x^3+2x^2-2x-12=0. tìm x
Những câu hỏi liên quan
Bài 13: Tìm x biết: a) (x-2)(x-3)-D0. b) (x-3)(x-4)-0. c) (x-7)(6-x)=0. d) (x-3)(x-13)=0. The Bài 14: Tìm x biết: a) (12-x)(2-x)=0. b) (x-33)(11-x)=0. c) (21-x)(12-x)=0. d) (50-x)(x-150) =0. Bài 15: Tìm x biết: a) 2x +x = 45. b) 2x +7x = 918. c) 2x+3x 60+5. d) 11x+22x 33.2.
13 tháng 12 2023 lúc 19:15
tìm x,biết:
a, 3(x-3)-6x=0
b, 2x(x-15)+2x
c, 2(x-3)+3x=9
d, x(x-11)+2(x-11)=0
e,x(x+2)+8=x^2
f, 8(x+1)+2x=-2
g,12-3(x+2)=0
a: \(3\left(x-3\right)-6x=0\)
=>\(3x-9-6x=0\)
=>-3x-9=0
=>3x+9=0
=>3x=-9
=>\(x=-\dfrac{9}{3}=-3\)
b: Đề thiếu vế phải rồi bạn
c: \(2\left(x-3\right)+3x=9\)
=>2x-6+3x=9
=>5x-6=9
=>5x=6+9=15
=>x=15/5=3
d: \(x\left(x-11\right)+2\left(x-11\right)=0\)
=>\(\left(x-11\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x-11=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-2\end{matrix}\right.\)
e: \(x\left(x+2\right)+8=x^2\)
=>\(x^2+2x+8=x^2\)
=>2x+8=0
=>2x=-8
=>x=-8/2=-4
f: \(8\left(x+1\right)+2x=-2\)
=>\(8x+8+2x=-2\)
=>10x=-2-8=-10
=>\(x=-\dfrac{10}{10}=-1\)
g: 12-3(x+2)=0
=>3(x+2)=12
=>x+2=12/3=4
=>x=4-2=2
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Tìm x biết
a) (x+2).(x+3) - (x-2).(x+5)=10
b) (3x+2). (2x+9) - (x+2). (8x+11)=(x+1).(3-2x)
c) 3.(2x-1).(3x-1)-(2x-3).(9x-1)=0
d) (5x-8).(4x-5)-(3x-4).(2x+12)=12
a)(x+2).(x+3)-(x-2).(x+5)=10
( x^2 +3x+2x+6)-(x^2 +5x-2x-10)=10
x^2 +3x+2x+6-x^2 -5x+2x+10-10=0
2x+6=0
2x=-6
x=-3
Tìm x:
a) x^2 - 25x=0
b) (x-3)^2 - 36x^2=0
c) 2x(3-x)+2x^2=12
d) x(x-2)-x+2=0
a) x2 - 25x = 0
=> x(x - 25) = 0
=> \(\orbr{\begin{cases}x=0\\x=25\end{cases}}\)
b) (x - 3)2 - 36x2 = 0
=> (x - 3)2 - (6x)2 = 0
=> \(\left(x+6x-3\right)\left(x-6x-3\right)=0\)
=> \(\orbr{\begin{cases}7x-3=0\\-5x-3=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{3}{7}\\x=-\frac{3}{5}\end{cases}}\)
c) 2x(3 - x) + 2x2 = 12
=> 6x - 2x2 + 2x2 = 12
=> 6x = 12
=> x = 2
d) x(x - 2) - x + 2 = 0
=> x(x - 2) - (x - 2) = 0
=> (x - 1)(x - 2) = 0
=> \(\orbr{\begin{cases}x=1\\x=2\end{cases}}\)
a. x2 - 25x = 0
\(\Leftrightarrow x\left(x-25\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-25=0\end{cases}}\)
\(\orbr{\begin{cases}x=0\\x=25\end{cases}}\)
Vậy ...
b.(x-3)2 - 36x2 = 0
\(\Leftrightarrow\left(x-3-6x\right)\left(x-3+6x\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}-5x-3=0\\7x-3=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{-3}{5}\\x=\frac{3}{7}\end{cases}}\)
Vậy...
c.2x(3-x)+2x2 = 12
<=> 6x - 2x2 + 2x2 = 12
<=> 6x = 12
<=> x = 2
d. x (x-2) - x + 2 =0
<=> x(x-2 ) - (x - 2 ) = 0
<=> ( x - 2 ) ( x - 1 ) = 0
\(\Leftrightarrow\orbr{\begin{cases}x-2=0\\x-1=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=2\\x=1\end{cases}}\)
Vậy...
Tìm x
3x - 4(2x-7)=3x-3
3(2x-7)-(4-2x) =2x-1
(x-2).(2x+12)=0
a) 3x - 4 (2x-7) = 3x - 3
3x - 8x + 28 = 3x - 3
=> 3x - 8x - 3x = -3 -28
-8x = -31
x = 31/8
b) 3(2x-7) - (4-2x) = 2x -1
6x - 7 - 4 + 2x = 2x -1
=> 6x + 2x - 2x = -1 + 7+4
6x = 10
x = 5/3
c) (x-2).(2x+12)=0
=> x -2 = 0 => x = 2
2x + 12 = 0 => 2x = -12 => x = -6
KL: x = 2 hoặc x = -6
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\(3x-4\left(2x-7\right)=3x-3\)
\(\Leftrightarrow3x-8x+28=3x-3\)
\(\Rightarrow3x-8x-3x=-3-28\)
\(\Leftrightarrow-8x=-31\)
\(\Leftrightarrow x=\frac{31}{8}\)
Mấy câu kia dễ đổi dấu tự làm :)
Tìm x,biết :
a) 2x^2-7x+5=0
b) x(2x-5) - 4x+10=0
c) (x-5)(x+5) - x(x-2)=15
d) x^2(2x-3) - 12+8x=0
e) x(x - 1)+5x - 5=0
f) (2x-3)^2 - 4x(x - 1)=5
g) x(5 - 2x)+2x(x - 1)=13
h)2(x+5)(2x - 5)+(x - 1)(5 - 2x)=0
\(2x^2-7x+5=0\)
\(2x^2-2x-5x+5=0\)
\(2x\left(x-1\right)-5\left(x-1\right)=0\)
\(\left(x-1\right)\left(2x-5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=\frac{5}{2}\end{array}\right.\)
\(x\left(2x-5\right)-4x+10=0\)
\(x\left(2x-5\right)-2\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(x-2\right)=0\)
\(\left[\begin{array}{nghiempt}x-2=0\\2x-5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\2x=5\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=2\\x=\frac{5}{2}\end{array}\right.\)
\(\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\)
\(x^2-25-x^2+2x=15\)
\(2x=15+25\)
\(2x=40\)
\(x=\frac{40}{2}\)
\(x=20\)
\(x^2\left(2x-3\right)-12+8x=0\)
\(x^2\left(2x-3\right)+4\left(2x-3\right)=0\)
\(\left(2x-3\right)\left(x^2+4\right)=0\)
\(2x-3=0\) (vì \(x^2\ge0\Rightarrow x^2+4\ge4>0\))
\(2x=3\)
\(x=\frac{3}{2}\)
\(x\left(x-1\right)+5x-5=0\)
\(x\left(x-1\right)+5\left(x-1\right)=0\)
\(\left(x-1\right)\left(x+5\right)=0\)
\(\left[\begin{array}{nghiempt}x-1=0\\x+5=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=1\\x=-5\end{array}\right.\)
\(\left(2x-3\right)^2-4x\left(x-1\right)=5\)
\(4x^2-12x+9-4x^2+4x=5\)
\(-8x=5-9\)
\(-8x=-4\)
\(x=\frac{4}{8}\)
\(x=\frac{1}{2}\)
\(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(5x-2x^2+2x^2-2x=13\)
\(3x=13\)
\(x=\frac{13}{3}\)
\(2\left(x+5\right)\left(2x-5\right)+\left(x-1\right)\left(5-2x\right)=0\)
\(\left(2x+10\right)\left(2x-5\right)-\left(x-1\right)\left(2x-5\right)=0\)
\(\left(2x-5\right)\left(2x+10-x+1\right)=0\)
\(\left(2x-5\right)\left(x+11\right)=0\)
\(\left[\begin{array}{nghiempt}2x-5=0\\x+11=0\end{array}\right.\)
\(\left[\begin{array}{nghiempt}2x=5\\x=-11\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{5}{2}\\x=-11\end{array}\right.\)
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\(a,2x^2-7x+5=0\Leftrightarrow2x^2-2x-5x+5=0\Leftrightarrow2x\left(x-1\right)-5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-1=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=2,5\end{matrix}\right.\)\(b,x\left(2x-5\right)-4x+10=0\Rightarrow x\left(2x-5\right)-2\left(2x-5\right)=0\Leftrightarrow\left(x-2\right)\left(2x-5\right)=0\Rightarrow\left[{}\begin{matrix}x-2=0\\2x-5=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\2x=5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x=2,5\end{matrix}\right.\)\(c,\left(x-5\right)\left(x+5\right)-x\left(x-2\right)=15\Leftrightarrow x^2-25-x^2+2x-15=0\Leftrightarrow2x-40=0\Rightarrow2x=40\Rightarrow x=20\)\(d,x^2\left(2x-3\right)-12+8x=0\Rightarrow2x^3-3x^2-12+8x=0\Leftrightarrow2x^3+8x-3x^2-12=0\Leftrightarrow2x\left(x^2+4\right)-2\left(x^2+4\right)=0\Leftrightarrow\left(2x-2\right)\left(x^2+4\right)=0\Rightarrow\left[{}\begin{matrix}2x-2=0\\x^2+4=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}2x=2\\x^2=-4\end{matrix}\right.\Rightarrow x=1\)
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Tìm x
a, ( x + 2 ) ³ + ( x- 2) ³ - 2x ( x ² + 3 ) = 27
b, 3x ( 2x - 1 ) - 24x + 12 = 0
c, x ² + x - 2019 . 2020 = 0
Giúp vs em cần gấp ạ
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Tìm x:
a)x(2x-4)-(x-2)(2x+3)
b)x(3x+2)+(x+1)^2-(2x-5)(2x+5)=-12
c)(x-1)(x^2+x+1)-x(x-3)^2=6x^2
d)x^2-x=0
e)(x+2)(x-3)-x-2=0
f)3x^3-27x
x(2x-4)-(x-2)(2x+3)
<=> (2x2 - 4x ) - (2x2 - 3x - 4x -6=0
<=> 2x2 - 4x -2x2 -3x - 4x + 6 =0
<=> -3x + 6 =0
<=> -3x = 6
<=> x = -(6/3) = -2
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Chứ bằng mấy, mk thấy ko có bằng nên làm mặc định là 0
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Tìm x
(2x-7)+17=6
12-2.(3-3x)=-2
-14+3.(-x+5)=-20
-90:5.(-3-2x)=6
(x+1).(x-3)=0
(2x-2).(x+4)=0
(22+4).(x+3)=0
(5-x).(6-2x)=0
3.(x+1)+5=x+8
-4.(2x+9)-(-8x+3)-(x+13)=0
(2x - 7) + 17 = 6
=> 2x - 7 = 6 - 17
=> 2x - 7 = -11
=> 2x = -11 + 7
=> 2x = -4
=> x = -4 : 2
=> x = -2
+) 12 -2(3 - 3x)= -2
=> 2(3 - 3x) = 12 + 2
=> 2(3 - 3x) = 14
=> 3 - 3x = 14 : 2
=> 3 - 3x = 7
=> 3x = 3 - 7
=> 3x = -4
=> x = -4/3
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\(\left(x+1\right)\left(x-3\right)=0\)
=> \(\orbr{\begin{cases}x+1=0\\x-3=0\end{cases}}\)
=> \(\orbr{\begin{cases}x=-1\\x=3\end{cases}}\)
Vậy...
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\(\left(2x-7\right)+17=6\)
\(\left(2x-7\right)=6-17\)
\(2x-7=-11\)
\(2x=-11+7\)
\(2x=-4\)
\(x=-4:2\)
\(\Rightarrow x=-2\)
\(V\text{ậy x = -2}\)
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