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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) \(\frac{2ax^2-4ax+2a}{5b-5bx^2}=\frac{2a(x^2-2x+1)}{5b(1-x^2)}=\frac{2a(x-1)^2}{5b(1-x)(1+x)}\)
\(=\frac{2a(x-1)}{5b(-1)(x+1)}=\frac{2a(1-x)}{5b(x+1)}\)
Tớ làm lôn nheé , không chép lại đề đâu
a) 15x3 - 6x2 - 3x
b) ĐKXĐ: x # 1
( x - 1)2 . \(\dfrac{1}{x-1}\)
= x - 1
c) ĐKXĐ: x # 1\(\dfrac{x^2-x-x+1}{x-1}=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\)
d)ĐKXĐ : x # 0 ; x # 5
\(\dfrac{\left(x-5\right)^2}{x\left(x-5\right)}.\dfrac{x-5}{x}=\dfrac{\left(x-5\right)^2}{x^2}\)
\(D=\frac{x^{2}-2x+2018}{x^{2}}\)
\(D=\frac{x^{2}-2*x*1+1+2017}{x^{2}}\)
\(D= \frac{(x-1)^{2}+2017}{x^{2}}\)
Nhận xét: Để D Đặt GTNN thì \((x-1)^{2} + 2017\) Đạt GTNN
Mà \((x-1)^{2} \geq 0\) . Nên:
\((x-1)^{2}+2017\)\(\geq 2017\). GTNN của \((x-1)^{2}+2017=2017 \) Khi x-1=0 => x=1
Thay x=1 vào D
GTNN D=2017
\(A=2030+\dfrac{8}{x}+\dfrac{1}{x^2}=\left(\dfrac{1}{x}\right)^2+8.\dfrac{1}{x}+16+2014\)
\(\Rightarrow A=\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)
\(\Rightarrow A_{min}=2014\) khi \(\dfrac{1}{x}+4=0\Rightarrow x=-\dfrac{1}{4}\)
\(A=\dfrac{2030x^2+8x+1}{x^2}\\ =\dfrac{2030x^2}{x^2}+\dfrac{8x}{x^2}+\dfrac{1}{x^2}\\ =2030+\dfrac{8}{x}+\dfrac{1}{x^2}\\ =\left(\dfrac{1}{x}\right)^2+2\cdot\dfrac{1}{x}\cdot4+16+2014\\ =\left(\dfrac{1}{x}+4\right)^2+2014\)
Do \(\left(\dfrac{1}{x}+4\right)^2\ge0,2014>0\)
\(\Rightarrow\left(\dfrac{1}{x}+4\right)^2+2014\ge2014\)
\(\Rightarrow Min\left(A\right)=2014\Leftrightarrow\dfrac{1}{x}+4=0\Rightarrow x=\dfrac{-1}{4}\)
\(P-\dfrac{2}{3}=\dfrac{x^2-6x+9}{3x^2}=\dfrac{\left(x-3\right)^2}{3x^2}\ge0\Rightarrow P\ge\dfrac{2}{3}\).
Dấu "=" xảy ra khi x = 3.