a) \(\sqrt{1-8x+16x^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{1^2-2\cdot4x\cdot1+\left(4x\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\sqrt{\left(4x-1\right)^2}=\dfrac{1}{3}\)

\(\Leftrightarrow\left|4x-1\right|=\dfrac{1}{3}\)

\(\Leftrightarrow\left[{}\begin{matrix}4x-1=\dfrac{1}{3}\left(ĐK:x\ge\dfrac{1}{4}\right)\\4x-1=\dfrac{1}{3}\left(ĐK:x< \dfrac{1}{4}\right)\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{4}{3}\\4x=\dfrac{2}{3}\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\left(tm\right)\\x=\dfrac{1}{6}\left(tm\right)\end{matrix}\right.\)

b) \(\sqrt{16x-32}+\sqrt{25x-50}=18+\sqrt{9x-18}\) (ĐK: \(x\ge2\)) 

\(\Leftrightarrow\sqrt{16\left(x-2\right)}+\sqrt{25\left(x-2\right)}=18+\sqrt{9\left(x-2\right)}\)

\(\Leftrightarrow4\sqrt{x-2}+5\sqrt{x-2}=18+3\sqrt{x-2}\)

\(\Leftrightarrow6\sqrt{x-2}=18\)

\(\Leftrightarrow\sqrt{x-2}=3\)

\(\Leftrightarrow x-2=9\)

\(\Leftrightarrow x=9+2\)

\(\Leftrightarrow x=11\left(tm\right)\)

\(1.\)

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

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

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

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

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

\(2.\)

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

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

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

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

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

\(3.\)

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

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

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

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

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

\(4.\)

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

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

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

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

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

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

\(5.\)

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

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

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

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

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

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

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

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

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

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

\(6.\)

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

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

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

\(7.\)

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

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

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

\(8.\)

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

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

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

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

\(9.\)

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

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

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

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

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

\(10.\)

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

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

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

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

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

Bài 2 :

\(A=4x^2-2.2x.2+4+1\)

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

Thấy : \(\left(2x-2\right)^2\ge0\)

\(A=\left(2x-2\right)^2+1\ge1\)

Vậy \(MinA=1\Leftrightarrow x=1\)

\(B=\left(5x\right)^2-2.5x.1+1-4\)

\(=\left(5x-1\right)^2-4\)

Thấy : \(\left(5x-1\right)^2\ge0\)

\(\Rightarrow B=\left(5x-1\right)^2-4\ge-4\)

Vậy \(MinB=-4\Leftrightarrow x=\dfrac{1}{5}\)

\(C=\left(7x\right)^2-2.7x.2+4-5\)

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

Thấy : \(\left(7x-2\right)^2\ge0\)

\(\Rightarrow C=\left(7x-2\right)^2-5\ge-5\)

Vậy \(MinC=-5\Leftrightarrow x=\dfrac{2}{7}\)

\(1.\)

\(A=-x^2-10x+1=-\left(x^2+10x-1\right)\)

\(=-\left(x^2+2.5x+5^2-5^2-1\right)=-\left[\left(x+5\right)^2-26\right]\)

\(=-\left(x+5\right)^2+26\le26\) dấu "=" xảy ra<=>x=-5

\(B=-4x^2-6x-5=-4\left(x^2+\dfrac{6}{4}x+\dfrac{5}{4}\right)\)

\(=-4\left(x^2+2.\dfrac{3}{4}x+\dfrac{9}{16}+\dfrac{11}{16}\right)\)\(=-4\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{11}{6}\right]\le-\dfrac{11}{4}\)

\(C=-16x^2+8x-1=-16\left(x^2-\dfrac{1}{2}x+\dfrac{1}{16}\right)\)

\(=-16\left(x^2-2.\dfrac{1}{4}x+\dfrac{1}{16}\right)=-16\left(x-\dfrac{1}{4}\right)^2\le0\)

dấu"=" xảy ra<=>x=1/4

Lời giải:

a. 

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

$=xy(xy+1)-2(xy+1)=(xy+1)(xy-2)$

b. Bạn xem lại đoạn $-16x^2$ là dấu - hay + vậy?

\(x^8+64\)

\(=x^8+16x^4+64-16x^4\)

\(=\left(x^4\right)^2+2.x^4.8+8^2-16x^4\)

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

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

5x²+11x+6

=5x²+5x+6x+6

=(5x²+5x)+(6x+6)

=5x(x+1)+6(x+1)

=(x+1)(5x+6)

x2-16x+15

= x²-x-15x+15

= (x²-x)-(15x-15)

= x(x-1)-15(x-1)

=(x-1)(x-15)