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\(-16x^2+8xy-y^2+49\)
\(=49-\left(16x^2-8xy+y^2\right)\)
\(=7^2-\left(4x-y\right)^2\)
\(=\left(7-4x+y\right)\left(7+4x-y\right)\)
a) Ta có: \(4x^2-28xy+49y^2\)
\(=\left(2x\right)^2-2\cdot2x\cdot7y+\left(7y\right)^2\)
\(=\left(2x-7y\right)^2\)
b) Ta có: \(x^2+8xy+16y^2\)
\(=x^2+2\cdot x\cdot4y+\left(4y\right)^2\)
\(=\left(x+4y\right)^2\)
c) Ta có: \(x^2-12x+36\)
\(=x^2-2\cdot x\cdot6+6^2\)
\(=\left(x-6\right)^2\)
\(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x-y\right)^2-4\left(4x-y\right)\)
\(=\left(4x-y\right)\left(4x-y-4\right)\)
a) \(16x^2+y^2+4y-16x-8xy\)
\(=\left(4x\right)^2-8xy+y^2+4\left(y-4x\right)\)
\(=\left(4x-y\right)^2+4\left(y-4x\right)\)
\(=\left(y-4x\right)^2+4\left(y-4x\right)=\left(y-4x\right)\left(y-4x+4\right)\)
a) 9x4+16y6-24x2y3
=(3x2)2-2.3x2.4y3+(4y3)2
=(3x2-4y3)2
b) 16x2-24xy+9y2
=(4x)2-2.4x.3y+(3y)2
=(4x-3y)2
c) 36x2-(3x-2)2
=(36x-3x+2)(36x+3x-2)
=(33x+2)(39x-2)
d) 27x3+54x2y+36xy2+8y3
=(3x)3+3.(3x)2.2y+3.3x.(2y)2+(2y)3
=(3x+2y)3
e) y9-9x2y6+27x4y3-27x6
=(y3)3-3.(y3)2.3x2+3.y3.(3x2)2-(3x2)3
=(y3-3x2)3
f) 64x3+1
= (4x)3+13
=(4x+1)[(4x)2-4x.1+12]
=(4x+1)(16x2-4x+1)
e) 27x6-8x3 *sửa đề*
=(3x2)3-(2x)3
=(3x2-2x)[(3x)2+3x2.2x+(2x)2]
=(3x2-2x)(9x2+6x3+4x2)
~~~
a. -\(-16x^2+8xy-y^2+49\)
= \(\left(-\left(4x\right)^2+8xy-y^2\right)+49\)
= \(-\left(\left(4x^2\right)-8xy+y^2\right)+49\)
= \(-\left(4x-y\right)^2+49\)
b. \(y^2\left(x^2+y\right)-zx^2-zy\)
= \(y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
= \(\left(x^2+y\right)\left(y^2-z\right)\)
_16x2+8xy_y2+49
=( _(4x)2+2 × 4 × xy _ y2 )+ 72
= _((4x)2_ 2×4×x × xy +y2)+72
= _(4x_y)2+72
=72_(4x_y)2
= (7_(4x_y))×(7+(4x_y))
= (7_4x+y)×(7+4x_y)
2)y2×(x2+y)_zx2_zy
=y×(x2+y)_z(x2+y)
= ( x2+y)×(y_z)
\(16x^2+y^2+4y-16y-8xy\)
\(=\left(16x^2-8xy+y^2\right)+4y-16y\)
\(=\left(4x+y\right)^2-12y\)
\(=\left(4x+y-\sqrt{12y}\right)\left(4x+y-\sqrt{12y}\right)\)
P/S : Sai thì thôi nha!
Kimetsu no YaibaNếu y âm thì căn thức vô nghĩa