\(\frac{6}{1+\sqrt{x}}\)- \(\frac{4}{1-\sqrt{x}}\)-\(\frac{100\sqrt{x}}{x-1}\)với x\(\ge\) 0 ; x\(\ne\)1
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TP
21 tháng 6 2019
\(B=\frac{9-x}{\sqrt{x}+3}-\frac{x-6\sqrt{x}+9}{\sqrt{x}-3}-6\)(đk: x ≥ 0 và x ≠ 9)
\(B=\frac{\left(3-\sqrt{x}\right)\left(3+\sqrt{x}\right)}{\sqrt{x}+3}-\frac{\left(\sqrt{x}-3\right)^2}{\sqrt{x}-3}-6\)
\(B=\left(3-\sqrt{x}\right)-\left(\sqrt{x}-3\right)-6\)
\(B=3-\sqrt{x}-\sqrt{x}+3-6\)
\(B=-2\sqrt{x}\)
TP
21 tháng 6 2019
\(A=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}+\frac{x}{36-x}\)(đk: x ≥ 0 và x ≠ 36)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
20 tháng 10 2023
a: \(A=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)+2\sqrt{x}\left(\sqrt{x}+2\right)-3x-4}{x-4}\)
\(=\dfrac{x-2\sqrt{x}+2x+4\sqrt{x}-3x-4}{x-4}\)
\(=\dfrac{2\sqrt{x}-4}{x-4}=\dfrac{2}{\sqrt{x}+2}\)
b: A=1/2
=>\(\sqrt{x}+2=4\)
=>\(\sqrt{x}=2\)
=>x=4(loại)
8 tháng 10 2019
c,C= \(\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\left(x\ge1\right)\)
=\(\sqrt{x-1+2\sqrt{x-1}+1}+\sqrt{x-1-2\sqrt{x-1}+1}\)
=\(\sqrt{x-1}+1+\left|\sqrt{x-1}-1\right|\) (1)
TH1: \(\sqrt{x-1}< 1\) hay \(1\le x< 2\)
Từ (1)=>C= \(\sqrt{x-1}+1+1-\sqrt{x-1}\)=2
TH2: \(\sqrt{x-1}\ge1\) hay \(x\ge2\)
Từ (1) =>C=\(\sqrt{x-1}+1+\sqrt{x-1}-1\)=\(2\sqrt{x-1}\)
d, D=\(\sqrt{13+30\sqrt{2}+\sqrt{9+4\sqrt{2}}}=\sqrt{13+30\sqrt{2}+\sqrt{8+2\sqrt{8}+1}}=\sqrt{13+30\sqrt{2}+\sqrt{\left(\sqrt{8}+1\right)^2}}\)
=\(\sqrt{13+30\sqrt{2}+\sqrt{8}+1}=\sqrt{14+30\sqrt{2}+2\sqrt{2}}\)
=\(\sqrt{14+32\sqrt{2}}\)
8 tháng 10 2019
a)\(\frac{x-y}{\sqrt{x}-\sqrt{y}}=\frac{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}-\sqrt{y}}=\sqrt{x}+\sqrt{y}\)
b)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
HT
10 tháng 6 2019
a/ ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne9\end{matrix}\right.\)
\(P=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(=\frac{x-\sqrt{x}-6}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}=\frac{\left(\sqrt{x}-3\right)\left(2+\sqrt{x}\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)\(=\frac{2+\sqrt{x}}{3+\sqrt{x}}\)
b/ i, \(x=\sqrt{4+4\sqrt{2}+2}+\sqrt{4-4\sqrt{2}+2}\)
\(=\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\)
\(=2+\sqrt{2}+2-\sqrt{2}=4\)
Thay vào P có:\(P=\frac{2+\sqrt{4}}{3+\sqrt{4}}=\frac{4}{5}\)
ii, \(x=\frac{\sqrt{2}+1-\sqrt{2}+1}{2-1}=2\)
Thay vào có:\(P=\frac{2+\sqrt{2}}{3+\sqrt{2}}=\frac{4+\sqrt{2}}{7}\)
10 tháng 6 2019
a) Ta có:
\(P=\frac{x-\sqrt{x}}{x-9}+\frac{\sqrt{x}-3}{x-9}-\frac{\sqrt{x}+3}{x-9}\)
\(P=\frac{x-\sqrt{x}+\sqrt{x}-3-\sqrt{x}-3}{x-9}\)
\(P=\frac{x-\sqrt{x}-6}{x-9}\)
12 tháng 10 2019
ta có\(x=\frac{1}{2}\left(\sqrt{6+4\sqrt{2}}+\sqrt{6-4\sqrt{2}}\right)\)
⇔\(x=\frac{1}{2}\left(\sqrt{2+4+4\sqrt{2}}+\sqrt{2+4-4\sqrt{2}}\right)\)(hằng dẳng thức)
⇔\(x=\frac{1}{2}\left(\sqrt{\left(2+\sqrt{2}\right)^2}+\sqrt{\left(2-\sqrt{2}\right)^2}\right)\)
⇔\(x=\frac{1}{2}\left(\left|2+\sqrt{2}\right|+\left|2-\sqrt{2}\right|\right)\)
⇔\(x=\frac{1}{2}\left(2+\sqrt{2}+2-\sqrt{2}\right)\)
⇔\(x=\frac{1}{2}\cdot4\)
⇔x=2
thay x=2 vào biểu thức ta có:
\(\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{\left(\sqrt{2}-1\right)^2}{2-1}=\left(\sqrt{2}-1\right)^2\)
=\(2-2\sqrt{2}+1\)
=\(3-2\sqrt{2}\)
\(=\frac{6\left(1-\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}-\frac{4\left(1+\sqrt{x}\right)}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}+\frac{100\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}\)
\(=\frac{6-6\sqrt{x}-4-4\sqrt{x}+100\sqrt{x}}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\frac{90\sqrt{x}-2}{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}\right)}=\frac{2\left(45\sqrt{x}-1\right)}{1-x}\)