Đưa thừa số ra ngoài dấu căn:
a.\(\sqrt{49.\text{3}\text{6}0}\)
b.\(\sqrt{\text{1}2\text{5}a^2}\) với a<0
c.\(-\sqrt{\text{5}00.\text{1}\text{6}2}\)
d.\(\frac{\text{1}}{\text{3}}\sqrt{22\text{5}a^2}\) với a>0
Đưa thừa số ra ngoài dấu căn :
a. \(\sqrt{9\text{6}}.\sqrt{\text{1}2\text{5}}\)
b.\(\sqrt{a^4.\text{6}^{\text{5}}}\)
c.\(\sqrt{a^{\text{6}}.b^{\text{1}\text{1}}}\)
d.\(\:\sqrt{a^{\text{3}}\left(\text{1}-a\right)^4}\)
a) Ta có: \(\sqrt{96}\cdot\sqrt{125}\)
\(=\sqrt{16}\cdot\sqrt{6}\cdot\sqrt{25}\cdot\sqrt{5}\)
\(=20\cdot\sqrt{30}\)
b) Ta có: \(\sqrt{a^4\cdot6^5}\)
\(=a^2\cdot36\cdot\sqrt{6}\)
c) Ta có: \(\sqrt{a^6\cdot b^{11}}\)
\(=\sqrt{a^6}\cdot\sqrt{b^{11}}\)
\(=\left|a^3\right|\cdot\left|b^5\right|\cdot\sqrt{b}\)
\(=a^3b^5\cdot\sqrt{b}\)
d) Ta có: \(\sqrt{a^3\left(1-a\right)^4}\)
\(=\sqrt{a^3}\cdot\sqrt{\left(1-a\right)^4}\)
\(=a\sqrt{a}\cdot\left(1-a\right)^2\)
đưa thừa số ra ngoài dấu căn
a)\(0,1\sqrt{20000}\)
b)\(-0,05\sqrt{28800}\)
c)\(\sqrt{7.63a^2}\)
d)\(\sqrt{72a^2b^4}v\text{ới}a< 0\)
a/ \(0,1\sqrt{2.10000=0,1\sqrt{ }2.100^{ }2=0,1\cdot100\sqrt{ }2=10\sqrt{ }2}\)
b/ \(-0,05\sqrt{28800}=-0,05\sqrt{288\cdot100=-0,05\cdot10\sqrt{ }288=6\sqrt{ }2}\)
c/\(\sqrt{7\cdot63}a^2=\sqrt{7\cdot9\cdot7}a^2=21a^2\)
\(\sqrt{72a^{ }2b\sqrt{ }4=\sqrt{ }6\cdot9\left|\right|ab^{ }2=-3\sqrt{ }6ab^{ }2}\)
\(\sqrt{72a^{ }2b^{ }4}=\sqrt{6.9}\left|ab^{ }2\right|=3\sqrt{6}-\left(ab^{ }2\right)=-3\sqrt{6}ab^2\)
mk sữa lại câu d
\(\sqrt{48.45}\) Đưa thừa số ra ngoài dấu căn:
\(\sqrt{225.17}\)
\(\sqrt{a^3b^7}với\) \(a\ge0;b\ge0\)
\(\sqrt{x^5\left(x-3\right)^2}\) với \(x>0\)
\(\sqrt{48\cdot45}=12\sqrt{15}\\ \sqrt{225\cdot17}=15\sqrt{17}\\ \sqrt{a^3b^7}=\left|ab^3\right|\sqrt{ab}=ab^3\sqrt{ab}\\ \sqrt{x^5\left(x-3\right)^2}=\left|x^2\left(x-3\right)\right|\sqrt{x}=x^2\left(x-3\right)\sqrt{x}\)
\(\sqrt{48\cdot45}=4\sqrt{3}\cdot3\sqrt{5}=12\sqrt{15}\)
\(\sqrt{225\cdot17}=15\sqrt{17}\)
Đưa một thừa số vào trong dấu căn.
a)$-\dfrac{2}{3} \sqrt{ab}$ với $a>0, b \geq 0 \text {; }$
b) $a \sqrt{\frac{3}{a}}$ với $a>0, b \geq 0 \text {; }$
c) $a\sqrt{7}$ với $\mathrm{a} \geq 0;$
d) $b \sqrt{3}$ với $b<0;$
e) $a b \sqrt{\dfrac{a}{b}}$ với $b \geq 0, a>0;$
f) $a b \sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$ với $a>0 , b>0$.
a, \(-\frac{2}{3}\sqrt{ab}=-\sqrt{\frac{4ab}{9}}\)
b, \(a\sqrt{\frac{3}{a}}=\sqrt{\frac{3a^2}{a}}=\sqrt{3a}\)
c, \(a\sqrt{7}=\sqrt{7a^2}\)
d, \(b\sqrt{3}=\sqrt{3b^2}\)
e, \(ab\sqrt{\frac{a}{b}}=\sqrt{\frac{a^3b^2}{b}}=\sqrt{a^3b}\)
f, \(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a^2b^2}{a}+\frac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)
a,\(-\sqrt{\dfrac{4}{9}ab}\)
d, trừ căn (3b^2)
b, căn 3a
c, căn 7a^2
e, căn của a^3b
f, căn của a^2b+ab^2
a) -√(4/9 *ab)
b) √3a
c) √7a²
d) √3b²
e) √a³b
f)√(ab²+a²b)
Đưa thừa số ra ngoài dấu căn:
a) $\sqrt{28 x^{4} y^{2}}$ với $y \leq 0$;
b) $\sqrt{63 a^{2} b^{4}}$ với $a \geq 0$;
c) $\sqrt{147(a-1)^{3}}$;
d) $\sqrt{192(y+2)^{5}}$.
a, -2x^2y căn 7
b, ab^2 căn 63
c, a-1 căn 147a-147
d, y+2 nhân căn [192 nhân (y+2)^3]
a)-2x²y√7
b) 3ab²√7
c) 7(a-1)√3(a-1)
d) 8(y+2)²√3(y+2)
1. CHỨNG MINH ĐẲNG THỨC
a. \(\text{[}3+2\sqrt{6}-\sqrt{33}\text{]}\cdot\text{[}\sqrt{22}+\sqrt{6}+4\text{]}=24\)
b. \(\text{[}\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\text{]}\cdot\text{[}15+2\sqrt{6}\text{]}\)
c.\(\text{[}\frac{4}{3}\cdot\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\text{]}\cdot\text{[}\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\text{]}=4\)
d. \(\sqrt{\text{[}1-\sqrt{1989}\text{]}^2}\cdot\sqrt{1990+2\sqrt{1989}}=1988\)
e. \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)với \(a>0;b>0\)và \(a\ne b\)
a) \(\left(3+1\sqrt{6}-\sqrt{33}\right)\left(\sqrt{22}+\sqrt{6}+4\right)\)
\(=\sqrt{3}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right).\sqrt{2}\left(\sqrt{11}+\sqrt{3}+2\sqrt{2}\right)\)
\(=\sqrt{6}\left(\sqrt{3}+2\sqrt{2}-\sqrt{11}\right)\left(\sqrt{3}+2\sqrt{2}+\sqrt{11}\right)\)
\(=\sqrt{6}\left[\left(\sqrt{3}+2\sqrt{2}\right)^2-11\right]=\sqrt{6}\left(11+4\sqrt{6}-11\right)=\sqrt{6}.4\sqrt{6}=6.4=24\)
b) \(\left(\frac{1}{5-2\sqrt{6}}+\frac{2}{5+2\sqrt{6}}\right)\left(15+2\sqrt{6}\right)=\left(\frac{5+2\sqrt{6}+10-4\sqrt{6}}{5^2-\left(2\sqrt{6}\right)^2}\right)\left(15+2\sqrt{6}\right)\)
\(=\left(15-2\sqrt{6}\right)\left(15+2\sqrt{6}\right)=15^2-24=201\)
C) \(\left(\frac{4}{3}.\sqrt{3}+\sqrt{2}+\sqrt{3\frac{1}{3}}\right)\left(\sqrt{1,2}+\sqrt{2}-4\sqrt{\frac{1}{5}}\right)\)
\(=\left(\frac{4}{\sqrt{3}}+\frac{\sqrt{6}}{\sqrt{3}}+\frac{\sqrt{10}}{\sqrt{3}}\right)\left(\frac{\sqrt{6}}{\sqrt{5}}+\frac{\sqrt{10}}{\sqrt{5}}-\frac{4}{\sqrt{5}}\right)\)
\(=\frac{1}{\sqrt{15}}\left(\sqrt{6}+\sqrt{10}+4\right)\left(\sqrt{6}+\sqrt{10}-4\right)=\frac{1}{\sqrt{15}}\left[\left(\sqrt{6}+\sqrt{10}\right)^2-16\right]\)
\(=\frac{1}{\sqrt{15}}\left(16+4\sqrt{15}-16\right)=\frac{4\sqrt{15}}{\sqrt{15}}=4\)
d) \(\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1990+2\sqrt{1989}}=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{1989+2\sqrt{1989}+1}\)
\(=\sqrt{\left(1-\sqrt{1989}\right)^2}.\sqrt{\left(\sqrt{1989}+1\right)^2}=\left(\sqrt{1989}-1\right)\left(\sqrt{1989}+1\right)=1989-1=1988\)
e) \(\frac{a-\sqrt{ab}+b}{a\sqrt{a}+b\sqrt{b}}-\frac{1}{a-b}=\frac{a-\sqrt{ab}+b}{\left(\sqrt{a}+\sqrt{b}\right)\left(a-\sqrt{ab}+b\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}}{\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)}-\frac{1}{a-b}=\frac{\sqrt{a}-\sqrt{b}-1}{a-b}\)
đưa thừa số ra ngoài dấu căn :
\(\sqrt{18b^3\left(1-2a\right)^2}\)( a≥\(\dfrac{1}{2}\); b ≥0)
\(\sqrt{18b^3\cdot\left(1-2a\right)^2}\)
\(=3\sqrt{2}\cdot b\sqrt{b}\cdot\left|1-2a\right|\)
\(=3\sqrt{2}\left(2a-1\right)\cdot b\sqrt{b}\)
Phân tích ra thừa số:
a) x - 9 với x > 0 ; \(\text{ b) x - 5\sqrt{x}+ 4 ;}\)
\(\text{c) 6√xy - 4x\sqrt{x} - 9y√y + 6xy ; }\) \(\text{ d) x - 2\sqrt{x-1} - a^2}\)
a)x-9=\(\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)\)
đưa thừa số ra ngoài dấu căn :
a) a2\(\sqrt{\dfrac{2}{3a}}\)( a > 0 )
b) \(\dfrac{x-3}{x}\)\(\sqrt{\dfrac{x^3}{9-x^2}}\)(0<x<3)
a: \(a^2\cdot\sqrt{\dfrac{2}{3a}}=a^2\cdot\dfrac{\sqrt{2}}{\sqrt{3}\cdot\sqrt{a}}=\dfrac{a\sqrt{2}}{\sqrt{3}}=\dfrac{a\sqrt{6}}{3}\)
b: \(\dfrac{x-3}{x}\cdot\sqrt{\dfrac{x^3}{9-x^2}}\)
\(=\dfrac{x-3}{x}\cdot\dfrac{x\sqrt{x}}{\sqrt{x-3}\cdot\sqrt{x+3}}\)
\(=\dfrac{\sqrt{x}\cdot\sqrt{x-3}}{\sqrt{x+3}}\)