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a, b, c, dung hang dang thuc A2-B2 =9A-B) (A+B)
d) a4 - 16 =(a2)2 -42= (a2-4) (a2+4) =(a-2)(a+2) (a2+4)
\(a..9-4x^2=3^2-\left(2x\right)^2=\left(3-2x\right)\left(3+2x\right)\)
\(b..16x^2-25=\left(4x\right)^2-5^2=\left(4x-5\right)\left(4x+5\right)\)
\(c..a^4-16=\left(a^2\right)^2-4^2=\left(a^2-4\right)\left(a^2+4\right)\)
\(d..\left(a+b\right)^2-1=\left(a+b\right)^2-1^2=\left(a+b-1\right)\left(a+b+1\right)\)
k mik đi
a. 25 - \(x^2\) = (5-x) (5+x)
b) -196 + 4\(x^2\) = 196 - 4\(x^2\) = (14- 2x) (14+2x)
c)\(5^4-81x^4\) = \(\left[\left(5^2\right)^2\right]-\left[\left(81x^2\right)^2\right]\) = (\(\left(5^2-81x^2\right)\left(5^2+81x^2\right)\)
\(a,25-e=\left(5-\sqrt{e}\right)\left(5+\sqrt{e}\right)\)
\(b,-196+g=-\left(196-g\right)=-\left(14-\sqrt{g}\right)\left(14+\sqrt{g}\right)\)
\(c,2^6-47^2=\left(2^3\right)^2-47^2=\left(2^3-47\right)\left(2^3+47\right)\)
\(d,5^4-81x^4=\left(5^2\right)^2-\left(9x^2\right)^2=\left(5^2-9x^2\right)\left(5^2+9x^2\right)=\left(25-9x^2\right)\left(25+9x^2\right)\)
\(i,\dfrac{25}{16}-9y^2=\left(\dfrac{5}{4}-3y\right)\left(\dfrac{5}{4}+3y\right)\)
a, \(x^2+10x+25=x^2+5x+5x+25\)
\(=\left(x+5\right)^2\)
b, \(x^2-12x+36=x^2-6x-6x+36\)
\(=\left(x-6\right)^2\)
c, \(9x^2+4+12x=9x^2+6x+6x+4\)
\(=3x\left(3x+2\right)+2\left(3x+2\right)=\left(3x+2\right)^2\)
d, \(x^2+49-14x=x^2-7x-7x+49\)
\(=\left(x-7\right)^2\)
e, \(9x^4+24x^2+16=9x^4+12x^2+12x^2+16\)
\(=3x^2\left(3x^2+4\right)+4\left(3x^2+4\right)=\left(3x^2+4\right)^2\)
g,\(4x^2-12xy+9y^2=4x^2-6xy-6xy+9y^2\)
\(=2x\left(2x-3y\right)-3y\left(2x-3y\right)=\left(2x-3y\right)^2\)
Chúc bạn học tốt!!!
\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x-3\right)\left(2x+3\right)\)
\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)
\(\left(a+b\right)^2-1=\left(a+b\right)^2-1^2=\left(a+b-1\right)\left(a+b-1\right)\)
\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b-m+n\right)\left(a+b+m-n\right)\)
\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+3^2\right)\)
\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2+\frac{4}{3}x+\frac{1}{9}\right)\)
Tham khảo~
\(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
\(4x^2-9=\left(2x\right)^2-3^2=\left(2x+3\right)\left(2x-3\right)\)
\(a^4-25=\left(a^2\right)^2-5^2=\left(a^2+5\right)\left(a^2-5\right)\)
\(\left(a+b\right)^2-1=\left(a+b+1\right)\left(a+b-1\right)\)
\(\left(a+b\right)^2-\left(m-n\right)^2=\left(a+b+m-n\right)\left(a+b-m+n\right)\)
\(x^3-27=x^3-3^3=\left(x-3\right)\left(x^2+3x+9\right)\)
\(64x^3+\frac{1}{27}=\left(4x\right)^3+\left(\frac{1}{3}\right)^3=\left(4x+\frac{1}{3}\right)\left(16x^2-\frac{4}{3}x+\frac{1}{9}\right)\)
a,16x2-9
=42x2-32
=(4x-3)(4x+3) HĐT thứ 3
b,9a2-25b2
=32a2-52b2
=(3a-5b)(3a+5b) HĐT thứ 3
c,81-y4
=32.32-y2.y2
=(32-y2)
=(3-y)(3+y) HĐT thứ 3
d,(2x+y)2-1
=(2x+y-1)(2x+y-1) HĐT thứ 3
e,(x+y+z)2-(x-y-z)2
cái này là HĐT thứ 8 mở rộng bạn lên mạng tìm nha
a) \(16x^2-9=\left(4x\right)^2-3^2=\left(4x-3\right).\left(4x+3\right)\)
b) \(9a^2-25b^4=\left(3a\right)^2-\left(5b^2\right)^2=\left(3a-5b^2\right).\left(3a+5b^2\right)\)
c) \(81-y^4=9^2-\left(y^2\right)^2=\left(9-y^2\right).\left(9+y^2\right)\)
d)\(\left(2x+y\right)^2-1=\left(2x+y\right)^2-1^2=\left(2x+y-1\right).\left(2x+y+1\right)\)
a , \(16x^2+8x+1=\left(4x\right)^2+2.4x.1+1^2=\left(4x+1\right)^2\)
b , \(x^2-x+\dfrac{1}{4}=x^2-2.x.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x-\dfrac{1}{2}\right)^2\)
a,(4x+1)2 e,\(\left(\dfrac{3}{2}x-\dfrac{2}{5}\right)^2\)
b,(x-\(\dfrac{1}{2}\))2 g,\(\left(xy+1\right)^2\)
c,(\(x+\dfrac{3}{2}\))2 h,\(\left(x+5\right)^2\)
d,\(\left(x-\dfrac{5}{4}\right)^2\) i,\(-\left(x-6\right)^2\)
k,\(-\left(2x+3\right)^2\)
a, \(A=\left(100+50\right)^2=22500\)
b, \(B=\left(127+73\right)^2=40000\)
c, \(C=-6x+25\)Thay x = 100 ta có :
\(C=-6.100+25=-600+25=-575\)
\(A=100^2+200.50+50^2\)
\(=100^2+2.100.5+50^2\)
\(=\left(100+50\right)^2=150^2\)
\(B=127^2+146.127+73^2\)
\(=127^2+2.73.127+73^2\)
\(=\left(127+73\right)^2=200^2\)
a) \(m^2-n^2=\left(m-n\right)\left(m+n\right)\)
b) \(\left(x^2+x-1\right)^2-\left(x^2+2x+3\right)^2\)
\(=\left(x^2+x-1+x^2+2x+3\right)\left(x^2+x-1-x^2-2x-3\right)\)
\(=\left(2x^2+3x+2\right)\left(-4-x\right)\)
c) \(-16+\left(x-3\right)^2=\left(x-3\right)^2-16=\left(x-3-4\right)+\left(x-3+4\right)=\left(x-7\right)\left(x+1\right)\)
d) \(64+16y+y^2\)
\(=8^2+2.8.y+y^2\)
\(=\left(8+y\right)^2\)
a) \(16-x^2\)
\(=\left(4\right)^2-x^2\)
\(=\left(4-x\right)\left(4+x\right)\)
b) \(4x^2-9y^2\)
\(=\left(2x\right)^2-\left(3y\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y\right)\)
c) \(a^4-25\)
\(=\left(a^2\right)^2-5^2\)
\(=\left(a^2-5\right)\left(a^2+5\right)\)
d) \(\left(a+b\right)^2-4\)
\(=\left(a+b\right)^2-2^2\)
\(=\left(a+b-2\right)\left(a+b+2\right)\)
a. \(16-x^2=4^2-x^2=\left(4-x\right)\left(4+x\right)\)
b. \(4x^2-9y^2=\left(2x\right)^2-\left(3y\right)^2=\left(2x-3y\right)\left(2x+3y\right)\)
c. \(a^4-25=\left(a^2\right)^2-5^2=\left(a^2-5\right)\left(a^2+5\right)\)
d. \(\left(a+b\right)^2-4=\left(a+b\right)^2-2^2=\left(a+b-2\right)\left(a+b+2\right)\)TK MIK NHA