Rút gọn các biểu thức sau:
a) \(\sqrt{\dfrac{16(4x-4\sqrt{x}+1)}{6x+3\sqrt{x}}}\) với \(x > 1\)
b) \(\dfrac{\sqrt{(x)^{2}}+\sqrt{4-4x+(x)^{2}}+1}{2x-1}\) với \(x > 2\sqrt{2}\)
c) \(\sqrt{(x)^{2}-8x+16}+\sqrt{36-12x+(x)^{2}}\) với \(4< x <6\)
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16 tháng 10 2021
a: Ta có: \(x=\sqrt{28-16\sqrt{3}}+2\sqrt{3}\)
\(=4-2\sqrt{3}+2\sqrt{3}\)
=4
Thay x=4 vào B, ta được:
\(B=\dfrac{2-4}{2}=-1\)
20 tháng 7 2021
a) Ta có: \(A=\left(\dfrac{1}{\sqrt{a}+2}+\dfrac{1}{\sqrt{a}-2}\right):\dfrac{\sqrt{a}}{a-4}\)
\(=\dfrac{\sqrt{a}-2+\sqrt{a}+2}{\left(\sqrt{a}-2\right)\left(\sqrt{a}+2\right)}\cdot\dfrac{\left(\sqrt{a}+2\right)\left(\sqrt{a}-2\right)}{\sqrt{a}}\)
=2
b) Ta có: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
5 tháng 7 2021
a) Pt \(\Leftrightarrow\sqrt{\left(x-2\right)^2}=5\Leftrightarrow\left|x-2\right|=5\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
Vậy...
b)Đk: \(x\ge-1\)
Pt \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
\(\Leftrightarrow4\sqrt{x+1}=16\)\(\Leftrightarrow x+1=16\)\(\Leftrightarrow x=15\) (tm)
Vậy...
\(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\) (a>0)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-\left(2\sqrt{a}+1\right)+1=a-\sqrt{a}\)
b) \(A=a-\sqrt{a}=a-2.\dfrac{1}{2}\sqrt{a}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
Dấu "=" xảy ra khi \(\sqrt{a}=\dfrac{1}{2}\Leftrightarrow a=\dfrac{1}{4}\left(tmđk\right)\)
Vậy \(A_{min}=-\dfrac{1}{4}\)
AT
5 tháng 7 2021
a) \(\sqrt{x^2-4x+4}=5\Rightarrow\sqrt{\left(x-2\right)^2}=5\Rightarrow\left|x-2\right|=5\)
\(\Rightarrow\left[{}\begin{matrix}x-2=5\\x-2=-5\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=7\\x=-3\end{matrix}\right.\)
b) \(\sqrt{16x+16}-3\sqrt{x+1}+\sqrt{4x+4}=16-\sqrt{x+1}\)
\(\Rightarrow\sqrt{16\left(x+1\right)}-3\sqrt{x+1}+\sqrt{4\left(x+1\right)}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)
\(\Rightarrow4\sqrt{x+1}=16\Rightarrow\sqrt{x+1}=4\Rightarrow x=15\)
a) \(A=\dfrac{a^2+\sqrt{a}}{a-\sqrt{a}+1}-\dfrac{2a+\sqrt{a}}{\sqrt{a}}+1\)
\(=\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{a-\sqrt{a}+1}-\dfrac{\sqrt{a}\left(2\sqrt{a}+1\right)}{\sqrt{a}}+1\)
\(=a+\sqrt{a}-2\sqrt{a}-1+1=a-\sqrt{a}\)
b) Ta có: \(a-\sqrt{a}=\left(\sqrt{a}\right)^2-2.\sqrt{a}.\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2-\dfrac{1}{4}\)
\(=\left(\sqrt{a}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(\Rightarrow A_{min}=-\dfrac{1}{4}\) khi \(a=\dfrac{1}{4}\)
16 tháng 8 2023
a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
H9
HT.Phong (9A5)
CTVHS
7 tháng 8 2023
a) \(x-2y-\sqrt{x^2-4xy+4y^2}\)
\(=x-2y-\sqrt{\left(x-2y\right)^2}\)
\(=x-2y-\left|x-2y\right|\)
TH1: \(x-2y--\left(x-2y\right)\)
\(=x-2y+x-2y\)
\(=2x-4y\)
TH2: \(x-2y-\left(x-2y\right)\)
\(=x-2y-x+2y\)
\(=0\)
b) \(x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\sqrt{\left(x^2-4\right)^2}\)
\(=x^2+\left|x^2-4\right|\)
TH1:
\(x^2+-\left(x^2-4\right)\)
\(=x^2-x^2+4\)
\(=4\)
TH2:
\(x^2+\left(x^2-4\right)\)
\(=x^2+x^2-4\)
\(=2x^2-4\)
c) \(2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\) (x>5)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}\)
\(=2x-1-\sqrt{x-5}\)
d) \(\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}\) (\(x>\sqrt{2}\))
\(=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}\)
\(=\sqrt{x^2-2}\)
e) \(\sqrt{\left(x^2-4\right)^2}+\dfrac{x-4}{\sqrt{x^2-8x+16}}\)
\(=\left|x^2-4\right|+\dfrac{x-4}{\sqrt{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+\sqrt{\dfrac{\left(x-4\right)^2}{\left(x-4\right)^2}}\)
\(=\left|x^2-4\right|+1\)
TH1:
\(x^2-4+1\)
\(=x^2-3\)
TH2:
\(-\left(x^2-4\right)+1\)
\(=-x^2+4+1\)
\(=-x^2+5\)
7 tháng 8 2023
a: \(A=x-2y-\sqrt{x^2-4xy+4y^2}\)
=x-2y-|x-2y|
Khi x>=2y thì A=x-2y-x+2y=0
Khi x<2y thì A=x-2y+x-2y=2x-4y
b: \(B=x^2+\sqrt{x^4-8x^2+16}\)
\(=x^2+\left|x^2-4\right|\)
TH1: x>=2 hoặc x<=-2
B=x^2+x^2-4=2x^2-4
TH2: -2<=x<=2
B=x^2+4-x^2=4
c: \(C=2x-1-\sqrt{\dfrac{x^2-10x+25}{x-5}}\)
\(=2x-1-\sqrt{\dfrac{\left(x-5\right)^2}{x-5}}=2x-1-\sqrt{x-5}\)
d: \(D=\sqrt{\dfrac{x^4-4x^2+4}{x^2-2}}=\sqrt{\dfrac{\left(x^2-2\right)^2}{x^2-2}}=\sqrt{x^2-2}\)
AT
10 tháng 7 2021
a) \(P=\dfrac{x^2+3x}{x^2-8x+16}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\right)\left(x\ne0,x\ne4\right)\)
\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\left(\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x\left(x-4\right)}\right)\)
\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{\left(x+4\right)\left(x-4\right)+x+19-x^2}{x\left(x-4\right)}\)
\(=\dfrac{x^2+3x}{\left(x-4\right)^2}:\dfrac{x+3}{x\left(x-4\right)}=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\dfrac{x\left(x-4\right)}{x+3}=\dfrac{x^2}{x-4}\)
b) \(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}+1-\sqrt{3}+1=2\)
\(\Rightarrow P=\dfrac{2^2}{2-4}=-2\)
10 tháng 7 2021
a)\(ĐKXĐ:\left\{{}\begin{matrix}x\left(x-4\right)\ne0\\\dfrac{x+4}{x}+\dfrac{1}{x-4}+\dfrac{19-x^2}{x^2-4x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne4\\x\ne0\\x\ne-3\end{matrix}\right.\)
\(P=\dfrac{x\left(x+3\right)}{\left(x-4\right)}:\left(\dfrac{x^2-16+x+19-x^2}{x\left(x-4\right)}\right)=\dfrac{x\left(x+3\right)}{\left(x-4\right)^2}.\left(\dfrac{x\left(x-4\right)}{x+3}\right)=\dfrac{x^2}{x-4}\)
b)\(x=\sqrt{4+2\sqrt{3}}-\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3+1}-\left(\sqrt{3}-1\right)=2\)
thay x=2 vào P ta có \(P=\dfrac{2^2}{2-4}=-2\)
XO
30 tháng 1 2023
b) ĐKXĐ : \(x\ne\pm1\)
\(P=\dfrac{x}{x-1}+\dfrac{3}{x+1}-\dfrac{6x-4}{x^2-1}\)
\(=\dfrac{x\left(x+1\right)+3\left(x-1\right)-\left(6x-4\right)}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{x^2-2x+1}{\left(x-1\right)\left(x+1\right)}=\dfrac{\left(x-1\right)^2}{\left(x-1\right)\left(x+1\right)}=\dfrac{x-1}{x+1}\)
c) ĐKXĐ : \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)
\(A=\dfrac{1}{x+\sqrt{x}}+\dfrac{2\sqrt{x}}{x-1}-\dfrac{1}{x-\sqrt{x}}\)
\(=\dfrac{1}{\sqrt{x}\left(\sqrt{x}+1\right)}+\dfrac{2\sqrt{x}}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}\)
\(=\dfrac{\sqrt{x}-1+2x-\left(\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{2\left(x-1\right)}{\sqrt{x}\left(x-1\right)}=\dfrac{2}{\sqrt{x}}\)
XO
30 tháng 1 2023
a) ĐKXĐ : \(x\ge0;x\ne16\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}+4}+\dfrac{4}{\sqrt{x-4}}\right):\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{\sqrt{x}\left(\sqrt{x}-4\right)+4\left(\sqrt{x}+4\right)}{x-16}:\dfrac{x+16}{\sqrt{x}+2}\)
\(=\dfrac{x+16}{x-16}:\dfrac{x+16}{\sqrt{x}+2}=\dfrac{\sqrt{x}+2}{x-16}\)
AH
Akai Haruma
Giáo viên
22 tháng 6 2023
1.
$x+3+\sqrt{x^2-6x+9}=x+3+\sqrt{(x-3)^2}=x+3+|x-3|$
$=x+3+(3-x)=6$
2.
$\sqrt{x^2+4x+4}-\sqrt{x^2}=\sqrt{(x+2)^2}-\sqrt{x^2}$
$=|x+2|-|x|=x+2-(-x)=2x+2$
3.
$\sqrt{x^2+2\sqrt{x^2-1}}-\sqrt{x^2-2\sqrt{x^2-1}}$
$=\sqrt{(\sqrt{x^2-1}+1)^2}-\sqrt{(\sqrt{x^2-1}-1)^2}$
$=|\sqrt{x^2-1}+1|+|\sqrt{x^2-1}-1|$
$=\sqrt{x^2-1}+1+|\sqrt{x^2-1}-1|$
AH
Akai Haruma
Giáo viên
22 tháng 6 2023
4.
$\frac{\sqrt{x^2-2x+1}}{x-1}=\frac{\sqrt{(x-1)^2}}{x-1}$
$=\frac{|x-1|}{x-1}=\frac{x-1}{x-1}=1$
5.
$|x-2|+\frac{\sqrt{x^2-4x+4}}{x-2}=2-x+\frac{\sqrt{(x-2)^2}}{x-2}$
$=2-x+\frac{|x-2|}{x-2}|=2-x+\frac{2-x}{x-2}=2-x+(-1)=1-x$
6.
$2x-1-\frac{\sqrt{x^2-10x+25}}{x-5}=2x-1-\frac{\sqrt{(x-5)^2}}{x-5}$
$=2x-1-\frac{|x-5|}{x-5}$
b: \(=\dfrac{\left|x\right|+\left|x-2\right|+1}{2x-1}=\dfrac{x+x-2+1}{2x-1}=\dfrac{2x-1}{2x-1}=1\)
c: \(=\left|x-4\right|+\left|x-6\right|\)
=x-4+6-x=2