tìm x: x^4 - 3x^2 + 1/x^4 - 3/x^2 = 2
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24 tháng 10 2021
d: ta có: \(x^2-4x+4=9\left(x-2\right)\)
\(\Leftrightarrow\left(x-2\right)\left(x-11\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=11\end{matrix}\right.\)
25 tháng 7 2021
a) (x-2)3+6(x+1)2-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6(x2+2x+1)-x3+12=0
\(\Rightarrow\)x3-6x2+12x-8+6x2+12x+6-x3+12=0
\(\Rightarrow\)24x+10=0
\(\Rightarrow\)24x=-10
\(\Rightarrow\)x=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
25 tháng 7 2021
b)(x-5)(x+5)-(x+3)2+3(x-2)2=(x+1)2-(x-4)(x+4)+3x2
\(\Rightarrow\)x2-25-(x2+6x+9)+3(x2-4x+4)=x2+2x+1-(x2-16)+3x2
\(\Rightarrow\)x2-25-x2-6x-9+3x2-12x+12=x2+2x+1-x2+16+3x2
\(\Rightarrow\)3x2-18x-22=3x2+2x+17
\(\Rightarrow\)3x2-18x-22-3x2-2x-17=0
\(\Rightarrow\)-20x-39=0
\(\Rightarrow\)-20x=39
\(\Rightarrow\)x=\(-\dfrac{39}{20}\)
15 tháng 10 2021
a) \(\Rightarrow9x^2+24x+16-9x^2+1=49\)
\(\Rightarrow24x=32\Rightarrow x=\dfrac{4}{3}\)
b) \(\Rightarrow x^2-13x+22=0\)
\(\Rightarrow\left(x-11\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=11\\x=2\end{matrix}\right.\)
c) \(\Rightarrow x^2-3x-10=0\)
\(\Rightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
8 tháng 9 2023
2:
a: =>x^2+3x-4x-12-(x^2-5x+x-5)=8
=>x^2-x-12-x^2+4x+5=8
=>3x-7=8
=>3x=15
=>x=5
b: =>3x^2+3x-2x-2-3x^2-21x=13
=>-20x=15
=>x=-3/4
c: =>x^2-25-x^2-2x=9
=>-2x=25+9=34
=>x=-17
d: =>x^3-1-x^3+3x=1
=>3x-1=1
=>3x=2
=>x=2/3
9 tháng 7 2019
b, \(\left(x-5\right)\left(x-4\right)-\left(x+1\right)\left(x-2\right)=7\)
\(\Rightarrow x^2-9x+20-x^2+x+2=7\)
\(\Rightarrow-8x+22=7\)
\(\Rightarrow-8x=-15\)
\(\Rightarrow x=\frac{15}{8}\)
c, \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)-3\)
\(\Rightarrow3x^2-10x+8=3x^2-27x-3\)
\(\Rightarrow3x^2-10x-3x^2+27x=\left(-3\right)+\left(-8\right)\)
\(\Rightarrow17x=-11\)
\(\Rightarrow x=-\frac{11}{17}\)
d, \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(5-x^2\right)=6x\)
\(\Rightarrow x^3+3x^2+9x-3x^2-9x-27+5x-x^3=6x\)
\(\Rightarrow6x=-27\)
\(\Rightarrow x=-\frac{27}{6}\)
\(\Rightarrow x=-\frac{9}{2}\)
e, \(\left(3x-5\right)\left(x+1\right)-\left(3x-1\right)\left(x+1\right)=x-4\)
\(\Rightarrow3x^2-2x-5-3x^2-2x+1=x-4\)
\(\Rightarrow-4=x-4\)
\(\Rightarrow x=0\)
9 tháng 7 2019
b) (x - 5)(x - 4) - (x + 1)(x - 2) = 7
<=> x2 - 9x + 20 - x2 + x + 2 - 7 = 0
<=> 8x - 15 = 0 <=> x = 15/8
c) (3x - 4)(x - 2) = 3x(x - 9) - 3
<=> 3x2 - 10x + 8 = 3x2 - 27x - 3
<=> 17x = -11 <=> x = -11/17
d) (x - 3)(x2 + 3x + 9) + x(5 - x2) = 6x
<=> x3 - 27 - x3 + 5x - 6x = 0
<=> x = -27
e) (3x - 5)(x + 1) - (3x - 1)(x + 1) = x - 4
<=> (x + 1)(3x - 5 - 3x + 1) - x + 4 = 0
<=> -4x - 4 - x + 4 = 0 <=> x = 0
8 tháng 9 2021
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
8 tháng 9 2021
Bài 4:
a: Ta có: \(3\left(2x-3\right)-2\left(x-2\right)=-3\)
\(\Leftrightarrow6x-9-2x+4=-3\)
\(\Leftrightarrow4x=2\)
hay \(x=\dfrac{1}{2}\)
b: Ta có: \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
hay \(x=\dfrac{13}{3}\)
c: Ta có: \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
hay x=1
H9
HT.Phong (9A5)
CTVHS
3 tháng 8 2023
a) \(\left(2x-3\right)\left(2x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=3\\2x=-3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b) \(\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
c) \(2x\left(3x-1\right)-3x\left(5+2x\right)=0\)
\(\Rightarrow x\left[2\left(3x-1\right)-3\left(5+2x\right)\right]=0\)
\(\Rightarrow x\left(6x-2-15-6x\right)\)
\(\Rightarrow-16x=0\)
\(\Rightarrow x=0\)
d) \(\left(3x-2\right)\left(3x+2\right)-4\left(x-1\right)=0\)
\(\Rightarrow9x^2-4-4x+4=0\)
\(\Rightarrow9x^2-4x=0\)
\(\Rightarrow x\left(9x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\9x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{4}{9}\end{matrix}\right.\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
3 tháng 8 2023
\(a,\left(2x-3\right)\left(2x+3\right)=0\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\2x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=-\dfrac{3}{2}\end{matrix}\right.\\ b,\left(x-4\right)\left(x-1\right)\left(x-2\right)=0\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=1\\x=2\end{matrix}\right.\)
16 tháng 9 2021
1) \(\left(x+1\right)\left(x+2\right)-3x\left(x-4\right)=x^2+3x+2-3x^2+12x=-2x^2+15x+2\)
2) \(\left(3x-4\right)\left(x-2\right)=3x\left(x-9\right)\)
\(\Leftrightarrow3x^2-10x+8=3x^2-27x\)
\(\Leftrightarrow17x=-8\Leftrightarrow x=-\dfrac{8}{17}\)
3) \(-3\left(x-4\right)\left(x-2\right)-x^2\left(-3x+18\right)+24x-25\)
\(=-3x^3+6x^2+12x^2-24x+3x^3-18x^2+24x-25=-25\)