cho biểu thức p=[\(\frac{x}{x-2}\)+ \(\frac{1}{x^2-4}\)] : \(\frac{x+1}{x+2}\)
a.rút gọn p
b.tính giá trị của p sau khi x=\(\frac{1}{2}\)
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1 tháng 1 2018
a)\(P=\left(\frac{1}{\sqrt{x}-2}+\frac{1}{\sqrt{x}+2}\right).\frac{\sqrt{x}-2}{2}\left(ĐK:x\ge0;x\ne4\right)\)
\(\Leftrightarrow P=\left(\frac{\sqrt{x}+2+\sqrt{x}-2}{x-4}\right).\frac{\sqrt{x}-2}{2}\)
\(\Leftrightarrow P=\left[\frac{2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right].\frac{\sqrt{x}-2}{2}\)
\(\Leftrightarrow P=\frac{\sqrt{x}}{\sqrt{x}+2}\)
b)Tại x=9 \(\Leftrightarrow\frac{\sqrt{9}}{\sqrt{9}+2}=\frac{3}{3+2}=\frac{3}{5}\)
1 tháng 1 2018
Ý c nàk
\(Q=P.\sqrt{x}=\sqrt{x}.\frac{\sqrt{x}}{\sqrt{x}+2}=\frac{x}{\sqrt{x}+2}=\frac{x-4+4}{\sqrt{x}+2}=\frac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)+4}{\sqrt{x}+2}\)
\(=\sqrt{x}-2+\frac{4}{\sqrt{x}+2}=\left(\sqrt{x}+2\right)+\frac{4}{\sqrt{x}+2}-4\)
Áp dụng bđt AM - GM ta có :
\(Q\ge2\sqrt{\left(\sqrt{x}+2\right).\frac{4}{\sqrt{x}+2}}-4=2.2-4=0\) có GTNN là 0
Dấu "=" xảy ra \(\Leftrightarrow x=0\)
14 tháng 2 2021
Sửa đề: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne\dfrac{9}{4}\end{matrix}\right.\)
a) Ta có: \(P=\left(2-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{2\cdot\left(2\sqrt{x}-3\right)}{2\sqrt{x}-3}-\dfrac{\sqrt{x}-1}{2\sqrt{x}-3}\right):\left(\dfrac{6\sqrt{x}+1}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}+\dfrac{\sqrt{x}\left(2\sqrt{x}-3\right)}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\right)\)
\(=\dfrac{4\sqrt{x}-6-\sqrt{x}+1}{2\sqrt{x}-3}:\dfrac{6\sqrt{x}+1+2x-3\sqrt{x}}{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}-3}\cdot\dfrac{\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)}{2x+3\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2x+2\sqrt{x}+\sqrt{x}+1}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{2\sqrt{x}\left(\sqrt{x}+1\right)+\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\left(3\sqrt{x}-5\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(2\sqrt{x}+1\right)}\)
\(=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\)
b) Ta có: \(x=\dfrac{3-2\sqrt{2}}{4}\)
\(\Leftrightarrow x=\dfrac{2-2\cdot\sqrt{2}\cdot1+1}{4}\)
\(\Leftrightarrow x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\)(thỏa ĐK)
Thay \(x=\dfrac{\left(\sqrt{2}-1\right)^2}{4}\) vào biểu thức \(P=\dfrac{3\sqrt{x}-5}{2\sqrt{x}+1}\), ta được:
\(P=\left(3\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}-5\right):\left(2\cdot\sqrt{\dfrac{\left(\sqrt{2}-1\right)^2}{4}}+1\right)\)
\(\Leftrightarrow P=\left(3\cdot\dfrac{\sqrt{2}-1}{2}-5\right):\left(2\cdot\dfrac{\sqrt{2}-1}{2}+1\right)\)
\(\Leftrightarrow P=\left(\dfrac{3\cdot\left(\sqrt{2}-1\right)}{2}-\dfrac{10}{2}\right):\left(\sqrt{2}-1+1\right)\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-3-10}{2}:\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{3\sqrt{2}-13}{2}\cdot\sqrt{2}\)
\(\Leftrightarrow P=\dfrac{6-13\sqrt{2}}{2}\)
Vậy: Khi \(x=\dfrac{3-2\sqrt{2}}{4}\) thì \(P=\dfrac{6-13\sqrt{2}}{2}\)
NL
20 tháng 4 2017
ĐKXĐ: x\(\ne\)1, x\(\ne\)-1
MTC (x-1)(x+1)
\(\Leftrightarrow\)(\(\frac{-\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}\)+ \(\frac{2\left(x-1\right)}{MTC}\)-\(\frac{-\left(5-x\right)}{MTC}\)) : \(\frac{1-2x}{MTC}\)
\(\Rightarrow\)\(\left[-\left(x+1\right)+2\left(x-1\right)+\left(5-x\right)\right]:\left(1-2x\right)\)
\(\Leftrightarrow\frac{-x-1+2x-2+5-x}{1-2x}\)
=\(\frac{-2x+2x+2}{1-2x}\)
=\(\frac{2}{1-2x}\)
b. mình chỉ biết \(x< \frac{1}{2}\) thôi chứ ko biết làm sao
hình như là giải Bất phương trình \(\frac{2}{1-2x}>0\)
MN
7 tháng 2 2020
\(ĐKXĐ:x\ne\pm1\)
a) \(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{4x^2}{1-x^2}\right):\frac{2x^2-2}{x^2-2x+1}\)
\(\Leftrightarrow A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}-\frac{4x^2}{x^2-1}\right):\frac{2\left(x^2-1\right)}{\left(x-1\right)^2}\)
\(\Leftrightarrow A=\frac{\left(x+1\right)^2-\left(x-1\right)^2-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{x^2+2x+1-x^2+2x-1}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{4x-4x^2}{x^2-1}.\frac{\left(x-1\right)^2}{2\left(x^2-1\right)}\)
\(\Leftrightarrow A=\frac{-4x\left(x-1\right)^3}{2\left(x-1\right)^2\left(x+1\right)^2}\)
\(\Leftrightarrow A=\frac{-2x\left(x-1\right)}{\left(x+1\right)^2}\)
b) Thay x = -3 vào A, ta được :
\(A=\frac{\left(-2\right)\left(-3\right)\left(-3-1\right)}{\left(-3+1\right)^2}\)
\(\Leftrightarrow A=\frac{6.\left(-4\right)}{2^2}\)
\(\Leftrightarrow A=-6\)
c) Để A > -1
\(\Leftrightarrow-2x\left(x-1\right)>-\left(x+1\right)^2\)
\(\Leftrightarrow2x\left(x-1\right)< \left(x+1\right)^2\)
\(\Leftrightarrow2x^2-2x< x^2+2x+1\)
\(\Leftrightarrow x^2-4x-1< 0\)
\(\Leftrightarrow\left(x-2\right)^2-5< 0\)
\(\Leftrightarrow\left(x-2\right)^2< 5\)
Đoạn này bạn tự tìm giá trị x thỏa mãn là xong (Chú ý ĐKXĐ)
TM
26 tháng 1 2017
\(A=\left(\frac{1}{x-1}-\frac{x}{1-x^3}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(A=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(1+x+x^2\right)}.\frac{x^2+x+1}{x+1}\right):\frac{2x+1}{x^2+2x+1}\)
\(A=\left(\frac{1}{x-1}-\frac{x}{\left(1-x\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+2x+1}\)
\(A=\left(\frac{-x-1}{\left(1-x\right)\left(x+1\right)}-\frac{x}{\left(1-x\right)\left(x+1\right)}\right):\frac{2x+1}{x^2+2x+1}\)
\(A=\frac{-2x-1}{\left(1-x\right)\left(x+1\right)}:\frac{2x+1}{x^2+2x+1}\)
\(A=\frac{-\left(2x+1\right)}{\left(1-x\right)\left(x+1\right)}.\frac{\left(x+1\right)^2}{2x+1}\)
\(A=\frac{-1}{1-x}.\frac{x+1}{1}\)
\(A=\frac{-x-1}{1-x}\)
a) \(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\left(x\ne\pm2\right)\)
\(=\left(\frac{x}{x-2}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right):\frac{x+1}{x+2}=\left(\frac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\frac{1}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{x+2}{x+1}\)
\(=\frac{x^2+2x+1}{\left(x-2\right)\left(x+2\right)}\cdot\frac{x+2}{x+1}=\frac{\left(x+1\right)^2\cdot\left(x+2\right)}{\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\frac{x+1}{x-2}\)
vậy \(P=\frac{x+1}{x-2}\left(x\ne\pm2\right)\)
b) ta có \(P=\frac{x+1}{x-2}\left(x\ne\pm2\right)\)
ta có x=\(\frac{1}{2}\left(tm\right)\)thay vào P ta được \(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=\frac{3}{2}:\left(\frac{-3}{2}\right)=\frac{3}{2}\cdot\frac{-2}{3}=-1\)
vậy P=-1 khi x=1/2
\(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\)
a) ĐKXĐ : \(x\ne\pm2\)
\(P=\left(\frac{x}{x-2}+\frac{1}{x^2-4}\right):\frac{x+1}{x+2}\)
\(P=\left(\frac{x}{x-2}+\frac{1}{\left(x+2\right)\left(x-2\right)}\right):\frac{x+1}{x+2}\)
\(P=\left(\frac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\frac{1}{\left(x+2\right)\left(x-2\right)}\right)\cdot\frac{x+2}{x+1}\)
\(P=\frac{x^2+2x+1}{\left(x+2\right)\left(x-2\right)}\cdot\frac{x+2}{x+1}\)
\(P=\frac{\left(x+1\right)^2\cdot\left(x+2\right)}{\left(x+2\right)\left(x-2\right)\left(x+1\right)}\)
\(P=\frac{x+1}{x-2}\)
b) Thế x = 1/2 vào P ta được :
\(P=\frac{\frac{1}{2}+1}{\frac{1}{2}-2}=\frac{\frac{3}{2}}{-\frac{3}{2}}=-1\)