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`@` `\text {Ans}`
`\downarrow`
`A= (2x - 3)^2 - (2x + 3)^2`
`= [(2x - 3) - (2x + 3)]*[(2x - 3) + (2x + 3)]`
`= (2x - 3 - 2x - 3) * (2x - 3 + 2x + 3)`
`= -6 * 4x`
`= -24x`
![](https://rs.olm.vn/images/avt/0.png?1311)
dài quá, làm từ từ nhé
1, \(\left(a-b\right)^2\left(2a-3b\right)-\left(b-a\right)^2\left(3a-5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(2a-3b-3a+5b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-b\right)^2\left(-a+2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=-\left(a-b\right)^2\left(a-2b\right)+\left(a+b\right)^2\left(a-2b\right)\)
\(=\left(a-2b\right)\left[\left(a+b\right)^2-\left(a-b\right)^2\right]\)
\(=\left(a-2b\right)\left(a+b-a+b\right)\left(a+b+a-b\right)\)
\(=4ab\left(a-2b\right)\)
2, \(x^4-4\left(x^2+5\right)-25=\left(x^2-25\right)-4\left(x^2+5\right)=\left(x^2-5\right)\left(x^2+5\right)-4\left(x^2+5\right)\)
\(=\left(x^2-9\right)\left(x^2+5\right)=\left(x-3\right)\left(x+3\right)\left(x^2+5\right)\)
3,\(\left(2-x\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)=\left(x-2\right)^2+\left(x-2\right)\left(x+3\right)-\left(4x^2-1\right)\)
\(=\left(x-2\right)\left(x-2+x+3\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2\right)\left(2x+1\right)-\left(2x-1\right)\left(2x+1\right)\)
\(=\left(x-2-2x+1\right)\left(2x+1\right)\)
\(=\left(-x-1\right)\left(2x+1\right)\)
4, câu này đề thiếu
5,\(16\left(xy+6\right)^2-\left(4x^2+y^2-25\right)^2=\left(4xy+24\right)^2-\left(4x^2+y^2-25\right)^2\)
\(=\left(4xy+24-4x^2-y^2+25\right)\left(4xy+24+4x^2+y^2-25\right)\)
\(=\left[49-\left(4x^2-4xy+y^2\right)\right]\left[\left(4x^2+4xy+y^2\right)-1\right]\)
\(=\left[49-\left(2x-y\right)^2\right]\left[\left(2x+y\right)^2-1\right]\)
\(=\left(7-2x+y\right)\left(7+2x-y\right)\left(2x+y-1\right)\left(2x+y+1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) 4x - 1 = 3x - 2
<=> 4x - 3x = -2 + 1
<=> x = -1
Vậy S = {-1}
b) x + 1 = 2. (x - 3)
<=> x + 1 = 2x - 6
<=> x - 2x = -6 -1
<=> -x = -7
<=> x = 7
Vậy:....
c) 2. (x + 1) + 3 = 2 - x
<=> 2x + 2 + 3 = 2 - x
<=> 2x + x = 2 - 2 - 3
<=> 3x = -3
<=> x = -1
Vậy:,,,,
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(x^2+7x+10-12x+9=x^2-10x+25\)
\(\Leftrightarrow5x=6\Leftrightarrow x=\dfrac{6}{5}\)
b, bạn ktra lại đề nhé
c, \(x^2-4+3x+3=3+x^2-x-2\)
\(\Leftrightarrow x^2+3x-1=x^2-x+1\Leftrightarrow4x=2\Leftrightarrow x=\dfrac{1}{2}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(5a^2-5b^2=5\left(a^2-b^2\right)=5\left(a+b\right)\left(a-b\right)\)
b, \(2ab^2-a^2b-b^3=b\left(2ab-a^2b-b^2\right)=-b\left(a^2-2ab+b^2\right)\)
\(=-b\left(a-b\right)^2\)
a) 5a2 - 5b2 = 5(a2-b2) = ( a+b) (a-b)
b) 2ab2 - a2b - b3 = b ( 2ab-a2-b2 ) = -b (a2-2ab+b2) = -b (a-b)2
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a)
$(3-4x)^2=16(x-3)^2=4^2(x-3)^2=(4x-12)^2$
$\Leftrightarrow [(3-4x)-(4x-12)][(3-4x)+(4x-12)]=0$
$\Leftrightarrow (15-8x)(-9)=0$
$\Rightarrow 15-8x=0\Rightarrow x=\frac{15}{8}$
b)
$(x^2+x+1)^2=(4x-1)^2$
$\Leftrightarrow (x^2+x+1)^2-(4x-1)^2=0$
$\Leftrightarrow (x^2+x+1-4x+1)(x^2+x+1+4x-1)=0$
$\Leftrightarrow (x^2-3x+2)(x^2+5x)=0$
$\Leftrightarrow (x-1)(x-2)x(x+5)=0$
\(\Rightarrow \left[\begin{matrix} x-1=0\\ x-2=0\\ x=0\\ x+5=0\end{matrix}\right.\Leftrightarrow \left[\begin{matrix} x=1\\ x=2\\ x=0\\ x=-5\end{matrix}\right.\)
a) 9 - 24x + 16x2 = 16(x2 - 6x + 9)
=> 16x2 - 24x + 9 = 16x2 - 96x + 144
=> -24x + 96x = 144 - 9
=> 72x = 135
=> x = \(\frac{15}{8}\)
b) (x2 + x + 1)2 = (4x - 1)2
=> x4 + x2 + 1 + 2x3 + 2x + 2x2 = 16x2 - 8x + 1
=> x4 + 2x3 + 3x2 + 2x + 1 = 16x2 - 8x + 1
=> x4 + 2x3 - 13x2 + 10x = 0
=> x4 - x3 + 3x3 - 3x2 - 10x2 + 10x = 0
=> (x3 + 3x2 - 10x)(x - 10) = 0
=> x(x2 + 3x - 10)(x - 10) = 0
=> x(x - 2)(x+5)(x-10) = 0
=> \(\left[{}\begin{matrix}x=0\\x=2\\x=-5\\x=10\end{matrix}\right.\)
click cho mình nha
a) 3-4x = 4.(x-3) hoặc 3-4x = -4.(x-3)
3-4x=4x-12 hoặc 3-4x = -4x +12
8x=15 hoặc -4x+4x=12-3
x=15/8
b) x^2+x+1=4x-1 hoặc x^2+x+1= -(4x-1)
x^2-3x+2=0 hoặc x^2+5x=0
TH1: x^2-3x+2=0
x^2-x-2x+2=0
(x^2-x)-(2x-2)=0
x(x-1)-2(x-1)=0
(x-1).(x-2)=0
x=1 hoặc x=2
TH2: x^2+5x=0
x.(x+5)=0
x=0 hoặc x=-5
Các bạn tự đáp số nhé