A) \(\frac{1-x\sqrt{x}}{1-\sqrt{x}}\) với x \(\ge\) 0; và x \(\ne\) 1
B) Giải phương trình: x(3-\(\sqrt{3x}-1\)) = \(\sqrt{3x^2+2x-1}-x\sqrt{x+1}+1\)
K
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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)\(\frac{x-2\sqrt{x}+1}{\sqrt{x}-1}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
NV
Nguyễn Việt Lâm
Giáo viên
17 tháng 10 2019
\(P=\left(\frac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\frac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\frac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\frac{2}{\sqrt{x}-1}\)
\(=\left(\frac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right).\left(\frac{2}{\sqrt{x}-1}\right)\)
\(=\frac{2\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)^2\left(x+\sqrt{x}+1\right)}=\frac{2}{x+\sqrt{x}+1}\)
Do \(x+\sqrt{x}+1=x+\sqrt{x}+\frac{1}{4}+\frac{3}{4}=\left(\sqrt{x}+\frac{1}{2}\right)^2+\frac{3}{4}>0\)
\(\Rightarrow P=\frac{2}{x+\sqrt{x}+1}>0\)
23 tháng 8 2019
a, \(A=\sqrt{\left(1-x\right)^2}-1=\left|1-x\right|-1=1-x-1\)(vì x<1)
<=> A=\(-x\)
b,B=\(\frac{3-\sqrt{x}}{x-9}\left(x\ge0,x\ne9\right)\)
=\(\frac{-\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=-\frac{1}{\sqrt{x}+3}\)
Vậy \(B=-\frac{1}{\sqrt{x}+3}\)
c, C=\(\frac{x-5\sqrt{x}+6}{\sqrt{x}-3}\left(x\ge0,x\ne9\right)\)
=\(\frac{x-2\sqrt{x}-3\sqrt{x}+6}{\sqrt{x}-3}\)=\(\frac{\sqrt{x}\left(\sqrt{x}-2\right)-3\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)=\(\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}{\sqrt{x}-3}\)=\(\sqrt{x}-2\)
Vậy C= \(\sqrt{x}-2\)
d, D=\(5-3x-\sqrt{25-10x+x^2}\left(x< 5\right)\)
= \(5-3x-\sqrt{\left(5-x\right)^2}\)=\(5-3x-\left|5-x\right|\)=\(5-3x-5+x\) (vì x<5)=-2x
Vậy D=-2x
e, E=\(\sqrt{3a}.\sqrt{27a}\) (đk \(a\ge0\))
=\(\sqrt{3.27.a^2}=\sqrt{3^4}.a=9a\)
Vậy E=9a
f, F=\(\frac{1}{a-1}\sqrt{9\left(a-1\right)^2}\) (đk :a>1)
= \(\frac{1}{a-1}.3\left|a-1\right|\)=\(\frac{1}{a-1}.3\left(a-1\right)\) (vì a>1)=3
Vậy F=3
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)
10 tháng 8 2019
\(A=\frac{x\sqrt{x}+26\sqrt{x}-19-2\sqrt{x}\left(\sqrt{x}+3\right)+\left(\sqrt{x}-3\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x\left(\sqrt{x}-1\right)+16\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{x+16}{\sqrt{x}+3}\)
+ \(A=\frac{x+16}{\sqrt{x}+3}=\frac{x-9+25}{\sqrt{x}+3}=\frac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)+25}{\sqrt{x}+3}\) \(=\sqrt{x}-3+\frac{25}{\sqrt{x}+3}\)
\(=\sqrt{x}+3+\frac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\frac{25}{\sqrt{x}+3}}-6=10-6=4\)
Dấu "=" \(\Leftrightarrow\sqrt{x}+3=\frac{25}{\sqrt{x}+3}\Leftrightarrow\sqrt{x}+3=5\Leftrightarrow x=4\)
Vậy \(A=\frac{x+16}{\sqrt{x}+3}\)
Min A = 4 \(\Leftrightarrow x=4\)
HT
28 tháng 6 2019
Sửa đề nha: \(\sqrt{x^3-1}\) thành \(\sqrt{x^3}-1\)
\(B=\left(\frac{2x+1-\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{1+\sqrt{x}}-\sqrt{x}\right)\)
\(B=\left(\frac{2x+1-x+\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right)\left(x-2\sqrt{x}+1\right)\)
\(B=\frac{\left(x+\sqrt{x}+1\right)\left(x-2\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-1}=\sqrt{x}-1\)
b/ Để B= 3\(\Leftrightarrow\sqrt{x}-1=3\Leftrightarrow x=16\)
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)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}}{\sqrt{x}-6}-\frac{3}{\sqrt{x}+6}-\frac{x}{x-36}\)
\(=\frac{\sqrt{x}\left(\sqrt{x}+6\right)-3\left(\sqrt{x-6}\right)-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{x+6\sqrt{x}-3\sqrt{x}+18-x}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3\sqrt{x}+18}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3(\sqrt{x}+6)}{(\sqrt{x}-6)\left(\sqrt{x}+6\right)}\)
\(=\frac{3}{\sqrt{x}-6}\)
A) Ta có: \(\frac{1-x\sqrt{x}}{1-\sqrt{x}}\)
\(=\frac{\left(1-\sqrt{x}\right)\left(1+\sqrt{x}+x\right)}{1-\sqrt{x}}\)
\(=x+\sqrt{x}+1\)