Tìm x biết:
\(x^2-4x+4=25\)
K
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29 tháng 4 2016
x2 - 4x + 4 = 25
x2 - 4x = 25 - 4
x2 - 4x = 21
x2 = 21 + 4x
x.x = 21 + 4x
\(\Rightarrow\) 3x = 21
x = 21 : 3
x = 7
4 tháng 4 2022
vì \(\left(4x^2-4x+1\right)^{2022}\ge0\left(\forall x\right)\),\(\left(y^2-\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}\ge0\left(\forall y\right)\),\(\left|x+y+z\right|\ge0\)
mà \(\left(4x^2-4x+1\right)^{2022}+\left(y^2+\dfrac{4}{5}y+\dfrac{4}{25}\right)^{2022}+\left|x+y-z\right|=0\)
=>\(\left\{{}\begin{matrix}4x^2-4x+1=0\\y^2+\dfrac{4}{5}y+\dfrac{4}{25}=0\\x+y-z=0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2x-1=0\\y+\dfrac{2}{5}=0\\x+y-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\\dfrac{1}{2}-\dfrac{2}{5}-z=0\end{matrix}\right.\)
<=>\(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
KL: vậy \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
BK
0
30 tháng 12 2021
a x = \(\dfrac{-1}{12}\)
b x = \(\dfrac{-4}{3}\)
c x = \(\dfrac{-1}{6}\)
d x = \(\dfrac{-1}{4}\)
30 tháng 12 2021
\(\left(4x+1\right)^2=\dfrac{4}{9}\)
\(\left(4x+1\right)=\perp\left(\dfrac{2}{3}\right)^2\)
\(\text{Vậy }4x+1=\dfrac{2}{3}\)
\(4x\) \(=\dfrac{2}{3}+\left(-1\right)=\dfrac{-1}{3}\)
\(x\) \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)
\(\text{hoặc }4x+1=\dfrac{-2}{3}\)
\(4x\) \(=\left(\dfrac{-2}{3}\right)+\left(-1\right)=\dfrac{-5}{3}\)
\(x\) \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)
\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)
\(\left(3x-1\right)^2=25\)
\(\left(3x-1\right)^2=\perp\left(5\right)^2\)
\(\text{Vậy }3x-1=5\)
\(3x\) \(=5+1=6\)
\(x\) \(=6:3=2\)
\(\text{hoặc }3x-1=-5\)
\(3x\) \(=\left(-5\right)+1=-4\)
\(x\) \(=\left(-4\right):3=\dfrac{-4}{3}\)
\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)
\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\left(x-\dfrac{1}{3}\right)^2=\perp\left(\dfrac{1}{2}\right)^2\)
\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)
\(x\) \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)
\(x\) \(=\left(\dfrac{-1}{2}\right)+\dfrac{1}{3}=\dfrac{-1}{6}\)
\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)
\(\left(4x-3\right)^2=16\)
\(\left(4x-3\right)=\perp\left(4\right)^2\)
\(\text{Vậy }4x-3=4\)
\(4x\) \(=4+3=7\)
\(x\) \(=7:4=\dfrac{7}{4}\)
\(\text{hoặc }4x-3=-4\)
\(4x\) \(=\left(-4\right)+3=-1\)
\(x\) \(=\left(-1\right):4=\dfrac{-1}{4}\)
\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)
BT
0
6 tháng 2 2022
a: =>25-4x=1
=>4x=24
hay x=6
b: =>2x-4=0
hay x=2
c: =>x-35=115
hay x=150
d: =>x-2=12
hay x=14
e: =>x-36=216
hay x=252
D
21 tháng 10 2016
8x2+30x+7=0
8x2+16x+14x+7=0
8x(x+2) +7(x+2)=0
(8x+7)(x+2)=0
=>\(\orbr{\begin{cases}8x+7=0\\x+2=0\end{cases}\Rightarrow\orbr{\begin{cases}x=-\frac{7}{8}\\x=-2\end{cases}}}\)
(x-2)2 = 25 => x- 2 = 5
=> x = 7
\(x^2-4x+4=25\)
\(\Leftrightarrow\left(x-2\right)^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}\)