登录
3. 计算:$(\sqrt {a^{3}b}-3ab+\sqrt {ab^{3}})÷(\sqrt {ab})(a＞0,b＞0).$
答案
$ 3. a + b - 3 \sqrt { a b } $
4. 计算.
(1)$(π-3)^{0}+|1-\sqrt {3}|+(-\frac {1}{2})^{-2};$
(2)$\sqrt {\frac {1}{2}}×\sqrt {12}-\sqrt {48}÷\sqrt {2}-3\sqrt {\frac {2}{3}}.$
(1)$(π-3)^{0}+|1-\sqrt {3}|+(-\frac {1}{2})^{-2};$
(2)$\sqrt {\frac {1}{2}}×\sqrt {12}-\sqrt {48}÷\sqrt {2}-3\sqrt {\frac {2}{3}}.$
答案
$ 4. $ (1) $ 4 + \sqrt { 3 } $ (2) $ - 2 \sqrt { 6 } $
5. 若$\sqrt {a+b-1}+(a-3)^{2}=\sqrt {x-3}· \sqrt {3-x}$,则$b^{a}+x^{a}$的值为
19
.答案
$ 5. 19 $
6. 阅读下列材料,并回答问题.
$\frac {1}{\sqrt {2}+1}=\frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}=\frac {\sqrt {2}-1}{(\sqrt {2})^{2}-1}=\sqrt {2}-1;$
$\frac {1}{\sqrt {3}+\sqrt {2}}=\frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}=\frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3})^{2}-(\sqrt {2})^{2}}=\sqrt {3}-\sqrt {2};$
$\frac {1}{\sqrt {4}+\sqrt {3}}=\frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}=\frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4})^{2}-(\sqrt {3})^{2}}=\sqrt {4}-\sqrt {3};$
……
(1)$\frac {1}{\sqrt {7}+\sqrt {6}}=$
(2)观察上述算式，请写出算式$\frac {1}{\sqrt {n+1}+\sqrt {n}}$(n 是正整数)的结果;
(3)试比较$\sqrt {12}-\sqrt {11}$与$\sqrt {11}-\sqrt {10}$的大小;
(4)计算:$\frac {1}{\sqrt {2}+1}+\frac {1}{\sqrt {3}+\sqrt {2}}+\frac {1}{\sqrt {4}+\sqrt {3}}+... +\frac {1}{\sqrt {2025}+\sqrt {2024}}$(提示:$45^{2}=2025$).
$\frac {1}{\sqrt {2}+1}=\frac {\sqrt {2}-1}{(\sqrt {2}+1)(\sqrt {2}-1)}=\frac {\sqrt {2}-1}{(\sqrt {2})^{2}-1}=\sqrt {2}-1;$
$\frac {1}{\sqrt {3}+\sqrt {2}}=\frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3}+\sqrt {2})(\sqrt {3}-\sqrt {2})}=\frac {\sqrt {3}-\sqrt {2}}{(\sqrt {3})^{2}-(\sqrt {2})^{2}}=\sqrt {3}-\sqrt {2};$
$\frac {1}{\sqrt {4}+\sqrt {3}}=\frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4}+\sqrt {3})(\sqrt {4}-\sqrt {3})}=\frac {\sqrt {4}-\sqrt {3}}{(\sqrt {4})^{2}-(\sqrt {3})^{2}}=\sqrt {4}-\sqrt {3};$
……
(1)$\frac {1}{\sqrt {7}+\sqrt {6}}=$
$\sqrt { 7 } - \sqrt { 6 }$
;(2)观察上述算式，请写出算式$\frac {1}{\sqrt {n+1}+\sqrt {n}}$(n 是正整数)的结果;
(3)试比较$\sqrt {12}-\sqrt {11}$与$\sqrt {11}-\sqrt {10}$的大小;
(4)计算:$\frac {1}{\sqrt {2}+1}+\frac {1}{\sqrt {3}+\sqrt {2}}+\frac {1}{\sqrt {4}+\sqrt {3}}+... +\frac {1}{\sqrt {2025}+\sqrt {2024}}$(提示:$45^{2}=2025$).
答案
$ 6. $ (1) $ \sqrt { 7 } - \sqrt { 6 } $ (2) $ \sqrt { n + 1 } - \sqrt { n } $ (3) $ \sqrt { 12 } - \sqrt { 11 } ＜ \sqrt { 11 } - \sqrt { 10 } $ (4) $ 44 $
7. (1)下列各组可以合并二次根式的是(
A.$\sqrt {2x}$与$\sqrt {2y}$
B.$\sqrt {\frac {8}{9}a^{3}b^{4}}$与$\sqrt {\frac {9}{2}a^{5}b^{8}}$
C.$\sqrt {mn}$与$\sqrt {n}$
D.$\sqrt {m+n}$与$\sqrt {m-n}$
(2)互为有理化因式:两个二次根式的乘积可以运用平方差公式$(a+b)(a-b)=a^{2}-b^{2}$求得,同时它们的积是有理数,不含有二次根式.如$x+1-\sqrt {x^{2}+2x}$与$x+1+\sqrt {x^{2}+2x}$就是互为有理化因式,$\sqrt {x}$与$\frac {1}{\sqrt {x}}$也是互为有理化因式.
$\sqrt {2}+\sqrt {3}$的有理化因式是
$x-\sqrt {y}$的有理化因式是
$-\sqrt {x+1}-\sqrt {x-1}$的有理化因式是
(3)分母有理化:把分母中的根号化去,通常在分子、分母上同乘以一个二次根式,达到化去分母中根号的目的.
把下列各式的分母有理化:
①$\frac {1}{\sqrt {5}-1}$; ②$\frac {1}{1+2\sqrt {3}}$; ③$\frac {\sqrt {2}}{\sqrt {6}-\sqrt {2}}.$
(4)规定:$a\otimes b=\sqrt {a}· \sqrt {b}+\sqrt {\frac {a}{b}},a※b=ab-b^{2}$.则$(2\otimes 4)※\sqrt {2}=$
B
)A.$\sqrt {2x}$与$\sqrt {2y}$
B.$\sqrt {\frac {8}{9}a^{3}b^{4}}$与$\sqrt {\frac {9}{2}a^{5}b^{8}}$
C.$\sqrt {mn}$与$\sqrt {n}$
D.$\sqrt {m+n}$与$\sqrt {m-n}$
(2)互为有理化因式:两个二次根式的乘积可以运用平方差公式$(a+b)(a-b)=a^{2}-b^{2}$求得,同时它们的积是有理数,不含有二次根式.如$x+1-\sqrt {x^{2}+2x}$与$x+1+\sqrt {x^{2}+2x}$就是互为有理化因式,$\sqrt {x}$与$\frac {1}{\sqrt {x}}$也是互为有理化因式.
$\sqrt {2}+\sqrt {3}$的有理化因式是
$\sqrt { 2 } - \sqrt { 3 }$
;$x-\sqrt {y}$的有理化因式是
$x + \sqrt { y }$
;$-\sqrt {x+1}-\sqrt {x-1}$的有理化因式是
$ - \sqrt { x + y } + \sqrt { x - 1 }$
.(3)分母有理化:把分母中的根号化去,通常在分子、分母上同乘以一个二次根式,达到化去分母中根号的目的.
把下列各式的分母有理化:
①$\frac {1}{\sqrt {5}-1}$; ②$\frac {1}{1+2\sqrt {3}}$; ③$\frac {\sqrt {2}}{\sqrt {6}-\sqrt {2}}.$
(4)规定:$a\otimes b=\sqrt {a}· \sqrt {b}+\sqrt {\frac {a}{b}},a※b=ab-b^{2}$.则$(2\otimes 4)※\sqrt {2}=$
3
.答案
$ 7. $ (1) $ \mathrm { B } $ (2) $ \sqrt { 2 } - \sqrt { 3 } ; x + \sqrt { y } ; - \sqrt { x + y } + \sqrt { x - 1 } $ (3) $ \frac { \sqrt { 5 } + 1 } { 4 } ; \frac { 2 \sqrt { 3 } - 1 } { 11 } ; \frac { \sqrt { 3 } + 1 } { 2 } $ (4) $ 3 $
