(=x\left(x+2\right)\left(x^2-2x+4-4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x-4\right)\)
\(d/\)
\(x^4-x^2+2x-1\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2+x-1\right)\left(x^2-x+1\right)\)
\(e/\)(Xem lại đề)
\(x^4+x^3+x^2+2x+1\)
\(=\left(x^4+x^3\right)+\left(x^2+2x+1\right)\)
\(=x^3\left(x+1\right)+\left(x+1\right)^2\)
\(=\left(x+1\right)\left(x^3+x+1\right)\)
\(f/\)
\(x^3-4x^2+4x-1\)
\(=x\left(x^2-4x+4\right)-1^2\)
\(=x\left(x-2\right)^2-1\)
\(=[\sqrt{x}\left(x-2\right)]^2-1\)
\(=[\sqrt{x}\left(x-2\right)-1][\sqrt{x}\left(x-2\right)+1]\)
12 tháng 8 2021
\(c/\)
\(x^3+x^2-4x-4\)
\(=\left(x^3-2x^2\right)+\left(3x^2-6x\right)+\left(2x-4\right)\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)[\left(x^2+x\right)+\left(2x+2\right)]\)
\(=\left(x-2\right)\left(x+1\right)\left(x+2\right)\)
JN
3
AH
Akai Haruma
Giáo viên
29 tháng 6 2023
Lời giải:
a. $\sqrt{x^2}=1$
$\Leftrightarrow |x|=1$
$\Leftrightarrow x=\pm 1$
b. $\sqrt{4x^2-4x+1}=3$
$\Leftrightarrow \sqrt{(2x-1)^2}=3$
$\Leftrightarrow |2x-1|=3$
$\Leftrightarrow 2x-1=\pm 3$
$\Leftrightarrow x=-1$ hoặc $x=2$
3. ĐKXĐ: $x^2\geq 4$
$\sqrt{x^2-4}+\sqrt{x^2+4x+4}=0$
Do $\sqrt{x^2-4}\geq 0; \sqrt{x^2+4x+4}\geq 0$ với mọi $x\in$ ĐKXĐ nên để tổng của chúng bằng $0$ thì:
$\sqrt{x^2-4}=\sqrt{x^2+4x+4}=0$
$\Leftrightarrow (x-2)(x+2)=(x+2)^2=0$
$\Leftrightarrow x=-2$
4.
PT \(\Leftrightarrow \left\{\begin{matrix} x-3\geq 0\\ x^2-4x+3=(x-3)^2\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ x^2-4x+3=x^2-6x+9\end{matrix}\right.\)
\(\Leftrightarrow \left\{\begin{matrix} x\geq 3\\ 2x=6\end{matrix}\right.\Leftrightarrow x=3\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 6 2023
Ý 1:
\(\sqrt{x^2}=1\\ \Leftrightarrow\left|x\right|=1\\ Vậy:x=1.hoặc.x=-1\\ S=\left\{\pm1\right\}\)
Ý 2:
\(\sqrt{4x^2-4x+1}=3\\ \Leftrightarrow\sqrt{\left(2x-1\right)^2}=3\\ \Leftrightarrow\left|2x-1\right|=3\\ \Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\\ Vậy:S=\left\{-1;2\right\}\)
14 tháng 10 2021
1: Ta có: \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=4x+17\)
\(\Leftrightarrow x^2+6x+9-x^2+4-4x=17\)
\(\Leftrightarrow x=2\)
3: Ta có: \(\left(2x+3\right)\left(x-1\right)+\left(2x-3\right)\left(1-x\right)=0\)
\(\Leftrightarrow2x^2-2x+3x-3+2x-2x^2-3+3x=0\)
\(\Leftrightarrow6x=6\)
hay x=1
PT
18 tháng 9 2018
a) \(4x-4y+x^2-2xy+y^2\)
\(=4\left(x-y\right)+\left(x-y\right)^2\)
\(=\left(x-y\right)\left(4+x-y\right)\)
b) \(x^4-4x^3-8x^2+8x\)
\(=x^4+2x^3-6x^3-12x^2+4x^2+8x\)
\(=x^3\left(x+2\right)-6x^2\left(x+2\right)+4x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^3-6x^2+4x\right)\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
c) \(x^3+x^2-4x-4\)
\(=x^3-2x^2+3x^2-6x+2x-4\)
\(=x^2\left(x-2\right)+3x\left(x-2\right)+2\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+3x+2\right)\)
\(=\left(x-2\right)\left(x^2+2x+x+2\right)\)
\(=\left(x-2\right)\left[x\left(x+2\right)+\left(x+2\right)\right]\)
\(=\left(x-2\right)\left(x+2\right)\left(x+1\right)\)
d) \(x^4-x^2+2x-1\)
\(=x^4-\left(x^2-2x+1\right)\)
\(=x^4-\left(x-1\right)^2\)
\(=\left(x^2\right)^2-\left(x-1\right)^2\)
\(=\left(x^2-x+1\right)\left(x^2+x-1\right)\)
e)Sửa đề \(x^4+x^3+x^2-1\)
\(=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)\)
\(=\left(x+1\right)\left(x^3+x-1\right)\)
f) \(x^3-4x^2+4x-1\)
\(=x^3-x^2-3x^2+3x+x-1\)
\(=x^2\left(x-1\right)-3x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2-3x+1\right)\)
Ta có:
3/4x-1/4=2(x-3)+1/4x
<=>3/4x-1/4x=2x-6+1/4
<=>x-2x=-23/4
<=>-x=-23/4
<=>x=23/4.
Vậy x=23/4