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\(A=x^2+3x+7\)
\(=x^2+2.1,5x+2,25+4,75\)
\(=\left(x+1,5\right)^2+4,75\ge4,75\)
Vậy \(A_{min}=4,75\Leftrightarrow x=-1,5\)
\(B=2x^2-8x\)
\(=2\left(x^2-4x\right)\)
\(=2\left(x^2-4x+4-4\right)\)
\(=2\left[\left(x-2\right)^2-4\right]\)
\(=2\left(x-2\right)^2-8\ge-8\)
Vậy \(B_{min}=-8\Leftrightarrow x=2\)
\(B=x^2-6x+y^2-2y+12=\left(x^2-6x+9\right)\left(y^2-2y+1\right)+2\)
\(B=\left(x-3\right)^2+\left(y-1\right)^2+2\text{ }\)
Ta thấy B lớn hơn hoặc bằng 2 suy ra GTNN của B là 2
Dấu = xảy ra khi x=3; y=1
\(C=2x^2-6x=\left(2x^2-6x+4,5\right)-4,5=2\left(x^2-3x+2,25\right)-4,5\)
\(C=2\left(x-1,5\right)^2-4,5\)
Ta thấy C luôn luôn lớn hơn hoặc bằng -4,5 nên GTNN của C là -4,5
Dấu = xảy ra khi x=1,5
Tối mình full cho còn giờ mình đi đá bóng đây
1) \(D=\frac{2016}{-4x^2+4x-5}\). Để D đạt giá trị nhỏ nhất suy ra \(-4x^2+4x-5\)đạt giá trị lớn nhất.
Ta có \(-4x^2+4x-5=-4x^2+4x-1-4=\left(-4x^2+4x-1\right)-4\)
\(-4\left(x^2-x+\frac{1}{4}\right)-4=-4\left(x-\frac{1}{2}\right)^2-4\).
Ta Thấy:\(-4\left(x-\frac{1}{2}\right)^2\) bé hơn hoặc bằng 0 nên \(-4\left(x-\frac{1}{2}\right)^2-4\)bé hơn hoặc bằng -4
nên ..... bạn tự kết luận
a)
\(\frac{x^2-16}{4x-x^2}=\frac{x^2-4^2}{x(4-x)}=\frac{(x-4)(x+4)}{x(4-x)}=\frac{x+4}{-x}\)
b) \(\frac{x^2+4x+3}{2x+6}=\frac{x^2+x+3x+3}{2(x+3)}=\frac{x(x+1)+3(x+1)}{2(x+3)}=\frac{(x+1)(x+3)}{2(x+3)}=\frac{x+1}{2}\)
c)
\(\frac{15x(x+y)^3}{5y(x+y)^2}=\frac{5.3.x(x+y)^2.(x+y)}{5y(x+y)^2}=\frac{3x(x+y)}{y}\)
d) \(\frac{5(x-y)-3(y-x)}{10(x-y)}=\frac{5(x-y)+3(x-y)}{10(x-y)}=\frac{8(x-y)}{10(x-y)}=\frac{8}{10}=\frac{4}{5}\)
e) \(\frac{2x+2y+5x+5y}{2x+2y-5x-5y}=\frac{7x+7y}{-3x-3y}=\frac{7(x+y)}{-3(x+y)}=\frac{-7}{3}\)
f) \(\frac{x^2-xy}{3xy-3y^2}=\frac{x(x-y)}{3y(x-y)}=\frac{x}{3y}\)
g) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
1) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)
2) \(\left(2x+3y\right)^2=4x^2+12xy+9y^2\)
3) \(\left(x+\frac{1}{3}\right)^4=\left[\left(x+\frac{1}{3}\right)^2\right]^2=\left(x^2+\frac{2}{3}x+\frac{1}{9}\right)^2=x^4+\frac{4}{9}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x+\frac{2}{9}x^2=x^4+\frac{2}{3}x^2+\frac{1}{81}+\frac{4}{3}x^3+\frac{4}{27}x\)
4) \(\left(2x+y^2\right)^3=8x^3+12x^2y^2+6xy^4+y^6\)
5) Sửa đề: \(\left(\frac{x}{2}-2y\right)^3=\frac{x^3}{8}-\frac{3x^2}{2}+6xy^2-8y^3\)
6) \(\left(\sqrt{2x-y}\right)^4=\left(2x-y\right)^2=4x^2-4xy+y^2\)
7) \(\left(x+1\right)\left(x^2-x+1\right)=x^3+1\)
8) \(\left(x-3\right)\left(x^2+3x+9\right)=x^3-27\)
1)
a) \(\dfrac{5x}{10}=\dfrac{x}{2}\)
b) \(\dfrac{4xy}{2y}=2x\left(y\ne0\right)\)
c) \(\dfrac{21x^2y^3}{6xy}=\dfrac{7xy^2}{2}\left(xy\ne0\right)\)
d) \(\dfrac{2x+2y}{4}=\dfrac{2\left(x+y\right)}{4}=\dfrac{x+y}{2}\)
e) \(\dfrac{5x-5y}{3x-3y}=\dfrac{5\left(x-y\right)}{3\left(x-y\right)}=\dfrac{5}{3}\left(x\ne y\right)\)
f) \(\dfrac{-15x\left(x-y\right)}{3\left(y-x\right)}=-5x\dfrac{x-y}{y-x}=-5x\dfrac{x-y}{-\left(x-y\right)}\)
\(=-5x.\left(-1\right)=5x\left(x\ne y\right)\)
2)
a) Nhớ ghi ĐK vào nhá, lười quá :V\(\dfrac{x^2-16}{4x-x^2}=-\dfrac{\left(x-4\right)\left(x+4\right)}{x^2-4x}=\dfrac{\left(x-4\right)\left(x+4\right)}{x\left(x-4\right)}=\dfrac{x+4}{x}\)
b) \(\dfrac{x^2+4x+3}{2x+6}=\dfrac{x^2+3x+x+3}{2\left(x+3\right)}=\dfrac{x\left(x+3\right)+\left(x+3\right)}{2\left(x+3\right)}\)
\(=\dfrac{\left(x+3\right)\left(x+1\right)}{2\left(x+3\right)}=\dfrac{x+1}{2}\)
c) \(\dfrac{15x\left(x+3\right)^3}{5y\left(x+y\right)^2}=\dfrac{3x\left(x+3\right)^3}{y\left(x+y\right)^2}\) ( câu này có gì đó sai sai )
d) \(\dfrac{5\left(x-y\right)-3\left(y-x\right)}{10\left(x-y\right)}=\dfrac{5\left(x-y\right)+3\left(x-y\right)}{10\left(x-y\right)}\)
\(=\dfrac{8\left(x-y\right)}{10\left(x-y\right)}=\dfrac{8}{10}=\dfrac{4}{5}\)
e) \(\dfrac{2x+2y+5x+5y}{2x+2y-5x-5y}=\dfrac{2\left(x+y\right)+5\left(x+y\right)}{2\left(x+y\right)-5\left(x+y\right)}\)
\(=\dfrac{7\left(x+y\right)}{-3\left(x+y\right)}=-\dfrac{7}{3}\)
\(a,3x^3y^3-15x^2y^2=3x^2y^2\left(xy-5\right)\)
\(b,5x^3y^2-25x^2y^3+40xy^4\)
\(=5xy^2\left(x^2-5xy+8y^2\right)\)
\(c,-4x^3y^2+6x^2y^2-8x^4y^3\)
\(=-2x^2y^2\left(2x-3+4x^2y\right)\)
\(d,a^3x^2y-\frac{5}{2}a^3x^4+\frac{2}{3}a^4x^2y\)
\(=a^3x^2\left(y-\frac{5}{2}x^2+\frac{2}{3}ay\right)\)
\(e,a\left(x+1\right)-b\left(x+1\right)=\left(x+1\right)\left(a-b\right)\)
\(f,2x\left(x-5y\right)+8y\left(5y-x\right)\)
\(=2x\left(x-5y\right)-8y\left(x-5y\right)=\left(x-5y\right)\left(2x-8y\right)\)
\(g,a\left(x^2+1\right)+b\left(-1-x^2\right)-c\left(x^2+1\right)\)
\(=\left(x^2+1\right)\left(a-b-c\right)\)
\(h,9\left(x-y\right)^2-27\left(y-x\right)^3\)
\(=9\left(x-y\right)^2+27\left(x-y\right)^3\)
\(=9\left(x-y\right)^2\left(1+3x-3y\right)\)
a,3x3y3−15x2y2=3x2y2(xy−5)a,3x3y3−15x2y2=3x2y2(xy−5)
b,5x3y2−25x2y3+40xy4b,5x3y2−25x2y3+40xy4
=5xy2(x2−5xy+8y2)=5xy2(x2−5xy+8y2)
c,−4x3y2+6x2y2−8x4y3c,−4x3y2+6x2y2−8x4y3
=−2x2y2(2x−3+4x2y)=−2x2y2(2x−3+4x2y)
d,a3x2y−52a3x4+23a4x2yd,a3x2y−52a3x4+23a4x2y
=a3x2(y−52x2+23ay)=a3x2(y−52x2+23ay)
e,a(x+1)−b(x+1)=(x+1)(a−b)e,a(x+1)−b(x+1)=(x+1)(a−b)
f,2x(x−5y)+8y(5y−x)f,2x(x−5y)+8y(5y−x)
=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)=2x(x−5y)−8y(x−5y)=(x−5y)(2x−8y)
g,a(x2+1)+b(−1−x2)−c(x2+1)g,a(x2+1)+b(−1−x2)−c(x2+1)
=(x2+1)(a−b−c)=(x2+1)(a−b−c)
h,9(x−y)2−27(y−x)3h,9(x−y)2−27(y−x)3
=9(x−y)2+27(x−y)3
\(A=\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x-4\right)\\ A=\left(x^2-5x+4\right)\left(x^2-5x+6\right)\\ A=\left(x^2-5x+5-1\right)\left(x^2-5x+5+1\right)\\ A=\left(x^2-5x+5\right)^2-1\ge-1\)
đẳng thức xảy ra khi :
\(x^2-5x+5=0\\ x^2-2.\dfrac{5}{2}x+\dfrac{25}{4}=\dfrac{25}{4}-5\\ \left(x-\dfrac{5}{2}\right)^2=\dfrac{5}{4}\\ \Rightarrow\left[{}\begin{matrix}x-\dfrac{5}{2}=\sqrt{\dfrac{5}{4}}\\x-\dfrac{5}{2}=-\sqrt{\dfrac{5}{4}}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)
vậy GTNN của A =-1 tại \(\left[{}\begin{matrix}x=\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{\sqrt{5}+5}{2}\\x=-\sqrt{\dfrac{5}{4}}+\dfrac{5}{2}=\dfrac{5-\sqrt{5}}{2}\end{matrix}\right.\)