x^3-64x=0
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26 tháng 8 2021
a) 9-64x^2=0
=> 64x^2 = 8
=> \(x^2=\frac{8}{64}=\frac{1}{8}\)
=> \(x=\frac{1}{\sqrt{8}}\)
b ) 25x^2 - 3 = 0
=> 25x^2 = 3
=> \(x^2=\frac{3}{25}\)
=> \(x=\frac{\sqrt{3}}{5}\)
C) 7 - 16x^2 =0
=> 16x^2 = 7
=> \(x^2=\frac{7}{16}\)
=> \(x=\frac{\sqrt{7}}{4}\)
d) 4x^2 - (x-4)^2 = 0
=> 4x^2 - x^2 + 8x - 16 =0
=> 3x^2 + 8x -16 = 0
=> ( 3x^2 + 12x ) - ( 4x +16 ) = 0
=> 3x( x + 4 ) - 4( x + 4 ) = 0
=>( x + 4 )( 3x - 4 ) = 0
=> \(\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}\)
e) ( 3x + 4 )^2 - ( 2x - 5 )^2 = 0
=> ( 3x + 4 + 2x - 5 )( 3x + 4 - 2x + 5 ) = 0
=> ( 5x -1 ) ( x + 9 ) = 0
=> \(\orbr{\begin{cases}5x-1=0\\x+9=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=\frac{1}{5}\\x=-9\end{cases}}\)
QA
26 tháng 8 2021
Trả lời:
a, \(9-64x^2=0\)
\(\Leftrightarrow\left(3-8x\right)\left(3+8x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3-8x=0\\3+8x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{8}\\x=-\frac{3}{8}\end{cases}}}\)
Vậy x = 3/8; x = - 3/8 là nghiệm của pt.
b, \(25x^2-3=0\)
\(\Leftrightarrow\left(5x-\sqrt{3}\right)\left(5x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}5x-\sqrt{3}=0\\5x+\sqrt{3}=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{3}}{5}\\x=-\frac{\sqrt{3}}{5}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{3}}{5}\)
c, \(7-16x^2=0\)
\(\Leftrightarrow\left(\sqrt{7}-4x\right)\left(\sqrt{7}+4x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}\sqrt{7}-4x=0\\\sqrt{7}+4x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{\sqrt{7}}{4}\\x=-\frac{\sqrt{7}}{4}\end{cases}}}\)
Vậy \(x=\pm\frac{\sqrt{7}}{4}\)
d, \(4x^2-\left(x-4\right)^2=0\)
\(\Leftrightarrow\left(2x-x+4\right)\left(2x+x-4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(3x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\3x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-4\\x=\frac{4}{3}\end{cases}}}\)
Vậy x = - 4; x = 4/3 là nghiệm của pt.
e, \(\left(3x+4\right)^2-\left(2x-5\right)^2=0\)
\(\Leftrightarrow\left(3x+4-2x+5\right)\left(3x+4+2x-5\right)=0\)
\(\Leftrightarrow\left(x+9\right)\left(5x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+9=0\\5x-1=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-9\\x=\frac{1}{5}\end{cases}}}\)
Vậy x = - 9; x = 1/5 là nghiệm của pt.
PT
2
23 tháng 11 2021
a) \(\Rightarrow x^2\left(x^2-64\right)=0\Rightarrow x^2\left(x-8\right)\left(x+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\\x=-8\end{matrix}\right.\)
b) \(\Rightarrow x^2\left(x-6\right)+3x\left(x-6\right)+21\left(x-6\right)=0\Rightarrow\left(x-6\right)\left(x^2+3x+21\right)=0\)
\(\Rightarrow x=6\)
FS
1
17 tháng 3 2019
\(x^6-6x^4-64x^3+12x^2-8=0\)
\(\Leftrightarrow\left(x^2-4x-2\right)\left(x^4+4x^3+12x^2-8x+4\right)=0\)
\(\Leftrightarrow\left(x^2-4x-2\right)\left[\left(x^4+4x^3+4x^2\right)+\left(8x^2-8x+\frac{8}{4}\right)+2\right]=0\)
\(\Leftrightarrow\left(x^2-4x-2\right)\left[\left(x^2+2x\right)^2+8\left(x-\frac{1}{2}\right)^2+2\right]=0\)
\(\Leftrightarrow x^2-4x-2=0\)
\(\Leftrightarrow x=2\pm\sqrt{6}\)
26 tháng 10 2021
a: \(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
26 tháng 10 2021
a, \(\sqrt{\left(2x-3\right)^2}=7\\ \Rightarrow\left|2x-3\right|=7\\ \Rightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c, \(\sqrt{x^2-9}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\sqrt{x+3}-3\sqrt{x-3}=0\\ \Rightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}\sqrt{x-3}=0\\\sqrt{x+3}-3=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x-3=0\\x+3=9\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
2 tháng 10 2021
a) \(\sqrt{\left(2x-3\right)^2}=7\)
\(\Leftrightarrow\left|2x-3\right|=7\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
b) \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\left(đk:x\ge-2\right)\)
\(\Leftrightarrow8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}=20\)
\(\Leftrightarrow5\sqrt{x+2}=20\)
\(\Leftrightarrow\sqrt{x+2}=4\Leftrightarrow x+2=16\Leftrightarrow x=14\left(tm\right)\)
c) \(\sqrt{x^2-9}-3\sqrt{x-3}=0\left(đk:x\ge3\right)\)
\(\Leftrightarrow\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}\left(\sqrt{x+3}-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\\sqrt{x+3}=3\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x+3=9\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=3\left(tm\right)\\x=6\left(tm\right)\end{matrix}\right.\)
HP
2 tháng 10 2021
a. \(\sqrt{\left(2x-3\right)^2}=7\)
<=> \(\left|2x-3\right|=7\)
<=> \(\left[{}\begin{matrix}2x-3=7\left(x\ge\dfrac{3}{2}\right)\\-2x+3=7\left(x< \dfrac{3}{2}\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}2x=10\\-2x=4\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=5\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
b. \(\sqrt{64x+128}-\sqrt{25x+50}+\sqrt{4x+8}=20\) ĐK: \(x\ge-2\)
<=> \(\sqrt{64\left(x+2\right)}-\sqrt{25\left(x+2\right)}+\sqrt{4\left(x+2\right)}-20=0\)
<=> \(8\sqrt{x+2}-5\sqrt{x+2}+2\sqrt{x+2}-20=0\)
<=> \(\sqrt{x+2}.\left(8-5+2\right)-20=0\)
<=> \(5\sqrt{x+2}=20\)
<=> \(\sqrt{x+2}=4\)
<=> \(\left(\sqrt{x+2}\right)^2=4^2\)
<=> \(\left|x+2\right|=16\)
<=> \(\left[{}\begin{matrix}x+2=16\left(x\ge-2\right)\\x+2=-16\left(x< -2\right)\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=14\left(TM\right)\\x=-18\left(TM\right)\end{matrix}\right.\)
c. \(\sqrt{x^2-9}-3\sqrt{x-3}=0\) ĐK: \(x\ge3\)
<=> \(\sqrt{\left(x-3\right)\left(x+3\right)}-3\sqrt{x-3}=0\)
<=> \(\sqrt{x-3}.\sqrt{x+3}-3\sqrt{x-3}=0\)
<=> \(\left(\sqrt{x+3}-3\right).\sqrt{x-3}=0\)
<=> \(\left[{}\begin{matrix}\sqrt{x+3}-3=0\\\sqrt{x-3}=0\end{matrix}\right.\)
<=> \(\left[{}\begin{matrix}x=6\\x=3\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
22 tháng 8 2023
Lời giải:
a.
$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$
b.
$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$
c.
$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$
d.
$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$
$=(x+y)(x^2-4xy+7y^2)$
H9
HT.Phong (9A5)
CTVHS
22 tháng 8 2023
a) \(64x^2-24y^2\)
\(=8\left(8x^2-3y^2\right)\)
b) \(64x^3-27y^3\)
\(=\left(4x\right)^3-\left(3y\right)^3\)
\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)
c) \(x^4-2x^3+x^2\)
\(=x^2\left(x^2-2x+1\right)\)
\(=x^2\left(x-1\right)^2\)
d) \(\left(x-y\right)^3+8y^3\)
\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)
\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)
VT
1
23 tháng 9 2017
pt<=>\(\sqrt{\left(x+6\right)^3}+\sqrt{x+6}=\left(x^2+4x\right)^3+x^2+4x\)
đặt\(\sqrt{x+6}=a;x^2+4x=b\)
H
4
30 tháng 6 2019
4x3 - 64x = 0
<=> 4x(x + 4)(x - 4) = 0
<=> \(\hept{\begin{cases}4x=0\\x+4=0\\x-4=0\end{cases}}\) <=> \(\hept{\begin{cases}x=0\\x=-4\\x=4\end{cases}}\)
=> x = 0 hoặc x = -4 hoặc x = 4
x^3-64x=0
<=>x^3-64x=(x-8)x(x+8)
=>(x-8)x(x+8)=0
Th1:x-8=0
=>x=8
Th2:x+8=0
=>x=-8
vậy pt có x=±8
x^3-64x=0
x.x.x-64x=0
=>x.x=64
Ta có:
x.x=64
x=\(\sqrt{64}\)
x=8