\(\sqrt{27.48\left(1-a\right)^2}vớia>1\) \(\dfrac{1}{a-b}.\sqrt{a^4\left(a-b\right)^2}\) với a > b
Rút gọn:
A = \(\sqrt{27.48\left(1-a^2\right)}\) với a > 1
B = \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a > b
C = \(\sqrt{5a}.\sqrt{45a}-3a\) với a ≥ 0
D = \(\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\) với a tùy ý
a) Ta có: \(\sqrt{27\cdot48\left(1-a^2\right)}\)
\(=\sqrt{3^4\cdot4^2\cdot\left(1-a^2\right)}\)
\(=36\sqrt{1-a^2}\)
c) Ta có: \(\sqrt{5a}\cdot\sqrt{45a}-3a\)
\(=15a-3a=12a\)
b) Ta có: \(B=\dfrac{1}{a-b}\cdot\sqrt{a^4\cdot\left(a-b\right)^2}\)
\(=\dfrac{1}{a-b}\cdot a^2\cdot\left(a-b\right)\)
\(=a^2\)
d) Ta có: \(D=\left(3-a\right)^2-\sqrt{0.2}\cdot\sqrt{180a^2}\)
\(=a^2-6a+9-\sqrt{36a^2}\)
\(=a^2-6a+9-\left|6a\right|\)
\(=\left[{}\begin{matrix}a^2-6a+9-6a\left(a\ge0\right)\\a^2-6a+9+6a\left(a< 0\right)\end{matrix}\right.\)
\(=\left[{}\begin{matrix}a^2-12a+9\\a^2+9\end{matrix}\right.\)
\(A=9.4\left|1-a\right|=36\left(a-1\right)\) (a>1)
\(B=\dfrac{a^2\left|a-b\right|}{a-b}=\dfrac{a^2\left(a-b\right)}{a-b}=a^2\) (a>b)
\(C=5.3\left|a\right|-3a=15a-3a=12a\)
\(D=9-6a+a^2-6\left|a\right|=\left[{}\begin{matrix}a^2-12a+9\left(a\ge0\right)\\a^2+9\left(a< 0\right)\end{matrix}\right.\)
Rút gọn :
a, A = \(\sqrt{27.48\left(1-a^2\right)}vớia>1\)
b, B = \(\frac{1}{a-b}\sqrt{a^4\left(a-b^2\right)}\) với a > b
c, C= \(\sqrt{5a}.\sqrt{45a}-3a\) với a >= 0
Rút gọn
\(A=\sqrt{27.48\left(1-a^2\right)}vớia>1\)
\(B=\frac{1}{a-b}.\sqrt{a^4.\left(a-b\right)^2}\)Với a>b
\(C=\sqrt{5a}.\sqrt{45a}-3a\)với a> hoặc bằng 0
\(D=\left(3-a\right)^2-\sqrt{0,2}.\sqrt{180a^2}\)Với a tùy ý
\(A=\sqrt{9.3.3.16\left(1-a^2\right)}=3.3.4.\left|1-a\right|=36\left(a-1\right)\)
\(B=\frac{1}{a-b}a^2.\left|a-b\right|=\frac{a^2\left(a-b\right)}{a-b}=a^2\)
\(C=\sqrt{5.45.a^2}-3a=\sqrt{5^2.3^2.a^2}-3a=15\left|a\right|-3a=15a-3a=12a\)
\(D=\left(3-a\right)^2-\sqrt{\frac{2.180}{10}a^2}=\left(3-a\right)^2-6\left|a\right|\)
a)\(\sqrt{4\left(a-3\right)^2}vớia\ge3\)
b)\(\sqrt{a^2\left(a+1\right)^2}vớia>0\)
c)\(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}vớia< 0,b\ne0\)
a) \(\sqrt{4\left(a-3\right)^2}=2\left(a-3\right)=2a-6\)
b) \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c) \(\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\dfrac{1}{\sqrt{8}\left|a\right|}=\dfrac{1}{-\sqrt{8}a}=\dfrac{-\sqrt{8}}{8a}\)
a: \(\sqrt{4\left(a-3\right)^2}=2\cdot\left(a-3\right)=2a-6\)
b: \(\sqrt{a^2\left(a+1\right)^2}=a\left(a+1\right)=a^2+a\)
c: \(\dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}=\sqrt{\dfrac{16a^4b^6}{128a^6b^6}}=\sqrt{\dfrac{1}{8a^2}}=\sqrt{\dfrac{2}{16a^2}}=-\dfrac{\sqrt{2}}{4a}\)
Rút gọn các biểu thức sau:
a. \(\sqrt{0,36a^2}\) với a < 0;
b. \(\sqrt{a^4\left(3-a\right)^2}\) với \(a\ge3;\)
c. \(\sqrt{27.48\left(1-a\right)^2}\) với a > 1.
d. \(\dfrac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a > b.
a) = = 0,6.│a│
Vì a < 0 nên │a│= -a. Do đó = -0,6a.
b) = . = ││.│3 - a│.
Vì ≥ 0 nên │b│= . Vì a ≥ 3 nên 3 - a ≤ 0, do đó │3 - a│= a - 3.
Vậy = (a - 3).
c) = = = √81.√16.
= 9.4.│1 - a│
Vì a > 1 nên 1 - a < 0. Do đó │1 - a│= a -1.
Vậy = 36(a - 1).
d) : = : ( = : (.│a - b│)
Vì a > b nên a -b > 0, do đó│a - b│= a - b.
Vậy : = : ((a - b)) = .
a) = = 0,6.│a│
Vì a < 0 nên │a│= -a. Do đó = -0,6a.
b) = . = ││.│3 - a│.
Vì ≥ 0 nên │b│= . Vì a ≥ 3 nên 3 - a ≤ 0, do đó │3 - a│= a - 3.
Vậy = (a - 3).
c) = = = √81.√16.
= 9.4.│1 - a│
Vì a > 1 nên 1 - a < 0. Do đó │1 - a│= a -1.
Vậy = 36(a - 1).
d) : = : ( = : (.│a - b│)
Vì a > b nên a -b > 0, do đó│a - b│= a - b.
Vậy : = : ((a - b)) = .
Rút gọn các biểu thức
a, \(\sqrt{27.48\left(1-a\right)^2}\) với a > 1
b, \(\frac{1}{a-b}.\sqrt{a^4\left(a-b\right)^2}\) với a > b
c, \(\sqrt{4+\sqrt{7}}+\sqrt{4-\sqrt{7}}\)
a/ \(=\sqrt{36^2\left(1-a\right)^2}=36.\left|1-a\right|=36\left(a-1\right)=36a-36\)
b/ \(=\frac{1}{a-b}.a^2\left|a-b\right|=\frac{1}{a-b}.a^2\left(a-b\right)=a^2\)
c/ \(=\frac{\sqrt{8+2\sqrt{7}}}{\sqrt{2}}+\frac{\sqrt{8-2\sqrt{7}}}{\sqrt{2}}=\frac{\sqrt{\left(\sqrt{7}+1\right)^2}+\sqrt{\left(\sqrt{7}-1\right)^2}}{\sqrt{2}}=\frac{\sqrt{7}+1+\sqrt{7}-1}{\sqrt{2}}=\frac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)
Rút gọn rồi tính giá trị của biểu thức
\(\sqrt{\frac{\sqrt{a}-1}{\sqrt{b}+1}}\div\sqrt{\frac{\sqrt{b}-1}{\sqrt{a}+1}}vớia=7,25;b=3,25\)
\(\frac{a-b}{\sqrt{a\times\left(a+2\times b\right)+b^2}}\div\sqrt{\frac{\left(a-b\right)^2}{a\times\left(a+b\right)}}vớia>b>0và\frac{a}{b}=\frac{9}{7}\)
\(\frac{x-1}{\sqrt{y}-1}\times\sqrt{\frac{\left(y-2\times\sqrt{y}+1\right)^2}{\left(x-1\right)^4}}vớix=\frac{-1}{2};y=121\); giúp mk vs
từ giả thiết, ta có \(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{zx}=1\)
đặt \(\left(\dfrac{1}{xy};\dfrac{1}{yz};\dfrac{1}{zx}\right)=\left(a;b;c\right)\Rightarrow a+b+c=1\) =>\(\left(\dfrac{ac}{b};\dfrac{ab}{c};\dfrac{bc}{a}\right)=\left(\dfrac{1}{x^2};\dfrac{1}{y^2};\dfrac{1}{z^2}\right)\)
ta có VT=\(\dfrac{1}{\sqrt{1+\dfrac{1}{x^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{y^2}}}+\dfrac{1}{\sqrt{1+\dfrac{1}{z^1}}}=\sqrt{\dfrac{1}{1+\dfrac{ac}{b}}}+\sqrt{\dfrac{1}{1+\dfrac{ab}{c}}}+\sqrt{\dfrac{1}{1+\dfrac{bc}{a}}}\)
=\(\dfrac{1}{\sqrt{\dfrac{b+ac}{b}}}+\dfrac{1}{\sqrt{\dfrac{a+bc}{a}}}+\dfrac{1}{\sqrt{\dfrac{c+ab}{c}}}=\sqrt{\dfrac{a}{\left(a+b\right)\left(a+c\right)}}+\sqrt{\dfrac{b}{\left(b+c\right)\left(b+a\right)}}+\sqrt{\dfrac{c}{\left(c+a\right)\left(c+b\right)}}\)
\(\le\sqrt{3}\sqrt{\dfrac{ac+ab+bc+ba+ca+cb}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=\sqrt{3}.\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
ta cần chứng minh \(\sqrt{\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\le\dfrac{3}{2}\Leftrightarrow\dfrac{2\left(ab+bc+ca\right)}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\le\dfrac{9}{4}\Leftrightarrow8\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
<=>\(8\left(a+b+c\right)\left(ab+bc+ca\right)\le9\left(a+b\right)\left(b+c\right)\left(c+a\right)\) (luôn đúng )
^_^
Rút gọn biểu thức
a) \(\sqrt{27.48\left(1-a\right)^2}\) với a>1
b) \(\frac{1}{a-b}\sqrt{a^4\left(a-b\right)^2}\) với a>b