Tìm x
a) (2x-5)2-(5+2x)=0
b) 27x3-54x2+36x=0
c)(x3+8)-(x+2)(x-4)=0
d)x6-1=0
Những câu hỏi liên quan
B1: Tìm x
1, 64x3 + 192x2 + 24x = 0 2, 8x3 - 24x2 +54x- 28= 0
3, 27x3 -54x2 +36x -7= 0 4, x3 +12x2+ 48x +65 =0
5, 27x3 -54x2 +36x -19 =0 6, 8x3 - 12x2 + 6x = 0
B2: Tính giá trị biểu thức
A= (a+ b - c)2 + (a- b + c)2 với a=1, b=3, c= -1
B= (a+b)2 - (a -b )2 + 8 với a=2 , b=5
Kíu Kíu?!!
Bài suốt ngày p đăng lại thôi khổ lắm.
Ai giúp mới!!!
Bài 2:
a: \(A=\left[a+\left(b-c\right)\right]^2+\left[a-\left(b-c\right)\right]^2\)
\(=a^2+2a\left(b-c\right)+\left(b-c\right)^2+a^2-2a\left(b-c\right)+\left(b-c\right)^2\)
\(=2a^2+2\left(b-c\right)^2\)
\(=2\cdot1^2+2\left(3+1\right)^2=2+32=34\)
b: \(B=a^2+2ab+b^2-a^2+2ab-b^2=4ab=4\cdot2\cdot5=40\)
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Tìm x
(x-5)2=(3+2x)2
27x3-54x2+36x=9
cho bt x-y=4 và xy=1 tính giá trị của các biểu thức A=x2+y2,B=x3-y3,C=x4+y4
a) \(\left(x-5\right)^2=\left(3+2x\right)^2\)
\(\Rightarrow\left(3+2x\right)^2-\left(x-5\right)^2=0\)
\(\Rightarrow\left(3+2x+x-5\right)\left(3+2x-x+5\right)=0\)
\(\Rightarrow\left(3x-2\right)\left(x+8\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}3x-2=0\\x+8=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{2}{3}\\x=-8\end{matrix}\right.\)
b) \(27x^3-54x^2+36x=9\)
\(\Rightarrow27x^3-54x^2+36x-9=0\)
\(\Rightarrow27x^3-54x^2+36x-8+8-9=0\)
\(\Rightarrow\left(3x-2\right)^3-1=0\)
\(\Rightarrow\left(3x-2-1\right)\left[\left(3x-2\right)^2+3x-2+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2\right)^2+3x-2+\dfrac{1}{4}-\dfrac{1}{4}+1\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-2+\dfrac{1}{2}\right)^2+\dfrac{3}{4}\right]=0\)
\(\Rightarrow\left(3x-3\right)\left[\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\left(1\right)\)
mà \(\left(3x-\dfrac{3}{2}\right)^2+\dfrac{3}{4}>0,\forall x\)
\(\left(1\right)\Rightarrow3x-3=0\Rightarrow3x=3\Rightarrow x=1\)
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(\(x-5\))2 = (3 +2\(x\))2 ⇒ \(\left[{}\begin{matrix}x-5=3+2x\\x-5=-3-2x\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=-8\\x=\dfrac{2}{3}\end{matrix}\right.\) vậy \(x\in\){-8; \(\dfrac{2}{3}\)}
27\(x^3\) - 54\(x^2\) + 36\(x\) = 9
27\(x^3\) - 54\(x^2\) + 36\(x\) - 8 = 1
(3\(x\) - 2)3 = 1 ⇒ 3\(x\) - 2 = 1 ⇒ \(x\) = 1
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Xem thêm câu trả lời
Tìm x thỏa mãn:a)
(
2
x
-
5
)
2
-
(
5
+
2
x
)
2
0; b) 27
x
3...
Đọc tiếp
Tìm x thỏa mãn:
a) ( 2 x - 5 ) 2 - ( 5 + 2 x ) 2 = 0; b) 27 x 3 - 54 x 2 + 36x = 8.
Tìm x biết:a)
x
6
+ 2
x
3
+1 0; b) x(x - 5) 4x - 20;c)
x
4
-2
x
2
8-4
x
2
; d) (
x
3
-
x
2
) - 4
x
2
+ 8x-4 0.
Đọc tiếp
Tìm x biết:
a) x 6 + 2 x 3 +1 = 0; b) x(x - 5) = 4x - 20;
c) x 4 -2 x 2 =8-4 x 2 ; d) ( x 3 - x 2 ) - 4 x 2 + 8x-4 = 0.
a) x = -1. b) x = 4 hoặc x = 5.
c) x = ± 2 . d) x = 1 hoặc x = 2.
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Tìm x, biết:a) 3x(x - 1) + x - 1 0;b) (x - 2)(
x
2
+ 2x + 7) + 2(
x
2
- 4) - 5(x - 2) 0;c)
(
2
x
-
1
)
2
- 25 0;d)
x
3
+ 27 + (x + 3)(x - 9) 0.
Đọc tiếp
Tìm x, biết:
a) 3x(x - 1) + x - 1 = 0;
b) (x - 2)( x 2 + 2x + 7) + 2( x 2 - 4) - 5(x - 2) = 0;
c) ( 2 x - 1 ) 2 - 25 = 0;
d) x 3 + 27 + (x + 3)(x - 9) = 0.
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
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Bài 6 : Tìm x, biết :
a, ( 2x-5)^2 - ( 5+2x)^2 = 0
b, 27x^3 - 54x^2 + 36x = 8
c, ( x^3 + 8 ) - ( x+2) ( x-4) = 0
d, x^6 - 1 = 0
Giúp mk vs ạ mk đang cần gấp
tìm x biết a) ( x + 3 )2 - ( 2x + 1 ).( x+3 ) = 0 ; b) x3 - 12x2 + 36x = 0
\(a,\Leftrightarrow\left(x+3\right)\left(x+3-2x-1\right)=0\\ \Leftrightarrow\left(x+3\right)\left(2-x\right)=0\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\end{matrix}\right.\\ b,\Leftrightarrow x\left(x^2-12x+36\right)=0\\ \Leftrightarrow x\left(x-6\right)^2=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=6\end{matrix}\right.\)
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a, (x+3)2 - ( 2x + 1 ).( x+3)=0 b, x3-12x2+36x =0
=> (x+3).(x+3-2x-1) => x(x2-12x+36) = 0
=>(x+3).(-x+2) => x(x-6)2 = 0
=> x+3=0 <=> x=-3 => x=0 <=> x=0
-x+2=0 <=> x=-2 x-6= 0 <=> x=6
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a) x2(x - 5) + 5 - x = 0; b) 3x4 - 9x3 = -9x2 + 27x;
c) x2(x + 8) + x2 = -8x; d) (x + 3)(x2 -3x + 5) = x2 + 3x.
e) 3x(x - 1) + x - 1 = 0;
f) (x - 2)(x2 + 2x + 7) + 2(x2 - 4) - 5(x - 2) = 0;
g) (2x - 1)2 - 25 = 0;
h) x3 + 27 + (x + 3)(x - 9) = 0.
i)8x3 - 50x = 0; k) 2(x + 3)-x2 - 3x = 0;
m)6x2 - 15x - (2x - 5)(2x + 5) =
a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
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tìm x biết: a)x2 + 3x 0 b) x3 – 4x 0 c) 5x(x-1) x-1 d) 2(x+5) - x2-5x 0 e) 2x(x-5)-x(3+2x)26 f) 5x.(x – 2012) – x + 2012 0
Đọc tiếp
tìm x biết:
a)x2 + 3x = 0 b) x3 – 4x = 0
c) 5x(x-1) = x-1 d) 2(x+5) - x2-5x = 0
e) 2x(x-5)-x(3+2x)=26 f) 5x.(x – 2012) – x + 2012 = 0
a) \(\Rightarrow x\left(x+3\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)
b) \(\Rightarrow x\left(x^2-4\right)=0\Rightarrow x\left(x-2\right)\left(x+2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=2\\x=-2\end{matrix}\right.\)
c) \(\Rightarrow\left(x-1\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{5}\end{matrix}\right.\)
d) \(\Rightarrow2\left(x+5\right)-x\left(x+5\right)=0\Rightarrow\left(x+5\right)\left(2-x\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=-5\\x=2\end{matrix}\right.\)
e) \(\Rightarrow2x^2-10x-3x-2x^2=26\)
\(\Rightarrow-13x=26\Rightarrow x=-2\)
f) \(\Rightarrow\left(x-2012\right)\left(5x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=2012\\x=\dfrac{1}{5}\end{matrix}\right.\)
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