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Bài 1
a) √81a - √36a - √144a = 9√a - 6√a - 12√a = -9√a
b) √75 - √48 - √300 = 5√3 - 4√3 - 10√3 = -9√3
Bài 2
a) √2x-3 = 7
⇒ 2x-3 = 49 ⇔ 2x = 52 ⇔ x =26
c) √16x - √9x = 2
⇔ 4√x - 3√x = 2 ⇔ √x = 2 ⇔ x = 4
Bài 3
a) √(2-√5)2 = l 2-√5 l = √5-2
b) (a - 3)2 + (a - 9)
= a2 - 6a + 9 + a - 9 = a2 - 5a
c) A=\(\dfrac{2\sqrt{x}}{\sqrt{x}+3}+\dfrac{\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+3}{x-9}:\left(\dfrac{2\sqrt{x}-2}{\sqrt{x}-3}-1\right)\)
=\(\left(\dfrac{2\sqrt{x}\left(\sqrt{x}-3\right)+\sqrt{x}\left(\sqrt{x}+3\right)-3x-3}{x-9}\right):\left(\dfrac{2\sqrt{x}-2-\sqrt{x}+3}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{2x-6\sqrt{x}+x+3\sqrt{x}-3x-3}{x-9}\right):\left(\dfrac{\sqrt{x}+1}{\sqrt{x}-3}\right)\)
=\(\left(\dfrac{-3\sqrt{x}-3}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\left(\dfrac{-3\left(\sqrt{x}+1\right)}{x-9}\right).\left(\dfrac{\sqrt{x}-3}{\sqrt{x}+1}\right)\)
=\(\dfrac{-3\sqrt{x}+9}{x-9}\)
a) \(\sqrt{3+2\sqrt{2}}+\sqrt{\left(\sqrt{2}-2\right)^2}\)
\(=\sqrt{\left(\sqrt{2}\right)^2+2\sqrt{2}\cdot1+1^2}+\left|\sqrt{2}-2\right|\)
\(=\sqrt{\left(\sqrt{2}+1\right)^2}-\left(\sqrt{2}-2\right)\)
\(=\left|\sqrt{2}+1\right|-\sqrt{2}+2\)
\(=\sqrt{2}+1-\sqrt{2}+2\)
\(=3\)
b) \(\dfrac{1}{5}\sqrt{50}-2\sqrt{96}-\dfrac{\sqrt{30}}{\sqrt{15}}+12\sqrt{\dfrac{1}{6}}\)
\(=\dfrac{1}{5}\cdot5\sqrt{2}-2\cdot4\sqrt{6}-\sqrt{\dfrac{30}{15}}+\sqrt{\dfrac{144}{6}}\)
\(=\sqrt{2}-8\sqrt{6}-\sqrt{2}+2\sqrt{6}\)
\(=-8\sqrt{6}+2\sqrt{6}\)
\(=-6\sqrt{6}\)
c) \(\left(\dfrac{5-\sqrt{5}}{\sqrt{5}}-2\right)\left(\dfrac{4}{1+\sqrt{5}}+4\right)\)
\(=\left[\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}-2\right]\left[\dfrac{4\left(1-\sqrt{5}\right)}{\left(1+\sqrt{5}\right)\left(1-\sqrt{5}\right)}+4\right]\)
\(=\left(\sqrt{5}-1-2\right)\left(\dfrac{4\left(1-\sqrt{5}\right)}{1-5}+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}-1+4\right)\)
\(=\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)\)
\(=\left(\sqrt{5}\right)^2-3^2\)
\(=-4\)
a) \(\sqrt[]{3+2\sqrt[]{2}}+\sqrt[]{\left(\sqrt[]{2}-2\right)^2}\)
\(=\sqrt[]{2+2\sqrt[]{2}.1+1}+\left|\sqrt[]{2}-2\right|\)
\(=\sqrt[]{\left(\sqrt[]{2}+1\right)^2}+\left(2-\sqrt[]{2}\right)\) \(\left(\left(\sqrt[]{2}\right)^2=2< 2^2=4\right)\)
\(=\left|\sqrt[]{2}+1\right|+2-\sqrt[]{2}\)
\(=\sqrt[]{2}+1+2-\sqrt[]{2}\)
\(=3\)
`a)sqrt{(sqrt7-4)^2}+sqrt7`
`=|sqrt7-4|+sqrt7`
`=4-sqrt7+sqrt7=4`
`b)\sqrt{81a}-sqrt{144a}+sqrt{36a}(a>=0)`
`=9sqrta-12sqrta+6sqrta=3sqrta`
a) Ta có: \(\sqrt{\left(\sqrt{7}-4\right)^2}+\sqrt{7}\)
\(=4-\sqrt{7}+\sqrt{7}\)
=4
b) Ta có: \(\sqrt{81a}-\sqrt{144a}+\sqrt{36a}\)
\(=9\sqrt{a}-12\sqrt{a}+6\sqrt{a}\)
\(=3\sqrt{a}\)
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
b: Ta có: \(B=\left(\dfrac{\sqrt{x}+2}{x+2\sqrt{x}+1}-\dfrac{\sqrt{x}-2}{x-1}\right)\cdot\left(\dfrac{x\sqrt{x}-1}{\sqrt{x}-1}+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)-\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)^2\cdot\left(\sqrt{x}-1\right)}\cdot\left(x+\sqrt{x}+1+\sqrt{x}\right)\)
\(=\dfrac{x+\sqrt{x}-2-x+\sqrt{x}+2}{\sqrt{x}-1}\)
\(=\dfrac{2\sqrt{x}}{\sqrt{x}-1}\)
Bài 6:
a: \(\Leftrightarrow\sqrt{x^2+4}=\sqrt{12}\)
=>x^2+4=12
=>x^2=8
=>\(x=\pm2\sqrt{2}\)
b: \(\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}=1\)
=>x+1=1
=>x=0
c: \(\Leftrightarrow3\sqrt{2x}+10\sqrt{2x}-3\sqrt{2x}-20=0\)
=>\(\sqrt{2x}=2\)
=>2x=4
=>x=2
d: \(\Leftrightarrow2\left|x+2\right|=8\)
=>x+2=4 hoặcx+2=-4
=>x=-6 hoặc x=2
\(A=-\dfrac{3+\sqrt{5}+3-\sqrt{5}}{\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)}\cdot\dfrac{\sqrt{5}}{5}\\ A=\dfrac{-6}{4}\cdot\dfrac{\sqrt{5}}{5}=\dfrac{-3\sqrt{5}}{10}\)
a, \(\sqrt{75}+\sqrt{48}-\sqrt{300}\)
\(=5\sqrt{3}+4\sqrt{3}-10\sqrt{3}\)
\(=-\sqrt{3}\)
b, \(\sqrt{81a}-\sqrt{36a}+\sqrt{144a}\)
\(=9\sqrt{a}-6\sqrt{a}+12\sqrt{a}\)
\(=15\sqrt{a}\)
c, \(\dfrac{4}{\sqrt{5}-2}-\dfrac{4}{\sqrt{5}+2}\)
\(=\dfrac{4\sqrt{5}+8-4\sqrt{5}+8}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{16}{5-4}=16\)
d, \(\dfrac{a\sqrt{b}-b\sqrt{a}}{\sqrt{a}-\sqrt{b}}=\dfrac{\sqrt{ab}\left(\sqrt{a}-\sqrt{b}\right)}{\sqrt{a}-\sqrt{b}}=\sqrt{ab}\)
Nguyễn Huy Tú anh sinh năm 2004 là lên lớp 8 mà sao lại tl được bài lớp 9