\(-x\)+\(x^2=8\)
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a) x8 : x2 = 16
x6 = 16 = ... ( chỗ này bn xem có số nào mũ 6 = 16 ko nha)
...
b) x3.x2.x-4 = 60
x3+2-4 = 60
x-1 = 60 = (1/60)-1
=> x = 1/60
a) \(\left(x-1\right)^3+3\left(x+1\right)^2=\left(x^2-2x+4\right)\)
\(\Leftrightarrow x^3+9x+2=x^3+8\)
\(\Leftrightarrow x^3+9x=x^3+8-2\)
\(\Leftrightarrow x^3+9x=x^3+6\)
\(\Leftrightarrow x^3+9x=x^3+6x-x^3\)
\(\Leftrightarrow\frac{2}{3}\)
b) \(x^2-4=8\left(x-2\right)\)
\(\Leftrightarrow x^2-4=8x-16\)
\(\Leftrightarrow x^4-4=8x-16+16\)
\(\Leftrightarrow x^2+12=8x\)
\(\Leftrightarrow x^2+12=8x-8x\)
\(\Leftrightarrow x^2-8x+12=0\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=6\end{cases}}\)
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
\(\left[{}\begin{matrix}x=-\dfrac{8029}{8}+\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\\x=-\dfrac{8029}{8}-\dfrac{1}{2}.\sqrt{\left(\dfrac{8029}{4}\right)^2-2017}\end{matrix}\right.\)
\(a,78-\left(x-3\right)=16\)
\(78-x+3=16\)
\(51-x=16\)
\(x=51-16=35\)
\(b,\left(2x+3\right):4=60\)
\(2x+3=60\cdot4=240\)
\(2x=240-3=237\)
\(x=\frac{237}{2}\)
\(c,\left(2x-8\right)=\left(2x-8\right)^7\)
\(\left(2x-8\right)-\left(2x-8\right)^7=0\)
Để mik nghĩ thêm tí r làm tiép
Để tui làm nốt ý C cho !!!
( 2x - 8 ) = ( 2x - 8 )7
=> ( 2x - 8 ) . 1 - ( 2x - 8 ) . ( 2x - 8 )6 = 0
=> ( 2x - 8 ) . [ 1 - ( 2x - 8 )6 ] = 0
=> 1 - ( 2x - 8 )6 = 0
=> ( 2x - 8 )6 = 1
=> ( 2x - 8 )6 = ( 16 )
\(\Rightarrow\orbr{\begin{cases}2x-8=1\\2x-8=-1\end{cases}\Rightarrow\orbr{\begin{cases}2x=9\\2x=7\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{9}{2}\\x=\frac{7}{2}\end{cases}}}\)
Vậy ........
a, \(\left(2x-6\right)\left(x-5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
b, \(2x+\frac{1}{2}=5\)
\(2x=5-\frac{1}{2}=\frac{9}{2}\)
\(x=\frac{9}{4}\)
c, bn viết thiếu đề rồi. Nếu đề là vậy thì như này :
\(8-x+\frac{1}{5}=\frac{41}{5}-x\)
a) \(\left(2\times x-6\right)\left(x-5\right)=0\)
=> \(\orbr{\begin{cases}2\times x-6=0\\x-5=0\end{cases}}\)
=> \(\orbr{\begin{cases}2\times x=6\\x=5\end{cases}}\)
=> \(\orbr{\begin{cases}x=3\\x=5\end{cases}}\)
Vậy x = 3,x = 5
b) \(2\times x+\frac{1}{2}=5\)
=> \(2\times x=5-\frac{1}{2}\)
=> \(2\times x=\frac{9}{2}\)
=> \(x=\frac{9}{2}:2=\frac{9}{2}\cdot\frac{1}{2}=\frac{9}{4}\)
Còn câu c thiếu
\(x\in B\left(8\right),x< 40\)
\(\Rightarrow B\left(8\right)=\left\{8;16;24;32;40;...\right\}\)
mà x < 40
\(\Rightarrow x=\left\{8;16;24;32\right\}\)
|7-x|=-13-5.(-8)
|7-x|=-13-(-40)
|7-x|=27
TH1:7-x=27
x=-20
TH2:7-x=-27
x=34
Vậy x=-20;x=34
Lời giải:
$-x+x^2=8$
$x^2-x-8=0$
$4x^2-4x-32=0$
$(2x-1)^2-33=0$
$(2x-1)^2=33$
$\Rightarrow 2x-1=\pm \sqrt{33}$
$\Rightarrow x=\frac{1\pm \sqrt{33}}{2}$