rút gọn biểu thức
a) \(\sqrt{\left(1-\sqrt{2}\right)^2}\)
b)\(\sqrt{\left(\sqrt{3}-2\right)^2}\)
c)\(\sqrt{\left(5-\sqrt{3}\right)^2}\)
Thực hiện phép tính (rút gọn biểu thức)
a)\(\sqrt{\left(3+\sqrt{2}\right)^2}\)-\(\sqrt{\left(3-2\sqrt{2}\right)^2}\)
b) \(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}\)-\(\sqrt{\left(\sqrt{7}+2\sqrt{2}\right)^2}\)
c)\(\sqrt{\left(3+\sqrt{5}\right)^2}\)+\(\sqrt{\left(3-\sqrt{5}\right)^2}\)
d) \(\sqrt{\left(2-\sqrt{3}\right)^2}\)-\(\sqrt{\left(2+\sqrt{3}\right)^2}\)
Lời giải:
a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$
$=3\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$
$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$
$=-2\sqrt{7}$
c.
$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$
d.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
rút gọn biểu thức chưa căn thức bậc hai:
1,\(\sqrt{\left(1-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{2}+3\right)^2}\)
2, \(\sqrt{\left(\sqrt{3}-2\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}\)
3,\(\sqrt{\left(\sqrt{5}-3\right)^2}+\sqrt{\left(\sqrt{5}-2\right)^2}\)
4,\(\sqrt{\left(3+\sqrt{2}\right)^2}+\sqrt{\left(3-\sqrt{2}\right)^2}\)
5,\(\sqrt{\left(2-\sqrt{3}\right)^2}-\sqrt{\left(2+\sqrt{3}\right)^2}\)
\(1,=\left|1-\sqrt{2}\right|+\left|\sqrt{2}+3\right|\\ =1-\sqrt{2}+3+\sqrt{2}\\ =4\\ 2,=\left|\sqrt{3}-2\right|+\left|\sqrt{3}-1\right|\\ =\sqrt{3}-2+\sqrt{3}-1\\ =2\sqrt{3}-3\\ 3,=\left|\sqrt{5}-3\right|+\left|\sqrt{5}-2\right|\\ =\sqrt{5}-3+\sqrt{5}-2\\ =2\sqrt{5}-5\\ 4,=\left|3+\sqrt{2}\right|+\left|3-\sqrt{2}\right|\\ =3+\sqrt{2}+\sqrt{3}-\sqrt{2}\\ =3+\sqrt{3}\\ 5,=\left|2-\sqrt{3}\right|-\left|2+\sqrt{3}\right|\\ =2-\sqrt{3}-\left(2+\sqrt{3}\right)\\ =2-\sqrt{3}-2-\sqrt{3}\\ =-2\sqrt{3}\)
Rút gọn các biểu thức sau:
A= \(3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
B= \(\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
C= \(3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
D= \(\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
E= \(\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(A=3\left(x+2\sqrt{x}\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x+6\sqrt{x}-\left(x-1\right)\)
\(=3x+6\sqrt{x}-x+1\)
\(=2x+6\sqrt{x}+1\)
\(B=\left(\sqrt{x}+1\right)\left(\sqrt{x}+3\right)-2\left(\sqrt{x}-1\right)^2\)
\(=x+3\sqrt{x}+\sqrt{x}+3-2\left(x-2\sqrt{x}+1\right)\)
\(=x+4\sqrt{x}+3-2x+4\sqrt{x}-2\)
\(=-x+8\sqrt{x}+1\)
\(C=3x-3\sqrt{x}-2+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=3x-3\sqrt{x}-2+\left(\sqrt{x^2}-1\right)\)
\(=3x-3\sqrt{x}-2+x-1\)
\(=4x-3\sqrt{x}-3\)
\(D=\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)-\left(2\sqrt{x}+1\right)\left(\sqrt{x}-2\right)\)
\(=x-9-\left(2x-3\sqrt{x}-2\right)\)
\(=x-9-2x+3\sqrt{x}+2\)
\(=-x+3\sqrt{x}-7\)
\(E=\left(\sqrt{x}+4\right)\left(\sqrt{x}-4\right)-2\left(2\sqrt{x}-1\right)\left(\sqrt{x}+2\right)\)
\(=\sqrt{x^2}-2^2-2\left(2x+4\sqrt{x}-\sqrt{x}-2\right)\)
\(=x-4-2\left(2x+3\sqrt{x}-2\right)\)
\(=x-4-4x-6\sqrt{x}+4\)
\(=-3-6\sqrt{x}\)
Rút gọn biểu thức:
\(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(B=3\sqrt{\left(1,5\right)^2}-4\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)
\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)
\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)
\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\)
Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
Khai triển và rút gọn các biểu thức sau:
a) \({\left( {2 + \sqrt 2 } \right)^4}\)
b) \({\left( {2 + \sqrt 2 } \right)^4} + {\left( {2 - \sqrt 2 } \right)^4}\)
c) \({\left( {1 - \sqrt 3 } \right)^5}\)
a) Áp dụng công thức nhị thức Newton, ta có
\(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\\ = \left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right] + \left[ {{{4.2}^3}.\left( {\sqrt 2 } \right) + 4.2.{{\left( {\sqrt 2 } \right)}^3}} \right]\\ = 68 + 48\sqrt 2 \end{array}\)
b) Áp dụng công thức nhị thức Newton, ta có
\({\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\)
\({\left( {2 - \sqrt 2 } \right)^4} = \left( {2 +(- \sqrt 2 )} \right)^4= {2^4} + {4.2^3}.\left( { - \sqrt 2 } \right) + {6.2^2}.{\left( { - \sqrt 2 } \right)^2} + 4.2.{\left( { - \sqrt 2 } \right)^3} + {\left( { - \sqrt 2 } \right)^4}\)
Từ đó,
\(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} + {\left( {2 - \sqrt 2 } \right)^4} = 2\left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right]\\ = 2\left( {16 + 48 + 4} \right) = 136\end{array}\)
c) Áp dụng công thức nhị thức Newton, ta có
\(\begin{array}{l}{\left( {1 - \sqrt 3 } \right)^5} = \left( {1 +(- \sqrt 3 )} \right)^5= 1 + 5.\left( { - \sqrt 3 } \right) + 10.{\left( { - \sqrt 3 } \right)^2} + 10.{\left( { - \sqrt 3 } \right)^3} + 5.{\left( { - \sqrt 3 } \right)^4} + 1.{\left( { - \sqrt 3 } \right)^5}\\ = \left[ {1 + 10.{{\left( { - \sqrt 3 } \right)}^2} + 5.{{\left( { - \sqrt 3 } \right)}^4}} \right] + \left[ {5.\left( { - \sqrt 3 } \right) + 10.{{\left( { - \sqrt 3 } \right)}^3} + 1.{{\left( { - \sqrt 3 } \right)}^5}} \right]\\ = 76 - 44\sqrt 3 \end{array}\)
rút gọn biểu thức A=\(\dfrac{\left(2-\sqrt{a}\right)-\left(\sqrt{a+3}\right)}{1+2\sqrt{a}}\) (với a>0) ; B=\(\dfrac{1}{1-\sqrt{2}+\sqrt{3}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\); C=\(\dfrac{1}{\sqrt{5-2}}+\dfrac{1}{\sqrt{5+\sqrt{2}}}\)
\(A=\dfrac{2-\sqrt{a}-\sqrt{a}-3}{2\sqrt{a}+1}=-1\)
\(B=\dfrac{1}{1-\sqrt{2+\sqrt{3}}}-\dfrac{1}{1-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}-1}-\dfrac{\sqrt{2}}{\sqrt{2}-\sqrt{3}+1}\)
\(=\dfrac{2-\sqrt{6}+\sqrt{2}-2+\sqrt{6}+\sqrt{2}}{5-2\sqrt{6}-1}\)
\(=\dfrac{2\sqrt{2}}{4-2\sqrt{6}}=\dfrac{1}{\sqrt{2}-\sqrt{3}}=-\sqrt{2}-\sqrt{3}\)
Rút gọn các biểu thức sau:
d) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\)
\(=2\left|3-\sqrt{2}\right|+\sqrt{18}-5.1=6-2\sqrt{2}+3\sqrt{2}-5\)
\(=1+\sqrt{2}\)
rút gọn biểu thức
\(D=\left(\sqrt{a}\right)^7.\left(\sqrt[3]{a}\right).\left(\sqrt[4]{a}\right)^7\) (a>0)
\(D=a^{\sqrt{2}-1}.\left(a^2\right)^{\sqrt{2}}.\left(a^3\right)^{1-\sqrt{2}}\)
\(D=a^{\dfrac{7}{2}}.a^{\dfrac{1}{3}}.a^{\dfrac{7}{4}}=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}=\sqrt[12]{a^{67}}\)
\(D=a^{\sqrt{2}-1}.a^{2\sqrt{2}}.a^{3-3\sqrt{2}}=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{3}}=a^2\)
\(D=\left(\sqrt{a}\right)^7\cdot\left(\sqrt[3]{a}\right)\left(\sqrt[4]{a}\right)^7\)
\(=a^{\dfrac{1}{2}\cdot7}\cdot a^{\dfrac{1}{3}}\cdot a^{\dfrac{1}{4}\cdot7}\)
\(=a^{\dfrac{7}{2}+\dfrac{1}{3}+\dfrac{7}{4}}=a^{\dfrac{67}{12}}\)
b: \(D=a^{\sqrt{2}-1}\cdot\left(a^2\right)^{\sqrt{2}}\cdot\left(a^3\right)^{1-\sqrt{2}}\)
\(=a^{\sqrt{2}-1}\cdot a^{2\sqrt{2}}\cdot a^{3-3\sqrt{2}}\)
\(=a^{\sqrt{2}-1+2\sqrt{2}+3-3\sqrt{2}}=a^2\)
rút gọn các biểu thức sau:
a) \(\sqrt{\left(2-\sqrt{3}\right)^2}\)
b) \(\sqrt{\left(3-\sqrt{11}\right)^2}\)
c) \(2\sqrt{a^2}\)với a ≥ 0
d) 3\(\sqrt{\left(a-2\right)^2}\)với a < 0
\(a,=\left|2-\sqrt{3}\right|=2-\sqrt{3}\\ b,=\left|3-\sqrt{11}\right|=\sqrt{11}-3\\ c,=2\left|a\right|=2a\\ d,=3\left|a-2\right|=3\left(2-a\right)\left(a< 0\Leftrightarrow a-2< 0\right)\)
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: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)