1.a. (x - 1/2)^2=0 b.(x - 2/5)^2=4/25
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16 tháng 11 2021
a: \(x\in\left\{0;25\right\}\)
c: \(x\in\left\{0;5\right\}\)
SB
1
6 tháng 8 2017
a) \(x^2-25-\left(x+5\right)=0\Leftrightarrow x^2-25-x-5=0\Leftrightarrow x^2-x-30=0\)
\(\Leftrightarrow x^2+5x-6x-30=0\Leftrightarrow x\left(x+5\right)-6\left(x+5\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+5\right)=0\Leftrightarrow\left\{{}\begin{matrix}x-6=0\\x+5=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\) vậy \(x=6;x=-5\)
b) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(2-4x=0\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{2}{4}=\dfrac{1}{2}\) vậy \(x=\dfrac{1}{2}\)
c) \(x^2\left(x^2+4\right)-x^2-4=0\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\Leftrightarrow\left\{{}\begin{matrix}x^2-1=0\\x^2+4=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x^2=1\\x^2=-4\left(vôlí\right)\end{matrix}\right.\)
ta có : \(x^2=1\Leftrightarrow x=\pm1\) vậy \(x=1;x=-1\)
SB
1
MK
6 tháng 8 2017
Tìm x:
a) \(x^2-25-\left(x+5\right)=0\)
\(\Leftrightarrow x^2-x-30=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=6\\ x=-5 \end{array} \right.\)
b) \(\left(2x-1\right)^2-\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
\(\Leftrightarrow2-4x=0\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
c) \(x^2\left(x^2+4\right)-x^2-4=0\)
\(\Leftrightarrow x^2\left(x^2+4\right)-\left(x^2+4\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+4\right)=0\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x^2-1=0\\ x^2+4=0 \end{array} \right.\)
\(\Leftrightarrow\)\(\left[\begin{array}{} x=1\\ x=-1 \end{array} \right.\)
CM
11 tháng 9 2019
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
DK
5 tháng 6 2021
Áp dụng BĐT phụ \(a^2+b^2\ge\dfrac{1}{2}\left(a+b\right)^2\Leftrightarrow\left(a-b\right)^2\ge0\)
\(A\ge\dfrac{1}{2}\left(x+y+\dfrac{1}{x}+\dfrac{1}{y}\right)^2\ge\dfrac{1}{2}\left(x+y+\dfrac{4}{x+y}\right)^2=\dfrac{1}{2}\left(1+\dfrac{4}{1}\right)^2=\dfrac{25}{2}\)
Dấu "=" \(x=y=\dfrac{1}{2}\)
16 tháng 8 2022
Bài 2:
a: \(\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\)
=>(x+5)(x-6)=0
=>x=-5 hoặc x=6
b: \(\Leftrightarrow4x^2-4x+1-4x^2+1=0\)
=>-4x+2=0
hay x=1/2
c: \(\Leftrightarrow\left(x^2+4\right)\left(x^2-1\right)=0\)
=>x=1 hoặc x=-1
KV
17 tháng 12 2020
a) x(2x - 1) - (x - 2)(2x + 3) = 5
2x2 - x - 2x2 - 3x + 4x + 6 = 5
0x = -1 (vô lý)
Vậy không tìm được x
b) (x - 3)2 - 25 = 0
(x - 3)2 - 52 = 0
(x - 3 - 5)(x - 3 + 5) = 0
(x - 8)(x + 2) = 0
\(\Rightarrow\) x - 8 = 0 hoặc x + 2 = 0
*) x - 8 = 0
x = 0 + 8
x = 8
*) x + 2 = 0
x = 0 - 2
x = -2
Vậy x = 8; x = -2
c) (x - 1)(2 - x) + (x + 3)2 = 4 - 2x
2x - x2 - 2 + x + x2 + 6x + 9 = 4 - 2x
9x + 7 = 4 - 2x
9x + 2x = 4 - 7
11x = -3
x = \(\dfrac{-3}{11}\)
Vậy x = \(\dfrac{-3}{11}\)
VH
0
21 tháng 8 2021
a) \(\left(x+3\right)^2-\left(x-2\right)^3=\left(x+5\right)\left(x^2-5x+25\right)-108\)
\(\Leftrightarrow x^2+6x+9-x^2+4x-4=x^3-5x^2+25x+5x^2-25x+125-108\)
\(\Leftrightarrow x^3-10x+12=0\Leftrightarrow\left(x-2\right)\left(x^2+2x+6\right)=0\)
\(\Leftrightarrow x=2\)( do \(x^2+2x+6=\left(x+1\right)^2+4\ge4>0\))
7 tháng 11 2021
b) x(x-4) - 2x+8 = 0
x(x-4) - 2(x-4) = 0
(x-2) (x-4) = 0
TH1: x-2=0 TH2: x-4=0
x=2 x=4
Vậy x\(\in\){2;4}
7 tháng 11 2021
\(b,\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=4\end{matrix}\right.\\ c,\Leftrightarrow\left(x-5\right)\left(x+5\right)-\left(x+5\right)=0\\ \Leftrightarrow\left(x+5\right)\left(x-6\right)=0\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-5\end{matrix}\right.\\ d,\Leftrightarrow\left(2x-1\right)^2-\left(2x-1\right)\left(2x+1\right)=0\\ \Leftrightarrow\left(2x-1\right)\left(2x-1-2x-1\right)=0\\ \Leftrightarrow x=\dfrac{1}{2}\\ e,\Leftrightarrow\left(3x-1-x-5\right)\left(3x-1+x+5\right)=0\\ \Leftrightarrow\left(2x-6\right)\left(4x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=3\end{matrix}\right.\\ f,\Leftrightarrow\left(x-2\right)\left(x^2+2x+4\right)-\left(x-2\right)\left(x-12\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x^2+x+16\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2\\\left(x+\dfrac{1}{2}\right)^2+\dfrac{63}{4}=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow x=2\)
a)\(\left(x-\frac{1}{2}\right)^2=0\)
\(\Rightarrow x-\frac{1}{2}=0\)
\(\Rightarrow x=\frac{1}{2}\)
b) \(\left(x-\frac{2}{5}\right)^2=\frac{4}{25}\)
\(\Rightarrow\orbr{\begin{cases}x-\frac{2}{5}=\frac{2}{5}\\x-\frac{2}{5}=-\frac{2}{5}\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{2}{5}+\frac{2}{5}=\frac{4}{5}\\x=-\frac{2}{5}+\frac{2}{5}=0\end{cases}}\)