Tìm x biết:4x²-4x+1=25
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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 \(\left\{{}\begin{matrix}x=\dfrac{1}{2}\\y=\dfrac{-2}{5}\\z=\dfrac{1}{10}\end{matrix}\right.\)
a x = \(\dfrac{-1}{12}\)
b x = \(\dfrac{-4}{3}\)
c x = \(\dfrac{-1}{6}\)
d x = \(\dfrac{-1}{4}\)
\(\left(4x+1\right)^2=\dfrac{4}{9}\)
\(\left(4x+1\right)=\perp\left(\dfrac{2}{3}\right)^2\)
\(\text{Vậy }4x+1=\dfrac{2}{3}\)
\(4x\) \(=\dfrac{2}{3}+\left(-1\right)=\dfrac{-1}{3}\)
\(x\) \(=\left(\dfrac{-1}{3}\right).\dfrac{1}{4}=\dfrac{-1}{12}\)
\(\text{hoặc }4x+1=\dfrac{-2}{3}\)
\(4x\) \(=\left(\dfrac{-2}{3}\right)+\left(-1\right)=\dfrac{-5}{3}\)
\(x\) \(=\left(\dfrac{-5}{3}\right).\dfrac{1}{4}=\dfrac{-5}{12}\)
\(\Rightarrow x\in\left\{\dfrac{-1}{12};\dfrac{-5}{12}\right\}\)
\(\left(3x-1\right)^2=25\)
\(\left(3x-1\right)^2=\perp\left(5\right)^2\)
\(\text{Vậy }3x-1=5\)
\(3x\) \(=5+1=6\)
\(x\) \(=6:3=2\)
\(\text{hoặc }3x-1=-5\)
\(3x\) \(=\left(-5\right)+1=-4\)
\(x\) \(=\left(-4\right):3=\dfrac{-4}{3}\)
\(\Rightarrow x\in\left\{2;\dfrac{-4}{3}\right\}\)
\(\left(x-\dfrac{1}{3}\right)^2=\dfrac{1}{4}\)
\(\left(x-\dfrac{1}{3}\right)^2=\perp\left(\dfrac{1}{2}\right)^2\)
\(\text{Vậy }x-\dfrac{1}{3}=\dfrac{1}{2}\)
\(x\) \(=\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
\(\text{hoặc }x-\dfrac{1}{3}=\dfrac{-1}{2}\)
\(x\) \(=\left(\dfrac{-1}{2}\right)+\dfrac{1}{3}=\dfrac{-1}{6}\)
\(\Rightarrow x\in\left\{\dfrac{5}{6};\dfrac{-1}{6}\right\}\)
\(\left(4x-3\right)^2=16\)
\(\left(4x-3\right)=\perp\left(4\right)^2\)
\(\text{Vậy }4x-3=4\)
\(4x\) \(=4+3=7\)
\(x\) \(=7:4=\dfrac{7}{4}\)
\(\text{hoặc }4x-3=-4\)
\(4x\) \(=\left(-4\right)+3=-1\)
\(x\) \(=\left(-1\right):4=\dfrac{-1}{4}\)
\(\Rightarrow x\in\left\{\dfrac{7}{4};\dfrac{-1}{4}\right\}\)
4x2 - 20x + 9 = 0
(4x^2 - 2x) - (18x - 9) = 0
2x(2x - 1) - 9(2x - 1) = 0
(2x - 1)(2x - 9) = 0
2x - 1 = 0 hoặc 2x - 9 = 0
x = 1/2 hoặc x = 9/2
Bài 1.
1) ( 2x + 1 )3 - ( 2x + 1 )( 4x2 - 2x + 1 ) - 3( 2x - 1 ) = 15
<=> 8x3 + 12x2 + 6x + 1 - [ ( 2x )3 - 13 ] - 6x + 3 = 15
<=> 8x3 + 12x2 + 4 - 8x3 + 1 = 15
<=> 12x2 + 15 = 15
<=> 12x2 = 0
<=> x = 0
2) x( x - 4 )( x + 4 ) - ( x - 5 )( x2 + 5x + 25 ) = 13
<=> x( x2 - 16 ) - ( x3 - 53 ) = 13
<=> x3 - 16x - x3 + 125 = 13
<=> 125 - 16x = 13
<=> 16x = 112
<=> x = 7
Bài 2.
A = ( x + 5 )( x2 - 5x + 25 ) - ( 2x + 1 )3 - 28x3 + 3x( -11x + 5 )
= x3 + 53 - ( 8x3 + 12x2 + 6x + 1 ) - 28x3 - 33x2 + 15x
= -27x3 + 125 - 8x3 - 12x2 - 6x - 1 - 33x2 + 15x
= -33x3 - 45x2 + 9x + 124 ( có phụ thuộc vào biến )
B = ( 3x + 2 )3 - 18x( 3x + 2 ) + ( x - 1 )3 - 28x3 + 3x( x - 1 )
= 27x3 + 54x2 + 36x + 8 - 54x2 - 36x + x3 - 3x2 + 3x - 1 - 28x3 + 3x2 - 3x
= 7 ( đpcm )
C = ( 4x - 1 )( 16x2 + 4x + 1 ) - ( 4x + 1 )3 + 12( 4x + 1 )3 + 12( 4x + 1 ) - 15
= ( 4x )3 - 13 - [ ( 4x + 1 )3 - 12( 4x + 1 )3 - 12( 4x + 1 ) ] - 15
= 64x3 - 1 - ( 4x + 1 )[ ( 4x + 1 )2 - 12( 4x + 1 )2 - 12 ] - 15
= 64x3 - 16 - ( 4x + 1 )[ 16x2 + 8x + 1 - 12( 16x2 + 8x + 1 ) - 12 ]
= 64x3 - 16 - ( 4x + 1 )( 16x2 + 8x - 11 - 192x2 - 96x - 12 )
= 64x3 - 16 - ( 4x + 1 )( -176x2 - 88x - 23 )
= 64x3 - 16 - ( -704x3 - 528x2 - 180x - 23 )
= 64x3 - 16 + 704x3 + 528x2 + 180x + 23
= 768x3 + 528x2 + 180x + 7 ( có phụ thuộc vào biến )
ĐKXĐ:\(2x-5\ne0\Rightarrow x\ne\dfrac{5}{2}\)
\(\dfrac{4x^2-25}{2x-5}=0\\ \Leftrightarrow\dfrac{\left(2x-5\right)\left(2x+5\right)}{2x-5}=0\\ \Leftrightarrow2x+5=0\\ \Leftrightarrow x=-\dfrac{5}{2}\left(tm\right)\)
\(4x^2-4x+1=25\)
\(4x^2-4x\) \(=25-1\)
\(4x^2-4x\) \(=24\)
Toi day thi hk p lam nua, nhung cuoi cung suy ra x =3
+) Ta có: 9 = 1 . 9 = 9.1 = 3.3
Lập bảng :
4x +1 | 1 | 9 | 3 |
y - 3 | 9 | 1 | 3 |
x | 0 | 2 | 1/2 |
y | 12 | 4 | 9 |
Vì x thuộc n nên ...
+) Ta có : 25 = 1.25 = 25.1 = 5.5
Lập bảng:
2x + 3 | 1 | 25 | 5 |
y + 4 | 25 | 1 | 5 |
x | -1 | 11 | 1 |
y | 21 | -3 | 1 |
vì x thuộc N nên...
a. 4x2-25=0
=> (2x)2-52=0
=> (2x-5)(2x+5)=0
=> 2x-5=0 hoặc 2x+5=0
=> 2x=5 hoặc 2x=-5
=> x=5:2 hoặc x=-5:2
=> x=2,5 hoặc x=-2,5
b. (x-1)(4x2-49)=0
=> (x-1)[(2x)2-72 ]=0
=> (x-1)(2x-7)(2x+7)=0
=> x-1=0 hoặc 2x-7=0 hoặc 2x+7=0
=> x=1 hoặc 2x=7 hoặc 2x=-7
=> x=1 hoặc x=7:2=3,5 hoặc x=-7:2=-3,5
4x(x-1)+1=25
4x(x-1)=24
x(x-1)=6
=>x=3