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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) $2\sqrt{3}.(\sqrt{3}-\sqrt{2})=6-2\sqrt{6}.$

b) $\frac{ab}{|ab|}\sqrt{1+a^2{b}^2}$. Rút gọn hơn, ta có kết quả

+) $ab>0$ thì $ab\sqrt{1+\frac{1}{a^2{b}^2}}=\sqrt{1+a^2{b}^2}$.

+) $ab<0$ thì $ab\sqrt{1+\frac{1}{a^2{b}^2}}=-\sqrt{1+a^2{b}^2}$.
c) $\frac{1}{b^2}\sqrt{ab+a}$.
d) Cách 1.

$\frac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\frac{(a+\sqrt{ab})(\sqrt{a}-\sqrt{b})}{(\sqrt{a}+\sqrt{b})(\sqrt{a}-\sqrt{b})}$.

$=\frac{a\sqrt{a}+\sqrt{a^{2}b}-a\sqrt{b}-\sqrt{{ab}^2}}{a-b}=\frac{\sqrt{a}(a-b)}{a-b}=\sqrt{a}$

+ Ta có:

2√6−√5=2(√6+√5)(√6−√5)(√6+√5)26−5=2(6+5)(6−5)(6+5)

                   =2(√6+√5)(√6)2−(√5)2=2(√6+√5)6−5=2(6+5)(6)2−(5)2=2(6+5)6−5

                   =2(√6+√5)1=2(√6+√5)=2(6+5)1=2(6+5).

+ Ta có:

3√10+√7=3(√10−√7)(√10+√7)(√10−√7)310+7=3(10−7)(10+7)(10−7)

                    =3(√10−√7)(√10)2−(√7)2=3(10−7)(10)2−(7)2=3(√10−√7)10−7=3(10−7)10−7

                    =3(√10−√7)3=√10−√7=3(10−7)3=10−7.

+ Ta có:

1√x−√y=1.(√x+√y)(√x−√y)(√x+√y)1x−y=1.(x+y)(x−y)(x+y)

=√x+√y(√x)2−(√y)2=√x+√yx−y=x+y(x)2−(y)2=x+yx−y

+ Ta có:

2ab√a−√b=2ab(√a+√b)(√a−√b)(√a+√b)2aba−b=2ab(a+b)(a−b)(a+b)

=2ab(√a+√b)(√a)2−(√b)2=2ab(√a+√b)a−b=2ab(a+b)(a)2−(b)2=2ab(a+b)a−b.

\(\frac{2}{\sqrt{6}-\sqrt{5}}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{\left(\sqrt{6}-\sqrt{5}\right)\left(\sqrt{6}+\sqrt{5}\right)}=\frac{2\left(\sqrt{6}+\sqrt{5}\right)}{6-5}=2\left(\sqrt{6}+\sqrt{5}\right)\)

\(\frac{3}{\sqrt{10}+\sqrt{7}}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{\left(\sqrt{10}-\sqrt{7}\right)\left(\sqrt{10}+\sqrt{7}\right)}=\frac{3\left(\sqrt{10}-\sqrt{7}\right)}{10-7}=\sqrt{10}-\sqrt{7}\)

\(\frac{1}{\sqrt{x}-\sqrt{y}}=\frac{\sqrt{x}+\sqrt{y}}{x-y}\)

\(\frac{2ab}{\sqrt{a}-\sqrt{b}}=\frac{2ab\left(\sqrt{a}+\sqrt{b}\right)}{a-b}\)

An chạy châm nhất !!!!!!!!!!!!!!!

làm bài 2 trước :v bài 1 nhìn k hiểu lắm ._.

a) Với m = -2 (1) <=> x² - 2x - 3 = 0

Dễ thấy (1) có a - b + c = 0 nên có hai nghiệm x₁ = -1 ; x₂ = -c/a = 3

Vậy với m = -2 thì (1) có hai nghiệm x₁ = -1 ; x₂ = 3

b) Δ = b² - 4ac = m² + 12 ≥ 12 > 0 ∀ m

hay (1) luôn có hai nghiệm phân biệt ( đpcm )

c) Theo Viète ta có : \(\hept{\begin{cases}x_1+x_2=-\frac{b}{a}=-m\\x_1x_2=\frac{c}{a}=-3\end{cases}}\)

\(\frac{1}{x_1}+\frac{1}{x_2}=-3\left(ĐKXĐ:x_1,x_2\ne0\right)\)

\(\Leftrightarrow\frac{x_1+x_2}{x_1x_2}=-3\)

\(\Rightarrow x_1+x_2=-3x_1x_2\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2+3x_1x_2=0\)

\(\Leftrightarrow\left(x_1+x_2\right)^2+x_1x_2=0\)

\(\Leftrightarrow m^2-3=0\Leftrightarrow m=\pm\sqrt{3}\)

Vậy ...

+ Ta có:

$\frac{3}{\sqrt{3}+1}=\frac{3(\sqrt{3}-1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=\frac{3\sqrt{3}-3.1}{(\sqrt{3}{)}^2-1^2}$

$=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}$.

+ Ta có:

$\frac{2}{\sqrt{3}-1}=\frac{2(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}=\frac{2(\sqrt{3}+1)}{(\sqrt{3}{)}^2-1^2}$

$=\frac{2(\sqrt{3}+1)}{3-1}=\frac{2(\sqrt{3}+1)}{2}=\sqrt{3}+1$.

+ Ta có:

$\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{(2+\sqrt{3}).(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}=\frac{(2+\sqrt{3}{)}^2}{2^2-(\sqrt{3}{)}^2}$

$=\frac{2^2+\mathrm{2.2.}\sqrt{3}+(\sqrt{3}{)}^2}{4-3}$$=\frac{4+4\sqrt{3}+3}{1}=\frac{(4+3)+4\sqrt{3}}{1}$

$=\frac{7+4\sqrt{3}}{1}=7+4\sqrt{3}$.

+ Ta có:

$\frac{b}{3+\sqrt{b}}=\frac{b(3-\sqrt{b})}{(3+\sqrt{b})(3-\sqrt{b})}$

$=\frac{b(3-\sqrt{b})}{3^2-(\sqrt{b}{)}^2}=\frac{b(3-\sqrt{b})}{9-b};(b\ne 9)$.

+ Ta có:

$\frac{p}{2\sqrt{p}-1}=\frac{p(2\sqrt{p}+1)}{(2\sqrt{p}-1)(2\sqrt{p}+1)}$

$=\frac{p(2\sqrt{p}+1)}{(2\sqrt{p}{)}^2-1^2}=\frac{p(2\sqrt{p}+1)}{4p-1}$$=\frac{2p\sqrt{p}+p}{4p-1}$

$\frac{\mathrm{Bài\; 51\; trang\; 30\; SGK\; Toán\; 9\; tập\; 1\; -\; loigiaihay.com}}{}$

#Ye Chi-Lien

\(\frac{3}{\sqrt{3}+1}=\frac{3\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\frac{3\sqrt{3}-3}{3-1}=\frac{3\sqrt{3}-3}{2}\)

\(\frac{2}{\sqrt{3}-1}=\frac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}=\frac{2\left(\sqrt{3}+1\right)}{3-1}=\sqrt{3}-1\)

\(\frac{2+\sqrt{3}}{2-\sqrt{3}}=\frac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=4-3}=\left(2+\sqrt{3}\right)^2=4+4\sqrt{3}+3=7+4\sqrt{3}\)

\(\frac{b}{3+\sqrt{b}}=\frac{b\left(3-\sqrt{b}\right)}{\left(3+\sqrt{b}\right)\left(3-\sqrt{b}\right)}=\frac{b\left(3-\sqrt{b}\right)}{9-b}\)

\(\frac{p}{2\sqrt{p}-1}=\frac{p\left(2\sqrt{p}+1\right)}{\left(2\sqrt{p}-1\right)\left(2\sqrt{b}+1\right)}=\frac{p\left(2\sqrt{b}+1\right)}{4p-1}\)

\(\frac{5}{\sqrt{10}}=\frac{5\sqrt{10}}{10}=\frac{\sqrt{10}}{2}\)

\(\frac{5}{2\sqrt{5}}=\frac{10\sqrt{5}}{20}=\frac{\sqrt{5}}{2}\)

\(\frac{1}{3\sqrt{20}}=\frac{3\sqrt{20}}{180}=\frac{\sqrt{20}}{60}=\frac{2\sqrt{5}}{60}=\frac{\sqrt{5}}{30}\)

\(\frac{2\sqrt{2}+2}{5\sqrt{2}}=\frac{10\sqrt{2}\left(\sqrt{2}+1\right)}{50}=\frac{20+10\sqrt{2}}{50}=\frac{10\left(2+\sqrt{2}\right)}{50}=\frac{2+\sqrt{2}}{5}\)

\(\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{y\left(\sqrt{y}+b\right)}{by}=\frac{\sqrt{y}+b}{b}\)

+ Ta có: 

$\frac{5}{\sqrt{10}}=\frac{5.\sqrt{10}}{\sqrt{10}.\sqrt{10}}=\frac{5\sqrt{10}}{(\sqrt{10}{)}^2}=\frac{5\sqrt{10}}{10}$

$=\frac{5.\sqrt{10}}{5.2}$$=\frac{\sqrt{10}}{2}$.

+ Ta có:

$\frac{5}{2\sqrt{5}}=\frac{5.\sqrt{5}}{2\sqrt{5}.\sqrt{5}}=\frac{5\sqrt{5}}{2.(\sqrt{5}.\sqrt{5})}=\frac{5\sqrt{5}}{2(\sqrt{5}{)}^2}$

$=\frac{5\sqrt{5}}{2.5}=\frac{\sqrt{5}}{2}$.

+ Ta có:

$\frac{1}{3\sqrt{20}}=\frac{1.\sqrt{20}}{3\sqrt{20}.\sqrt{20}}=\frac{\sqrt{20}}{3.(\sqrt{20}.\sqrt{20})}=\frac{\sqrt{20}}{3.(\sqrt{20}{)}^2}$

              $=\frac{\sqrt{20}}{3.20}=\frac{\sqrt{2^2.5}}{60}=\frac{2\sqrt{5}}{60}=\frac{2\sqrt{5}}{2.30}=\frac{\sqrt{5}}{30}$.

+ Ta có:

$\frac{(2\sqrt{2}+2)}{5.\sqrt{2}}=\frac{(2\sqrt{2}+2).\sqrt{2}}{5\sqrt{2}.\sqrt{2}}=\frac{2\sqrt{2}.\sqrt{2}+2.\sqrt{2}}{5.(\sqrt{2}{)}^2}$

                    $=\frac{2.2+2\sqrt{2}}{5.2}=\frac{2(2+\sqrt{2})}{5.2}=\frac{2+\sqrt{2}}{5}$.

+ Ta có:

 $\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{(y+b\sqrt{y}).\sqrt{y}}{b\sqrt{y}.\sqrt{y}}=\frac{y\sqrt{y}+b\sqrt{y}.\sqrt{y}}{b.(\sqrt{y}{)}^2}$

                    $=\frac{y\sqrt{y}+b(\sqrt{y}{)}^2}{by}=\frac{y\sqrt{y}+by}{by}$

                    $=\frac{y(\sqrt{y}+b)}{b.y}=\frac{\sqrt{y}+b}{b}$.

Cách khác:

$\frac{y+b\sqrt{y}}{b\sqrt{y}}=\frac{{\left(\sqrt{y}\right)}^2+b\sqrt{y}}{b\sqrt{y}}$$=\frac{\sqrt{y}\left(\sqrt{y}+b\right)}{b\sqrt{y}}=\frac{\sqrt{y}+b}{b}$

Nguồn : Bài 50 trang 30 SGK Toán 9 tập 1 - loigiaihay.com

#Ye Chi-Lien

(do xy > 0 (gt) nên đưa thừa số xy vào trong căn để khử mẫu)

#Học tốt!!!

\(ab\cdot\sqrt{\dfrac{a}{b}}=a\cdot\sqrt{ab}\)

\(\dfrac{a}{b}\cdot\sqrt{\dfrac{b}{a}}=\dfrac{\sqrt{a\cdot b}}{b}\)

\(\sqrt{\dfrac{1}{b}+\dfrac{1}{b^2}}=\dfrac{\sqrt{b+1}}{b}\)

\(\sqrt{\dfrac{9\cdot a^3}{36\cdot b}}=\dfrac{\sqrt{a^3\cdot b}}{2\cdot b}\)

\(3\cdot x\cdot y\cdot\sqrt{\dfrac{2}{x\cdot y}}=3\cdot\sqrt{2\cdot x\cdot y}\)