B1: Rút gọn
a) \(\sqrt{(3-a)^2.a^4}\)với a ≥3
b) \(\sqrt{27.48.(1-a)^2}\)với a>1
c) \(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3x}{8}}\)với a≥0
d) \(\sqrt{13a}.\sqrt{\frac{52}{x}}với\:a\:>0\)
Mn giúp mình với ạ
1. Rút gọn biểu thức:
a) \(\sqrt{27\cdot48\cdot\left(1-a\right)^2}\)với a>1
b) \(\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}\) với a>b
c) \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}\)với \(a\ge0\)
d) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}\)với a>0
e) \(\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
Rút gọn các biểu thức sau:
a) $\sqrt{9a^4}$
b) 2$\sqrt{a^{2}}$- 5a (với a<0)
c) $\sqrt{16(1+4x+4x^2)}$ với x $\geq$ $\frac{1}{2}$
d) $\frac{1}{a-3}$$\sqrt{9(a^2-3a+9)}$ với a<3
a) \(\sqrt{9a^4}=\sqrt{\left(3a^2\right)^2}=\left|3a^2\right|=3a^2\)
b) \(2\sqrt{a^2}-5a=2\left|a\right|-5a=-2a-5a=-7a\)
c) \(\sqrt{16\left(1+4x+4x^2\right)}=\sqrt{\left[4\left(1+2x\right)\right]^2}=\left|4\left(1+2x\right)\right|=4\left(1+2x\right)\)
1)\(\frac{1}{a-b}.\sqrt{a^4.\left(a-b\right)^2}\) với a>b
2)\(\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}\)với a>=0 ( a lớn hơn hoặc bằng 0)
3.\(\sqrt{13}a.\sqrt{\frac{52}{a}}\)với a>0
1) \(\frac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}=\frac{1}{a-b}\cdot a^2\cdot\left|a-b\right|=a^2\)(Vì a > b => a - b > 0 và a^2 luôn dương với mọi a)
2) \(\sqrt{\frac{2a}{3}}\cdot\sqrt{\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\frac{a}{2}\)(vì \(a\ge0\))
3) \(\sqrt{13}a\cdot\sqrt{\frac{52}{a}}=\frac{a\cdot\sqrt{13}\cdot\sqrt{4\cdot13}}{\sqrt{a}}=\frac{2a\cdot\sqrt{13\cdot13}}{\sqrt{a}}=26\sqrt{a}\)(vì a > 0)
Rút gọn
\(B=\frac{2}{x^2-y^2}\sqrt{\frac{9\left(x^2+2xy+y^2\right)}{4}}\) với x > -y
\(C=\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}\) với a >hoặc= 0
\(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a > 0
\(B=\frac{2}{x^2-y^2}\cdot\sqrt{\frac{9\left(x^2+2xy+y^2\right)}{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\sqrt{\frac{9\left(x+y\right)^2}{4}}\)
\(=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{\sqrt{9\left(x+y\right)^2}}{\sqrt{4}}=\frac{2}{\left(x-y\right)\left(x+y\right)}\cdot\frac{3\left(x+y\right)}{2}\)(vì x > -y <=> x + y > 0)
\(=\frac{3}{x-y}\)
\(C=\sqrt{\frac{2a}{3}}.\sqrt{\frac{3a}{8}}=\sqrt{\frac{2a}{3}\cdot\frac{3a}{8}}=\sqrt{\frac{6a^2}{24}}=\sqrt{\frac{a^2}{4}}=\frac{a}{2}\)(vì a > = 0)
\(D=\frac{1}{a-b}\cdot\sqrt{a^4\left(a-b\right)^2}=\frac{1}{a-b}\cdot a^2\left(a-b\right)=a^2\)(a > b > 0)
câu cuối điều kiện là a>b
\(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}=\frac{a^2\left|a-b\right|}{a-b}=\frac{a^2\left(a-b\right)}{a-b}=a^2\) (vì a>b)
Rút gọn biểu thức:
a, \(\frac{2}{a}\sqrt{\frac{16a^2}{9}}\) với a < 0
b, \(\frac{3}{a-1}\sqrt{\frac{4a^2-8a+4}{25}}\) với a > 1
c, \(\frac{3\sqrt{18a^2b^4}}{\sqrt{2a^2b^2}}\) với a ≠ b
d, \(\left(1+\frac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\frac{a-\sqrt{a}}{\sqrt{a}-1}\right)\) với a ≠ 1, a ≥ 0
a/ \(\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{-8a}{3a}=-\frac{8}{3}\)
b/ \(\frac{3}{a-1}\sqrt{\frac{4\left(a-1\right)^2}{25}}=\frac{3}{\left(a-1\right)}.\frac{2\left|a-1\right|}{5}=\frac{6\left(a-1\right)}{5\left(a-1\right)}=\frac{6}{5}\)
c/ \(\frac{3\sqrt{9a^2b^4}}{\sqrt{a^2b^2}}=\frac{9.\left|a\right|.b^2}{\left|a\right|\left|b\right|}=9\left|b\right|\)
d/ \(\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
a/ \(=\frac{2}{a}.\frac{4\left|a\right|}{3}=\frac{2}{a}.\frac{-4a}{3}=\frac{-8}{3}\)
b/ \(=\frac{3}{a-1}.\frac{\left|2a-2\right|}{5}=\frac{3}{a-1}.\frac{2\left(a-1\right)}{5}=\frac{6}{5}\)
c/ \(=\sqrt{\frac{162a^2b^4}{2a^2b^2}}=\sqrt{81b^2}=9\left|b\right|\)
d/ \(=\left(1+\frac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\frac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)=1-a\)
rút gọn giúp mình nha
Q=\(\frac{x-y}{\sqrt{x}-\sqrt{y}}\)\(-\frac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\)với x ≥ 0, y ≥ 0 và x 6= y.
R=\(\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\)\(\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)với a ≥ 0 và a 6= 1.
\(Q=\frac{x-y}{\sqrt{x}-\sqrt{y}}-\frac{\sqrt{x^3}-\sqrt{y^3}}{x-y}\)
\(Q=\frac{\left(\sqrt{x}+\sqrt{y}\right)\left(x-y\right)-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{x\sqrt{x}-y\sqrt{x}+x\sqrt{y}-y\sqrt{y}-x\sqrt{x}+y\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(Q=\frac{\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\)
\(R=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(R=\left[\frac{\left(1-\sqrt{a}\right)\left(1+\sqrt{a}+a\right)}{1-\sqrt{a}}+\sqrt{a}\right].\frac{\left(1-\sqrt{a}\right)^2}{\left(1-a\right)^2}\)
\(R=\left(1+\sqrt{a}+a\right).\frac{\left(1-\sqrt{a}\right)^2}{\left(1-\sqrt{a}\right)^2.\left(1+\sqrt{a}\right)^2}\)
\(=\left(1+\sqrt{a}\right)^2.\frac{1}{\left(1+\sqrt{a}\right)^2}=1\)
BT. RÚT GỌN
1.( với a >=3 )\(\sqrt{\frac{24}{3}}\times\sqrt{\frac{3a}{8}}\)
2. ( với a>3 )\(\sqrt{13a}\times\sqrt{\frac{52}{a}}\)
3. ( với a >=0 )\(\sqrt{5a}\times\sqrt{45a}-3a\)
4. ( 3-a )2-\(\sqrt{0,2}\times\sqrt{180a^2}\)
GIÚP MÌNH VỚI!!!!!!!HHHH
1) \(\sqrt{\frac{24}{3}}\cdot\sqrt{\frac{3a}{8}}=\sqrt{\frac{72a}{24}}=\sqrt{3a}\)
2) \(\sqrt{13a}\cdot\sqrt{\frac{52}{a}}=\sqrt{\frac{13a\cdot52}{a}}=\sqrt{676}=26\)
3) \(\sqrt{5a}\cdot\sqrt{45a}-3a=\sqrt{225a^2}-3a=15a-3a=12a\)
4) \(\left(3-a\right)^2-\sqrt{0,2}\cdot\sqrt{180a^2}=a^2-6a+9-\sqrt{36a^2}=a^2-6a+9-6a=a^2-12a+9\)
Cho \(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\left(ĐK:0< a< 1\right)\)
a, Rút gọn Q ( câu này viết kq với hướng làm thôi cũng được !)
b, so sánh Q với Q3
Giúp mình với !!! UHUHUHU
a) ĐK: \(0< a< 1\)
\(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}\right).\sqrt{\left(1-a\right)^2}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right).\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\left(\sqrt{1+a}+\sqrt{1-a}\right)^2}{\left(\sqrt{1+a}-\sqrt{1-a}\right)\left(\sqrt{1+a}+\sqrt{1-a}\right)}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{2+2\sqrt{1-a^2}}{2a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1-a^2}+1}{a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{-a^2\left(1-a\right)}{a^2}=a-1\)
\(Q=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a^2}-1+a}\right)\left(\sqrt{\frac{1}{a^2}-1}-\frac{1}{a}\right)\sqrt{a^2-2a+1}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{1-a}{\sqrt{1-a}\left(\sqrt{1+a}-\sqrt{1-a}\right)}\right)\left(\frac{\sqrt{1-a^2}}{a}-\frac{1}{a}\right).\sqrt{\left(1-a\right)^2}\)
\(=\left(\frac{\sqrt{1+a}}{\sqrt{1+a}-\sqrt{1-a}}+\frac{\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}\right).\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1+a}+\sqrt{1-a}}{\sqrt{1+a}-\sqrt{1-a}}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{2+2\sqrt{1-a^2}}{2a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{\sqrt{1-a^2}+1}{a}.\frac{\sqrt{1-a^2}-1}{a}.\left(1-a\right)\)
\(=\frac{-a^2\left(1-a\right)}{a^2}=a-1\)
b) Xét: \(Q^3-Q=\left(a-1\right)^3-\left(a-1\right)=\left(a-1\right)^2\left(a-1-1\right)=\left(a-1\right)^2\left(a-2\right)\)
Do \(a< 1\)=> \(a-2< 0\) và \(a-1< 0\)
nên \(\left(a-1\right)^2\left(a-2\right)< 0\)
=> \(Q^3-Q< 0\)
<=> \(Q^3< Q\)
xin lỗi nhé, câu b mk sai, sửa lại:
\(Q^3-A=\left(a-1\right)^3-\left(a-1\right)=\left(a-1\right)\left[\left(a-1\right)^2-1\right]\)
\(=\left(a-1\right)\left(a-1-1\right)\left(a-1+1\right)=\left(a-2\right)\left(a-1\right)a\)
Do \(0< a< 1\)nên \(a-2< 0;\)\(a-1< 0\)
=> \(\left(a-2\right)\left(a-1\right)a>0\)
=> \(Q^3-Q>0\)
<=> \(Q^3>Q\)
1,Rút gọn C với C bằng (\(\frac{\sqrt{a}}{\sqrt{a}-1}\)-- \(\frac{\sqrt{a}}{\sqrt{a}-\sqrt{a}}\)) : \(\frac{\sqrt{a}+1}{a-1}\)
2, a, Rút gọn A với A = \(\frac{\sqrt{a}+1}{x+4\sqrt{x}+4}\): ( \(\frac{x}{x+2\sqrt{x}}+\frac{x}{\sqrt{x}+2}\)) với x > 0
b, Tìm tất cả các giá trị của x để A >= \(\frac{1}{3\sqrt{x}}\)