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toàn hằng đẳng thức (1) và (2) thôi mà bạn, đọc SGK 8 tập 1 là hiểu ngay. Có gì khó hiểu hỏi nhé!
a, x2-6x +9 = (x-3)2
b, 4x2+4x +1 = (2x)2+2.2x.1 +12=(2x+1)2
c, 9x2 -12x +4 = (3x-2)2
d, 25x2 -10x +1= (5x -1)2
e, x4-4x2+4 = (x2 -2)2
f, x2 +8x +16 = (x+4)2
Làm bài 1 thôi !! Mấy bài kia tương tự . Tìm nhân tử chung ra .
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-x-4\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3+4\right)\left(x-3-4\right)=x\left(x-7\right)\)
d) \(64+16y+y^2=\left(y+8\right)\left(y+8\right)\)
a) \(x^2-6x+9=x^2-2.3.x+3^2=\left(x-3\right)^2\)
b)\(x^2+4x+4=x^2+2.2.x+2^2=\left(x+2\right)^2\)
c)\(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
d)\(4x^2+12xy+9y^2=\left(2x\right)^2+2.2x.3y+\left(3y\right)^2=\left(2x+3y\right)^2\)
e)\(x^2-8x+16=x^2-2.4.x+4^2=\left(x-4\right)^2\)
a) x2 -6x +9 = (x-3)2
b) x2+4x +4= (x+2)2
c) 4x2+4x+1= (2x+1)2
d) 4x2+12xy+9y2 = (2x+3y)2
e) x2-8x+16 = (x-4)2
Đây chính là hằng đẳng thức nhé bn....
bình phương tổng chứ
b, B= x^2+ 2xy+y^2 +4y+4
= x^2+2xy+y^2+y^2+4y+4
=(x+y)^2+(y+2)^2
c, C= 2x^2+6xy+9y^2+2x+1
= x^2+6xy+9y^2+x^2+2x+1
= (x+3)^2+(x+1)^2
d, D= x(x+2) +(x+1)(x+3) +2
= x^2+2x+x^2+3x+x+3+2
= x^2+2x+1+x^2+4x+4
= (x+1)^2+(x+2)^2
e, E= x^2-2xy+2y^2+2y+1
= x^2-2xy+y^2+y^2+2y+1
= (x-y)^2+(y+1)^2
f, F= 4x^2-12xy+10y^2+4y+4
=4x^2-12xy+9y^2+y^2+4y+4
=(2x-3y)^2+(y+2)^2
g, G=2x^2+4xy+4y^2+4x+4
=x^2+4xy+4y^2+x^2+4x+4
=(x+2y)^2+(x+2)^2
Xong r.... dài quá...mới hè lớp 7 nên có j bỏ qua ak
a) \(4x^2-12x+9=\left(2x\right)^2-2.2x.3+3^2=\left(2x-3\right)^2\)
b) \(4x^2+4x+1=\left(2x\right)^2+2.2x.1+1^2=\left(2x+1\right)^2\)
c) \(1+12x+36x^2=1^2+2.6x.1+\left(6x\right)^2=\left(1+6x\right)^2\)
d) \(9x^2-24xy+16y^2=\left(3x\right)^2-2.3x.4y+\left(4y\right)^2=\left(3x-4y\right)^2\)
f) \(-x^2+10x-25=-\left(x^2-10x+25\right)=-\left(x-5\right)^2\)
g) \(-16a^4b^6-24a^5b^5-9a^6b^4=-\left(16a^4b^6+24a^5b^5+9a^6b^4\right)\)
\(=-\left[\left(4a^2b^3\right)^2+2.4a^2b^3.3a^3b^2+\left(3a^3b^2\right)^2\right]\)
\(=-\left(4a^2b^3+3a^3b^2\right)^2\)
h) \(25x^2-20xy+4y^2=\left(5x\right)^2-2.5x.2y+\left(2y\right)^2\) \(=\left(5x-2y\right)^2\)
i) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
a/ 9x2-12xy+4y2 = (3x - 2y)2
b/ 25x2-10x+1 = (5x - 1)2
c/ 9x2-12x+4 = (3x - 2)2
d/ 4x2+20x+25 = (2x + 5)2
e/ x4-4x2+4 = (x2 - 2)2
\(1,a,\left(12x-5\right)^2=12^2x^2-2.12.5x+5^2\)
\(b,\left(4x^2-y\right)^3=\left(4x^2\right)^3-3.\left(4x^2\right)^2y+3.4x^2.y^2-y^3=4x^6-3.16x^4y+12x^2y^2\)
\(c,\left(7x+8\right)^3=\left(7x\right)^3+3.\left(7x\right)^28+3.7x.8+8^3\)
\(a,x^3-16x=x\left(x^2-16\right)=x\left(x^2-4^2\right)=x\left(x-4\right)\left(x+4\right)\)
\(b,x^2-12x+36=x^2-2.x.6+6^2=\left(x-6\right)^2\)
\(1-8x^3=1^3-\left(2x\right)^3=\left(1-2x\right)\left(1^2+1.2x+\left(2x\right)^2\right)=\left(1-2x\right)\left(1+2x+4x^2\right)\)
\(d,\dfrac{1}{25}x^2-\dfrac{1}{64}y^2=\left(\dfrac{1}{5}x\right)^2-\left(\dfrac{1}{8}x\right)^2=\left(\dfrac{1}{5}x-\dfrac{1}{8}x\right)\left(\dfrac{1}{5}x+\dfrac{1}{8}x\right)=x\left(\dfrac{1}{5}-\dfrac{1}{8}\right)\left(\dfrac{1}{5}+\dfrac{1}{8}\right)\)
Bài 1:tìm x ,biết:
a) (2x - 1)(3x + 2) - 6x(x + 1) = 0
\(\Leftrightarrow6x^2+x-2-6x^2-6x=0\)
\(\Leftrightarrow-5x=2\)
\(\Leftrightarrow x=\frac{-2}{5}\)
b) \(\left(4x-1\right)^2-\left(2x+1\right)\left(8x-3\right)=0\)
\(\Leftrightarrow16x^2-8x+1-16x^2-2x+3=0\)
\(\Leftrightarrow-10x=-4\)
\(\Leftrightarrow x=\frac{2}{5}\)
c) \(4x^2-1=2\left(2x+1\right)\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-1\right)-2\left(2x+1\right)=0\)
\(\Leftrightarrow\left(2x+1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=\frac{3}{2}\end{cases}}\)
2a) \(4x^2-9y^2-6y-1=4x^2-\left(3y+1\right)^2\)
\(=\left(2x-3y-1\right)\left(2x+3y+1\right)\)
b) \(4x^2-1-2x\left(2x-1\right)=\left(2x-1\right)\left(2x+1\right)-2x\left(2x-1\right)\)
\(=1.\left(2x-1\right)\)
c) \(x^2-8x-4y^2+16=\left(x-4\right)^2-4y^2\)
\(=\left(x-4-2y\right)\left(x-4+2y\right)\)
d) \(9x^2-12x-y^2+4=\left(3x-2\right)^2-y^2\)
\(=\left(3x-2-y\right)\left(3x-2+y\right)\)
e) \(4x^2+10x-5=4x^2+2.2.\frac{5}{2}x+\frac{25}{4}-\frac{25}{4}-5\)
\(=\left(2x+\frac{5}{2}\right)^2-\frac{45}{4}\)
\(=\left(2x+\frac{5+3\sqrt{5}}{2}\right)\left(2x+\frac{5-3\sqrt{5}}{2}\right)\)
a) \(=\left(x-2\right)^2\)
b) \(=\left(3x-2\right)^2\)
c) \(=\left(x-3y\right)^2\)
d) \(=\left(\dfrac{x}{2}+1\right)^2\)
e) \(=\left(x-4\right)^2\)
f) \(=\left(\dfrac{1}{2}xy^2+1\right)^2\)
g) \(=\left(x-1\right)\left(x+1\right)\)
h) \(=\left(5x-4\right)\left(5x+4\right)\)
thank you bạn nha