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6 tháng 10 2018
\(4x\left(x-1\right)=8\left(x+1\right)\)
\(\Rightarrow4x^2-4x=8x+8\)
\(\Rightarrow4x^2-4x-8x-8=0\)
\(\Rightarrow4x^2-12x-8=0\)
\(\Rightarrow\left(2x\right)^2-2.2x.3+9-17=0\)
\(\Rightarrow\left(2x-3\right)^2=17\)
\(\Rightarrow\orbr{\begin{cases}2x-3=\sqrt{17}\\2x-3=-\sqrt{17}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{\sqrt{17}+3}{2}\\x=\frac{-\sqrt{17}+3}{2}\end{cases}}}\)
Chúc bạn học tốt.
NO
2
8 tháng 12 2017
a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy \(x=1;-3\)
b) \(x^2-4x+8=2x-1\)
\(\Leftrightarrow x^2-4x+8-2x+1=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy x=3
9 tháng 12 2017
a) \(\left(x-1\right)\left(2x+3\right)-x\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+3-x\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+3\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-1=0\\x+3=0\end{cases}\Rightarrow\orbr{\begin{cases}x=1\\x=-3\end{cases}}}\)
Vậy \(x=1;-3\)
b) \(x^2-4x+8=2x-1\)
\(\Leftrightarrow x^2-4x+8-2x+1=0\)
\(\Leftrightarrow x^2-6x+9=0\)
\(\Leftrightarrow\left(x-3\right)^2=0\)
\(\Leftrightarrow x-3=0\)
\(\Leftrightarrow x=3\)
Vậy \(x=3\)
BT
0
P
1
6 tháng 10 2016
a) 4x(x + 1) = 8(x + 1)
=> 4x(x + 1) - 8(x + 1) = 0
=> 4(x + 1).(x - 2) = 0
=> (x + 1)(x - 2) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x+1=0\\x-2=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=-1\\x=2\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=-1\\x=2\end{array}\right.\)
b) x2(x - 2) + 2 - x = 0
=> x2.(x - 2) - (x - 2) = 0
=> (x - 2).(x2 - 1) = 0
=> (x - 2).(x - 1).(x + 1) = 0
\(\Rightarrow\left[\begin{array}{nghiempt}x-2=0\\x-1=0\\x+1=0\end{array}\right.\)\(\Rightarrow\left[\begin{array}{nghiempt}x=2\\x=1\\x=-1\end{array}\right.\)
Vậy \(\left[\begin{array}{nghiempt}x=2\\x=1\\x=-1\end{array}\right.\)
25 tháng 8 2019
a) \(\left(x-3\right)^2-4=0\)
\(\left(x-3\right)^2=0+4\)
\(\left(x-3\right)^2=4\)
\(\left(x-3\right)^2=\pm4\)
\(\left(x-3\right)^2=\pm2^2\)
\(\orbr{\begin{cases}x-3=2\\x-3=-2\end{cases}}\)
\(\orbr{\begin{cases}x=5\\x=1\end{cases}}\)
b) \(\left(2x+3\right)^2-\left(2x+1\right)\left(2x-1\right)=22\)
\(4x^2+12x+9-4x^2+1=22\)
\(12x+10=22\)
\(12x=22-10\)
\(12x=12\)
\(x=1\)
c) \(\left(4x+3\right)\left(4x-3\right)-\left(4x-5\right)^2=16\)
\(16x^2-9-16x^2+40x-25=16\)
\(-34+40x=16\)
\(40x=16+34\)
\(40x=50\)
\(x=\frac{50}{40}=\frac{5}{4}\)
d) \(x^3-9x^2+27x-27=-8\)
\(x^3-9x^2+27x-27+8=0\)
\(x^3-9x^2+27x-19=0\)
\(\left(x^2-8x+19\right)\left(x-1\right)=0\)
Vì \(\left(x^2-8x+19\right)>0\) nên:
\(x-1=0\)
\(x=1\)
e) \(\left(x+1\right)^3-x^2\left(x+3\right)=2\)
\(x^3+2x^2+x+x^2+2x+1-x^2-3x^2=2\)
\(3x+1=2\)
\(3x=2-1\)
\(3x=1\)
\(x=\frac{1}{3}\)
29 tháng 11 2023
a: \(x^3-4x^2-x+4=0\)
=>\(\left(x^3-4x^2\right)-\left(x-4\right)=0\)
=>\(x^2\left(x-4\right)-\left(x-4\right)=0\)
=>\(\left(x-4\right)\left(x^2-1\right)=0\)
=>\(\left[{}\begin{matrix}x-4=0\\x^2-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=4\\x^2=1\end{matrix}\right.\Leftrightarrow x\in\left\{2;1;-1\right\}\)
b: Sửa đề: \(x^3+3x^2+3x+1=0\)
=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)
=>\(\left(x+1\right)^3=0\)
=>x+1=0
=>x=-1
c: \(x^3+3x^2-4x-12=0\)
=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)
=>\(x^2\cdot\left(x+3\right)-4\left(x+3\right)=0\)
=>\(\left(x+3\right)\left(x^2-4\right)=0\)
=>\(\left(x+3\right)\left(x-2\right)\left(x+2\right)=0\)
=>\(\left[{}\begin{matrix}x+3=0\\x-2=0\\x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=2\\x=-2\end{matrix}\right.\)
d: \(\left(x-2\right)^2-4x+8=0\)
=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)
=>\(\left(x-2\right)^2-4\left(x-2\right)=0\)
=>\(\left(x-2\right)\left(x-2-4\right)=0\)
=>(x-2)(x-6)=0
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)
4x * ( x + 1 ) = 8 * ( x + 1 )
Nên 4x : 4 * ( x + 1 ) = 8 : 4 * ( x + 1 )
x * ( x + 1 ) = 2 * ( x + 1 )
\(\Rightarrow\)x = 2
Ta có: 4x*(x+1)=8*(x+1)
=>4x*(x+1)-8*(x+1)=0
=>(x+1)(4x-8)=0
=>x+1=0 hoặc 4x-8=0
=>x=-1 hoặc x=2
Vậy .......