1)\(\sqrt{9a}\) - \(\sqrt{16a}\) +\(\sqrt{49a}\) (a \(\ge\) 0)
2)2\(\sqrt{40\sqrt{12}}\) - 2\(\sqrt{\sqrt{75}}\) - 3\(\sqrt{5\sqrt{80}}\)
3)\(\dfrac{\sqrt{X^3}-1}{\sqrt{X}-1}\)
Rút gọn:
2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\) với a \(\ge\) 0
4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\) với b \(\ge\) 0
2) \(\sqrt{98}-\sqrt{72}+0,5\sqrt{8}\)
\(=7\sqrt{2}-6\sqrt{2}+\sqrt{2}\)
\(=\left(7-6+1\right)\sqrt{2}\)
\(=2\sqrt{2}\)
3) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\)
\(=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
\(=\left(3-4+7\right)\sqrt{a}\)
\(=6\sqrt{a}\)
4) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)
\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
\(=4\sqrt{b}-5\sqrt{10b}\)
Bài 7: Rút Gọn Các Biểu Thức Sau
a. 5\(\sqrt{25^2}\) - 25x Với X<O
B \(\sqrt{49a^2}\) + 3a Với a \(\ge\) 0
C \(\sqrt{16a^4}\) + 6a\(^2\) Với a Bất Kì
d 3\(\sqrt{9a^6}\) - 6a\(^3\) với a bất kì
e 3\(\sqrt{9a^6}\) - 6a\(^3\) Với a\(\ge\) 0
f \(\sqrt{16a^{10}}\) + 6a\(^5\) với a \(\le0\)
b: B=căn 49a^2+3a
=|7a|+3a
=7a+3a(a>=0)
=10a
c: C=căn16a^4+6a^2
=4a^2+6a^2
=10a^2
d: \(D=3\cdot3\cdot\sqrt{a^6}-6a^3=6\cdot\left|a^3\right|-6a^3\)
TH1: a>=0
D=6a^3-6a^3=0
TH2: a<0
D=-6a^3-6a^3=-12a^3
e: \(E=3\sqrt{9a^6}-6a^3\)
\(=3\cdot\sqrt{\left(3a^3\right)^2}-6a^3\)
=3*3a^3-6a^3(a>=0)
=3a^3
f: \(F=\sqrt{16a^{10}}+6a^5\)
\(=\sqrt{\left(4a^5\right)^2}+6a^5\)
=-4a^5+6a^5(a<=0)
=2a^5
1, rút gọn
g, \(\sqrt{5a}\) - \(\sqrt{16a}\) + \(\sqrt{49a}\) (a>=0)
m, \(\dfrac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
g: \(=\sqrt{5a}-4\sqrt{a}+7\sqrt{a}\)
\(=\sqrt{5a}+3\sqrt{a}\)
b: \(=\dfrac{40}{6+2\sqrt{5}+2\cdot\sqrt{2+2\sqrt{5}}}\)
\(=\dfrac{40}{\left(\sqrt{5}+1\right)^2+\sqrt{2}\cdot\sqrt{4+4\sqrt{5}}}\)
\(=\dfrac{40}{\left(\sqrt{5}+1\right)^2+2\sqrt{2}\cdot\sqrt{\sqrt{5}+1}}\)
\(=\dfrac{40}{\left(\sqrt{\sqrt{5}+1}\right)\left[\left(\sqrt{\sqrt{5}+1}\right)^3+2\sqrt{2}\right]}\)
1. rút gọn
g, \(\sqrt{54a}\)+ \(\sqrt{16a}\)+ \(\sqrt{49a}\) (a>0)
m, \(\dfrac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
Rút Gọn Biểu Thức
\(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)
\(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\) (a ≥ 0)
\(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}\)
\(\left(a\sqrt{\dfrac{a}{b}+2\sqrt{ab}+b\sqrt{\dfrac{a}{b}}}\right)\sqrt{ab}\)
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}\right)\)
a: \(\dfrac{2}{\sqrt{3}-1}-\dfrac{2}{\sqrt{3}+1}\)
\(=\dfrac{2\left(\sqrt{3}+1\right)-2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\dfrac{2\sqrt{3}+2-2\sqrt{3}+2}{2}=\dfrac{4}{2}=2\)
b: \(\dfrac{\sqrt{12}-\sqrt{6}}{\sqrt{30}-\sqrt{15}}\)
\(=\dfrac{\sqrt{6}\left(\sqrt{2}-1\right)}{\sqrt{15}\left(\sqrt{2}-1\right)}\)
\(=\dfrac{\sqrt{6}}{\sqrt{15}}=\sqrt{\dfrac{6}{15}}=\sqrt{\dfrac{2}{5}}=\dfrac{\sqrt{10}}{5}\)
c: \(\sqrt{9a}+\sqrt{81a}+3\sqrt{25a}-16\sqrt{49a}\)
\(=3\sqrt{a}+9\sqrt{a}+3\cdot5\sqrt{a}-16\cdot7\sqrt{a}\)
\(=27\sqrt{a}-112\sqrt{a}=-85\sqrt{a}\)
d: \(\dfrac{ab-bc}{\sqrt{ab}-\sqrt{bc}}=\dfrac{\left(\sqrt{ab}-\sqrt{bc}\right)\left(\sqrt{ab}+\sqrt{bc}\right)}{\sqrt{ab}-\sqrt{bc}}\)
\(=\sqrt{ab}+\sqrt{bc}\)
e: \(a\left(\sqrt{\dfrac{a}{b}+2\sqrt{ab}+b\cdot\sqrt{\dfrac{a}{b}}}\right)\cdot\sqrt{ab}\)
\(=a\cdot\sqrt{\dfrac{a}{b}\cdot ab+2\sqrt{ab}\cdot ab+b\cdot\sqrt{\dfrac{a}{b}}\cdot ab}\)
\(=a\cdot\sqrt{a^2+2\cdot ab\cdot\sqrt{ab}+a\sqrt{a}\cdot b\sqrt{b}}\)
\(=a\cdot\sqrt{a^2+3\cdot a\cdot\sqrt{a}\cdot b\cdot\sqrt{b}}\)
e: ĐKXĐ: a>=0 và a<>1
\(\left(\dfrac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}\)
\(=\left(\dfrac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\dfrac{\left(\sqrt{a}+1\right)\left(a-\sqrt{a}+1\right)}{\sqrt{a}+1}\)
\(=\left(1+\sqrt{a}+\sqrt{a}+a\right)\cdot\left(a-\sqrt{a}+1\right)\)
\(=\left(\sqrt{a}+1\right)^2\cdot\left(a-\sqrt{a}+1\right)\)
1) thực hiện phép tính và rút gọn biểu thức
a) A=\(\sqrt{9a}-\sqrt{16a}-\sqrt{49a}\) với \(a\ge0\)
b) B=\(\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
lm nhanh giúp mk nhé mk đang cần gấp
a) \(A=\sqrt{9a}-\sqrt{16a}-\sqrt{49a}=3\sqrt{a}-4\sqrt{a}-7\sqrt{a}=-8\sqrt{a}\)
b) \(B=\dfrac{3+2\sqrt{3}}{\sqrt{3}}+\dfrac{2+\sqrt{2}}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=\dfrac{\sqrt{3}\left(2+\sqrt{3}\right)}{\sqrt{3}}+\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}}-\left(\sqrt{3}+\sqrt{2}\right)\)
\(=2+\sqrt{3}+\sqrt{2}+1-\sqrt{3}-\sqrt{2}=3\)
1. rút gọn
a, \(\sqrt{54a}\) - \(\sqrt{16a}\) + \(\sqrt{49a}\) (a>0)
m, \(\dfrac{20}{3+\sqrt{5}+\sqrt{2+2\sqrt{5}}}\)
nếu câu a sai thì hãy làm câu b nhé
Bài 1: Thực hiện phép tính
a) \(\dfrac{1}{2}\sqrt{48}-\sqrt{32}-\sqrt{75}\)\(-\dfrac{1}{5}\sqrt{50}\)
b) \(\dfrac{3+\sqrt{3}}{3-\sqrt{3}}+\dfrac{3-\sqrt{3}}{3+\sqrt{3}}\)
c) \(4\sqrt{\dfrac{3}{2}}-\dfrac{5}{2}\sqrt{24}+\dfrac{1}{2}\sqrt{50}\)
d) \(\left(2\sqrt{5}+5\sqrt{2}\right).\sqrt{5}-\sqrt{250}\)
Bài 2: Rút gọn biểu thức sau
\(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}\) với \(a\ge0\)
Bài 3: Cho biểu thức sau
A=\(\left(\dfrac{\sqrt{x}}{\sqrt{x}-a}+\dfrac{\sqrt{x}}{\sqrt{x}+2}\right).\dfrac{4-x}{2\sqrt{x}}\)với \(x>0\)và \(x\ne4\)
a) Rút gọn A b) Tìm x để A=-3
Bài 4: Rút gọn biểu thức sau
A=\(\left(\dfrac{1}{\sqrt{x}-1}+\dfrac{1}{1+\sqrt{x}}\right):\dfrac{1}{x-1}\) với \(x\ge0\) và \(x\ne1\)
Bài 5: Cho biểu thức
C= \(\left(\dfrac{2+\sqrt{a}}{2-\sqrt{a}}-\dfrac{2-\sqrt{a}}{2+\sqrt{a}}-\dfrac{4a}{a-4}\right):\left(\dfrac{2}{2-\sqrt{a}}-\dfrac{\sqrt{a}+3}{2\sqrt{a}-a}\right)\)
a) Rút gọn C b) Timg giá trị của a để C>0 c) Tìm giá trị của a để C=-1
Bài 6: Giải phương trình
a) \(2\sqrt{3}-\sqrt{4+x^2}=0\\\)
b) \(\sqrt{16x+16}-\sqrt{9x+9}=1\)
c) \(3\sqrt{2x}+5\sqrt{8x}-20-\sqrt{18x}=0\)
d) \(\sqrt{4\left(x+2\right)^2}=8\)
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
Rút gọn các biểu thức sau:
\(a.A=2\sqrt{3}-\sqrt{75}+2\sqrt{12}\)
\(b.B=\sqrt{\left(2-\sqrt{5}\right)^2}+\sqrt{\left(3-\sqrt{5}\right)^2}\)
\(c.C=\left(\dfrac{x+2\sqrt{x}}{x-2\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right).\dfrac{1}{\sqrt{x}+1}\) (x > 0;x ≠ 4)