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12 tháng 10 2020
a) 2x (x-5) -(x2-10x +25)=0
\(\Leftrightarrow\)2x(x-5)-(x-5)2=0
\(\Leftrightarrow\)(x-5)(2x-x+5)=0
\(\Leftrightarrow\)(x-5)(x+5)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x-5=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
b) x2 - 9 +3x(x+3) = 0
\(\Leftrightarrow\)(x2 - 9) +3x(x+3) =0
\(\Leftrightarrow\)(x-3)(x+3)+3x(x+3)=0
\(\Leftrightarrow\)(x+3)(x-3+3x)=0
\(\Leftrightarrow\)(x+3)(4x-3)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x+3=0\\4x-3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=-3\\4x=3\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=\frac{3}{4}\end{matrix}\right.\)
c) x3 - 16x = 0
\(\Leftrightarrow\)x(x2-16)=0
\(\Leftrightarrow\)x(x-4)(x+4)=0
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
d) (2x+3)(x-2) - (x2 -4x+4) = 0
\(\Leftrightarrow\)(2x+3)(x-2) -(x-2)2=0
\(\Leftrightarrow\)(x-2)(2x+3-x+2)=0
\(\Leftrightarrow\)(x-2)(x+5)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+5=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=2\\x=-5\end{matrix}\right.\)
e) 9x2 -(x2 -2x +1)=0
\(\Leftrightarrow\)(3x)2-(x-1)2=0
\(\Leftrightarrow\)(3x-x+1)(3x+x-1)=0
\(\Leftrightarrow\)(2x+1)(4x-1)=0
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x+1=0\\4x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}2x=-1\\4x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{-1}{2}\\x=\frac{1}{4}\end{matrix}\right.\)
f)x3-4x2 -9x +36 = 0
\(\Leftrightarrow\)(x3-9x)-(4x2-36)=0
\(\Leftrightarrow\)x(x2-9)-4(x2-9)=0
\(\Leftrightarrow\)(x-4)(x2-9)=0
\(\Leftrightarrow\)(x-4)(x-3)(x+3)=0
\(\Leftrightarrow\left[{}\begin{matrix}x-4=0\\x-3=0\\x+3=0\end{matrix}\right.\)
\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=4\\x=3\\x=-3\end{matrix}\right.\)
g) 3x - 6 = (x-1).(x-2)
\(\Leftrightarrow\)3(x-2)=(x-1)(x-2)
\(\Leftrightarrow\)x-1=3
\(\Leftrightarrow\)x=4
i) (x-2).(x+2) +(2x+1)2 =-5x.(x-3) =5 (?? đề sao vậy ??)
k) x2 -1 = (x-1).(2x+3)
\(\Leftrightarrow\)(x-1)(x+1)=(x-1)(2x+3)
\(\Leftrightarrow\)x+1=2x+3
\(\Leftrightarrow\)x-2x=3-1
\(\Leftrightarrow\)-x=2
\(\Leftrightarrow\)x=-2
l) (2x-1)2 +(x+3).(x-3) -5x(x-2)=6
\(\Leftrightarrow\)4x2-4x+1+x2-9-5x2+10x=6
\(\Leftrightarrow\)6x-8=6
\(\Leftrightarrow\)6x=14
\(\Leftrightarrow\)x=\(\frac{7}{3}\)
HP
14 tháng 8 2016
a) \(\left(y-1\right)^2=9\)
\(\Rightarrow\left(y-1\right)^2=3^2=\left(-3\right)^2\)
\(\Rightarrow x-1=3\Rightarrow x=4\)
\(\Rightarrow x-1=-3\Rightarrow x=-2\)
Vậy: \(x=4\) hoặc \(-2\)
PH
2
21 tháng 9 2017
a) \(x^3-4x^2-9x+36=0\Leftrightarrow x^3-7x^2+12x+3x^2-21x+36=0\) \(x\left(x^2-7x+12\right)+3\left(x^2-7x+12\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-7x+12\right)=0\) \(\Leftrightarrow\left(x+3\right)\left(x^2-7x+12\right)=0\Leftrightarrow\left(x+3\right)\left(x^2-3x-4x+12\right)=0\) \(\Leftrightarrow\left(x+3\right)\left(x\left(x-3\right)-4\left(x-3\right)\right)=0\Leftrightarrow\left(x+3\right)\left(x-4\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+3=0\\x-4=0\\x-3=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-3\\x=4\\x=3\end{matrix}\right.\) vậy \(x=-3;x=4;x=3\)
b) \(5x^2-4\left(x^2-2x+1\right)-5=0\) \(\Leftrightarrow5x^2-4x^2+8x-4-5=0\)
\(\Leftrightarrow x^2+8x-9=0\Leftrightarrow x^2-x+9x-9=0\)
\(\Leftrightarrow x\left(x-1\right)+9\left(x-1\right)\Leftrightarrow\left(x+9\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+9=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=-9\\x=1\end{matrix}\right.\) vậy \(x=-9;x=1\)
c) đề có sai o bn
d) \(x^3-3x+2=0\Leftrightarrow x^3+x^2-2x-x^2-x+2=0\)
\(\Leftrightarrow x\left(x^2+x-2\right)-\left(x^2+x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x^2+x-2\right)\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-x+2x-2\right)=0\Leftrightarrow\left(x-1\right)\left(x\left(x-1\right)+2\left(x-1\right)\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+2\right)\left(x-1\right)=0\) \(\Leftrightarrow\left\{{}\begin{matrix}x-1=0\\x+2=0\\x-1=0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=1\\x=-2\\x=1\end{matrix}\right.\)
vậy \(x=1;x=-2\)
LH
21 tháng 9 2017
1. \(x^3-4x^2-9x+36=0\)
\(\Rightarrow x^2.\left(x-4\right)-9\left(x-4\right)=0\)
\(\Rightarrow\left(x^2-9\right)\left(x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^2-9=0\Rightarrow x\in\left\{3;-3\right\}\\x-4=0\Rightarrow x=4\end{matrix}\right.\)
Vậy ..........
2. \(5x^2-4\left(x^2-2x+1\right)-5=0\)
\(\Rightarrow5x^2-4\left(x^2-1\right)-5=0\)
\(\Rightarrow5x^2-4x^2+4-5=0\)
\(\Rightarrow x^2-1=0\)
\(\Rightarrow x^2=1\)
\(\Rightarrow x=\pm1\)
Vậy .......
3. \(x^3-3x+2=0\)
\(\Rightarrow x^3-4x+x+2=0\)
\(\Rightarrow x.\left(x^2-4\right)+x+2=0\)
\(\Rightarrow x.\left(x-2\right).\left(x+2\right)+x+2=0\)
\(\Rightarrow\left(x+2\right).\left(x^2-2x+1\right)=0\)
\(\Rightarrow\left(x+2\right)\left(x-1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}\left(x+2\right)=0\\\left(x-1\right)^2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-2\\x=1\end{matrix}\right.\)
Vậy .......
11 tháng 3 2020
Đề bài là giải các phương trình nha :Đ
11 tháng 3 2020
\(b,\left(2x+1\right)^2=9\)
\(\left(2x+1\right)^2=3^2\)
\(\Rightarrow\orbr{\begin{cases}2x+1=3\\2x+1=-3\end{cases}\Rightarrow\orbr{\begin{cases}2x=2\\2x=-4\end{cases}\Rightarrow}\orbr{\begin{cases}x=1\\x=-2\end{cases}}}\)
\(c,x^3+5x^2-4x-20=0\)
\(x^2\left(x+5\right)-4\left(x+5\right)=0\)
\(\left(x^2-4\right)\left(x+5\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x^2-4=0\\x+5=0\end{cases}\Rightarrow\orbr{\begin{cases}x^2=4\\x=5\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=-2\end{cases};x=5}\)
ko phải mk lười đâu , cái này bn làm nó mới có ý nghĩa , cố gắng nốt nha !
U
21 tháng 7 2018
e, x(x - 2) + x - 2 = 0
=> (x-1)(x-2) = 0
=> x - 1 = 0 hoặc x - 2 = 0
=> x = 1 hoặc x = 2
vậy_
b, x2 + 3x = 0
=> x(x + 3) = 0
=> x = 0 hoặc x + 3 = 0
=> x = 0 hoặc x = -3
vậy_
2x2 - 5x + 3 = 0
=> 2.x.x - 5.x = -3
=> x(2x - 5) = -3
đoạn này lập bảng
d) 4x2 - 9x + 5 = 0
=> 4.x.x - 9.x = -5
=> x(4x - 9) = -5
đến đây cx lập bảng
TT
1
1 tháng 10 2016
a) \(x\left(x-5\right)-4x+20=0\)
\(\Leftrightarrow x\left(x-5\right)-4\left(x-5\right)=0\)
\(\Leftrightarrow\left(x-4\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-4=0\\x-5=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=4\\x=5\end{array}\right.\)
b) \(x\left(x+6\right)-7x-42=0\)
\(\Leftrightarrow x\left(x+6\right)-7\left(x+6\right)=0\)
\(\Leftrightarrow\left(x+6\right)\left(x-7\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x+6=0\\x-7=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=-6\\x=7\end{array}\right.\)
d) \(x^2-9x+8=0\)
\(\Leftrightarrow x^2-x-8x+8=0\)
\(\Leftrightarrow x\left(x-1\right)-8\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-8\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\x-8=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=8\end{array}\right.\)
g) \(3x^2-5x+2=0\)
\(\Leftrightarrow3x^2-3x-2x+2=0\)
\(\Leftrightarrow3x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-2\right)=0\)
\(\Leftrightarrow\left[\begin{array}{nghiempt}x-1=0\\3x-2=0\end{array}\right.\)\(\Leftrightarrow\left[\begin{array}{nghiempt}x=1\\x=\frac{2}{3}\end{array}\right.\)
LT
3
5 tháng 7 2019
a) \(9x^2-12x+4=0\)
\(\Leftrightarrow\left(3x\right)^2-3x.2.2+2^2=0\)
\(\Leftrightarrow\left(3x-2\right)^2=0\)
\(\Leftrightarrow3x-2=0\)
\(\Leftrightarrow3x=2\)
\(\Leftrightarrow x=\frac{2}{3}\)
Vậy ...
b) \(\left(x-2\right)^2-25=0\)
\(\Leftrightarrow\left(x-2\right)^2=25\)
\(\Leftrightarrow\left(x-2\right)^2=5^2=\left(-5\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x-2=5\\x-2=-5\end{cases}\Leftrightarrow\orbr{\begin{cases}x=7\\x=-3\end{cases}}}\)
Vậy ...
5 tháng 7 2019
c) \(x^3+3x^2+3x+1=0\)
\(\Leftrightarrow\left(x+1\right)^3=0\)
\(\Leftrightarrow x+1=0\)
\(\Leftrightarrow x=-1\)
Vậy ...