Tìm x
\(\sqrt{7}\)X - \(\sqrt{28}\)= \(\sqrt{63}+\sqrt{7}\)
\(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\) (ĐK x>0; x\(\ne9\))
a)Rút gọn A và B
b) Tìm các giá trị của x để giá trị biểu thức A lớn hơn giá trị biểu thức B
a) \(A=\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=2\sqrt{7}-3\sqrt{7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=-\sqrt{7}+\sqrt{7}+1-\sqrt{7}-1=-\sqrt{7}\)
\(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}=\dfrac{2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}.\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
\(=\dfrac{8}{\sqrt{x}-3}\)
b) \(A>B\Rightarrow-\sqrt{7}>\dfrac{8}{\sqrt{x}-3}\Rightarrow\dfrac{8}{\sqrt{x}-3}+\sqrt{7}< 0\)
\(\Rightarrow\dfrac{\sqrt{7x}+8-3\sqrt{7}}{\sqrt{x}-3}< 0\)
Ta có: \(\left\{{}\begin{matrix}8=\sqrt{64}\\3\sqrt{7}=\sqrt{63}\end{matrix}\right.\Rightarrow8-3\sqrt{7}>0\Rightarrow8-3\sqrt{7}+\sqrt{7x}>0\)
\(\Rightarrow\sqrt{x}-3< 0\Rightarrow\sqrt{x}< 3\Rightarrow x< 9\Rightarrow0< x< 9\)
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
\(\dfrac{8}{\sqrt{x}-3}< -\sqrt{7}\)
\(\Leftrightarrow8< -\sqrt{7x}+3\sqrt{7}\)
\(\Leftrightarrow x< \dfrac{\left(3\sqrt{7}-8\right)^2}{7}\)
\(\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(\sqrt{28}-\sqrt{63}+\dfrac{7+\sqrt{7}}{\sqrt{7}}-\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(=\sqrt{4.7}-\sqrt{9.7}+\dfrac{\sqrt{7}\left(\sqrt{7}+1\right)}{\sqrt{7}}-\left|\sqrt{7}+1\right|\)
\(=2\sqrt{7}-3\sqrt{7}+\left(\sqrt{7}+1\right)-\left(\sqrt{7}+1\right)=-\sqrt{7}\)
a) Rút gọn biểu thức: $A=\sqrt{28}+\sqrt{63}-2 \sqrt{7}$
b) Chứng minh rằng: $\dfrac{x \sqrt{y}+y \sqrt{x}}{\sqrt{x y}}: \dfrac{1}{\sqrt{x}-\sqrt{y}}=x-y$ với $x>0 ; y>0 ; x \neq y$
a) \(A=2\sqrt{7}+3\sqrt{7}-2\sqrt{7}\\ A=3\sqrt{7}\)
b) \(\Leftrightarrow\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ x-y=x-y\)
a) \(A=2\sqrt{7}+3\sqrt{7}-2\sqrt{7}\\ A=3\sqrt{7}\)
b) \(\Leftrightarrow\dfrac{\left(x\sqrt{y}+y\sqrt{x}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}+xy-xy-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ \dfrac{x\sqrt{xy}-y\sqrt{xy}}{\sqrt{xy}}=x-y\\ x-y=x-y\)
rút gọn
A=\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)
B=\(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}}\)
C=\(\left(\sqrt{7}-\sqrt{10}\right)^2+\sqrt{280}\)
D=\(\dfrac{\sqrt{99}}{\sqrt{11}}+\sqrt{7}\cdot\sqrt{63}-\sqrt{\sqrt{81}}\)
E=\(\sqrt{27}\left(s-\sqrt{5}\right)^2\cdot\left(3\sqrt{48}\right)\)
giải chi tiết ra giúp mik nha,cảm ơn nhiều
\(2\sqrt{112}-\dfrac{7}{6}\sqrt{252}-5\sqrt{63+3\sqrt{28}}\)
\(=2.4\sqrt{7}-\dfrac{7}{6}.6\sqrt{7}-5.3\sqrt{7}+3.2\sqrt{7}\)
\(=8\sqrt{7}-7\sqrt{7}-15\sqrt{7}+6\sqrt{7}\)
\(=-8\sqrt{7}\)
tính
1.\(\sqrt{147}+\sqrt{54}-4\sqrt{27}\)
2.\(\sqrt{28}-4\sqrt{63}+7\sqrt{112}\)
3.\(\sqrt{49}-5\sqrt{28}+\dfrac{1}{2}\sqrt{63}\)
4.\(\left(2\sqrt{6}-4\sqrt{3}-\dfrac{1}{4}\sqrt{8}\right).3\sqrt{6}\)
5.(\(2\sqrt{1\dfrac{9}{16}}-5\sqrt{5\dfrac{1}{16}}\)):\(\sqrt{16}\)
6.\(\left(\sqrt{48}-3\sqrt{27}-\sqrt{147}\right):\sqrt{3}\)
7.\(\left(\sqrt{50}-3\sqrt{49}\right):\sqrt{2}-\sqrt{162}:\sqrt{2}\)
8.\(\left(2\sqrt{1\dfrac{9}{10}}-\sqrt{5\dfrac{1}{10}}\right):\sqrt{10}\)
9.\(2\sqrt{\dfrac{16}{3}}-3\sqrt{\dfrac{1}{27}}-6\sqrt{\dfrac{4}{75}}\)
10.\(2\sqrt{27}-6\sqrt{\dfrac{4}{3}}+\dfrac{3}{5}\sqrt{75}\)
11.\(\dfrac{\sqrt{18}}{\sqrt{2}}-\dfrac{\sqrt{12}}{\sqrt{3}}\)
12.\(\dfrac{\sqrt{27}}{\sqrt{3}}+\dfrac{\sqrt{98}}{\sqrt{2}}-\sqrt{175}:\sqrt{7}\)
13.\(\left(\dfrac{\sqrt{8}}{\sqrt{2}}-\dfrac{\sqrt{180}}{\sqrt{5}}\right).\sqrt{5}-\sqrt{\dfrac{81}{11}}.\sqrt{11}\)
14.\(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)
15.\(\left(3\sqrt{2}-2\sqrt{3}\right)\left(3\sqrt{2}+2\sqrt{3}\right)\)
16.\(\left(1+\sqrt{5}-\sqrt{3}\right)\left(1+\sqrt{5}+\sqrt{3}\right)\)
cho hai biểu thức
A = \(\sqrt{63}-\sqrt{28}-\sqrt{\left(\sqrt{7}-1\right)^2}\)
B = \(\left(\frac{1}{\sqrt{x}+3}+\frac{1}{\sqrt{x}-3}\right).\frac{4.\left(\sqrt{x}+3\right)}{\sqrt{x}}\)
rút gọn biểu thức A và B
tìm giá trị của x để giá trị biểu thức B bằng giá trị biểu thức A
\(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
a: \(2\sqrt{8\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\sqrt{12}}\)
\(=2\sqrt{4\cdot2\sqrt{3}}-\sqrt{2\sqrt{3}}-\sqrt{9\cdot2\sqrt{3}}\)
\(=4\sqrt{2\sqrt{3}}-\sqrt{2\sqrt{3}}-3\sqrt{2\sqrt{3}}\)
=0
b: \(\sqrt{3}+\sqrt{7-4\sqrt{3}}\)
\(=\sqrt{3}+\sqrt{\left(2-\sqrt{3}\right)^2}\)
\(=\sqrt{3}+\left|2-\sqrt{3}\right|\)
\(=\sqrt{3}+2-\sqrt{3}\)
=2
c: \(\sqrt{\left(\sqrt{7}-4\right)^2}-\sqrt{28}+\sqrt{63}\)
\(=\left|\sqrt{7}-4\right|-2\sqrt{7}+3\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
d: \(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
\(=\dfrac{\sqrt{10}\left(15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\right)}{\sqrt{10}}\)
\(=15\sqrt{5}+5\sqrt{20}-3\sqrt{45}\)
\(=15\sqrt{5}+5\cdot2\sqrt{5}-3\cdot3\sqrt{5}\)
\(=16\sqrt{5}\)
e: \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
\(=\sqrt{3}-2\cdot4\sqrt{3}+3\cdot5\sqrt{3}-4\cdot6\sqrt{3}\)
\(=\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
\(=-16\sqrt{3}\)