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\(\frac{25}{x^2}-\frac{5}{2}+\frac{x^2}{16}=\left(\frac{5}{x}-\frac{x}{4}\right)^2\)
\(x^3-4x^2-8x+8\)
\(\Leftrightarrow\left(x^3-4x^2\right)-\left(8x-8\right)\)
\(\Leftrightarrow x^2\left(x-4\right)-4\left(x-4\right)\)
\(\Leftrightarrow\left(x-4\right)\left(x^2-4\right)\)
\(-4x^2+x+1\)
\(=-\left(4x^2-x-1\right)\)
\(=-\left(\left(2x\right)^2-2.2x.\frac{1}{4}+\frac{1}{16}-\frac{17}{16}\right)\)
\(=\frac{17}{16}-\left(2x-\frac{1}{4}\right)^2\le\frac{17}{16}\)
Vậy MAX P = \(\frac{17}{16}< =>2x-\frac{1}{4}=0=>x=\frac{1}{8}\)
P= -(4x2 - x -1)
P=- [(2x)2 - 2.2x.\(\frac{1}{4}\)+\(\frac{1}{16}\)- \(\frac{17}{16}\)]
P=-(2x-\(\frac{1}{4}\))2 + \(\frac{17}{16}\)
vậy P <= 17/16
P=\(^{-4x^2}\)+2.2.1/4.x-1/16+17/16
=\(-\left(2x-\frac{1}{4}\right)\)^2+\(\frac{17}{16}\)<=17/16
vay max P=17/16 khi do x=1/8
Q=-(x^2+4x+4)+6
=-(x+2)^2+6nho hon hoac=6
vay max Q=6khi do x=-2
a, <=> (x-1)^2-4=0
<=> (x-1-2).(x-1+2)=0
<=> (x-3).(x+1)=0
<=> x-3=0 hoặc x+1=0
<=> x=3 hoặc x=-1
b, <=> x^2-x+2x-2=0
<=> x^2+x-2=0
<=> (x^2-x)+(2x-2)=0
<=> (x-1).(x+2)=0
<=> x-1=0 hoặc x+2=0
<=> x=1 hoặc x=-2
c, <=> (2x+1)^2=x^2
<=> 2x+1=x hoặc 2x+1=-x
<=> x=-1 hoặc x=-1/3
d, <=> (x^2-2x)-(3x-6)=0
<=> (x-2).(x-3)=0
<=> x-2=0 hoặc x-3=0
<=> x=2 hoặc x=3
Tk mk nha
a,\(\left(x^2-2x+1\right)-4=0\)
\(\Leftrightarrow\left(x-1\right)^2-4=0\)
\(\Leftrightarrow\left(x-1-2\right)\left(x-1+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+1\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x=3\\x=-1\end{cases}}\)
câu 2 :
\(100+10x+\frac{1}{4x^2}\)
<=> \(\left(10\right)^2+2.10.\frac{1}{2x}+\left(\frac{1}{2x}\right)^2\)
<=> \(\left(10+\frac{1}{2}x\right)^2\)
câu 1 :
\(\frac{25}{x^2}-\frac{5}{2}+\frac{x^2}{16}\)
\(\Leftrightarrow\)\(\left(\frac{5}{x}\right)^2-2.\frac{5}{x}.\frac{x}{4}+\left(\frac{x}{4}\right)^2\)
\(\Leftrightarrow\)\(\left(\frac{5}{x}+\frac{x}{4}\right)^2\) k mình nhé bạn !!