rút gọn
\(a,\left(\sqrt{\dfrac{7-\sqrt{7}}{7}-2}\right)\left(\dfrac{6}{\sqrt{7}+1}+4\right)\)
\(b,\sqrt{\dfrac{5}{12}-\dfrac{1}{\sqrt{6}}}+\dfrac{1}{\sqrt{6}}\)
b1,tính
a,\(\sqrt{\left(\sqrt{7}-4\right)^2}+\sqrt{8-2\sqrt{7}}\)
b,\(\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{6+2\sqrt{5}}\)
b2,rút gọn các biểu thức sau
a,\(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{2}\sqrt{20}+\sqrt{5}\)
b,\(\sqrt{\dfrac{1}{2}}+\sqrt{4,5}+\sqrt{12,5}\)
c,\(\sqrt{20}-\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
d,\(0,1\times\sqrt{200}+2\times\sqrt{0,08}+0,4\times\sqrt{50}\)
Bài 1:
a/
$\sqrt{(\sqrt{7}-4)^2}+\sqrt{8-2\sqrt{7}}$
$=|\sqrt{7}-4|+\sqrt{7+1-2\sqrt{7}}=|\sqrt{7}-4|+\sqrt{(\sqrt{7}-1)^2}$
$=4-\sqrt{7}+|\sqrt{7}-1|=4-\sqrt{7}+\sqrt{7}-1=3$
b/
\(\sqrt{(\sqrt{5}-2)^2}+\sqrt{6+2\sqrt{5}}\\ =|\sqrt{5}-2|+\sqrt{5+1+2\sqrt{5}}\\ =\sqrt{5}-2+\sqrt{(\sqrt{5}+1)^2}\\ =\sqrt{5}-2+|\sqrt{5}+1|=\sqrt{5}-2+\sqrt{5}+1=2\sqrt{5}-1\)
Bài 2:
a. $=\sqrt{5}+\sqrt{5}+\sqrt{5}=3\sqrt{5}$
b. $=\frac{\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}+\frac{5\sqrt{2}}{2}$
$=\frac{\sqrt{2}+3\sqrt{2}+5\sqrt{2}}{2}=\frac{9\sqrt{2}}{2}$
c.
$=2\sqrt{5}-3\sqrt{5}+9\sqrt{2}+6\sqrt{2}$
$=-\sqrt{5}+15\sqrt{2}$
d.
$=0,1.10\sqrt{2}+2.\frac{\sqrt{2}}{5}+0,4.5\sqrt{2}$
$=\sqrt{2}+0,4\sqrt{2}+2\sqrt{2}$
$=\sqrt{2}(1+0,4+2)=3,4\sqrt{2}$
Rút gọn các biểu thức:
\(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{x-4}{3\sqrt{x}}\)
\(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}+\dfrac{6-7\sqrt{a}}{a-4}\right).\left(\sqrt{a}+2\right)\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
Rút gọn:
1) \(\dfrac{16-6\sqrt{7}}{\sqrt{7}-3}\)
2) \(\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2+4\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)
3) \(\dfrac{\left(\sqrt{3}+2\sqrt{5}\right)^2-8\sqrt{15}}{\sqrt{6}-2\sqrt{10}}\)
Giúp em với ạ. Help mee !!!
Câu 1,2 bạn đã đăng và có lời giải rồi
Câu 3:
\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)
* Rút gọn biểu thức
a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
- Rút gọn biểu thức
a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
a) Ta có: \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)
\(=10\sqrt{5}\)
b) Ta có: \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
\(=3\left(2-\sqrt{3}\right)+4+\sqrt{3}+2\sqrt{3}\)
\(=6-2\sqrt{3}+4+3\sqrt{3}\)
\(=10+\sqrt{3}\)
c) Ta có: \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)
=7-5=2
d) Ta có: \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=-3+2\sqrt{3}\)
a. \(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
\(=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}\)
\(=10\sqrt{5}\)
b. \(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
\(=\dfrac{3\left(2-\sqrt{3}\right)}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}+\dfrac{13\left(4+\sqrt{3}\right)}{\left(4-\sqrt{3}\right)\left(4+\sqrt{3}\right)}+\dfrac{6\sqrt{3}}{\sqrt{3}.\sqrt{3}}\)
\(=\dfrac{3\left(2-\sqrt{3}\right)}{4-3}+\dfrac{13\left(4+\sqrt{3}\right)}{16-3}+\dfrac{6\sqrt{3}}{3}\)
\(=3\left(2-\sqrt{3}\right)+\dfrac{13\left(4+\sqrt{3}\right)}{13}+2\sqrt{3}\)
\(=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)
\(=10\)
c. \(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
\(=\left(\dfrac{\sqrt{7}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}+\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=\left(\sqrt{7}+\sqrt{5}\right).\left(\sqrt{7}-\sqrt{5}\right)\)
\(=7-5=2\)
d. \(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
\(=\left|2+\sqrt{3}\right|-\sqrt{5^2-2.5.\sqrt{3}+\left(\sqrt{3}\right)^2}\)
\(=\left|2+\sqrt{3}\right|-\left(5-\sqrt{3}\right)^2\)
\(=\left|2+\sqrt{3}\right|-\left|5-\sqrt{3}\right|\)
\(=2+\sqrt{3}-\left(5-\sqrt{3}\right)\) (vì \(\left|2+\sqrt{3}\right|\ge0,\left|5-\sqrt{3}\right|\ge0\))
\(=2+\sqrt{3}-5+\sqrt{3}\)
\(=2\sqrt{3}-3\)
rút gọn biểu thức
a.\(2\sqrt{80}+3\sqrt{45}-\sqrt{245}\)
b.\(\dfrac{3}{2+\sqrt{3}}+\dfrac{13}{4-\sqrt{3}}+\dfrac{6}{\sqrt{3}}\)
c.\(\left(\dfrac{\sqrt{14}-\sqrt{7}}{\sqrt{2}-1}+\dfrac{\sqrt{15}-\sqrt{5}}{\sqrt{3}-1}\right):\dfrac{1}{\sqrt{7}-\sqrt{5}}\)
d.\(\sqrt{\left(2+\sqrt{3}\right)^2}-\sqrt{28-10\sqrt{3}}\)
Lời giải:
a.
\(=2\sqrt{4^2.5}+3\sqrt{3^2.5}-\sqrt{7^2.5}=2.4\sqrt{5}+3.3\sqrt{5}-7\sqrt{5}\)
\(=8\sqrt{5}+9\sqrt{5}-7\sqrt{5}=10\sqrt{5}\)
b.
\(=\frac{3(2-\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}+\frac{13(4+\sqrt{3})}{(4-\sqrt{3})(4+\sqrt{3})}+\frac{6\sqrt{3}}{3}\)
\(=\frac{6-3\sqrt{3}}{1}+\frac{13(4+\sqrt{3})}{13}+2\sqrt{3}=6-3\sqrt{3}+4+\sqrt{3}+2\sqrt{3}\)
\(=10\)
c.
\(=\left[\frac{\sqrt{7}(\sqrt{2}-1)}{\sqrt{2}-1}+\frac{\sqrt{5}(\sqrt{3}-1)}{\sqrt{3}-1}\right].(\sqrt{7}-\sqrt{5})\)
\(=(\sqrt{7}+\sqrt{5})(\sqrt{7}-\sqrt{5})=7-5=2\)
d.
\(=|2+\sqrt{3}|-\sqrt{5^2-2.5\sqrt{3}+3}=|2+\sqrt{3}|-\sqrt{(5-\sqrt{3})^2}\)
\(=|2+\sqrt{3}|-|5-\sqrt{3}|=2+\sqrt{3}-(5-\sqrt{3})=-3+2\sqrt{3}\)
Cho hai biểu thức:
A= \(\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)}^2\)
B= \(\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\left(x>0;x\ne9\right)\)
a) Rút gọn A,B
b) Tìm các giá trị của x để A>B?
Help !!!
a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{2^2\cdot7}-\sqrt{3^2\cdot7}+\dfrac{\sqrt{7}\cdot\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=2\sqrt{7}-3\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1\)
\(=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\left[\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{2\sqrt{x}}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{2\cdot4}{\sqrt{x}-3}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\) khi
\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)
\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)
\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)
Rút gọn các biểu thức sau:
a. \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}\) - \(\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
b.\(\dfrac{1}{4-3\sqrt{2}}\) - \(\dfrac{1}{4+3\sqrt{2}}\)
c.\(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right)\): \(\sqrt{28}\)
d.\(\dfrac{3}{\sqrt{6}-\sqrt{3}}\)+\(\dfrac{4}{\sqrt{7}+\sqrt{3}}\)
a: Ta có: \(\dfrac{8}{\left(\sqrt{5}+\sqrt{3}\right)^2}-\dfrac{8}{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\dfrac{8}{8+2\sqrt{15}}-\dfrac{8}{8-2\sqrt{15}}\)
\(=\dfrac{64-16\sqrt{15}-64-16\sqrt{15}}{4}\)
\(=\dfrac{-32\sqrt{15}}{4}=-8\sqrt{15}\)
b: Ta có: \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}\)
\(=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{-2}\)
\(=-\dfrac{6\sqrt{2}}{2}=-3\sqrt{2}\)
b) \(\dfrac{1}{4-3\sqrt{2}}-\dfrac{1}{4+3\sqrt{2}}=\dfrac{4+3\sqrt{2}-4+3\sqrt{2}}{\left(4-3\sqrt{2}\right)\left(4+3\sqrt{2}\right)}=\dfrac{6\sqrt{2}}{-2}=-3\sqrt{2}\)
c) \(\left(\dfrac{\sqrt{7}+3}{\sqrt{7}-3}-\dfrac{\sqrt{7}-3}{\sqrt{7}+3}\right):\sqrt{28}=\dfrac{\left(\sqrt{7}+3\right)^2-\left(\sqrt{7}-3\right)^2}{\left(\sqrt{7}-3\right)\left(\sqrt{7}+3\right)}:\sqrt{28}=\dfrac{16+6\sqrt{7}-16+6\sqrt{7}}{7-9}=\dfrac{12\sqrt{7}}{-2}=-6\sqrt{7}\)