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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\\ b,=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\\ c,=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=3\sqrt{2}\left(\sqrt{3}-\sqrt{2}\right)-3\sqrt{6}\)
=3căn 6-6-3căn 6=-6
b: \(=\dfrac{a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}-\sqrt{a}\)
\(=\dfrac{a+\sqrt{ab}-a+\sqrt{ab}}{\sqrt{a}-\sqrt{b}}=\dfrac{2\sqrt{ab}}{\sqrt{a}-\sqrt{b}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\\ b,=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
LG a
√18(√2−√3)2;18(2−3)2;
Phương pháp giải:
+ √ab=√a.√bab=a.b, với a, b≥0a, b≥0.
+ |a|=a|a|=a, nếu a≥0a≥0
|a|=−a|a|=−a nếu a<0a<0.
+ Sử dụng định lí so sánh hai căn bậc hai số học: Với hai số a, ba, b không âm, ta có:
a<b⇔√a<√ba<b⇔a<b
Lời giải chi tiết:
Ta có:
√18(√2−√3)2=√18.√(√2−√3)218(2−3)2=18.(2−3)2
=√9.2.|√2−√3|=√32.2.|√2−√3|=9.2.|2−3|=32.2.|2−3|
=3√2.|√2−√3|=3√2(√3−√2)=32.|2−3|=32(3−2)
=3√2.3−3(√2)2=32.3−3(2)2
=3√6−3.2=3√6−6=36−3.2=36−6.
(Vì 2<3⇔√2<√3⇔√2−√3<02<3⇔2<3⇔2−3<0
Do đó: |√2−√3|=−(√2−√3)=−√2+√3|2−3|=−(2−3)=−2+3=√3−√2=3−2).
LG b
ab√1+1a2b2ab1+1a2b2
Phương pháp giải:
+ √ab=√a.√bab=a.b, với a, b≥0a, b≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ |a|=a|a|=a, nếu a≥0a≥0
|a|=−a|a|=−a nếu a<0a<0.
Lời giải chi tiết:
Ta có:
ab√1+1a2b2=ab√a2b2a2b2+1a2b2=ab√a2b2+1a2b2ab1+1a2b2=aba2b2a2b2+1a2b2=aba2b2+1a2b2
=ab√a2b2+1√a2b2=ab√a2b2+1√(ab)2=aba2b2+1a2b2=aba2b2+1(ab)2
=ab√a2b2+1|ab|=aba2b2+1|ab|
Nếu ab>0ab>0 thì |ab|=ab|ab|=ab
⇒ab√a2b2+1|ab|=ab√a2b2+1ab=√a2b2+1⇒aba2b2+1|ab|=aba2b2+1ab=a2b2+1.
Nếu ab<0ab<0 thì |ab|=−ab|ab|=−ab
⇒ab√a2b2+1|ab|=ab√a2b2+1−ab=−√a2b2+1⇒aba2b2+1|ab|=aba2b2+1−ab=−a2b2+1.
LG c
√ab3+ab4ab3+ab4
Phương pháp giải:
+ √ab=√a.√bab=a.b, với a, b≥0a, b≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ |a|=a|a|=a, nếu a≥0a≥0
|a|=−a|a|=−a nếu a<0a<0.
Lời giải chi tiết:
Ta có:
√ab3+ab4=√a.bb3.b+ab4=√abb4+ab4ab3+ab4=a.bb3.b+ab4=abb4+ab4
=√ab+ab4=√ab+a√(b2)2=√ab+a|b2|=√ab+ab2=ab+ab4=ab+a(b2)2=ab+a|b2|=ab+ab2.
(Vì b2>0b2>0 với mọi b≠0b≠0 nên |b2|=b2|b2|=b2).
LG d
a+√ab√a+√ba+aba+b
Phương pháp giải:
+ √ab=√a.√bab=a.b, với a, b≥0a, b≥0.
+ √ab=√a√bab=ab, với a≥0, b>0a≥0, b>0.
+ |a|=a|a|=a, nếu a≥0a≥0
|a|=−a|a|=−a nếu a<0a<0.
Lời giải chi tiết:
Ta có:
a+√ab√a+√b=(√a)2+√a.√b√a+√b=√a(√a+√b)√a+√ba+aba+b=(a)2+a.ba+b=a(a+b)a+b
=√a=a.
Cách khác:
a+√ab√a+√b=(a+√ab)(√a−√b)(√a+√b)(√a−√b)=a√a−a√b+√ab.√a−√ab.√b(√a)2−(√b)2=a√a−a√b+a√b−b√aa−b=a√a−b√aa−b=√a(a−b)a−b=√a
a)
b) . Rút gọn hơn, ta có kết quả
+) thì .
+) thì .
c) .
d) Cách 1.
.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\left(2+\sqrt{2}\right)\left(\sqrt{2}-1\right)}{2-1}=2\sqrt{2}-2+2-\sqrt{2}=\sqrt{2}\)
\(\dfrac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}=\dfrac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}=-\sqrt{5}\)
\(\dfrac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}=\dfrac{\sqrt{6}}{2}\)
\(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{\left(a-\sqrt{a}\right)\left(1+\sqrt{a}\right)}{1-a}=\dfrac{a+a\sqrt{a}-\sqrt{a}-a}{1-a}=\dfrac{\sqrt{a}\left(a-1\right)}{1-a}=-\sqrt{a}\)
\(\dfrac{p-2\sqrt{p}}{\sqrt{p}-2}=\dfrac{\sqrt{p}\left(\sqrt{p}-2\right)}{\sqrt{p}-2}=\sqrt{p}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(B=\left(\dfrac{4x}{\sqrt{x}-1}-\dfrac{\sqrt{x}-2}{x-3\sqrt{x}+2}\right)\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{\sqrt{x}-1}\cdot\dfrac{\sqrt{x}-1}{x^2}\)
\(=\dfrac{4x-1}{x^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
a) ĐS:
.
b) ĐS: Nếu
thì ![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?ab%5Csqrt%7B1+%5Cfrac%7B1%7D%7Ba%5E%7B2%7Db%5E%7B2%7D%7D%7D%3D%5Csqrt%7B%7Ba%5E%7B2%7Db%5E%7B2%7D+1%7D%7D.)
Nếu ab
c) ĐS:![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Csqrt%7Bab+a%7D%7D%7Bb%5E%7B2%7D%7D.)
d)![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cfrac%7Ba+%5Csqrt%7Bab%7D%7D%7B%5Csqrt%7Ba%7D+%5Csqrt%7Bb%7D%7D%3D%5Cfrac%7B%28a+%5Csqrt%7Bab%7D%29%28%5Csqrt%7Ba%7D-%5Csqrt%7Bb%7D%29%7D%7Ba-b%7D%3D%5Cfrac%7Ba%5Csqrt%7Ba%7D-a%5Csqrt%7Bb%7D+%5Csqrt%7Bab%7D%5Csqrt%7Ba%7D-%5Csqrt%7Bab%7D%5Csqrt%7Bb%7D%7D%7Ba-b%7D)
Nhận xét. Nhận thấy rằng để
có nghĩa thì
Do đó
. Vì thế có thể phân tích tử thành nhân tử.
a) ĐS:
.
b) ĐS: Nếu
thì ![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?ab%5Csqrt%7B1+%5Cfrac%7B1%7D%7Ba%5E%7B2%7Db%5E%7B2%7D%7D%7D%3D%5Csqrt%7B%7Ba%5E%7B2%7Db%5E%7B2%7D+1%7D%7D.)
Nếu ab
c) ĐS:![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cfrac%7B%5Csqrt%7Bab+a%7D%7D%7Bb%5E%7B2%7D%7D.)
d)![This is the rendered form of the equation. You can not edit this directly. Right click will give you the option to save the image, and in most browsers you can drag the image onto your desktop or another program.](http://latex.codecogs.com/gif.latex?%5Cfrac%7Ba+%5Csqrt%7Bab%7D%7D%7B%5Csqrt%7Ba%7D+%5Csqrt%7Bb%7D%7D%3D%5Cfrac%7B%28a+%5Csqrt%7Bab%7D%29%28%5Csqrt%7Ba%7D-%5Csqrt%7Bb%7D%29%7D%7Ba-b%7D%3D%5Cfrac%7Ba%5Csqrt%7Ba%7D-a%5Csqrt%7Bb%7D+%5Csqrt%7Bab%7D%5Csqrt%7Ba%7D-%5Csqrt%7Bab%7D%5Csqrt%7Bb%7D%7D%7Ba-b%7D)
Nhận xét. Nhận thấy rằng để
có nghĩa thì
Do đó
. Vì thế có thể phân tích tử thành nhân tử.