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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) \(\left(y-3\right)\left(y+3\right)\left(y^2+9\right)-\left(y^2+2\right)\left(y^2-2\right)\)
\(=\left(y^2-9\right)\left(y^2+9\right)-\left(y^4-4\right)\)
\(=y^4-81-y^4+4\)
\(=-77\)
1) \(B=5\left(2x-1\right)^2+4\left(x-1\right)\left(x+3\right)+2\left(5-3x\right)^2\)
\(=5\left(4x^2-4x+1\right)+\left(4x-4\right)\cdot\left(x+3\right)+2\left(25-30x+9x^2\right)\)
\(=20x^2-20x+5+4x^2+12x-4x-12+50-60+18x^2\)
\(=42x^2-72x+43\)
2) \(C=\left(2a^2+2a+1\right)\left(2a^2-2a+1\right)-\left(2a+1\right)^2\)
\(=4a^4-4a^3+2a^2+4a^3-4a^2+2a+2a^2-2a+1-\left(4a^2+4a+1\right)\)
\(=4a^4+2a^2-4a^2+2a^2+1-4a^2-4a-1\)
\(=4a^4-4a^2-4a\)
3) Sky Sơn Tùng làm đúng rồi nhé.
4) \(E=\left(x^2-5x+1\right)^2+2\left(5x-1\right)\left(x^2-5x+1\right)\left(5x-1\right)^2\)
\(=x^4+27x^2+1-10x^3+250x^5-1400x^4+1030x^3-302x^2+40x-2\)
\(=-1399x^4-275x^2-1+1020x^3+250x^5+40x\)
5) \(F=\left(a^2+b^2-c^2\right)^2-\left(a^2-b^2+c^2\right)^2\)
\(=\left[a^2+b^2-c^2-\left(a^2-b^2+c^2\right)\right]\cdot\left(a^2+b^2-c^2+a^2-b^2+c^2\right)\)
\(=\left(a^2+b^2-c^2-a^2+b^2-c^2\right)\cdot2a^2\)
\(=\left(2b^2-2c^2\right)\cdot2a^2\)
\(=2\left(b^2-c^2\right)\cdot2a^2\)
\(=2\left(b-c\right)\left(b+c\right)\cdot2a^2\)
\(=2\cdot2a^2\cdot\left(b-c\right)\left(b+c\right)\)
\(=4a^2\cdot\left(b-c\right)\left(b+c\right)\)
6) \(G=\left(a+b+c\right)^2+\left(a+b-c\right)^2-2\left(a+b\right)^2\)
\(=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+\left(-c\right)^2+2ab-2ac-2bc-2\left(a^2+2ab+b^2\right)\)
\(=a^2+b^2+c^2+2ab+a^2+b^2+\left(-c\right)^2+2ab-2a^2-4ab-2b^2\)
\(=0+0+c^2+0+c^2\)
\(=2c^2\)
7) \(H=\left(a+c\right)\left(a-c\right)-\left(a-b-c\right)\left(a-b+c\right)+b\left(b-2x\right)\)
\(=a^2-c^2-\left[\left(a-b\right)^2-c^2\right]+b^2-2bx\)
\(=a^2-c^2-\left(a^2-2ab+b^2-c^2\right)+b^2-2bx\)
\(=a^2-b^2-a^2+2ab-b^2+c^2+b^2-2bx\)
\(=2ab-2bx\)
\(D=\left(9x-1\right)^2+\left(1-5x\right)^2+2\left(9x-1\right)\left(1-5x\right)=\left(9x-1+1-5x\right)^2=\left(4x\right)^2=16x^2\)
Phần a? phải là \(4a^2-4a+1\)chứ
a) \(4a^2-4a+1=\left(2a\right)^2+2.2a+1\)
\(=\left(2a+1\right)^2\)
b) \(9x^2-25y^2=\left(3x\right)^2-\left(5y\right)^2\)
\(=\left(3x-5y\right)\left(3x+5y\right)\)
c) \(1-2x+a^2=\left(1-a\right)^2\)
d) \(\left(2x+1\right)-2.\left(2x+1\right)\left(3x-y\right)+\left(3x-y\right)^2\)
\(=\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)
nếu có sai thì bn thông cảm
1.
b) nó là hằng đẳng thức rồi bn nhá
c) \(1-2a+a^2\)= \(1^2-2a1+a^2\)=\(\left(1-a\right)^2\)
d)\(\left[\left(2x+1\right)-\left(3x-y\right)\right]^2\)=\(\left(2x+1-3x+y\right)^2\)=\(\left(1-x+y\right)^2\)
2.
a)\(\left(\frac{1}{2}x\right)^2-\left(3y\right)^2\)=\(\left(\frac{x}{2}-3y\right)\left(\frac{x}{2}+3y\right)\)
b) Ko khai triển đc
c) \(4x^2+2xy+\frac{1}{4}y^2\)
- Viết 7 hằng đẳng thức đáng nhớ :
\(\left(A+B\right)^2=A^2+2AB+B^2\)
\(\left(A-B\right)^2=A^2-2AB+B^2\)
\(A^2-B^2=\left(A-B\right)\left(A+B\right)\)
\(\left(A+B\right)^3=A^3+3A^2B+3AB^2+B^3\)
\(\left(A-B\right)^3=A^3-3A^2B+3AB^2-B^3\)
\(A^3-B^3=\left(A-B\right)\left(A^2+AB+B^2\right)\)
\(A^3+B^3=\left(A+B\right)\left(A^2-AB+B^2\right)\)
- Áp dụng :
\(a,\left(x+2y\right)^2=x^2+4xy+4y^2\)
\(b,\left(\dfrac{5x-1}{2}\right)^2=\dfrac{\left(5x-1\right)^2}{2^2}=\dfrac{25x^2-10x+1}{4}\)
\(c,\left(\dfrac{1}{3x-3}\right)\left(\dfrac{1}{3x+3}\right)=\dfrac{1.1}{\left(3x-3\right)\left(3x+3\right)}=\dfrac{1}{9x^2-9}\)
\(d,\left(2x+3\right)^3=8x^3+36x^2+54x+27\)
\(e,\left(\dfrac{1}{4y-2x}\right)^2=\dfrac{1}{\left(4y-2x\right)^2}=\dfrac{1}{16y^2-16xy+4x^2}\)
\(f,\left(2x-y\right)\left(4x^2+2xy+y^2\right)=\left(2x\right)^3-y^3=8x^3-y^3\)
\(g,\left(x+3\right)\left(x^2-3x+9\right)=x^3+27\)
a) \(x^2-y^2-5x-5y\)
\(=\left(x^2-y^2\right)-\left(5x+5y\right)\)
\(=\left(x-y\right)\left(x+y\right)-5\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-5\right)\)
b) \(5x^3-5x^2y-10x^2+10xy\)
\(=\left(5x^3-5x^2y\right)-\left(10x^2-10xy\right)\)
\(=5x^2\left(x-y\right)-10x\left(x-y\right)\)
\(=\left(x-y\right)\left(5x^2-10x\right)\)
\(=5x\left(x-y\right)\left(x-2\right)\)
c) \(x^3-2x^2-x+2\)
\(=\left(x^3-2x^2\right)-\left(x-2\right)\)
\(=x^2\left(x-2\right)-\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2-1\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\)
d) \(-y^2+2xy-x^2+3x-3y\)
\(=-\left(y^2-2xy+x^2\right)+\left(3x-3y\right)\)
\(=-\left(y-x\right)^2+3\left(x-y\right)\)
\(=-\left(x-y\right)^2+3\left(x-y\right)\)
\(=\left(x-y\right)\left[-\left(x-y\right)+3\right]\)
\(=\left(x-y\right)\left(-x+y+3\right)\)
g) \(4x^2-8x+3\)
\(=4x^2-6x-2x+3\)
\(=\left(4x^2-6x\right)-\left(2x-3\right)\)
\(=2x\left(2x-3\right)-\left(2x-3\right)\)
\(=\left(2x-3\right)\left(2x-1\right)\)
h) \(2x^2-5x-7\)
\(=2x^2+2x-7x-7\)
\(=\left(2x^2+2x\right)-\left(7x+7\right)\)
\(=2x\left(x+1\right)-7\left(x+1\right)\)
\(=\left(x+1\right)\left(2x-7\right)\)
k) \(x^4+4\)
\(=x^4+4x^2+4-4x^2\)
\(=\left[\left(x^2\right)^2+2.x^2.2+2^2\right]-4x^2\)
\(=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
mik cần gấp
2\(\left(3a-1\right)^2=9a^2-6a+1\)
\(\left(a-2\right)^2=a^2-4a+4\)
\(\left(1-5a\right)^2=1-10a+25a^2\)
\(\left(3a-2b\right)^2=9a^2-12ab+4a^2\)
\(\left(4-3a\right)^2=16-24a+9a^2\)
\(\left(5a-4b\right)^2=25a^2-40ab+16b^2\)
\(\left(5a-3b\right)\left(5a+3b\right)=25a^2-9b^2\)
\(\left(3x+1\right)\left(3x-1\right)=9x^2-1\)
\(\left(5x^2-2\right)\left(5x^2+2\right)=25x^4-4\)
\(\left(2a+\dfrac{1}{2}\right)\left(2a-\dfrac{1}{2}\right)=4a^2-\dfrac{1}{4}\)
\(\left(3x^2-y\right)\left(3x^2+y\right)=9x^4-y^2\)
\(\left(\dfrac{1}{2}x-1\right)\left(\dfrac{1}{2}x+1\right)=\dfrac{1}{4}x^2-1\)
\(\left(\dfrac{3}{4}x+2\right)\left(\dfrac{3}{4}x-2\right)=\dfrac{9}{16}x^2-4\)
\(\left(5x-\dfrac{3}{2}\right)\left(5x+\dfrac{3}{2}\right)=25x^2-\dfrac{9}{4}\)
\(\left(2a^2-7\right)\left(2a^2+7\right)=4a^2-49\)
\(a,=x^2+x+\dfrac{1}{4}\\ b,=4x^2+2x+\dfrac{1}{4}\\ c,=x^2-2+\dfrac{1}{x^2}\\ d,=4x^2+\dfrac{8}{3}x+\dfrac{4}{9}x^2\\ e,=a^2-1\\ f,=25x^4-4\)
\(a,\left(x+\dfrac{1}{2}\right)^2=x^2+x+\dfrac{1}{4}\)
\(b,\left(2x+\dfrac{1}{2}\right)^2=4x^2+2x+\dfrac{1}{4}\)
\(c,\left(x-\dfrac{1}{x}\right)^2=x^2-2+\dfrac{1}{x^2}\)
\(d,\left(\dfrac{2x+2}{3x}\right)^2=\dfrac{\left(2x+2\right)^2}{9x^2}=\dfrac{4x^2+8x+4}{9x^2}\)
\(e,\left(a-1\right).\left(a+1\right)=a^2-1\)
\(f,\left(5x^2-2\right).\left(5x^2+2\right)=25x^4-4\)