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(x - 2)3 - (x - 3)(x2 + 3x + 9) + 6(x + 1)2 = 49
<=>x3-6x2+12x-8-(x3-27)+6(x2+2x+1)=49
<=>x3-6x2+12x-8-x3+27+6x2+12x+6=49
<=>24x+25=49
<=>24x=24
<=>x=1 x(x + 5)(x - 5) - (x + 2)(x2 - 2x + 4) = 42
<=>x(x2-25)-(x3+8)=42
<=>x3-25x-x3-8=42
<=>-25x-8=42
<=>-25x=50
<=>x=-2
a) \(2x-6=0\)
\(\Leftrightarrow2x=6\)
\(\Leftrightarrow x=\dfrac{6}{2}=3\)
b) \(x^2-4x=0\)
\(\Leftrightarrow x\left(x-4\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-4=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=4\end{matrix}\right.\)
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
Vậy \(x=5\)hoặc \(x=1\)
\(b,x^2-2x=24\)
\(\Leftrightarrow x^2-2x+1-1=24\)
\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)
\(\Leftrightarrow x-1=\pm5\)
\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
Vậy \(x=6\) hoặc \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6
=> (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = 9 => x = 0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x -3x + 1) - (2x . 9x - 2x -3 . 9x + 3) = 0
=> 3(6x2 - 5x +1) - (18x2 - 29x + 3) = 0
=> (18x2 -15x + 1) -(18x2 - 29x +3) = 0
=> 18x2 - 15x +1 -18x2 + 29x - 3 = 0
=> 14x = 0 => x = 0
a)(x+2)(x+3)-(x-2)(x+5)=6
x(x+3)+2(x+3)-x(x+5)+2(x+5)=6
x2+3x+2x+6-x2-5x+2x+10=6
(x2-x2)+(3x+2x-5x+2x)+(10+6)=6
2x+16=6
2x=6-16
2x=-10
x=-10/2
x=-5. Vậy x=-5
b)3x(2x+9)+2(2x+9)-x(6x+1)-2(6x+1)=x+1-x+6
6x2+27x+4x+18-6x2-x-12x-2=7
(6x2-6x2)+(27x+4x-x-12x)+(18-2)=7
18x+16=7
18x=7-16
x=-9/18=-1/2. Vậy x=-1/2
c)[3(3x-1)](2x-1)-(2x-3)(9x-1)=0
(9x-3)(2x-1)-(2x-3)(9x-1)=0
9x(2x-1)-3(2x-1)-2x(9x-1)+3(9x-1)=0
18x2-9x-6x+3-18x2+2x+27x-3=0
(18x2-18x2)+(27x+2x-6x-9x)+(3-3)=0
14x=0
x=0/14
x=0. Vậy x=0
a) (x + 2) . (x + 3) - (x - 2) . (x + 5) = 6 => (x . x + 3x + 2x + 2 . 3) - (x . x + 5x - 2x - 2 . 5) = 6
=> (x2 + 5x + 6) - (x2 + 3x - 10) = 6
=> x2 + 5x + 6 - x2 - 3x + 10 = 6
=> 2x +16 = 6 => 2x = -10 => x = -5
b) (3x + 2) . (2x + 9) - (x + 2) . (6x + 1) = (x + 1) - (x - 6)
=> (3x . 2x + 3x . 9 + 2 . 2x + 2 . 9) - (x . 6x + 1x + 2 . 6x + 2 .1) = x + 1 - x + 6
=> (6x2 + 31x + 18) - (6x2 + 13x + 2) = 7
=> 6x2 + 31x + 18 - 6x2 - 13x - 2 = 7
=> 18x + 16 = 7 => 18x = -9 => x = -0,5
c) 3 . (2x - 1) . (3x - 1) - (2x - 3) . (9x - 1) = 0
=> 3(2x . 3x - 2x - 3x + 1) - (2x . 9x - 2x - 3. 9x + 3) = 0
=> 3(6x2 - 5x + 1) - (18x2 - 29x + 3) = 0
=> 18x2 - 15x + 3 - 18x2 + 29x -3 = 0
=> 14x = 0 => x = 0.
\(\left(x+3\right)^3-x\left(3x+1\right)^2+\left(2x+1\right)\left(4x^2-2x+1\right)=28\)
\(\Leftrightarrow x^3+9x^2+27x+27-9x^3-6x^2-x+8x^3+1=28\)
\(\Leftrightarrow3x^2+26x+28=28\)
\(\Leftrightarrow3x^2+26x=0\)\(\Leftrightarrow x\left(3x+26\right)=0\)
Suy ra x=0 hoặc x=-26/3
a) (x-2)2 -(x-3)(x-3)=6
=>x2 -4x+4-x2+3=6
=>7-4x=6
=>4x=1 =>x=\(\frac{1}{4}\)
b)4(x-3)2 -(2x-1)(2x+1)=10
=>4(x2-6x+9)-4x2+1=10
=>4x2-24x+36-4x2+1=10
=>37-24x=10 =>24x=27 =>x=\(\frac{9}{8}\)
c)x2-16-3(x+4)=0
=>(x-4)(x+4)-3(x+4)=0
=>(x-7)(x+4)=0
=>\(\orbr{\begin{cases}x-7=0\\x+4=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-4\end{cases}}}\)
=>x\(\in\left\{-4;7\right\}\)
d)(x-4)2-(x-2)(x+2)=6
=>x2-8x+16-x2+4=6
=>20-8x=6
=>8x=14 =>x=\(\frac{4}{7}\)
e) 9(x+1)2-(3x-2)(3x+2)=10
=>9(x2 +2x+1)-9x2+4=10
=>9x2+18x+9-9x2+4=10
=>18x+13=10
=>18x=-3
=>x=\(\frac{-1}{6}\)
mình chỉ làm bài 1 nha
nhớ chon mk đúng nha
a, \(3x+2\left(x-5\right)=6-\left(5x-1\right)\)
\(\Leftrightarrow3x+2x-10=6-5x+1\)
\(\Leftrightarrow-15\ne0\)Vậy phương trình vô nghiệm
b, \(x^3-3x^2-x+3=0\)
\(\Leftrightarrow x\left(x^2-1\right)-3\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x+1\right)=0\Leftrightarrow x=3;\pm1\)
Vậy tập nghiệm của phương trình là S = { 1 ; -1 ; 3 }
c, \(\frac{1}{x-3}+\frac{x}{x+3}=\frac{2}{x^2-9}ĐK:x\ne\pm3\)
\(\Leftrightarrow\frac{x+3}{\left(x-3\right)\left(x+3\right)}+\frac{x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}=\frac{2}{\left(x-3\right)\left(x+3\right)}\)
\(\Leftrightarrow x+3+x^2-3x-2=0\)
\(\Leftrightarrow x^2-2x+1=0\Leftrightarrow\left(x-1\right)^2=0\Leftrightarrow x=1\)thỏa mãn
Vậy ...
giải phá các ngoặc ra rút gọn x rồi tìm
ko ghi lại đề nha !
a) \(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+x\left(2^2-x^2\right)=0\)
\(\Leftrightarrow x^3+3x^2+9x-3x^2-9x-27+4x-x^3=0\)
\(\Leftrightarrow-27+4x=0\)
\(\Leftrightarrow4x=27\)
\(\Leftrightarrow x=6,75\)
b)\(\Leftrightarrow\left(x^3+3x^2+3x+1\right)-\left(x^3-3x^2+3x-1\right)-6\left(x^2-2x+1\right)=-10\)
\(\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6=-10\)
\(\Leftrightarrow6x^2+2-6x^2+12x-6=-10\)
\(\Leftrightarrow12x-4=-10\)
\(\Leftrightarrow12x=-6\)
\(\Leftrightarrow x=-0,5\)