Tìm x: a) 2x(x-5)-x(3+2x)=26
b) (x+4)2-(x+1)(x-1)=16
c) (2x-1)2-4(x+7)(x-7)=0
d) x(x-5)-4x+20=0
Tìm x , Biết
a) (x-4) x - (x-3)^2=0
b) 3x-6 = x^2-16
c) (2x-3)^2 - 49=0
d) 2x (x-5) - 7 (5-x)=0
a) \(\Leftrightarrow x^2-4x-x^2+6x-9=0\\ \Leftrightarrow2x=9\\ \Leftrightarrow x=4,5\)
b) \(\Leftrightarrow x^2-3x-10=0\\ \Leftrightarrow\left(x^2+2x\right)-\left(5x+10\right)=0\\ \Leftrightarrow x\left(x+2\right)-5\left(x+2\right)=0\\ \left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
c) \(\Leftrightarrow\left(2x-3-7\right)\left(2x-3+7\right)=0\\ \Leftrightarrow\left(2x-10\right)\left(2x+4\right)=0\\ \Leftrightarrow\left(x-5\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
d) \(\Leftrightarrow\left(2x+7\right)\left(x-5\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{7}{2}\\x=5\end{matrix}\right.\)
Bài 1: Tìm x , Biết
a) (x-4) x - (x-3)^2=0
b) 3x-6 = x^2-16
c) (2x-3)^2 - 49=0
d) 2x (x-5) - 7 (5-x)=0
Bài 2: Tìm m để đa thức
A(x)= 2x^3 + x^2 - 4x + m chia hết cho đa thức B(x)= 2x-1
Bài 3 : Phân tích đa thức thành nhân tử
a) x^2 - 8x
b) x^2 - xy - 6x + 6y
Bài 1:
b: \(3x-6=x^2-16\)
\(\Leftrightarrow x^2-3x-10=0\)
\(\Leftrightarrow\left(x-5\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)
a)2x-3/4-4x-5/3=5-x/6
b)3(x-1^2)=16
c)(x+7)(x-4)=2(x-4)
d)4x^2-3x-1=0
`a)[2x-3]/4-[4x-5]/3=[5-x]/6`
`<=>3(2x-3)-4(4x-5)=2(5-x)`
`<=>6x-9-16x+20=10-2x`
`<=>8x=1`
`<=>x=1/8`
(`b->c` mk đã làm r).
Tìm x: a) 2x(x-3)-x(3+2x)=26
b) (x+4)2-(x+1)(x-1)=16
c) (2x-1)2-4(x+7)(x-7)=0
d) x(x-5)-4x+20=0
mk làm lun nha
a, 2x^2-6x-3x-2x^2=26
-9x=26
x=-26/9
b,x^2+2.x.4+16-(x^2-1)=16
x^2+8x+16-x^2+1=16
8x=-1
x=-1/8
c,(2x)^2-2.2x.1+1-4(x^2-7^2)=0
4x^2-4x+1-4x^2+196=0
-4x=-197
x=197/4
d,x^2-5x-4x+20=0
-9x=-20
x=20/9
**** cho mk nha
Bài 4: Tìm x, biết.
a) 4x(x - 7) - 4x2 = 56
b) 12x(3x - 2) - (4 - 6x) = 0
c) 4(x - 5) - (5 - x)2 = 0
d) x(x +1) - x(x - 3) = 0
e) - 6x + 8 = 0 f) 2 + 2x + = 0
c: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\end{matrix}\right.\)
1. Tìm x,y:
a) (x+2)2 + (x-3)2 = 2x ( x+ 7)
b) x3- 3x2 + 3x - 126 = 0
c) x2 + y2 - 2x + 4y + 5 = 0
d) 2x2 - 2xy + y2 + 4x + 4 = 0
\(a.\left(x^2+4x+4\right)+\left(x^2-6x+9\right)=2x^2+14x\)
\(x^2+4x+4+x^2-6x+9-2x^2-14x=0\)
\(-18x+13=0\)
\(x=\dfrac{13}{18}\)
Vậy \(S=\left\{\dfrac{13}{18}\right\}\)
\(b.\left(x-1\right)^3-125=0\)
\(\left(x-1\right)^3=125\)
\(x-1=5\)
\(x=6\)
Vậy \(S=\left\{6\right\}\)
\(c.\left(x-1\right)^2+\left(y +2\right)^2=0\)
\(Do\left(x-1\right)^2\ge0\forall x;\left(y+2\right)^2\ge0\forall y\)
\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\forall x,y\)
Mà \(\left(x-1\right)^2+\left(y+2\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-1\right)^2=0\\\left(y+2\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\y+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)
Vậy \(S=\left\{1;-2\right\}\)
\(d.x^2-4x+4+x^2-2xy+y^2=0\)
\(\left(x-2\right)^2+\left(x-y\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\\left(x-y\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x-y=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\y=2\end{matrix}\right.\)
Vậy \(S=\left\{2;2\right\}\)
Giúp với mn ơi ;-;a) 9(2x+1)^2-4(x+1)^2=0b) (x+1)^2+2(x+1)+1=0c) (x-1)(x^2-9)+x+3=0d) (7-x)^2__________ -(x+5)^2=0 4e) 4x^2+(x-1)^2-(2x+1)^2=0f) x^3+1=(x+1)(2-x)
Rối quá bn ơi
Bn ko dùng dấu enter để xuống dòng à
Mk sẽ giúp bn 3 câu đầu thôi còn 3 câu sau thì..... chịu vì ko biết làm sao
a) \(9\left(2x+1\right)^2-4\left(x+1\right)^2=0\)
(=)\(\left[9\left(2x+1\right)-4\left(x+1\right)\right]\left[9\left(2x+1\right)+4\left(x+1\right)\right]=0\)
(=)\(\left(18x+9-4x-4\right)\left(18x+9+4x+4\right)=0\)
(=)\(\left(14x+5\right)\left(22x+12\right)=0\)
(=)\(\left(14x+5\right)=0\) hoặc \(\left(22x+12\right)=0\)
1) \(14x+5=0\Rightarrow x=-\dfrac{5}{14}\)
2) \(22x+12=0\Rightarrow x=-\dfrac{6}{11}\)
Vậy........
b)\(\left(x+1\right)^2+2\left(x+1\right)+1=0\)
(=)\(\left(x+1\right)^2+2\left(x+1\right).1+1^2=0\)
(=) \(\left(x+2\right)^2=0\)
(=) \(x+2=0\Rightarrow x=-2\)
Vậy........
c) \(\left(x-1\right)\left(x^2-9\right)+x+3=0\)
(=) \(\left(x-1\right)\left(x-3\right)\left(x+3\right)+x+3=0\)
(=) \(\left(x+3\right)\left(x-3+1\right)\left(x-1\right)=0\)
(=) \(\left(x+3\right)\left(x-2\right)\left(x-1\right)=0\)
(=) \(x+3=0\) hoặc \(x-2=0\) hoặc \(x-1=0\)
1)\(x+3=0\Rightarrow x=-3\)
2)\(x-2=0\Rightarrow x=2\)
3) \(x-1=0\Rightarrow x=1\)
Vậy.........
tìm x
1,(2x+3)^2+(2x+1)(2x-1)=22
2,(4x+3)(4x-3)=46+(4x-5)^2
3,(2x-1)^3-4x^2(2x-3)=5
4,(x+4)^2-(x-1)(x+1)=16
5,(2x-1)^2+(x+3)^2=5(x+7)(x-7)+325
6,(x^2+1)^2-(x^4+x^2+1)(x^2-1)=0
1)4x-20=0 ; 2) 5x+15=0 ; 3) 3x-5=7x+2 ; 4) 4x-(x-1)=2(1+x) ; 5) x2 -2x=0 ; 6) 2(3x-5)-3(x-2)=3(x+4) ; 7) (x+3)(2x-7)=0
8) 5x(x-3)+2x-6=0 ; 9) (3x-1)(2x-1)-(3x-1)(x+2)=0
10)|2x-1|+1=8 ; 11) |x-2|=3x+1 ; 12) |2x|=21-x
Giải các phương trình nha mọi người ^_^