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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) \(25x^4-10x^2y+y^2=\left(5x^2\right)^2-2.5x^2.y+y^2=\left(5x^2-y\right)^2\)
Bài 8:
b. 1+8x6y3 = 13+23(x2)3y3 = 13+(2x2y)3
= (1+2x2y)(1-2x2y+4x4y2)
e. 27x3+\(\dfrac{y^3}{8}\)\(=\left(3x\right)^3+\left(\dfrac{y}{2}\right)^3\)
= (3x+\(\dfrac{y}{2}\))(9x2-\(\dfrac{3xy}{2}\)+\(\dfrac{y^2}{4}\))
Bài 9:
c. 1- 9x +27x2 -27x3 = 13-3.12.3x+3.(3x)2-(3x)3
= (1-3x)3
d. x3+\(\dfrac{3}{2}x^2\)+\(\dfrac{3}{4}x+\dfrac{1}{8}\) = x3+\(3x^2.\dfrac{1}{2}\)+\(3x.\dfrac{1}{4}+\left(\dfrac{1}{2}\right)^3\)
= (x+\(\dfrac{1}{2}\))3
f. x2 - 2xy +y2 -4m2 +4m.n - n2 = (x2 - 2xy +y2)-((2m)2 -2.2m.n + n2)
= (x-y)2-(2m-n)2 = (x-y-2m+n)(x-y+2m-n)
Bạn khá hiểu bài rồi đó. Đúng hết 4 câu đầu luôn.
Bổ sung thêm vào câu 3 một chút (nối tiếp theo sau nhé):
\(\Rightarrow\left(m-n\right)\left(x-\sqrt{3}y\right)\left(x+\sqrt{3}y\right)\)
Bổ xung thêm vào câu 4:
\(\Rightarrow\left(x-y\right)\left(2x-3y\right)\left(2x+3y\right)\)
Sửa lại câu 5:
\(10x^2\left(a-2b\right)^2-\left(x^2+2\right)\left(2b-a\right)^2\)
\(=-10x^2\left(2b-a\right)^2-\left(x^2+2\right)\left(2b-a\right)^2\)
\(=\left[-10x^2-\left(x^2+2\right)\right]\left(2b-a\right)^2\)
\(=\left(-10x^2-x^2-2\right)\left(2b-a\right)^2\)
\(=\left(-11x^2-2\right)\left(4b^2-4ab+a^2\right)\)
a) Mình không hiểu đề cho lắm 
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x\left(x-4\right)\)
\(=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)
\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)\)
\(=2\left(2x+5\right)^2+3\left(4x+1\right)\left(4x-1\right)\)
\(=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3\)
\(=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)\)
\(=x\left(x^2-16\right)-\left(x^4-1\right)\)
\(=x^3-16x-x^4+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4\)
\(=-77\)
a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)\)
\(=x^2+y^2\)
b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)
\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)
\(=7x-9y\)
c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)
\(=\left(x+y\right)^3:\left(x+y\right)\)
\(=\left(x+y\right)^2=x^2+2xy+y^2\)
d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3:\left(x-y\right)^2\)
\(=\left(x-y\right)\)
e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)
Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)
\(=\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{2x+1}\)
\(=4x^2-2x+1\)
f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\frac{\left(2x-1\right)\left(4x^2+2x+1\right)}{4x^2+2x+1}\)
\(=2x-1\)
a, (x4 + 2x2y2 + y4) : (x2 + y2)
= (x2 + y2)2 : (x2 + y2)
= x2 + y2
b, (49x2 - 81y2) : (7x + 9y)
= (7x - 9y)(7x + 9y) : (7x + 9y)
= 7x - 9y
c, (x3 + 3x2y + 3xy2 + y3) : (x + y)
= (x + y)3 : (x + y)
= (x + y)2
d, (x3 - 3x2y + 3xy2 - y3) : (x2 - 2xy + y2)
= (x - y)3 : (x - y)2
= x - y
Phần e thiếu thì phải
f, (8x3 - 1) : (4x2 + 2x + 1)
= (2x - 1)(4x2 + 2x + 1) : (4x2 + 2x + 1)
= 2x - 1
Chúc bn học tốt!
Bài 14:Tìm x
a,\(x-3=\left(3-x\right)^2\)
\(\Rightarrow\left(x-3\right)-\left(3-x\right)^2=0\)
\(\Rightarrow\left(x-3\right)+\left(x-3\right)^2=0\)
\(\Rightarrow\left(x-3\right)\left(1+x-3\right)=0\)
\(\Rightarrow\left(x-3\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x-3=0\\x-2=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=2\end{matrix}\right.\)
b,\(\left(2x-5\right)-\left(5+2x\right)^2=0\)
\(\Rightarrow\left(2x-5\right)+\left(2x-5\right)^2=0\)
\(\Rightarrow\left(2x-5\right)\left(1+2x-5\right)=0\)
\(\Rightarrow\left(2x-5\right)\left(2x-4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}2x-5=0\\2x-4=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x=5\\2x=4\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{5}{2}\\x=2\end{matrix}\right.\)
\(4x^2+4x+1\)
\(=\left(2x\right)^2+2.2x.1+1\)
\(=\left(2x+1\right)^2\)
\(1+12x+36x^2\)
\(=1+2.6x+\left(6x\right)^2\)
\(=\left(1+6x\right)^2\)
b) \(3x\left(x-1\right)^2-2x\left(x+3\right)\left(x-3\right)+4x\left(x-4\right)\)
\(=3x\left(x^2-2x+1\right)-2x\left(x^2-9\right)+4x^2-16x=3x^3-6x^2+3x-2x^3+18x+4x^2-16x\)\(=x^3-2x^2+5x\)
c) \(2\left(2x+5\right)^2-3\left(4x+1\right)\left(1-4x\right)=2\left(4x^2+20x+25\right)+3\left(16x^2-1\right)\)
\(=8x^2+40x+50+48x^2-3=56x^2+40x+47\)
d) \(x\left(x+4\right)\left(x-4\right)-\left(x^2+1\right)\left(x^2-1\right)=x\left(x^2-16\right)-x^4+1=x^3-x^4-16x+1\)
e) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)=\left(y^2-9\right)\left(y^2+9\right)-y^4+4=y^4-81-y^2+4=-77\)
a. \(x^2-10x+25=\left(x-5\right)^2\)
b.\(4-4x^2+x^4=\left(2-x^2\right)^2\)
c. \(x^2-6y+9y^2=\left(x-3y\right)^2\)
d. \(\left(2x+y^2\right)\left(2x-y^2\right)=4x^2-y^4\)
a) x2 - 10x + 25x = ( x - 5)2
b) 4 - 4x2 + x4 = ( 2 - x2 )2
c) x2 - 6xy + 9y2 = (x - 3y )2
d_ (2x + y2 ). (2x - y2 ) = 4x2 - y4