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AD
31 tháng 7 2021
a)\(\left(4-x\right)^2-16=0\)
\(\Leftrightarrow\left(4-x\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=4\\4-x=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)
b) \(25-\left(3-x\right)^2=0\)
\(\Leftrightarrow\left(3-x\right)^2=25\)
\(\Leftrightarrow\orbr{\begin{cases}3-x=5\\3-x=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)
c)\(3x^2-6x+3-27=0\)
\(\Leftrightarrow3x^2-6x-24=0\)
\(\Leftrightarrow\left(3x+6\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x+6=0\\x-4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=4\end{cases}}}\)
#H
31 tháng 7 2021
1.(4-x)2-16 =0
<=> 16 -8x+x2 -16 =0
<=> -x(8-x) =0
<=> TH1: x=0
. TH2: 8-x=0
. => x= -8
2. 25 - (3-x)2 = 0
<=> 25 - (9-6x+x2) = 0
<=> 25 - 9+6x-x2 = 0
<=> -x2+6x+16 = 0
<=> -(x-8)(x+2) = 0 (bước này bạn nhập phương trình trên mtinh là nó sẽ ra nghiệm nhe)
<=> TH1:x-8=0
. x= 8
. TH2:x+2=0
. x=-2
3.(bạn tự làm nhé, giải bth thui)
QA
2 tháng 8 2021
Trả lời:
\(1,\left(4x-x\right)^2-16=0\)
\(\Leftrightarrow\left(3x\right)^2-16=0\)
\(\Leftrightarrow\left(3x-4\right)\left(3x+4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}3x-4=0\\3x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{4}{3}\\x=-\frac{4}{3}\end{cases}}}\)
Vậy x = 4/3; x = - 4/3 là nghiệm của pt.
\(2,25-\left(3-x\right)^2=0\)
\(\Leftrightarrow\left(5-3+x\right)\left(5+3-x\right)=0\)
\(\Leftrightarrow\left(2+x\right)\left(8-x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}2+x=0\\8-x=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=-2\\x=8\end{cases}}}\)
Vậy x = - 2; x = 8 là nghiệm của pt.
\(3,3x^2-6x+3-27=0\)
\(\Leftrightarrow3x^2-6x-24=0\)
\(\Leftrightarrow3\left(x^2-2x-8\right)=0\)
\(\Leftrightarrow x^2-2x-8=0\)
\(\Leftrightarrow x^2-4x+2x-8=0\)
\(\Leftrightarrow x\left(x-4\right)+2\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x+2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-4=0\\x+2=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}\)
Vậy x = 4; x = - 2 là nghiệm của pt.
21 tháng 9 2020
a) ( x - 3 )2 - 4 = 0
<=> ( x - 3 )2 - 22 = 0
<=> ( x - 3 - 2 )( x - 3 + 2 ) = 0
<=> ( x - 5 )( x - 1 ) = 0
<=> x = 5 hoặc x = 1
b( 2x + 3 )2 - ( 2x + 1 )( 2x - 1 ) = 22
<=> 4x2 + 12x + 9 - ( 4x2 - 1 ) = 22
<=> 4x2 + 12x + 9 - 4x2 + 1 = 22
<=> 12x + 10 = 22
<=> 12x = 12
<=> x = 1
c) ( 4x + 3 )( 4x - 3 ) - ( 4x - 5 )2 = 16
<=> 16x2 - 9 - ( 16x2 - 40x + 25 ) = 16
<=> 16x2 - 9 - 16x2 + 40x - 25 = 16
<=> 40x - 34 = 16
<=> 40x = 50
<=> x = 50/40 = 5/4
d) x3 - 9x2 + 27x - 27 = -8
<=> ( x - 3 )3 = -8
<=> ( x - 3 )3 = (-2)3
<=> x - 3 = -2
<=> x = 1
e) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
f) ( x - 2 )3 - x( x - 1 )( x + 1 ) + 6x2 = 5
<=> x3 - 6x2 + 12x - 8 - x( x2 - 1 ) + 6x2 = 5
<=> x3 + 12x - 8 - x3 + x = 5
<=> 13x - 8 = 5
<=> 13x = 13
<=> x = 1
21 tháng 9 2020
a) \(\left(x-3\right)^2-4=0\)
=> \(\left(x-3\right)^2-2^2=0\)
=> \(\left(x-3-2\right)\left(x-3+2\right)=0\)
=> \(\left(x-5\right)\left(x-1\right)=0\)
=> \(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
=> \(\left(2x+3\right)^2-\left[\left(2x\right)^2-1^2\right]=22\)
=> \(\left(2x+3\right)^2-\left(4x^2-1\right)=22\)
=> \(\left(2x\right)^2+2\cdot2x\cdot3+3^2-4x^2+1=22\)
=> \(4x^2+12x+9-4x^2+1=22\)
=> \(12x+9+1=22\)
=> \(12x+10=22\)
=> 12x = 12
=> x = 1
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
=> \(\left(4x\right)^2-3^2-\left[\left(4x\right)^2-2\cdot4x\cdot5+5^2\right]=16\)
=> \(16x^2-9-\left(16x^2-40x+25\right)=16\)
=> \(16x^2-9-16x^2+40x-25=16\)
=> \(-9+40x-25=16\)
=> \(40x=16+25-\left(-9\right)=16+25+9=50\)
=> x = 50/40 = 5/4
d) \(x^3-9x^2+27x-27=-8\)
=> \(x^3-3\cdot x^2\cdot3+3\cdot x\cdot3^2-3^3=8\)
=> \(\left(x-3\right)^3=-8\)
=> \(\left(x-3\right)^3=\left(-2\right)^3\)
=> x - 3 = -2 => x = 1
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
=> \(x^3+3x^2+3x+1-x^3-3x^2=2\)
=> \(3x+1=2\)
=> \(3x=1\)=> x = 1/3
f) \(\left(x-2\right)^3-x\left(x-1\right)\left(x+1\right)+6x^2=5\)
=> \(x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3-x\left(x^2-1\right)+6x^2=5\)
=> \(x^3-6x^2+12x-8-x^3+x+6x^2=5\)
=> \(\left(12x+x\right)-8=5\)
=> 13x = 13
=> x = 1
13 tháng 7 2021
a, ( 2x - 3 )2- (2x + 1)2 = -3
4x2-12x+9-4x2+4x-1=-3
-8x-1=-3
-8x=-2
x=\(\frac{1}{4}\)
b, (5x - 1) 2 - (5x + 4)(5x - 4) = 7
25x2-10x+1-25x2+16=7
-10x+17=7
-10x=-10
x=1
c, ( x- 5)2 + (x-3)(x+3) - 2(x + 1)2=0
x2-10x+25+x2-9-2x2-4x-2=0
-14x+14=0
-14(x-1)=0
=>x-1=0
x=1
AD
13 tháng 7 2021
a) \(\left(2x-3\right)^2-\left(2x+1\right)^2=-3\)
\(\Leftrightarrow4x^2-12x+9-4x^2-4x-1=-3\)
\(\Leftrightarrow-16x+8=-3\)
\(\Leftrightarrow-16x=-11\)
\(\Leftrightarrow x=\frac{11}{16}\)
b)\(\left(5x-1\right)^2-\left(5x+4\right)\left(5x-4\right)=7\)
\(\Leftrightarrow25x^2-10x+1-25x^2+16=7\)
\(\Leftrightarrow-10x+17=7\)
\(\Leftrightarrow-10x=-10\)
\(\Leftrightarrow x=1\)
c)\(\left(x-5\right)^2+\left(x-3\right)\left(x+3\right)-2\left(x+1\right)^2=0\)
\(\Leftrightarrow x^2-10x+25+x^2-9-2\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow2x^2-10x-16-2x^2-4x-2=0\)
\(\Leftrightarrow-14x-18=0\)
\(\Leftrightarrow-14x=18\)
\(\Leftrightarrow x=-\frac{9}{7}\)
#H
YN
4 tháng 10 2021
1, \(3x\left(x-7\right)+2x-14=0\)
\(\Rightarrow3x\left(x-7\right)+2\left(x-7\right)=0\)
\(\Rightarrow\left(x-7\right)\left(3x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=7\\x=\frac{-2}{3}\end{cases}}\)
2, \(x^3+3x^2-\left(x+3\right)=0\)
\(\Rightarrow x^2\left(x+3\right)-\left(x+3\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x^2-1\right)=0\)
\(\Rightarrow\left(x+3\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\pm1\end{cases}}\)
3, \(15x-5+6x^2-2x=0\)
\(\Rightarrow\left(15x-5\right)+\left(6x^2-2x\right)=0\)
\(\Rightarrow5\left(3x-1\right)+2x\left(3x-1\right)=0\)
\(\Rightarrow\left(3x-1\right)\left(5+2x\right)=0\)
\(\Rightarrow\orbr{\begin{cases}3x-1=0\\5+2x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{3}\\x=\frac{-5}{2}\end{cases}}\)
4, \(5x-2-25x^2+10x=0\)
\(\Rightarrow\left(5x-25x^2\right)-\left(2-10x\right)=0\)
\(\Rightarrow5x\left(1-5x\right)-2\left(1-5x\right)=0\)
\(\Rightarrow\left(1-5x\right)\left(5x-2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}1-5x=0\\5x-2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{1}{5}\\x=\frac{2}{5}\end{cases}}\)
8 tháng 8 2023
a. x mũ 2 - 2x + 1 = 25
= x^2 + 2.x.1 + 1^2
= ( x + 1 ) ^2
ko bt có đúng ko nữa, mấy câu kia tui ko bt lm
6 tháng 8 2021
e, \(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\Leftrightarrow\left(x^2-1\right)\left(x-4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)\left(x-4\right)=0\Leftrightarrow x=\pm1;x=4\)
f, \(2x^3-242x=0\Leftrightarrow2x\left(x^2-121\right)=0\)
\(\Leftrightarrow2x\left(x-11\right)\left(x+11\right)=0\Leftrightarrow x=\pm11;x=0\)
g, \(x^5-9x=0\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow x\left(x^2-3\right)\left(x^2+3>0\right)=0\Leftrightarrow x=\pm\sqrt{3};x=0\)
PT
2
12 tháng 7 2018
\(a,25x^2-9=0\)
\(\Leftrightarrow\left(5x-3\right)\left(5x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-3=0\\5x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{5}\\x=-\dfrac{3}{5}\end{matrix}\right.\)
\(b,\left(x+4\right)^2-\left(x+1\right)\left(x-1\right)=16\)
\(\Leftrightarrow2x=255\Leftrightarrow x=\dfrac{255}{2}\)\(\Leftrightarrow x^2+8x+16-x^2+1=16\)
\(\Leftrightarrow8x=-1\)
\(\Leftrightarrow x=-\dfrac{1}{8}\)
\(c,\left(2x-1\right)^2+\left(x+3\right)^2-5\left(x+7\right)\left(x-7\right)=0\)
\(\Leftrightarrow4x^2-4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow2x=-255\Leftrightarrow x=-\dfrac{255}{2}\)
TS
12 tháng 7 2018
\(a,25x^2-9=0\)
\(25x^2=9\)
\(x^2=\dfrac{9}{25}\)
\(x=\dfrac{3}{5}\)
( 4 - x ) mũ 2 - 16 = 0
<=>(4-x)^2=4^2
<=>4-x=4
=>x=0
\(\left(4-x\right)^2-16=0\)
\(\Leftrightarrow\left(4-x\right)^2=16\)
\(\Leftrightarrow\orbr{\begin{cases}4-x=4\\4-x=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=8\end{cases}}}\)
#H