Tìm x biết
(1-3x)2 - (x-2)(9x+1) = (3x-4)(3x+4) - 9(x+3)2
Giúp mình đi :))
Những câu hỏi liên quan
tìm x biết :
(1-3x2) - (x-2)(9x+1) = (3x-4)(3x+4)-9(x+3)2
(1-3x2)-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
⇒1-3x2-(9x2+x-18x-2)=9x2-16-9(x2+6x+9)
⇒1-3x2-(9x2-17x-2)= -56x-97
⇒1-3x2-9x2+17x+2=-56x-97
⇒3-12x2+17x=-56x-97
⇒3-12x2+17x+56x+97=0
⇒-12x2+73x+100=0
⇒-(12x2-73x-100)=0
Đúng 0
Bình luận (0)
Tìm x, biết:
a) (5x – 1)(x + 1) – 2(x – 3)2 = (x + 2)(3x – 1) – (x + 4)2 + (x2 – x).
b) (1 – 3x2) – (x – 2)(9x + 1) = (3x – 4)(3x + 4) – 9(x + 3)2.
Tìm x :
( 1 - 3x2 ) - ( x - 2 )( 9x + 1 ) = ( 3x - 4 )( 3x + 4 ) - 9( x + 3 )2
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
<=> \(1-6x+9x^2-\left(9x^2-17x-2\right)=\left(9x^2-4\right)-\left[3\left(x+3\right)\right]^2\)
<=> \(1-6x+9x^2-9x^2+17x+2=9x^2-4-\left(3x+9\right)^2\)
<=> \(3+11x=\left(3x-3x-9\right)\left(3x+3x+9\right)-4\)
<=> \(3+4+11x=-9\left(6x+9\right)\)
<=> \(7+11x=-9.3\left(2x+3\right)\)
<=> \(7+11x=-27\left(2x+3\right)\)
<=> \(7+11x+27\left(2x+3\right)=0\)
<=> \(7+11x+54x+81=0\)
<=> \(65x=-88\)
<=> \(x=-\frac{88}{65}\)
Đúng 0
Bình luận (0)
I, thực hiện phép tính:
a,( 9/ x^3 - 9x + 1/ x+3) : (x-3/ x^2+3x -x/3x+9x)
b, (3x/1-3x +2x/ 3x+1) : (6x^2+10x/1-6x+9x^2)
c, a^2-b^2/ a^2 * a^4 / (a+b)^2
d,( 3-3x/ (1+ x )^2 : (6x^2- 6 / x+1)
e, x^2 -1/x+10 * x/ x+2 + x^2-1/ x+10 * 1- x/x -2
II, tìm ĐKXĐ của các phân thức sau:
a, x^2-4/ 9x^2- 16
b, 2x-1/ x^2 -4x +4
c, x^2 -4/ x^2+1
d, 5x-3/ 2x^2 -x
cần gấp chiều 2h đi học
Tìm x, biết :
a) (x-2)3 +6(x+1)2-x3+12=0
b) (x-5) (x+5) - (x+3)2+3(x-2)2=(x+1)2- (x+4)(x-4)+3x2
c) (2x+3)2 +(x-1)(x+1)=5(x+2)2-(x-5)(x+1)+(x+4)
d) (1-3x)2-(x-2)(9x+1)=(3x-4)(3x+4)-9(x+3)2
Giúp mk với ạ, mk cảm ơn !
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=\(\dfrac{-10}{24}=\dfrac{-5}{12}\)
Đúng 3
Bình luận (0)
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=\(-\dfrac{39}{20}\)
Đúng 2
Bình luận (0)
c) (2x+3)2 +(x-1)(x+1)=5(x+2)2-(x-5)(x+1)+(x+4)
⇒4x2+12x+9+x2-1=5(x2+4x+4)-(x2+x-5x-5)+x+4
⇒5x2+12x+8=5x2+20x+20-x2-x+5x+5+x+4
⇒5x2+12x+8-5x2-20x-20+x2+x-5x-5-x-4=0
⇒x2-13x-21=0
Đúng 10
Bình luận (0)
Xem thêm câu trả lời
Tìm x biết
(1-3x)2 - (x-2)(9x+1) = (3x-4)(3x+4) - 9(x+3)2
(1 - 3x)2 - (x - 2)(9x + 1) = (3x - 4)(3x + 4) - 9(x + 3)2
⇌ 1 - 6x + 9x2 - 9x2 - x + 18x + 2 = 9x2 - 16 - 9x2 - 54x - 81 ⇌ 65x = -100 ⇌ x = \(-\frac{20}{13}\)
Tìm x biết
(1-3x)2 - (x-2)(9x+1) = (3x-4)(3x+4) - 9(x+3)2
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Rightarrow1-6x+9x^2-9x^2+18x-x-2=9x^2-16-9x^2-6x-9\)
\(\Rightarrow\left(-6x+18x-x+6x\right)+\left(9x^2-9x^2-9x^2+9x^2\right)=-1+2-16-9\)
\(\Rightarrow17x=-24\)
\(\Rightarrow x=-\dfrac{24}{17}.\)
Vậy \(x=-\dfrac{24}{17}.\)
Đúng 0
Bình luận (0)
\(\left(1-3x\right)^2-\left(x-2\right)\left(9x+1\right)=\left(3x-4\right)\left(3x+4\right)-9\left(x+3\right)^2\)
\(\Rightarrow1-6x+9x^2-x\left(9x+1\right)+2\left(9x+1\right)=9x^2-16-9\left(x^2+6x+9\right)\)\(\Rightarrow1-6x+9x^2-9x^2-x-18x-2=9x^2-16-9x^2-54x-81\)\(\Rightarrow-1-24x=97-54x\)
\(\Rightarrow-1-24x-97+54x=0\)
\(\Rightarrow-98x+20x=0\)
\(\Rightarrow x=\dfrac{49}{10}\)
Đúng 0
Bình luận (0)
\(\left(1-3x^{ }\right)^2-\left(x-2\right).\left(9x=1\right)=\left(3x-4\right).\left(3x+4\right)-9\left(x+3\right)^2\)
\(1-6x+9x^2-9x^2+18x-x-2=9x^2-16-9x^2-6x-9\)
\(\left(-6x+18x-x+6x\right)+\left(9x^2-9x^2-9x^2=9x^2\right)=-1+2=16-9\)
\(17x=-24\)
\(=>x=-\dfrac{24}{17}\)
Vậy \(x=-\dfrac{24}{17}\)
Đúng 0
Bình luận (0)
Tìm x,biết
b) |3/4x-5|-2/3=|-1/4|
a) (3x-1) (-1/4 - 5x) (-2/7x + 3)=0
c)|-2/5 - 3x| - |-7/9 + 2x|=0
d) |-3/7x - 1| + |5+2/3x|=0
e)|3x-1|+|9x^2 - 1|=0
f) x+2/2018 + x+4/2016 = x+6/2014 - 1
Mọi người giải giúp e với ạ chiều e phải nộp r
ngu thế à bạn
tìm x;y: 1) 9x^2-6x=9
2) (2x+3)^2 + 2(2x+3)(x-2)+(2-x)^2=4
3)(x+3)(3-x)=5
4)(3x+2)(9x^2-3x+1)=2
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(2-x\right)^2=4\)
\(\left(2x+3\right)^2+2\left(2x+3\right)\left(x-2\right)+\left(x-2\right)^2=4\)
\(\left(2x+3+x-2\right)^2=\left(\pm2\right)^2\)
\(\left(3x+1\right)^2=\left(\pm2\right)^2\)
\(\left[\begin{array}{nghiempt}3x+1=2\\3x+1=-2\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=2-1\\3x=-2-1\end{array}\right.\)
\(\left[\begin{array}{nghiempt}3x=1\\3x=-3\end{array}\right.\)
\(\left[\begin{array}{nghiempt}x=\frac{1}{3}\\x=-1\end{array}\right.\)
***
\(\left(x+3\right)\left(3-x\right)=5\)
\(3^2-x^2=5\)
\(x^2=9-5\)
\(x^2=4\)
\(x^2=\left(\pm2\right)^2\)
\(x=\pm2\)
***
\(\left(3x+1\right)\left(9x^2-3x+1\right)=2\)
\(27x^3+3=2\)
\(27x^3=2-3\)
\(\left(3x\right)^3=-1\)
\(3x=-1\)
\(x=-\frac{1}{3}\)
Đúng 0
Bình luận (0)
