tìm x , biết :
a, ( x mũ 3 - 4 x mũ 2 ) - ( x -4 ) = 0
b, x mũ 5 - 9x = 0
c, ( x mxu 3 - x mũ 2 ) mũ 2 - 4 x mũ 2 + 8x - 4 = 0
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9 tháng 8 2021
1, \(x^3+4x^2+4x=0\Leftrightarrow x\left(x^2+4x+4\right)=0\)
\(\Leftrightarrow x\left(x+2\right)^2=0\Leftrightarrow x=-2;x=0\)
2, \(\left(x+3\right)^2-4=0\Leftrightarrow\left(x+3-2\right)\left(x+3+2\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=1\)
3, \(x^4-9x^2=0\Leftrightarrow x^2\left(x^2-9\right)=0\)
\(\Leftrightarrow x^2\left(x-3\right)\left(x+3\right)=0\Leftrightarrow x=0;\pm3\)
4, \(x^2-6x+9=81\Leftrightarrow\left(x-3\right)^2=9^2\)
\(\Leftrightarrow\left(x-3-9\right)\left(x-3+9\right)=0\Leftrightarrow\left(x-12\right)\left(x+6\right)=0\Leftrightarrow x=-6;x=12\)
5, em xem lại đề nhé
9 tháng 8 2021
à lag tý @@
5, \(x^3+6x^2+9x-4x=0\Leftrightarrow x^3+6x^2+5x=0\)
\(\Leftrightarrow x\left(x^2+6x+5\right)=0\Leftrightarrow x\left(x^2+x+5x+5\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+5\right)=0\Leftrightarrow x=-5;x=-1;x=0\)
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 2022
b \(\Leftrightarrow x^3\left(x-2\right)+10x\left(x-2\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
=>x=8 hoặc x=-2/3
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
=>x=2 hoặc x=1
e: \(\Leftrightarrow x\left(x^2-11x+30\right)=0\)
=>x(x-5)(x-6)=0
hay \(x\in\left\{0;5;6\right\}\)
6 tháng 8 2022
b: \(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left(x-2\right)=0\)
hay \(x\in\left\{0;2\right\}\)
c: \(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
=>(x-8)(3x+2)=0
hay \(x\in\left\{8;-\dfrac{2}{3}\right\}\)
d: \(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
=>x=1 hoặc x=2
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
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, \(x^5-9x=0\Leftrightarrow x\left(x^4-9\right)=0\)
\(\Leftrightarrow x\left(x^2-3\right)\left(x^2+3>0\right)=0\Leftrightarrow x=\pm\sqrt{3};x=0\)
a/
\(x^3-4x^2-\left(x-4\right)=0\)
\(\Leftrightarrow x^2\left(x-4\right)-\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x+1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=-1\end{matrix}\right.\)
b/
\(x^5-9x=0\)
\(\Leftrightarrow x\left(x^4-9\right)=x\left(x^2-3\right)\left(x^2+3\right)=0\)
\(\Leftrightarrow x\left(x-\sqrt{3}\right)\left(x+\sqrt{3}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\sqrt{3}\\x=-\sqrt{3}\end{matrix}\right.\)
c/
\(\left(x^3-x^2\right)^2-4x^2+8x-4=0\)
\(\Leftrightarrow x^4\left(x-1\right)^2-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^4-4\right)=0\)
\(\Leftrightarrow\left(x-1\right)^2\left(x^2-2\right)\left(x^2+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x^2-2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\pm\sqrt{2}\end{matrix}\right.\)
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