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KM
2
22 tháng 10 2018
a) \(P=x^2+\frac{1}{2}x+\frac{1}{16}\)
\(P=x^2+2\cdot x\cdot\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(P=\left(x+\frac{1}{4}\right)^2\)
b) Thay x = 49,75 vào P ta có :
\(P=\left(49,75+\frac{1}{4}\right)^2\)
\(P=50^2\)
\(P=2500\)
Vậy với x = 49,75 thì P = 2500
KM
1
22 tháng 10 2018
\(P=x^2+\frac{1}{2}x+\frac{1}{16}\)
\(=x^2+2.x.\frac{1}{4}+\left(\frac{1}{4}\right)^2\)
\(=\left(x+\frac{1}{4}\right)^2=\left(x+0,25\right)^2\)
b, Với x = 49,75 thì:
\(P=\left(x+0,25\right)^2=\left(49,75+0,25\right)^2=50^2=2500\)
MN
4 tháng 2 2020
\(ĐKXĐ:x\ne1\)
a) \(A=\left(1+\frac{x^2}{x^2+1}\right):\left(\frac{1}{x-1}-\frac{2x}{x^3+x-x^2-1}\right)\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{x\left(x^2+1\right)-\left(x^2+1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\left[\frac{1}{x-1}-\frac{2x}{\left(x^2+1\right)\left(x-1\right)}\right]\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x^2+1-2x}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{\left(x-1\right)^2}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x^2+1}:\frac{x-1}{x^2+1}\)
\(\Leftrightarrow A=\frac{\left(2x^2+1\right)\left(x^2+1\right)}{\left(x^2+1\right)\left(x-1\right)}\)
\(\Leftrightarrow A=\frac{2x^2+1}{x-1}\)
b) Thay \(x=-\frac{1}{2}\)vào A, ta được :
\(A=\frac{2\left(-\frac{1}{2}\right)^2+1}{-\frac{1}{2}-1}\)
\(\Leftrightarrow A=\frac{\frac{3}{2}}{-\frac{3}{2}}\)
\(\Leftrightarrow A=-1\)
c) Để A < 1
\(\Leftrightarrow2x^2+1< x-1\)
\(\Leftrightarrow2x^2-x+2< 0\)
\(\Leftrightarrow2\left(x^2-\frac{1}{2}x+\frac{1}{16}\right)+\frac{15}{8}< 0\)
\(\Leftrightarrow2\left(x-\frac{1}{4}\right)^2+\frac{15}{8}< 0\)
\(\Leftrightarrow x\in\varnothing\)
Vậy để \(A< 1\Leftrightarrow x\in\varnothing\)
d) Để A có giá trị nguyên
\(\Leftrightarrow2x^2+1⋮x-1\)
\(\Leftrightarrow2x^2-2x+2x-2+3⋮x-1\)
\(\Leftrightarrow2x\left(x-1\right)+2\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow2\left(x+1\right)\left(x-1\right)+3⋮x-1\)
\(\Leftrightarrow3⋮x-1\)
\(\Leftrightarrow x-1\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)
\(\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
Vậy để \(A\inℤ\Leftrightarrow x\in\left\{2;0;4;-2\right\}\)
23 tháng 12 2017
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(P=\left(\frac{8}{\left(x+4\right)\left(x-4\right)}+\frac{x-4}{\left(x-4\right)\left(x+4\right)}\right)\cdot\frac{x^2-2x-8}{1}\)
\(P=\left(\frac{x+4}{\left(x+4\right)\left(x-4\right)}\right)\cdot x^2-2x-8\)
\(P=\frac{1}{x-4}\cdot x^2-2x-8\)
P\(P=\frac{x^2+2x-4x+8}{x-4}\)
\(P=\frac{x\left(x+2\right)-4\left(x+2\right)}{x-4}\)
\(P=\frac{\left(x-4\right)\left(x+2\right)}{x-4}\)
\(P=x+2\)
14 tháng 1 2018
2 ,\(x^2-9x+20=0\)
\(\Rightarrow x^2-4x-5x+20=0\)
\(\Rightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Rightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}}\)\(\Rightarrow\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\orbr{\begin{cases}x=5\Rightarrow\\x=4\Rightarrow\end{cases}}\orbr{\begin{cases}P=7\\P=6\end{cases}}\)
9 tháng 7 2020
a) A = \(\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\left(x-2+\frac{10-x^2}{x+2}\right)\)
A = \(\left[\frac{x}{\left(x-2\right)\left(x+2\right)}-\frac{2\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right]:\left[\frac{\left(x-2\right)\left(x+2\right)}{x+2}+\frac{10-x^2}{x+2}\right]\)
A = \(\left[\frac{x-2x-4+x-2}{\left(x-2\right)\left(x+2\right)}\right]:\left[\frac{x^2-4+10-x^2}{x+2}\right]\)
A = \(-\frac{6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)
A = \(-\frac{6\left(x+2\right)}{6\left(x-2\right)\left(x+2\right)}\)
A = \(-\frac{6}{6\left(x-2\right)}\)
A = \(-\frac{1}{x-2}\)
b) |x| = \(\hept{\begin{cases}x=\frac{1}{2}\\x=-\frac{1}{2}\end{cases}}\)
+) với x = 1/2, ta có:
A = \(-\frac{1}{\frac{1}{2}-2}=\frac{2}{3}\)
+) với x = -1/2, ta có:
A = \(-\frac{1}{\left(-\frac{1}{2}\right)-2}=\frac{2}{5}\)
MN
2 tháng 2 2020
a) \(ĐKXĐ:x\ne\pm4;x\ne-2\)
\(P=\left(\frac{8}{x^2-16}+\frac{1}{x+4}\right):\frac{1}{x^2-2x-8}\)
\(\Leftrightarrow P=\left(\frac{8}{\left(x-4\right)\left(x+4\right)}+\frac{1}{x+4}\right):\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{8+x-4}{\left(x-4\right)\left(x+4\right)}:\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{x+4}{\left(x-4\right)\left(x+4\right)}:\frac{1}{\left(x-4\right)\left(x+2\right)}\)
\(\Leftrightarrow P=\frac{1}{x-4}.\left(x-4\right)\left(x+2\right)\)
\(\Leftrightarrow P=\frac{\left(x-4\right)\left(x+2\right)}{\left(x-4\right)}\)
\(P=x+2\)
b) Ta có :
\(x^2-9x+20=0\)
\(\Leftrightarrow x^2-4x-5x+20=0\)
\(\Leftrightarrow x\left(x-4\right)-5\left(x-4\right)=0\)
\(\Leftrightarrow\left(x-5\right)\left(x-4\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-5=0\\x-4=0\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=5\\x=4\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}P=x+2=5+2=7\\P=x+2=4+2=6\end{cases}}\)
Vậy \(P\in\left\{7;6\right\}\)
13 tháng 12 2018
\(A=\frac{1}{x^2-x}+\frac{1}{x^2+x+1}+\frac{2x}{1-x^3}\)
\(A=\frac{1}{x.\left(x-1\right)}+\frac{1}{x^2+x+1}+\frac{2x}{\left(1-x\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x\left(x-1\right)}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1}{x.\left(x-1\right)\left(x^2+x+1\right)}+\frac{x^2-x}{x.\left(x-1\right)\left(x^2+x+1\right)}-\frac{2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{x^2+x+1+x^2-x-2x^2}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\frac{1}{x.\left(x^3-1\right)}\)
Với x=10
\(\Rightarrow A=\frac{1}{10.\left(10^3-1\right)}\)
\(A=\frac{1}{10.999}\)
\(A=\frac{1}{9990}\)
Vậy \(A=\frac{1}{9990}\)tại x=10
\(P=x^2+\frac{1}{2}x+\frac{1}{16}\)
\(\Rightarrow\)\(16P=\left(4x\right)^2+2.4.x+1^2=\left(4x+1\right)^2\)
\(\Rightarrow\)\(P=\frac{\left(4x+1\right)^2}{16}\)
b,
Gía trị của P tại x = 49.75 là \(\frac{\left(49,75.4+1\right)^2}{16}=\frac{200^2}{16}=\frac{40000}{16}=2500\)