4x^2-4x-36x^2+1
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AH
Akai Haruma
Giáo viên
24 tháng 9 2023
Lời giải:
a/
PT $\Leftrightarrow \sqrt{(2x-1)^2}=3$
$\Leftrightarrow |2x-1|=3\Leftrightarrow 2x-1=\pm 3$
$\Leftrightarrow x=2$ hoặc $x=-1$ (đều tm)
b/ ĐKXĐ: $x\geq 1$
PT $\Leftrightarrow \sqrt{49(x-1)}-\sqrt{36(x-1)}=3\sqrt{2}$
$\Leftrightarrow 7\sqrt{x-1}-6\sqrt{x-1}=3\sqrt{2}$
$\Leftrightarrow \sqrt{x-1}=\sqrt{18}$
$\Leftrightarrow x-1=18$
$\Leftrightarrow x=19$ (tm)
EM
2
9 tháng 7 2018
\(A=\sqrt{4x^2-4x+1}+\sqrt{4x^2-36x+81}\)
\(=\sqrt{\left(2x\right)^2-2.2x.1+1^2}+\sqrt{\left(2x\right)^2-2.2x.9+9^2}\)
\(=\sqrt{\left(2x-1\right)^2}+\sqrt{\left(2x-9\right)^2}\)
\(=\left|2x-1\right|+\left|2x-9\right|\)
\(=2x-1+9-2x=8\)
10 tháng 7 2021
a,\(\sqrt{\left(3x-1\right)^2}=5=>|3x-1|=5=>\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b, \(\sqrt{4x^2-4x+1}=3=\sqrt{\left(2x-1\right)^2}=3=>\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\)
\(=>\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c, \(\sqrt{x^2-6x+9}+3x=4=>|x-3|=4-3x\)
TH1: \(|x-3|=x-3< =>x\ge3=>x-3=4-3x=>x=1,75\left(ktm\right)\)
TH2 \(|x-3|=3-x< =>x< 3=>3-x=4-3x=>x=0,5\left(tm\right)\)
Vậy x=0,5...
d, đk \(x\ge-1\)
=>pt đã cho \(< =>9\sqrt{x+1}-6\sqrt{x+1}+4\sqrt{x+1}=12\)
\(=>7\sqrt{x+1}=12=>x+1=\dfrac{144}{49}=>x=\dfrac{95}{49}\left(tm\right)\)
10 tháng 7 2021
a) Ta có: \(\sqrt{\left(3x-1\right)^2}=5\)
\(\Leftrightarrow\left|3x-1\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}3x-1=5\\3x-1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3x=6\\3x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{4}{3}\end{matrix}\right.\)
b) Ta có: \(\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow\left|2x-1\right|=3\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=3\\2x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=4\\2x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
c) Ta có: \(\sqrt{x^2-6x+9}+3x=4\)
\(\Leftrightarrow\left|x-3\right|=4-3x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=4-23x\left(x\ge3\right)\\x-3=23x-4\left(x< 3\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x+23x=4+3\\x-23x=4+3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{24}\left(loại\right)\\x=\dfrac{-4}{22}=\dfrac{-2}{11}\left(loại\right)\end{matrix}\right.\)
IL
2
20 tháng 10 2020
a) a3 - a2c + a2b - abc
= a( a2 - ac + ab - bc )
= a[ ( a2 + ab ) - ( ac + bc ) ]
= a( a( a + b ) - c( a + b ) ]
= a( a + b )( a - c )
b) ( x2 + 1 )2 - 4x2
= ( x2 + 1 )2 - ( 2x )2
= ( x2 - 2x + 1 )( x2 + 2x + 1 )
= ( x - 1 )2( x + 1 )2
c) x2 - 10x - 9y2 + 25
= ( x2 - 10x + 25 ) - 9y2
= ( x - 5 )2 - ( 3y )2
= ( x - 3y - 5 )( x + 3y - 5 )
d) 4x2 - 36x + 56
= 4x2 - 8x - 28x + 56
= 4x( x - 2 ) - 28( x - 2 )
= 4( x - 2 )( x - 7 )
TQ
1
H9
HT.Phong (9A5)
CTVHS
6 tháng 11 2023
a) \(4x^3-36x=0\)
\(\Leftrightarrow4x\left(x^2-9\right)=0\)
\(\Leftrightarrow4x\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x+3=0\\x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=3\end{matrix}\right.\)
b) \(\left(x-2\right)^2-4x+8=0\)
\(\Leftrightarrow\left(x-2\right)^2-\left(4x-8\right)=0\)
\(\Leftrightarrow\left(x-2\right)^2-4\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-2-4\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(x-6\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
c) \(x^3+\left(x+3\right)\left(x-9\right)=-27\)
\(\Leftrightarrow\left(x^3+27\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-9\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-2x\right)=0\)
\(\Leftrightarrow x\left(x+3\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\\x-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\\x=2\end{matrix}\right.\)
KL
0
NV
1
ND
Nguyễn Đức Trí
VIP
22 tháng 8 2023
2) \(x^4-5x^2+4\)
\(=x^4-x^2-4x^2+4\)
\(=x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
\(=\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(4x^2-4x-36x^2+1\\ =\left(4x^2-4x+1\right)-36x^2\\ =\left(2x-1\right)^2-\left(6x\right)^2\\ =\left(2x-1-6x\right)\left(2x-1+6x\right)\\ =-\left(4x+1\right)\left(8x-1\right)\)
Cảm ơn ạ