Tìm x biết :
\(x^2-6x+9=\left(2x+1\right)^2\)
TÌM X, BIẾT
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=x^2-x-6x-6\)
\(\Leftrightarrow18x+16=x^2-7x-6\)
\(\Leftrightarrow x^2-7x-18x=16+6\)
\(\Leftrightarrow x^2-15x=22\)
\(\Leftrightarrow x^2-15x-22=0\)
......
Tìm x biết:
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
\(\Leftrightarrow\left(6x^2+27x+4x+18\right)-\left(6x^2+x+12x+2\right)=x-1-x+6\)
\(\Leftrightarrow6x^2+31x+18-6x^2-x-12x-2=7\)
\(\Leftrightarrow18x+16=7\)
\(\Leftrightarrow18x=-9\)
\(\Leftrightarrow x=\frac{-1}{2}\)
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
\(3x\left(2x+9\right)+2\left(2x+9\right)-x\left(6x+1\right)-2\left(6x+1\right)=x-1-x+6\)
\(6x^2+27x+4x+18-6x^2-x-12x-2=5\)
\(6x^2+\left(27x+4x\right)+18-6x^2-\left(12x+x\right)-2=5\)
\(6x^2+31x+18-6x^2-13x-2=5\)
\(\left(6x^2-6x^2\right)+\left(31x-13x\right)+\left(18-2\right)=5\)
\(18x+16=5\)
\(18x=5+16\)
\(18x=21\)
\(x=21:18\)
\(x=\frac{7}{6}\)
Vậy \(x=\frac{7}{6}\)
P/s: Mình mới lớp 6 nên hi vọng bn xem bài của mik thật kĩ xem có sai sót không,cảm ơn.
Tìm x biết : \(\left(2x+5\right)^2+\left(4x+10\right)\left(3-x\right)+x^2-6x+9=0\)
(2x + 5 )2 + (4x+10)(3-x) + x2-6x+9=0
=>(2x+5)2_ 2(2x+5)(x-3) + (x-3)2=0
=>[(2x+5)-(x-3)]2 =0
=>(x+8)2=0
=> x+8=0
=> x=-8
Tìm x, biết
a)\(\left|x+1\right|\) +2x=-2
b)x^2-6x+9=1
a, |x+1| +2x \(=\)-2
\(\Leftrightarrow\) |x+1|\(=\)-2-2x (*)
TH1: Nếu x+1\(\ge0\)\(\Leftrightarrow x\ge-1\)thì \(\left|x+1\right|=x+1\)
Thay vào (*) ta có:
\(x+1=-2-2x\)
\(\Leftrightarrow x+2x=-2-1\)
\(\Leftrightarrow3x=-3\)
\(\Leftrightarrow x=-1\)(TMĐK)
TH2: Nếu \(x+1< 0\Leftrightarrow x< -1\Rightarrow\left|x+1\right|=-\left(x+1\right)\)
Thay vào (*), ta có:
\(-x-1=\)\(-2-2x\)
\(\Leftrightarrow-x+2x=-2+1\)
\(\Leftrightarrow x=-1\)(kTMĐK)
Vậy S\(=\){-1}
b, \(x^2-6x+9=1\)
\(\Leftrightarrow x^2-2.3.x+3^2\)
\(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left(x-3\right)^2-1^2=0\)
\(\Leftrightarrow\left(x-3-1\right)\left(x-3+1\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Vậy S\(=\){4;2}
a) |x + 1| + 2x = 2
\(\Rightarrow\) |3x + 1| = 2
\(\Rightarrow\) |3x| = 1
\(\Rightarrow\) |x| = 1 : 3 = \(\dfrac{1}{3}\)
\(\Rightarrow\) x = \(\dfrac{1}{3}\) hoặc \(-\dfrac{1}{3}\)
Tìm x, biết
a)\(\left|x+1\right|\) +2x=-2
b)x^2-6x+9=1
a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\\left(-2x-2\right)^2-\left(x+1\right)^2=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x< =-1\\3\left(x+1\right)^2=0\end{matrix}\right.\Leftrightarrow x=-1\)
b: \(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x=2\end{matrix}\right.\)
Tìm x biết:
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
Giải:
\(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x-1\right)-\left(x-6\right)\)
\(\Leftrightarrow6x^2+4x+27x+18-\left(6x^2+12x+x+2\right)=x-1-x+6\)
\(\Leftrightarrow6x^2+4x+27x+18-6x^2-12x-x-2=5\)
\(\Leftrightarrow16+18x=5\)
\(\Leftrightarrow18x=-11\)
\(\Leftrightarrow x=-\dfrac{11}{18}\)
Vậy ...
Tìm x biết
a) \(\left|x+1\right|\) +2x=-2
b) x^2-6x+9=1
a: \(\Leftrightarrow\left|x+1\right|=-2x-2\)
\(\Leftrightarrow\left[{}\begin{matrix}\left(-2x-2\right)^2-\left(x+1\right)^2=0\\-2x-2>=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}3\left(x+1\right)^2=0\\x< =-1\end{matrix}\right.\Leftrightarrow x=-1\)
b: \(\Leftrightarrow\left(x-3\right)^2=1\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=1\\x-3=-1\end{matrix}\right.\Leftrightarrow x\in\left\{4;2\right\}\)
Tìm x, biết
a) \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
b) \(3\left(2x-1\right)\left(3x-1\right)-\left(2x-3\right)\left(9x-1\right)=0\)
a) (3x+2)(2x+9) - (x+2)(6x+1) = (x+1) - (x-6)
<=> 6x2 + 27x + 4x + 18 - 6x2 - x - 12x - 2 = x+1 - x+6
<=> 18x + 16 = 7
<=> 18x = -9
<=> x = \(-\dfrac{1}{2}\)
b) 3(2x-1)(3x-1) - (2x-3)(9x-1) = 0
<=> 3.(6x2-2x-3x+1) - (18x2-2x-27x+3) = 0
<=> 3.(6x2-5x+1) - 18x2+29x-3 = 0
<=> 18x2-15x+3 - 18x2+29x - 3 = 0
<=> 14x = 0
<=> x = 0
b) \(\left(3x+2\right)\left(2x+9\right)-\left(x+2\right)\left(6x+1\right)=\left(x+1\right)-\left(x-6\right)\)
\(\Leftrightarrow\left(3x+2\right)2x+\left(3x+2\right)9-x.\left(6x+1\right)-2.\left(6x+1\right)=x+1-x+6\)
\(\Leftrightarrow6x^2+4x+27x+18-6x^2-x-12x-2=7\)
\(\Leftrightarrow18x+16=7\)
\(\Leftrightarrow18x=-9\)
\(\Leftrightarrow x=\frac{-1}{2}\)
Vậy ...