Câu hỏi là j?

ko biet cau hoi

m)\(x^4-256=\left(x^2\right)^2-16^2\)

\(=\left(x^2+16\right)\left(x^2-16\right)\)

\(=\left(x^2+16\right)\left(x+4\right)\left(x-4\right)\)

h) \(15x\left(x-y\right)-25x+25y\)

\(=15x\left(x-y\right)-25\left(x-y\right)\)

\(=\left(x-y\right)\left(15x-25\right)\)

\(=5\left(x-y\right)\left(3x-5\right)\)

k) \(-y^2+\frac{1}{9}=\frac{1}{9}-y^2\)

\(=\left(\frac{1}{3}\right)^2-y^2\)

\(=\left(y+\frac{1}{3}\right)\left(\frac{1}{3}-y\right)\)

\(1.\)

\(x^3z+x^2yz-x^2z^2-xyz^2\)

\(=x^3z-x^2z^2+x^2yz-xyz^2\)

\(=x^2z\left(x-z\right)-xyz\left(x-z\right)\)

\(=\left(x^2z-xyz\right)\left(x-z\right)\)

\(=xz\left(x-y\right)\left(x-z\right)\)

\(2.\)

\(x^2-\left(a+b\right)xy+aby^2\)

\(=x^2-axy-bxy+aby^2\)

\(=x^2-bxy-axy+aby^2\)

\(=x\left(x-by\right)-ay\left(x-by\right)\)

\(=\left(x-ay\right)\left(x-by\right)\)

\(3.\)

\(ab\left(x^2+y^2\right)+xy\left(x^2+y^2\right)\)

\(=abx^2+aby^2+a^2xy+b^2xy\)

\(=abx^2+b^2xy+a^2xy+aby^2\)

\(=bx\left(ax+by\right)+ay\left(ax+by\right)\)

\(=\left(ax+by\right)\left(bx+ay\right)\)

\(4.\)

\(\left(xy+ab\right)^2+\left(ay-bx\right)^2\)

\(=x^2y^2+2abxy+a^2b^2+a^2y^2-2aybx+b^2x^2\)

\(=x^2y^2+a^2b^2+a^2y^2+b^2x^2\)

\(=x^2y^2+b^2x^2+a^2b^2+a^2y^2\)

\(=x^2\left(b^2+y^2\right)+a^2\left(b^2+y^2\right)\)

\(=\left(a^2+x^2\right)\left(b^2+y^2\right)\)

\(5.\)

\(a^2\left(b-c\right)+b^2\left(c-a\right)+c^2\left(a-b\right)\)

\(=a^2b-a^2c+b^2c-ab^2+ac^2-bc^2\)

\(=a^2b-ab^2-a^2c-b^2c+ac^2-bc^2\)

\(=ab\left(a-b\right)-c\left(a^2-b^2\right)+c^2\left(a-b\right)\)

\(=ab\left(a-b\right)-c\left(a-b\right)\left(a+b\right)+c^2\left(a-b\right)\)

\(=\left(a-b\right)\left(ab-ac-bc+c^2\right)\)

\(=\left(a-b\right)\left(ab-bc-ac+c^2\right)\)

\(=\left(a-b\right)\left[b\left(a-c\right)-c\left(a-c\right)\right]\)

\(=\left(a-c\right)\left(b-c\right)\left(a-c\right)\)

\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\)

\(6.\)

\(16x^2-40xy+2y^2\)

\(=\left(4x\right)^2-2\cdot4\cdot5xy+\left(5y\right)^2\)

\(=\left(4x-5y\right)^2\)

\(7.\)

\(25x^4-10x^2y+y^2\)

\(=\left(5x^2\right)^2-2\cdot5x^2y+y^2\)

\(=\left(5x^2+y\right)^2\)

\(8.\)

\(-16x^4y^6-24x^5y^5-9x^6y^4\)

\(=-\left(4^2x^4y^6+2\cdot4\cdot3x^5y^5+3^2x^6y^4\right)\)

\(=-\left[\left(4x^2y^3\right)^2+2\left(4x^2y^3\right)\left(3x^3y^2\right)+\left(3x^3y^2\right)^2\right]\)

\(=\left(4x^2y^3+3x^3y^2\right)^2\)

\(9.\)

\(16x^2-4y^2-8x+1\)

\(=\left(4x\right)^2-\left(2y\right)^2-8x+1\)

\(=\left(4x\right)^2-8x+1-\left(2y\right)^2\)

\(=\left(4x+1\right)^2-\left(2y\right)^2\)

\(=\left(4x-2y+1\right)\left(4x+2y+1\right)\)

\(10.\)

\(49x^2-25+42xy+9y^2\)

\(=\left(7x\right)^2-5^2+2\cdot7\cdot3xy+\left(3y\right)^2\)

\(=\left(7x\right)^2+2\cdot7\cdot3xy+\left(3y\right)^2-5^2\)

\(=\left(7x+3y\right)^2-5^2\)

\(=\left(7x+5y+5\right)\left(7x+3y-5\right)\)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

a² + 2ab + b² = (a + b)²

4x² + 4xy + y² = (2x)² + 2.2x.y + y² = (2x + y)²

25x² + 10xy + y² = (5x)² + 2.5x.y + y² = (5x + y)²

a^2+2ab+b^2=(a+b)^2

4x^2+4xy+y^2=(2x+y)^2

25x^2+10xy+y^2=(5x+y)^2

Bài 1:

a) \(2x^2y\left(3xy-4xy^2-\frac{3}{2xy^3}\right)\) \(=6x^3y^2-8x^2y^3-\frac{3x}{y^2}\)

b) \(\left(15x^4y^2-36x^3y^4+21x^2y^5\right):3x^2y^2\)\(=5x^2-12xy^2+7y^3\)

Bài 2:

a) \(x^2-4xy+4y^2-16\) \(=\left(x-2y\right)^2-16=\left(x-2y-4\right)\left(x-2y+4\right)\)

b) \(2x+xy-x^2-xy\) \(=x\left(2-x\right)\)

c)\(16x^2-25y^2=\left(4x-5y\right)\left(4x+5y\right)\)

Lời giải:
$A=a^2+ab+b^2-3b-3a+3$

$4A=4a^2+4ab+4b^2-12a-12b+12$

$=(4a^2+4ab+b^2)-12a-12b+3b^2+12$

$=(2a+b)^2-6(2a+b)+9+(3b^2-6b+3)$

$=(2a+b-3)^2+3(b-1)^2\geq 0+3.0=0$

Vậy $A_{\min}=0$. Giá trị này đạt tại $2a+b-3=b-1=0$

$\Leftrightarrow b=1; a=1$

Câu B tương tự câu A nhé. Chỉ khác mỗi đặt tên biến.

---------------

$C=x^2+5y^2-4xy+2y-3$

$=(x^2-4xy+4y^2)+(y^2+2y)-3$

$=(x-2y)^2+(y^2+2y+1)-4$

$=(x-2y)^2+(y+1)^2-4\geq 0+0-4=-4$

Vậy $C_{\min}=-4$. Giá trị này đạt tại $x-2y=y+1=0$

$\Leftrightarrow y=-1; x=-2$