tìm x biêt:
(x+2)^3-(2x+3)^2+(2x+3)(2x-3)=(x-2)(x^2+2x+4)-6x(x+2)
làm phép chia :
a) (x^4 -2x^3 + 2x -1) : (x^2 - 1)
b) (x^3 -8) : (x^2 + 2x +4)
c) (x^6 - 2x^5 + 2x^4 + 6x^3 - 4x^2)n: 6x^2
d) (-2x^5 + 3x^2 - 4x^3) :2x^2
e) (15x^3 - 10x^2 + x - 2) : (x - 2)
f) (2x^4 - 3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
b: =x-2
d: \(=-x^3+\dfrac{3}{2}-2x\)
Bài 4: Tìm x, biết:
a) 3(2x – 3) + 2(2 – x) = –3 ; b) x(5 – 2x) + 2x(x – 1) = 13 ;
c) 5x(x – 1) – (x + 2)(5x – 7) = 6 ; d) 3x(2x + 3) – (2x + 5)(3x – 2) = 8 ;
e) 2(5x – 8) – 3(4x – 5) = 4(3x – 4) + 11; f) 2x(6x – 2x 2 ) + 3x 2 (x – 4) = 8.
\(a,3\left(2x-3\right)+2\left(2-x\right)=-3\\ \Leftrightarrow6x-9+4-2x=-3\\ \Leftrightarrow4x=2\\ \Leftrightarrow x=\dfrac{1}{2}\\ b,x\left(5-2x\right)+2x\left(x-1\right)=13\\ \Leftrightarrow5x-2x^2+2x^2-2x=13\\ \Leftrightarrow3x=13\\ \Leftrightarrow x=\dfrac{13}{3}\\ c,5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\\ \Leftrightarrow5x^2-5x-5x^2-3x+14=6\\ \Leftrightarrow-8x=-8\\ \Leftrightarrow x=1\\ d,3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\\ \Leftrightarrow6x^2+9x-6x^2-11x+10=8\\ \Leftrightarrow-2x=-2\\ \Leftrightarrow x=1\)
\(e,2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\\ \Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ f,2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\\ \Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3-8=0\\ \Leftrightarrow-\left(x^3+8\right)=0\\ \Leftrightarrow-\left(x+2\right)\left(x^2-2x+4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-2\\x\in\varnothing\left(x^2-2x+4=\left(x-1\right)^2+3>0\right)\end{matrix}\right.\)
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
a/ \(3\left(2x-3\right)+2\left(2-x\right)=-3\)
\(\Leftrightarrow6x-9+4-2x=-3\)
\(\Leftrightarrow4x=2\)
\(\Leftrightarrow x=\dfrac{1}{2}\)
Vậy: \(x=\dfrac{1}{2}\)
===========
b/ \(x\left(5-2x\right)+2x\left(x-1\right)=13\)
\(\Leftrightarrow5x-2x^2+2x^2-2x=13\)
\(\Leftrightarrow3x=13\)
\(\Leftrightarrow x=\dfrac{13}{3}\)
Vậy: \(x=\dfrac{13}{3}\)
==========
c/ \(5x\left(x-1\right)-\left(x+2\right)\left(5x-7\right)=6\)
\(\Leftrightarrow5x^2-5x-5x^2+7x-10x+14=6\)
\(\Leftrightarrow-8x=-8\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
d/ \(3x\left(2x+3\right)-\left(2x+5\right)\left(3x-2\right)=8\)
\(\Leftrightarrow6x^2+9x-6x^2+4x-15x+10=8\)
\(\Leftrightarrow-2x=-2\)
\(\Leftrightarrow x=1\)
Vậy: \(x=1\)
==========
e/ \(2\left(5x-8\right)-3\left(4x-5\right)=4\left(3x-4\right)+11\)
\(\Leftrightarrow10x-16-12x+15=12x-16+11\)
\(\Leftrightarrow-14x=-4\)
\(\Leftrightarrow x=\dfrac{2}{7}\)
Vậy: \(x=\dfrac{2}{7}\)
==========
f/ \(2x\left(6x-2x^2\right)+3x^2\left(x-4\right)=8\)
\(\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\)
\(\Leftrightarrow-x^3=8\)
\(\Leftrightarrow x=-2\)
Vậy: \(x=-2\)
Tìm x Toán đại 8 Hằng đẳng thức đáng nhớ?
Tìm x:
1. (x-1)^3+3.(x-3)^2-(x+2).(x^2-2x+4) = (x+2)^3-(x-3).(x^2+9)-6x^2+5
2.(3+2x)^3-(6x-1).(6x+1) = (2x-1)^3+(x+4)^2-x^3+(x+1).(x^2+x+1)
1. (x - 1)^3 + 3.(x - 3)^2 - (x + 2).(x^2 - 2x + 4) = (x + 2)^3 - (x - 3).(x^2 + 9) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3(x^2 - 6x + 9) - (x^3 + 2^3)
= x^3 + 6x^2 + 12x + 8 - (x^3 - 3x^2 + 9x -27) - 6x^2 + 5
<=> x^3 - 3x^2 + 3x - 1 + 3x^2 - 18x + 27 - x^3 - 8
= x^3 + 6x^2 + 12x + 8 - x^3 + 3x^2 - 9x + 27 - 6x^2 + 5
<=> 3x - 18x -12x - 3x^2 + 9x = 27 + 5 + 8 + 8 + 1 - 27
<=> - 3x^2 - 18x - 22 = 0
<=> 3x^2 + 18x + 22 = 0
Nửa chu vi mảnh đất là:
120 : 2 = 60 (m)
Chiều dài hơn chiều rộng là:
5 + 5 = 10 (m)
Chiều rộng là:
( 60 - 10 ) : 2 = 25 (m)
Chiều dài là:
25 + 10 = 35 (m)
Diện tích là:
25 35 = 875 ( )
2x ^3 -5x^2+4x-1) : (2x+1)
(x63 -2x+4) ; (x+2)
(6x^3 - 19x^2+23x-12):(2x-3)
(x^4 - 2 x ^3 - 1+ 2 x ):(x^2 - 1)
(6x^3 - 5x^2 + 4x -1 ) : (2x^2-x+1)
(x^4 -5x^2+4):(x^2-3x+2)
d: \(\dfrac{x^4-2x^3+2x-1}{x^2-1}\)
\(=\dfrac{\left(x^2-1\right)\left(x^2+1\right)-2x\left(x^2-1\right)}{x^2-1}\)
\(=x^2-2x+1\)
\(=\left(x-1\right)^2\)
Tìm x
(2x-3).(x+1)-2x^2+6x=0
(X^2-x+1).(x-3)-x^3+4x^2=0
(X^2-2).(x^2+2)-x^4-2x+5=0
(X-3).(x^2-3x+2)-(x^2-2x-7).(x-2)+2x^2-2x=0
( 2x - 3 )( x + 1 ) - 2x2 + 6x = 0
<=> 2x2 - x - 3 - 2x2 + 6x = 0
<=> 5x - 3 = 0
<=> 5x = 3
<=> x = 3/5
( x2 - x + 1 )( x - 3 ) - x3 + 4x2 = 0
<=> x3 - 4x2 + 4x - 3 - x3 + 4x2 = 0
<=> 4x - 3 = 0
<=> 4x = 3
<=> x = 3/4
( x2 - 2 )( x2 + 2 ) - x4 - 2x + 5 = 0
<=> ( x2 )2 - 4 - x4 - 2x + 5 = 0
<=> x4 + 1 - x4 - 2x = 0
<=> 1 - 2x = 0
<=> 2x = 1
<=> x = 1/2
( x - 3 )( x2 - 3x + 2 ) - ( x2 - 2x - 7 )( x - 2 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - ( x3 - 4x2 - 3x + 14 ) + 2x2 - 2x = 0
<=> x3 - 6x2 + 11x - 6 - x3 + 4x2 + 3x - 14 + 2x2 - 2x = 0
<=> 12x - 20 = 0
<=> 12x = 20
<=> x = 20/12 = 5/3
a, \(\left(2x-3\right)\left(x+1\right)-2x^2+6x=0\)
\(\Leftrightarrow2x^2+2x-3x-3-2x^2+6x=0\Leftrightarrow5x-3=0\Leftrightarrow x=\frac{3}{5}\)
b, \(\left(x^2-x+1\right)\left(x-3\right)-x^3+4x^2=0\)
\(\Leftrightarrow x^3-3x^2-x^2+3x+x-3-x^3+4x^2=0\Leftrightarrow4x-3=0\Leftrightarrow x=\frac{3}{4}\)
c ; d tương tự nhé !
làm phép chia
1) (x^6 - 2x^4 + 6x^3 - 4x^2) : 6x^2
2) (-2x^5 = 3x^2 - 4x^3) : 2x^2
3) (15x^3 - 10x^2 + x - 2) : (x - 2)
4) (2x^4 -3x^3 - 3x^2 + 6x - 2) : (x^2 - 2)
Tìm x
(2x+3)(4x^2-6x+9)-(2x-1)^3=(x+4)^3-(x-2)(x^2+2x+4)
pt <=> \(8x^3+27-\left(8x^3-12x^2+6x-1\right)=\left(x^3+12x^2+48x+64\right)-\left(x^3-8\right)\)
<=> \(8x^3+27-8x^3+12x^2-6x+1=x^3+12x^2+48x+64-x^3+8\)
<=> \(12x^2-6x+28=12x^2+48x+72\)
<=> \(54x+44=0\)
<=> \(x=-\frac{22}{27}\)
các bạn giúp mình với nha, bài tìm x, biêt:
6x^2-(2x-3)x(3x+2)=0
Tìm đkxđ của: 1, 3x/ 4x-8 2, 2x/ x²-9 3, 6/x³+1 4, 6x²/x²-2x+1 5, x-2/x²+3 6, 2x/x²+3+2
1) \(\dfrac{3x}{4x-8}\)
\(ĐKXĐ:4x-8\ne0\Leftrightarrow x\ne2\)
2) \(\dfrac{2x}{x^2-9}\)
\(ĐKXĐ:x^2-9\ne0\Leftrightarrow\)\(\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)
3) \(\dfrac{6}{x^3+1}=\dfrac{6}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(ĐKXĐ:\)\(x+1\ne0\Leftrightarrow x\ne-1\)
(do \(x^2-x+1=\left(x-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
4) \(\dfrac{6x^2}{x^2-2x+1}=\dfrac{6x^2}{\left(x-1\right)^2}\)
\(ĐKXĐ:x-1\ne0\Leftrightarrow x\ne1\)
5) \(\dfrac{x-2}{x^2+3}\)
Do \(x^2+3>0\forall x\in R\)
Vậy biểu thức trên xác định với mọi x
6) \(\dfrac{2x}{x^2+3x+2}=\dfrac{2x}{\left(x+1\right)\left(x+2\right)}\)
\(ĐKXĐ:\)\(\left\{{}\begin{matrix}x+1\ne0\\x+2\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x\ne-1\\x\ne-2\end{matrix}\right.\)