Ôn tập chương 1: Căn bậc hai. Căn bậc ba

-2x^2-9<=-9<0 với mọi x

=>\(\sqrt{-2x^2-9}\) không xác định với mọi x

Do đó: Phương trình này vô nghiệm

giải gúp em bài 12 với

12:

a: =>\(\sqrt{\left(x-3\right)^2}=3\)

=>|x-3|=3

=>x-3=3 hoặc x-3=-3

=>x=6 hoặc x=0

b: \(\Leftrightarrow\left|x-4\right|=x+2\)

=>(x-4)^2=(x+2)^2 và x+2>=0

=>x>=-2 và x^2-8x+16=x^2+4x+4

=>x>=-2 và -8x-4x=4-16=-12

=>x=1(nhận)

c: =>|x+3|=3x-6

=>3x-6>=0và (3x-6)^2=(x+3)^2

=>x>=2 và (3x-6-x-3)(3x-6+x+3)=0

=>x>=2 và (2x-9)(4x-3)=0

=>x=9/2

d: =>|x-2|=2x-5

=>x>=5/2 và (2x-5)^2=(x-2)^2

=>x>=5/2 và (2x-5-x+2)(2x-5+x-2)=0

=>x>=5/2 và (x-3)(3x-7)=0

=>x=3

Bài 11:

a) \(\sqrt{x-1}+\sqrt{4x-4}-\sqrt{25x-25}+2=0\) (ĐK: \(x\ge1\))

\(\Leftrightarrow\sqrt{x-1}+2\sqrt{x-1}-5\sqrt{x-1}+2=0\)

\(\Leftrightarrow-2\sqrt{x-1}+2=0\)

\(\Leftrightarrow2\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{x-1}=1\)

\(\Leftrightarrow x-1=1\)

\(\Leftrightarrow x=2\left(tm\right)\)

b) \(\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}=16\) (ĐK: \(x\ge-1\))

\(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}=16\)

\(\Leftrightarrow4\sqrt{x+1}=16\)

\(\Leftrightarrow\sqrt{x+1}=4\)

\(\Leftrightarrow x+1=16\)

\(\Leftrightarrow x=15\left(tm\right)\)

c) \(\sqrt{4x+20}+\sqrt{x+5}-\dfrac{1}{3}\sqrt{9x+45}=4\) (ĐK: \(x\ge-5\)

\(\Leftrightarrow2\sqrt{x+5}+\sqrt{x+5}-\dfrac{1}{3}\cdot3\sqrt{x+5}=4\)

\(\Leftrightarrow3\sqrt{x+5}-\sqrt{x+5}=4\)

\(\Leftrightarrow2\sqrt{x+5}=4\)

\(\Leftrightarrow\sqrt{x+5}=2\)

\(\Leftrightarrow x+5=4\)

\(\Leftrightarrow x=-1\left(tm\right)\)

d) \(\dfrac{1}{3}\sqrt{2x}-\sqrt{8x}+\sqrt{18x}-10=2\) (ĐK: \(x\ge0\))

\(\Leftrightarrow\dfrac{1}{3}\sqrt{2x}-2\sqrt{2x}+3\sqrt{2x}=2+10\)

\(\Leftrightarrow\left(\dfrac{1}{3}-2+3\right)\sqrt{2x}-12\)

\(\Leftrightarrow\dfrac{4}{3}\sqrt{2x}=12\)

\(\Leftrightarrow\sqrt{2x}=9\)

\(\Leftrightarrow2x=81\)

\(\Leftrightarrow x=\dfrac{81}{2}\left(tm\right)\)

giải giúp em bài 10+11

8:

c: \(=2\cdot\dfrac{4}{\sqrt{3}}-3\cdot\dfrac{1}{3\sqrt{3}}-6\cdot\dfrac{2}{5\sqrt{3}}\)

\(=\dfrac{8}{\sqrt{3}}-\dfrac{1}{\sqrt{3}}-\dfrac{12}{5\sqrt{3}}\)

\(=\dfrac{7}{\sqrt{3}}-\dfrac{12\sqrt{3}}{15}\)

\(=\dfrac{35\sqrt{3}-12\sqrt{3}}{15}=\dfrac{23\sqrt{3}}{15}\)

d: \(=\sqrt{\dfrac{\left(2-\sqrt{3}\right)^2}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}}+\sqrt{\dfrac{\left(2+\sqrt{3}\right)^2}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}}\)

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

\(=2-\sqrt{3}+2+\sqrt{3}\)

=4

9:

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

\(=\dfrac{3\sqrt{5}+5-3-\sqrt{5}}{2\left(\sqrt{5}+1\right)}=\dfrac{2\sqrt{5}+2}{2\left(\sqrt{5}+1\right)}\)

=1

b: \(=\dfrac{4\left(\sqrt{3}+1\right)}{2}+\dfrac{\sqrt{3}+1}{2}-\dfrac{6\left(3+\sqrt{3}\right)}{6}\)

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

\(=2\sqrt{3}+2-\dfrac{1}{2}\sqrt{3}-\dfrac{5}{2}\)

\(=\dfrac{3}{2}\sqrt{3}-\dfrac{1}{2}\)

c: \(=\sqrt{4-2\sqrt{2}-1}=\sqrt{3-2\sqrt{2}}=\sqrt{2}-1\)

d: \(=\dfrac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)

\(=\dfrac{\sqrt{2}}{2+\sqrt{3}+1}+\dfrac{\sqrt{2}}{2-\sqrt{3}+1}\)

\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}=\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)+\sqrt{2}\left(3+\sqrt{3}\right)}{9-3}\)

\(=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{6}\)

\(=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

giúp em câu c và d bài 8 và bài 9

8:

a: \(=\dfrac{\sqrt{20}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-\dfrac{8}{\sqrt{5}-1}\)

\(=\sqrt{20}-\dfrac{8\left(\sqrt{5}+1\right)}{4}\)

\(=2\sqrt{5}-2\left(\sqrt{5}+1\right)=2\sqrt{5}-2\sqrt{5}-2=-2\)

b: \(=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\)

\(=\dfrac{-2}{\sqrt{6}}-\dfrac{1}{\sqrt{6}}=\dfrac{-3}{\sqrt{6}}=\dfrac{-3\sqrt{6}}{6}=\dfrac{-\sqrt{6}}{2}\)

giúp em bài 8 câu a và b

7:

a: \(\dfrac{3}{\sqrt{5}}=\dfrac{3\sqrt{5}}{5}\)

b: \(\dfrac{2\sqrt{3}}{\sqrt{2}}=\dfrac{2\sqrt{3}\cdot\sqrt{2}}{2}=\sqrt{6}\)

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

d: \(\dfrac{1}{\sqrt{3}+\sqrt{2}}=\dfrac{\sqrt{3}-\sqrt{2}}{3-2}=\sqrt{3}-\sqrt{2}\)

e: \(\dfrac{\sqrt{2}+1}{\sqrt{2}-1}=\dfrac{\left(\sqrt{2}+1\right)^2}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}=\left(\sqrt{2}+1\right)^2=3+2\sqrt{2}\)

g: \(\dfrac{3\sqrt{2}}{\sqrt{3}+1}=\dfrac{3\sqrt{2}\left(\sqrt{3}-1\right)}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{3\sqrt{6}-3\sqrt{2}}{2}\)

giúp em làm bài 7 bài 8 câu a và b với ạ thanks

6:

a: \(=\dfrac{20\cdot2\sqrt{3}-15\cdot3\sqrt{3}}{5\sqrt{3}}=\dfrac{40\sqrt{3}-45\sqrt{3}}{5\sqrt{3}}\)

\(=-\dfrac{5\sqrt{3}}{5\sqrt{3}}=-1\)

b: \(=\dfrac{5\sqrt{3}+9\sqrt{3}-4\sqrt{3}}{\sqrt{3}}=5+9-4=10\)

c: \(\left(\sqrt{2}+1\right)^2+\left(\sqrt{2}-1\right)^2\)

\(=3+2\sqrt{2}+3-2\sqrt{2}\)

=6

d: \(=\left(2\sqrt{7}-2\sqrt{3}+\sqrt{7}\right)\cdot\sqrt{7}+2\sqrt{21}\)

\(=\left(3\sqrt{7}-2\sqrt{3}\right)\cdot\sqrt{7}+2\sqrt{21}\)

\(=21-2\sqrt{21}+2\sqrt{21}=21\)

giải giúp em bài 6 ạ

6. a) -1

    b) 10

    c) 6

    d) 21

\(B=tan^267^0-cot^223^0+2\cdot\left(sin^216^0+cos^216^0\right)-2\)

\(=0+2\cdot1-2=0\)

\(A=cot67\cdot tan67-2\left(\dfrac{\sqrt{2}}{2}\cdot sin64\right)^2-2\cdot\dfrac{sin23}{3\cdot sin23}-sin^226^0\)

\(=1-2\cdot\dfrac{1}{2}\cdot sin^264^0-\dfrac{2}{3}-sin^226^0\)

\(=1-1-\dfrac{2}{3}=-\dfrac{2}{3}\)

\(\sqrt{4x-5}+3=0\)

=>\(\sqrt{4x-5}=-3\)(vô lý)

Vậy: Phương trình vô nghiệm