\(\sqrt{4x^2-4x+1}+2=3x\)
Rút gọn
\(7\sqrt{a}-5b\sqrt{16a^3}+4a\sqrt{25ab^2}-3\sqrt{16a}\) với a>0, b>0
rút gọn các biểu thức sau
c,\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}\) d,\(5\sqrt{16a}-4\sqrt{25a}-2\sqrt{100a}+\sqrt{169a}\) với a ≥ 0
e,\(5\sqrt{4a}-4\sqrt{a^2}-\sqrt{100a}\) với a ≥ 0 f,\(3\sqrt{4a^6}-5^3\) với a ≤ 0
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) \(\sqrt{16b}+2\sqrt{40b}-3\sqrt{90b}\)
\(=4\sqrt{b}+4\sqrt{10b}-9\sqrt{10b}\)
\(=4\sqrt{b}-5\sqrt{10b}\)
Cho A = \(\sqrt{16a}\) - 5b \(\sqrt{25a^3}\) + 5a\(\sqrt{25ab^2}\) + 5a\(\sqrt{25ab^2}\) - 6\(\sqrt{a}\) Yêu cầu: a, rút gọn biểu thức. b, xác định giá trị a để A = 4
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
B=\(\dfrac{2\left(x+4\right)}{x-3\sqrt{x}-4}+\dfrac{\sqrt{x}}{\sqrt{x}+1}-\dfrac{8}{\sqrt{x}-4}\) Với x≥0,x≠16
a)Rút gọn B
Với x >= 0 ; x khác 16
\(B=\dfrac{2x+8+x-4\sqrt{x}-8\sqrt{x}-8}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3x-12\sqrt{x}}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-4\right)}=\dfrac{3\sqrt{x}}{\sqrt{x}+1}\)
Câu 1. Rút gọn các biểu thức sau:
a/\(\sqrt{4a^2}\)(với a<0)
b/\(\sqrt{4x^2-12x+9}\)(với x<3/2)
a) \(\sqrt{4a^2}=2\left|a\right|=-2a\) ( do a<0)
b) \(\sqrt{4x^2-12x+9}=\sqrt{\left(2x-3\right)^2}=\left|2x-3\right|=3-2x\)(do \(x< \dfrac{3}{2}\Leftrightarrow2x-3< 0\))
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
\(2x-\sqrt{4x^2+4x+1}\)với x bé hơn -1/2
\(3x+2-\sqrt{9x^2-12x+4}\)
\(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}vớia\ge0\)
\(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}vớia\ge0\)
\(\frac{3}{a^2-b^2}\sqrt{\frac{2a^2-4ab+2b^2}{9}}\)
\(\left(1-\frac{x-3\sqrt{x}}{x-9}\right)\div\frac{\sqrt{a}+1}{a}\)
a) \(2x-\sqrt{4x^2+4x+1}=2x-\sqrt{\left(2x+1\right)^2}=2x-\left|2x+1\right|\)
Vì \(x< -\frac{1}{2}\)nên \(\left|2x+1\right|=-\left(2x+1\right)\)
\(\Rightarrow2x+2x+1=4x+1\)
b) \(3x+2-\sqrt{9x^2-12x+4}=3x+2-\sqrt{\left(3x-2\right)^2}=3x+2-\left|3x-2\right|\)
Khi \(x\ge\frac{2}{3}\)thì \(\left|3x-2\right|=3x-2\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-3x+2=4\)
Khi \(x< \frac{2}{3}\) thì \(\left|3x-2\right|=2-3x\)
\(\Leftrightarrow3x+2-\left|3x-2\right|=3x+2-\left(2-3x\right)=6x\)
c) \(\sqrt{9a}-\sqrt{16a}+\sqrt{49a}=3\sqrt{a}-4\sqrt{a}+7\sqrt{a}\)
Đặt \(\sqrt{a}=x\) ta được : \(3x-4x+7x=6x\)\(=6\sqrt{a}\)( Do \(a\ge0\))
d) \(\sqrt{160a}+2\sqrt{40a}-3\sqrt{90a}=4\sqrt{10a}+4\sqrt{10a}-9\sqrt{10a}\)\(=-\sqrt{10}\)
TK NKA !!!