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a) \(\left(x+\frac{1}{4}\right)^2\)
\(=x^2+\frac{1}{2}x+\frac{1}{8}\)
b) HĐT 5
c) HĐT 5
d) HĐT 7
e) HĐT 7
f) HĐT 7
1.=(x-y)(5x+1)
2.=(x+3)(2x+1)
3.=(3x-2y)2(1+1)=2(3x-2y)2
4.bạn chép sai hay sao ý
5.=(2x-3y)2
6. = -(x+y)2
7. = -(a-5)2
1. 5x(x-y)-(y-x)
= 5x(x-y)+(x-y)
= (x-y)(5x+1)
2. 2x(x+3)+(3+x)
= (x+3)(2x+1)
3. (3x-2y)2-(2x-3y)2
= (3x-2y-2x+3y)(3x-2y+2x-3y)
=(x+y)(5x-5y)
=5(x+y)(x-y)
4. 4-(a-b)2
= 22-(a-b)2
= (2-a+b)(2+a-b)
5. 4x2-12xy+9y2
= (2x-3y)2
6. -x2-2xy-y2
= -(x+y)2
7. 10a-a2-25
= -a2+10a-25
= -(a-5)2
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
\(a,36x^2-\left(3x-2\right)^2=\left(6x-3x+2\right)\left(6x+3x-2\right)\)
\(=\left(3x+2\right)\left(9x-2\right)\)
phần b,c,d lm tg tự
\(e,16x^2-24xy+9y^2=\left(4x-3y\right)^2\)
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
2) \(1-9x^2=\left(1-3x\right)\left(1+3x\right)\)
3) \(\frac{x^2}{9}-\frac{y^2}{16}=\left(\frac{x}{3}-\frac{y}{4}\right)\left(\frac{x}{3}+\frac{y}{4}\right)\)
4) \(a^4-b^4=\left(a^2-b^2\right)\left(a^2+b^2\right)=\left(a-b\right)\left(a+b\right)\left(a^2+b^2\right)\)
5) \(\left(a-b\right)^2-1=\left(a-b+1\right)\left(a-b-1\right)\)
6) \(4-\left(a-b\right)^2=\left(2-a+b\right)\left(2+a-b\right)\)
7) \(\left(x-y\right)^2-\left(m+n\right)^2=\left(x-y-m-n\right)\left(x-y+m+n\right)\)
8) \(\left(3x-2y\right)^2-\left(2x-3y\right)^2=\left(3x-2y-2x+3y\right)\left(3x-2y+2x-3y\right)\)
\(=\left[3\left(x+y\right)-2\left(x+y\right)\right]\left[3\left(x-y\right)+2\left(x-y\right)\right]=5\left(x+y\right)\left(x-y\right)\)
9) \(4x^2-12xy+9y^2=\left(2x-3y\right)^2\)
10) \(\left(x^4+2x^2+1\right)=\left(x^2+1\right)^2\)
11) \(\left(a^4+4-4x^2\right)=\left(a^2-2\right)^2\)
\(a,x^2-5x\)
\(=x\left(x-5\right)\)
\(b,5x\left(x+5\right)+4x+20\)
\(=5x\left(x+5\right)+4\left(x+5\right)\)
\(=\left(5x+4\right)\left(x+5\right)\)
\(c,7x\left(2x-1\right)-4x+2\)
\(=7x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(7x-2\right)-\left(2x-1\right)\)
\(d,x^2-16+2\left(x+4\right)\)
\(=x^2-16+2x+8\)
\(=x\left(x-2\right)-8\) ( Ý này thì k chắc lắm, sai thông cảm :)) )
\(e,x^2-10x+9\)
\(=x^2-x-9x+9\)
\(=x\left(x-1\right)-9\left(x-1\right)\)
\(=\left(x-9\right)\left(x-1\right)\)
\(f,\left(2x-1\right)^2-\left(x-3\right)^2=0\) ( mk đoán bài này là tìm x, sai thì bảo mk để mk sửa nhé )
\(\Rightarrow\left(2x-1\right)^2=\left(x-3\right)^2\)
\(\Leftrightarrow\pm\left(2x-1\right)=\pm\left(x-3\right)\)
\(\Rightarrow\hept{\begin{cases}2x-1=x-3\\-\left(2x-1\right)=-\left(x-3\right)\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2x-1-x+3=0\\-2x+1-x+3=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x+2=0\\-3x+4=0\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}x=\left(-2\right)\\x=\frac{4}{3}\end{cases}}\)
Vậy ...
1 Điền vào chỗ chấm để có hằng đẳng thức thích hợp :
a) x2 - 4y2 = ........(x-2y)(x+2y)......................
b) 1/42 + 2xy + 4y2 =.............(1/4+2y)2..........................
c) 64x3 - 1 =..................(4x-1)(16x2+4x+1)................
d) 25 + 10y + y2 =........(5+y)2....................
a) \(\left(x+4\right)\left(x^2-4x+16\right)\)
\(x^3-4x^2+16x+4x^2-16x+64\)
\(=x^3+64\)
\(=x^3+4^3\)
\(=\left(x+4\right)\left(x^2-4x+16\right)\)
b) \(\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
\(=\frac{1}{27}x^3-\frac{2}{9}x^2y+\frac{4}{3}xy^2+\frac{2}{9}x^2y-\frac{4}{3}xy^2+8y^3\)
\(=\frac{1}{27}x^3+8y^3\)
\(=\left(\frac{1}{3}x\right)^3+\left(2y\right)^3\)
\(=\left(\frac{1}{3}x+2y\right)[\left(\frac{1}{3}x\right)^2-(\frac{1}{3}x.2y)+\left(2y\right)^2]\)
\(=\left(\frac{1}{3}x+2y\right)\left(\frac{1}{9}x^2-\frac{2}{3}xy+4y^2\right)\)
Câu c và d tương tự .
cảm ơn bạn nhé