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mình k cho bạn rồi nha, tích lại cho mình, số điểm của mình là -159 điểm
c)\(x^2+x+\dfrac{1}{4}=\left(x+\dfrac{1}{2}\right)^2\)
d)\(\dfrac{a^2}{4}-2a+4=\left(\dfrac{a}{2}-2\right)^2\)
e) \(4y^2-9x^2=\left(2y-3x\right)\left(2y+3x\right)\)
f)\(9y^2-\dfrac{1}{4}=\left(3y-\dfrac{1}{2}\right)\left(3y+\dfrac{1}{2}\right)\)
g)\(8x^3+8a^3=\left(2x+2a\right)\left(4x^2-4xa+4a^2\right)\)
a) Ta có: \(\left(x^4+2x^2y^2+y^4\right):\left(x^2+y^2\right)\)
\(=\left(x^2+y^2\right)^2:\left(x^2+y^2\right)\)
\(=x^2+y^2\)
b) Ta có: \(\left(49x^2-81y^2\right):\left(7x+9y\right)\)
\(=\frac{\left(7x+9y\right)\left(7x-9y\right)}{7x+9y}\)
\(=7x-9y\)
c) Ta có: \(\left(x^3+3x^2y+3xy^2+y^3\right):\left(x+y\right)\)
\(=\left(x+y\right)^3:\left(x+y\right)\)
\(=\left(x+y\right)^2=x^2+2xy+y^2\)
d) Ta có: \(\left(x^3-3x^2y+3xy^2-y^3\right):\left(x^2-2xy+y^2\right)\)
\(=\left(x-y\right)^3:\left(x-y\right)^2\)
\(=\left(x-y\right)\)
e)Sửa đề: \(\left(8x^3+1\right):\left(2x+1\right)\)
Ta có: \(\left(8x^3+1\right):\left(2x+1\right)\)
\(=\frac{\left(2x+1\right)\left(4x^2-2x+1\right)}{2x+1}\)
\(=4x^2-2x+1\)
f) Ta có: \(\left(8x^3-1\right):\left(4x^2+2x+1\right)\)
\(=\frac{\left(2x-1\right)\left(4x^2+2x+1\right)}{4x^2+2x+1}\)
\(=2x-1\)
a, (x4 + 2x2y2 + y4) : (x2 + y2)
= (x2 + y2)2 : (x2 + y2)
= x2 + y2
b, (49x2 - 81y2) : (7x + 9y)
= (7x - 9y)(7x + 9y) : (7x + 9y)
= 7x - 9y
c, (x3 + 3x2y + 3xy2 + y3) : (x + y)
= (x + y)3 : (x + y)
= (x + y)2
d, (x3 - 3x2y + 3xy2 - y3) : (x2 - 2xy + y2)
= (x - y)3 : (x - y)2
= x - y
Phần e thiếu thì phải
f, (8x3 - 1) : (4x2 + 2x + 1)
= (2x - 1)(4x2 + 2x + 1) : (4x2 + 2x + 1)
= 2x - 1
Chúc bn học tốt!
Ik mk nha, hôm nay ngày mai, ngày kia mk ik 3 lần lại cho bạn (thành 9 lần)
Nhớ kb với mìn lun nha!! Mk rất vui đc làm quen vs bạn, cảm ơn mn nhìu lắm
a) \(A=x^2-8x+17=\left(x-4\right)^2+1\ge1\)
Vậy MIN A = 1 khi x = 4
b) \(T=x^2-4x+7=\left(x-2\right)^2+3\ge3\)
Vậy MIN T = 3 khi x = 2
c) \(H=3x^2+6x-1=3\left(x+1\right)^2-4\ge-4\)
Vậy MIN H = -4 khi x = -1
d) \(E=x^2+y^2-4\left(x+y\right)+16=\left(x-2\right)^2+\left(y-2\right)^2+8\ge8\)
Vậy MIN E = 8 khi x = y = 2
e) \(K=4x^2+y^2-4x-2y+3=\left(2x-1\right)^2+\left(y-1\right)^2+1\ge1\)
Vậy MIN K = 1 khi x = 1/2; y = 1
f) \(M=\frac{3}{2}x^2+x+1=\frac{3}{2}\left(x+\frac{1}{3}\right)^2+\frac{5}{6}\ge\frac{5}{6}\)
Vậy MIN M = 5/6 khi x = -1/3
\(1-2y+y^2=\left(1-y\right)^2\)
\(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right)\)
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
\(27+27x+9x^2=9\left(3+3x+x^2\right)\)
\(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
\(3x^2-6xy+9y^2=3\left(x^2-2xy+3y^2\right)\)
\(x^2+4x+3=x^2+x+3x+3=x\left(x+1\right)+3\left(x+1\right)=\left(x+1\right)\left(x+3\right)\)
\(x^2-4x-5=x^2+x-5x-5=x\left(x+1\right)-5\left(x+1\right)=\left(x-5\right)\left(x+1\right)\)
a ) \(1-2y+y^2=y^2-2y+1=\left(y-1\right)^2\)
b ) \(\left(x+1\right)^2-25=\left(x+1-5\right)\left(x+1+5\right)=\left(x-4\right)\left(x+6\right).\)
c ) \(1-4x^2=\left(1-2x\right)\left(1+2x\right).\)
d ) \(27+27x+9x^2=9\left(3+3x+x\right)=9\left(3+4x\right).\)
e ) \(8x^3-12x^2y+6xy^2-y^3=\left(2x-y\right)^3\)
f ) \(3x^2-6xy+9y^2=3\left(x^2-2xy+3y^2\right).\)
g ) \(x^2+4x+3==x^2+3x+x+3=\left(x+1\right)\left(x+3\right)\)
h ) \(x^2-4x-5=x^2+x-5x-5=\left(x-5\right)\left(x+1\right).\)
\(A=49x^2-28x+25\)
\(A=\left(7x\right)^2-2.7x.2+4-4+25\)
\(A=\left(7x-2\right)^2+21\)
Vì \(\left(7x-2\right)^2\ge0\) với mọi x
\(\Rightarrow\left(7x-2\right)^2+21\ge21\) với mọi x
\(\Rightarrow Amin=21\Leftrightarrow7x-2=0\)
\(\Rightarrow7x=2\)
\(\Rightarrow x=\dfrac{2}{7}\)
Vậy \(Amin=21\Leftrightarrow x=\dfrac{2}{7}\)
\(B=8x^2-28x-1\)
\(B=2\left(4x^2-14x-\dfrac{1}{2}\right)\)
\(B=2\left[\left(2x\right)^2-2.2x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\left(\dfrac{7}{2}\right)^2-\dfrac{1}{2}\right]\)
\(B=2\left[\left(2x\right)^2-2.2x.\dfrac{7}{2}+\left(\dfrac{7}{2}\right)^2-\dfrac{51}{4}\right]\)
\(B=2\left(2x-\dfrac{7}{2}\right)^2-\dfrac{51}{2}\)
Vì \(2\left(2x-\dfrac{7}{2}\right)^2\ge0\) với mọi x
\(\Rightarrow2\left(2x-\dfrac{7}{2}\right)^2-\dfrac{51}{2}\ge-\dfrac{51}{2}\)
\(\Rightarrow Bmin=-\dfrac{51}{2}\Leftrightarrow2x-\dfrac{7}{2}=0\)
\(\Rightarrow2x=\dfrac{7}{2}\)
\(\Rightarrow x=\dfrac{7}{4}\)
Vậy \(Bmin=-\dfrac{51}{2}\Leftrightarrow x=\dfrac{7}{4}\)
\(C=\left(2x^2+5\right)^2+10\)
Vì \(\left(2x^2+5\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x^2+5\right)^2+10\ge10\) với mọi x
\(\Rightarrow Cmin=10\Leftrightarrow2x^2+5=0\)
\(\Rightarrow2x^2=-5\)
\(\Rightarrow x^2=-\dfrac{5}{2}\)
\(\Rightarrow\) Không tồn tại x thỏa mãn
Vậy C không có giá trị nhỏ nhất
P/s: Câu c mình làm không có chắc nha, thấy nó sao sao ấy, không biết có sai đề không? 
\(D=3x^2-8x+7\)
\(D=3\left(x^2-\dfrac{8}{3}x+\dfrac{7}{3}\right)\)
\(D=3\left(x^2-2.x.\dfrac{4}{3}+\dfrac{16}{9}-\dfrac{16}{9}+\dfrac{7}{3}\right)\)
\(D=3\left(x^2-2.x.\dfrac{4}{3}+\dfrac{16}{9}+\dfrac{5}{9}\right)\)
\(D=3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\)
Vì \(3\left(x-\dfrac{4}{3}\right)^2\ge0\) với mọi x
\(\Rightarrow3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\ge\dfrac{5}{3}\)
\(\Rightarrow Dmin=\dfrac{5}{3}\Leftrightarrow x-\dfrac{4}{3}=0\)
\(\Rightarrow x=\dfrac{4}{3}\)
Vậy \(Dmin=\dfrac{5}{3}\Leftrightarrow x=\dfrac{4}{3}\)
\(E=x^4-2x^2+12\)
\(E=\left(x^2\right)^2-2x^2+1+11\)
\(E=\left(x^2-1\right)^2+11\)
Vì \(\left(x^2-1\right)^2\ge0\) với mọi x
\(\Rightarrow\left(x^2-1\right)^2+11\ge11\) với mọi x
\(\Rightarrow Emin=11\Leftrightarrow x^2-1=0\)
\(\Rightarrow x^2=1\)
\(\Rightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(Emin=11\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
\(F=4x^2+15x+2\)
\(F=\left(2x\right)^2+2.2x.\dfrac{15}{4}+\left(\dfrac{15}{4}\right)^2-\left(\dfrac{15}{4}\right)^2+2\)
\(F=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{225}{16}+\dfrac{32}{16}\)
\(F=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\)
Vì \(\left(2x+\dfrac{15}{4}\right)^2\ge0\) với mọi x
\(\Rightarrow\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\ge-\dfrac{193}{16}\)
\(\Rightarrow Fmin=-\dfrac{193}{16}\Leftrightarrow2x+\dfrac{15}{4}=0\)
\(\Rightarrow2x=-\dfrac{15}{4}\)
\(\Rightarrow x=-\dfrac{15}{4}.\dfrac{1}{2}\)
\(\Rightarrow x=-\dfrac{15}{8}\)
Vậy \(Fmin=-\dfrac{193}{16}\Leftrightarrow x=-\dfrac{15}{8}\)
\(H=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(H=\left(x^2+4x-5\right)\left(x^2+4x+5\right)\)
\(H=\left(x^2+4x\right)^2-5^2\)
\(H=\left(x^2+4x\right)^2-25\)
Vì \(\left(x^2+4x\right)^2\ge0\)
\(\Rightarrow\left(x^2+4x\right)^2-25\ge-25\) với mọi x
\(\Rightarrow Hmin=-25\Leftrightarrow x^2+4x=0\)
\(\Rightarrow x\left(x+4\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
Vậy \(Hmin=-25\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-4\end{matrix}\right.\)
\(I=\left(x^6+6\right)^2\)
Vì \(\left(x^6+6\right)^2\ge0\)
\(\Rightarrow Imin=0\Leftrightarrow x^6+6=0\)
\(\Rightarrow\left(x^3\right)^2=-6\)
\(\Rightarrow\) Không tồn tại x
Vậy I không có giá trị nhỏ nhất
\(A=49x^2-28x+25=\left(49x^2-28x+1\right)+24=\left(7x-1\right)^2+24\ge24\)
Vậy GTNN của A là 24 khi x = \(\dfrac{1}{7}\)
\(B=8x^2-28x-1=8\left(x^2-\dfrac{7}{2}x+\dfrac{49}{16}\right)-\dfrac{51}{2}=8\left(x-\dfrac{7}{4}\right)^2-\dfrac{51}{2}\ge-\dfrac{51}{2}\)
Vậy GTNN của B là \(-\dfrac{51}{2}\) khi x = \(\dfrac{7}{4}\)
\(C=\left(2x^2+5\right)^2+10=4x^4+20x^2+35\ge35\)
Vậy GTNN của C là 35 khi x = 0
\(D=3x^2-8x+7=3\left(x^2-\dfrac{8}{3}x+\dfrac{16}{9}\right)+\dfrac{5}{3}=3\left(x-\dfrac{4}{3}\right)^2+\dfrac{5}{3}\ge\dfrac{5}{3}\)
Vậy GTNN của D là \(\dfrac{5}{3}\) khi x = \(\dfrac{4}{3}\)
\(E=x^4-2x^2+12=\left(x^4-2x^2+1\right)+11=\left(x^2-1\right)^2+11\ge11\)
Vậy GTNN của E là 11 khi x = 1 hoặc x = -1
\(F=4x^2+15x+2=\left(4x^2+15x+\dfrac{225}{16}\right)-\dfrac{193}{16}=\left(2x+\dfrac{15}{4}\right)^2-\dfrac{193}{16}\ge-\dfrac{193}{16}\)
Vậy GTNN của F là \(-\dfrac{193}{16}\) khi x = \(-\dfrac{15}{8}\)
\(G=8\left(a+2\right)^3-\left(2a+1\right)^3\)
\(G=36a^2+90a+63\)
\(G=9\left(4a^2+10a+7\right)\)
\(G=9\left(4a^2+10a+\dfrac{25}{4}\right)+\dfrac{27}{4}\)
\(G=9\left(2a+\dfrac{5}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}\)
Vậy GTNN của G là \(\dfrac{27}{4}\) khi x = \(-\dfrac{5}{4}\)
\(H=\left(x-1\right)\left(x+5\right)\left(x^2+4x+5\right)\)
\(H=x^4+8x^3+16x^2-25\)
\(H=\left(x^2+4x\right)^2-25\ge-25\)
Vậy GTNN của H là -25 khi x = -4 hoặc x = 0
\(I=\left(x^6+6\right)^2=x^{12}+12x^6+36\ge36\)
Vậy GTNN của I là 36 khi x = 0
a) \(a^2x+a^2y-9x-9y\)
\(=\left(a^2x+a^2y\right)-\left(9x+9y\right)\)
\(=a^2\left(x+y\right)-9\left(x+y\right)\)
\(=\left(x+y\right)\left(a^2-9\right)\)
\(=\left(x+y\right)\left(a-3\right)\left(a+3\right)\)
b) \(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2-4x\right)+\left(3x-12\right)\)
\(=x\left(x-4\right)+3\left(x-4\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
c) \(x^2\left(x-3\right)+12-4x\)
\(=x^2\left(x-3\right)-\left(4x-12\right)\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-3\right)\left(x^2-4\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
d) \(4x\left(x-y\right)+6y\left(x-y\right)\)
\(=\left(x-y\right)\left(4x+6y\right)\)
\(=2\left(x-y\right)\left(2x+3y\right)\)
e) \(5\left(x+y\right)-xy-y^2\)
\(=5\left(x+y\right)-\left(xy+y^2\right)\)
\(=5\left(x+y\right)-y\left(x+y\right)\)
\(=\left(x+y\right)\left(5-y\right)\)
a) \(x^2-25+y^2+2xy=\left(x-y\right)^2-5^2\)
= \(\left(x-y-5\right)\left(x-y+5\right)\)
b) \(x^2-2x-4y^2-4y\)
= \(\left(x^2-2x+1\right)-\left(4y^2+4y+1\right)\)
= \(\left(x-1\right)^2-\left(2y+1\right)^2\)
= \(\left(x-1-2y-1\right)\left(x-1+2y+1\right)\)
= \(\left(x-2y-2\right)\left(x+2y\right)\)
c) Câu này chắc là sai đề nên sửa luôn:
\(16x^3+0,25y^3z^3\)
= \(0,25\left(64x^3+y^3z^3\right)\)
= \(0,25\left(4x+yz\right)\left(16x^2-4xyz+y^2z^2\right)\)
d) \(x^3-x^2-x+1\)
= \(x^2\left(x-1\right)-\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\)
e) \(x^4+x^3+x^2-1\)
= \(x^3\left(x+1\right)+\left(x+1\right)\left(x-1\right)\)
= \(\left(x+1\right)\left(x^3+x-1\right)\)
f) \(x^4+6x^2y+9y^2-1\)
= \(\left(x^2+3y-1\right)\left(x^2+3y+1\right)\)
g) \(x^2+4x-y^2+4\)
= \(\left(x^2+4x+4\right)-\left(y^2-4y+4\right)\)
= \(\left(x+2\right)^2-\left(y-2\right)^2\)
= \(\left(x+2-y+2\right)\left(x+2+y-2\right)\)
= \(\left(x-y+4\right)\left(x+y\right)\)
h) \(x^3+3x^2-3x-1\)
= \(x^3-x^2+4x^2-4x+x-1\)
= \(x^2\left(x-1\right)+4x\left(x-1\right)+\left(x-1\right)\)
= \(\left(x-1\right)\left(x^2+4x+1\right)\)
\(a,x^2-25+y^2+2xy=\left(x^2+2xy+y^2\right)-25=\left(x+y\right)^2-5^2=\left(x+y+5\right)\left(x+y-5\right)\)\(b,x^2-2x-4y^2-4y=\left(x^2-2x+1\right)-\left(4y^2-4y+1\right)=\left(x-1\right)^2-\left(2y-1\right)^2=\left(x-1-2y+1\right)\left(x-1+2y-1\right)=\left(x-2y\right)\left(x+2y-2\right)\)\(c,16x^3+0,25yz^3=0,25\left(64x^3+yz^3\right)\)
\(d,x^3-x^2-x+1=x^2\left(x-1\right)-\left(x-1\right)=\left(x^2-1\right)\left(x-1\right)=\left(x-1\right)^2\left(x+1\right)\)\(e,x^4+x^3+x^2-1=x^3\left(x+1\right)+\left(x-1\right)\left(x+1\right)=\left(x+1\right)\left(x^3+x-1\right)\)\(f,x^4+6x^2y+9y^2-1=\left(x^2+3y\right)^2-1=\left(x^2+3y+1\right)\left(x^2+3y-1\right)\)\(g,x^2-4x+y^2+4=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)\(h,x^3+3x^2-3x-1=x^3-1+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+4x+1\right)\)
a,\(2x^2-8x+y^2+2y+9=0\)
\(\Rightarrow2\left(x^2-4x+4\right)+\left(y^2+2y+1\right)=0\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2=0\)
Mà \(2\left(x-2\right)^2\ge0\forall x\); \(\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow2\left(x-2\right)^2+\left(y+1\right)^2\ge0\forall x;y\)
Dấu "=" xảy ra<=> \(\hept{\begin{cases}2\left(x-2\right)^2=0\\\left(y+1\right)^2=0\end{cases}\Rightarrow\hept{\begin{cases}x=2\\y=-1\end{cases}}}\)
Vậy x=2;y=-1
a.
$x^2+20x+100=(x+10)^2$
b.
$16x^2+24xy+9y^2=(4x+3y)^2$
c.
$y^2-14y+49=(y-7)^2$
d.
$9x^2-42xy+49y^2=(3x-7y)^2$
e.
$4x^2-9y^2=(2x-3y)(2x+3y)$
f.
$16-x^2=(4-x)(4+x)$
g.
$49x^2-1=(7x-1)(7x+1)$
h.
$16x^2-25=(4x-5)(4x+5)$
i.
$8x^3+24x^2y+54xy^2+27y^3=(2x+3y)^3$
k.
$x^3-6x^2y+12xy^2-8y^3=(x-2y)^3$
l.
$(2a+b)(4a^2-2ab+b^2)=(2a)^3+b^3=8a^3+b^3$
m.
$(3x-4y)(9x^2+12xy+16y^2)=(3x)^3-(4y)^3=27x^3-64y^3$