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1,=> 36x^2-12x-36x^2+27x=30
=>15x =30
=> x =2
2,=>5x-2x^2+2x^2-2x=15
=>3x =15
=>x =5
\(a,\left(x-3\right)^2-4=0\)
\(\Leftrightarrow\left(x-3\right)^2=4\)
\(\Rightarrow x-3=\pm2\)
\(\hept{\begin{cases}x-3=2\Rightarrow x=5\\x-3=-2\Rightarrow x=1\end{cases}}\)
Vậy \(x=5\)hoặc \(x=1\)
\(b,x^2-2x=24\)
\(\Leftrightarrow x^2-2x+1-1=24\)
\(\Leftrightarrow\left(x-1\right)^2=24+1=25\)
\(\Leftrightarrow x-1=\pm5\)
\(\hept{\begin{cases}x-1=5\Rightarrow x=6\\x-1=-5\Rightarrow x=-4\end{cases}}\)
Vậy \(x=6\) hoặc \(x=-4\)
\(c,\left(2x+1\right)^2+\left(x+3\right)^2-5\left(x-7\right)\left(x+7\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5\left(x^2-49\right)=0\)
\(\Leftrightarrow4x^2+4x+1+x^2+6x+9-5x^2+245=0\)
\(\Leftrightarrow10x+255=0\)
\(\Leftrightarrow10x=-255\)
\(\Leftrightarrow x=\frac{-51}{2}\)
\(d,\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27+x\left(2x-x^2+4-2x\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
\(\Leftrightarrow4x-27=1\)
\(\Leftrightarrow4x=28\)
\(\Leftrightarrow x=7\)
\(\left(x+2\right)^2=\left(2x-1\right)^2\\ \Leftrightarrow\left(x+2\right)^2-\left(2x-1\right)^2=0\\\Leftrightarrow\left[x+2-\left(2x-1\right)\right]\left[x+2+2x-1\right]=0\\ \Leftrightarrow\left(x+2-2x+1\right)\left(x+2+2x-1\right)=0\\ \Leftrightarrow\left(-x+3\right)\left(3x+1\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}-x+3=0\\3x+1=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}-x=-3\\3x=-1\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
\(\left(x+2\right)^2=\left(2x-1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}x+2=2x-1\\x+2=-\left(2x-1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2x=-1-2\\x+2=-2x+1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}-x=-3\\x+2x=1-2\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\3x=-1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-\dfrac{1}{3}\end{matrix}\right.\)
a/ Ta có : \(49.x^2-4=0\)
\(\Rightarrow49x^2=4\)
\(\Rightarrow x^2=\frac{4}{49}\Rightarrow\orbr{\begin{cases}x=\frac{-2}{7}\\x=\frac{2}{7}\end{cases}}\)
b/ \(\left(x+3\right)^2-\left(x+2\right)\left(x-2\right)=11\)
\(\left(x+3\right)\left(x+3\right)-\left(x+2\right)\left(x-2\right)=11\)
\(\Rightarrow\left(x^2+2.3.x+3^2\right)-\left(x^2-2^2\right)=11\)
\(\Rightarrow x^2+6x+9-x^2+4=11\)
\(\Rightarrow6x+13=11\)
\(\Rightarrow6x=11-13\)
\(\Rightarrow x=\frac{-2}{6}=\frac{-1}{3}\)
c/ \(\left(2x+1\right)^2-\left(x-3\right)^2-3\left(x+5\right)\left(x-5\right)=5\)
\(\Rightarrow\left(2x+1\right)\left(2x+1\right)-\left(x-3\right)\left(x-3\right)-3\left[\left(x+5\right)\left(x-5\right)\right]=5\)
\(\Rightarrow\left(4x^2+2.2x+1\right)-\left(x^2-2.3x+9\right)-3\left(x^2-25\right)\)\(=5\)
\(\Rightarrow\left(4x^2+4x+1\right)-\left(x^2-6x+9\right)-\left(3x^2-75\right)=5\)
\(\Rightarrow4x^2+4x+1-x^2+6x-9-3x^2+75=5\)
\(\Rightarrow\left(4x^2-x^2-3x^2\right)+\left(4x+6x\right)+\left(1-9+75\right)=5\)
\(\Rightarrow10x+67=5\)
\(\Rightarrow10x=5-67=-62\)
\(\Rightarrow x=\frac{-62}{10}=\frac{-31}{5}\)
d/ \(\left(3x+1\right)\left(3x-1\right)=8\)
\(\Rightarrow9x^2-1=8\)
\(\Rightarrow9x^2=8+1=9\)
\(\Rightarrow x^2=\frac{9}{9}=1\Leftrightarrow\orbr{\begin{cases}x=-1\\x=1\end{cases}}\)
Ai đó bấm hộ mình cái nút đúng đi!
Ta có : 49x2 - 4 = 0
=> 49x2 = 4
=> x2 = 196
=> x2 = 142 ; (-14)2
=> x = 14 ; -14
áp dụng các hằng đẳng thức thôi mà :)
a)\(x^2-2x+1=25\)
=>\(\left(x-1\right)^2=25\)
=>\(\orbr{\begin{cases}x-1=-5\\x-1=5\end{cases}}\)
b)\(3\left(x-1\right)^2-3x\left(x-5\right)=1\)
=>\(3\left[\left(x-1\right)^2-x\left(x-5\right)\right]=1\)
=>\(3\left(x^2-2x+1-x^2+5x\right)=1\)
=>\(3\left(3x+1\right)=1\)
=>\(3x+1=\frac{1}{3}\)
=>\(3x=\frac{-2}{3}\)
=>\(x=\frac{-2}{9}\)
c)\(\left(5-2x\right)^2-16=0\)
=>\(\left(5-2x\right)^2-4^2=0\)
=>\(\left(5-2x-4\right)\left(5-2x+4\right)=0\)
=>\(\orbr{\begin{cases}5-2x-4=0\\5-2x+4=0\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{1}{2}\\x=\frac{9}{2}\end{cases}}}\)
1) (x + 2)(x - 2) - (x + 3)(x + 1)
= x^2 - 4 - (x - 3)(x + 1)
= x^2 - 4 - x^2 + 2x + 3
= 2x - 1
2) a) 5(x - y) - 3x(y - x)
= 5x - 5y - 3x(y - x)
= 5x - 5y - 3xy + 3x2
b) 5x^2 - 16 + 3
= (5x^2 - x) + (-15x + 3)
= x(5x - 1) - 3(5x - 1)
= (5x - 1)(x - 3)
3) a) 2x(x + 3) + 12 - 2x^2 = 0
<=> 2x(x + 3) + 12 - 2x^2 = 0 - 12
<=> 2x(x + 3) - 2x^2 = -12
<=> x = -2
b) x^3 - 16x = 0
<=> x(x + 4)(x - 4) = 0
<=> x = 0
<=> x = 0; x = +- 4
c) (2x - 1)^2 = (x + 3)^2
<=> 4x^2 - 4x + 1 = x^2 + 6x + 9
<=> 4x^2 - 4x + 1 = x^2 + 6x + 9 - 9
<=> 4x^2 - 4x - 8 = x^2 + 6x
<=> 4x^2 - 4x - 8 = x^2 + 6x - 6x
<=> 4x^2 -10x - 8 = x^2
<=> 3x^2 - 10x - 8 = 0
<=> x = 4, x = -2/3
d) x^2 - x - 6 = 0
<=> x = -2; x = 3
\(\left(x+2\right)\left(x-2\right)-\left(x+3\right)\left(x+1\right)\)
\(=x^2-4-\left(x^2+4x+3\right)\)
\(=x^2-4-x^2-4x-3\)
\(=-4x-7\)
(2x - 1)^2 + (x + 3)^2 - 5(x + 7)(x - 7) = 0
<=>4x^2-4x+1+x^2+6x+9-5x^2+245=0
<=>2x+255=0
<=>2x=-255
<=>x=-255/2
Có trên google ( ghi nguồn đầy đủ )
\(a,=x^2-1-\left(x^2+4x+4\right)=x^2-1-x^2-4x-4=11\)
\(\Leftrightarrow-5x=15\)
\(\Leftrightarrow x=-3\)
Vậy ...
\(b,=\left(x-3-2x+5\right)\left(x-3+2x-5\right)=0\)
\(\Leftrightarrow\left(-x+2\right)\left(3x-8\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\dfrac{8}{3}\end{matrix}\right.\)
Vậy ...
a) \(\left(x-1\right)\left(x+1\right)-\left(x+2\right)^2=11\)
\(\Rightarrow x^2-1-x^2-4x-4-11=0\)
=> -4x - 16 = 0
=> -4x = 16
=> x = -4
b) \(\left(x-3\right)^2-\left(2x-5\right)^2=0\)
=> (x - 3 + 2x - 5).(x - 3 - 2x + 5) = 0
=> (3x - 8).(-x + 2) = 0
=> x = 8/3 hoặc x = 2