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a: \(A=\dfrac{1}{x-1}\cdot5\sqrt{3}\cdot\left|x-1\right|\cdot\sqrt{x-1}\)
\(=\dfrac{5\sqrt{3}}{x-1}\cdot\left(x-1\right)\cdot\sqrt{x-1}=5\sqrt{3}\cdot\sqrt{x-1}\)
b: \(B=10\sqrt{x}-3\cdot\dfrac{10\sqrt{x}}{3}-\dfrac{4}{x}\cdot\dfrac{x\sqrt{x}}{2}\)
\(=10\sqrt{x}-10\sqrt{x}-\dfrac{4\sqrt{x}}{2}=-2\sqrt{x}\)
c: \(C=x-4+\left|x-4\right|\)
=x-4+x-4
=2x-8
Bài 2:
\(x=\sqrt{4+2\sqrt{3}}=\sqrt{3}+1\)
Ta có: \(P=x^2-2x+2020\)
\(=4+2\sqrt{3}-2\left(\sqrt{3}-1\right)+2020\)
\(=4+2\sqrt{3}-2\sqrt{3}+2+2020\)
=2026
Bài 1:
\(A=-\dfrac{3}{4}\cdot\sqrt{9-4\sqrt{5}}\cdot\sqrt{\left(-8\right)^2\cdot\left(2+\sqrt{5}\right)^2}\)
\(=\dfrac{-3}{4}\cdot8\cdot\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)\)
=-6
1)\(=\sqrt{\left(\sqrt{5}-2\right)^2}+\sqrt{26^2}=\sqrt{5}-2+26=24-\sqrt{5}\)
2) \(=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
3) \(=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{\left|x-1\right|}{x-1}\)\(=\left[{}\begin{matrix}1\left(x>1\right)\\-1\left(x< 1\right)\end{matrix}\right.\)
4) \(=\sqrt{\left(\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}\right)^2}-\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{1}{2}}\right)^2}=\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{\dfrac{7}{2}}+\sqrt{\dfrac{1}{2}}=2\sqrt{\dfrac{1}{2}}=\sqrt{2}\)
2. \(\dfrac{x^2-5}{x+\sqrt{5}}=\dfrac{x^2-\left(\sqrt{5}\right)^2}{x+\sqrt{5}}=\dfrac{\left(x-\sqrt{5}\right)\left(x+\sqrt{5}\right)}{x+\sqrt{5}}=x-\sqrt{5}\)
3. \(\dfrac{\sqrt{x^2-2x+1}}{x-1}=\dfrac{\sqrt{x^2-2.x.1+1^2}}{x-1}=\dfrac{\sqrt{\left(x-1\right)^2}}{x-1}=\dfrac{|x-1|}{x-1}=\left[{}\begin{matrix}x-1>0\left(x>1\right)\\x-1< 0\left(x< 1\right)\end{matrix}\right.=\left[{}\begin{matrix}=1\\=\dfrac{x+1}{x-1}\end{matrix}\right.\)
\(1,\sqrt{4\left(a-4\right)^2}\left(dkxd:a\ge4\right)\)
\(=\sqrt{4}.\sqrt{\left(a-4\right)^2}\)
\(=\sqrt{2^2}.\left|a-4\right|\)
\(=2\left(a-4\right)\)
\(=2a-8\)
\(2,\sqrt{9\left(b-5\right)^2}\left(dkxd:b< 5\right)\)
\(=\sqrt{9}.\sqrt{\left(b-5\right)^2}\)
\(=\sqrt{3^2}.\left|b-5\right|\)
\(=3\left(-b+5\right)\)
\(=-3b+15\)
\(A=3\sqrt{4x^6}-3x^3=3\sqrt{\left(2x^3\right)^2}-3x^3\\=3\left|2x^3\right|-3x^3=3.\left(-2x^3\right)-3x^3\left(Do:x\le0\right)\\ =-6x^3-3x^3=-9x^3\\ B=\left(a-3\right)b^3.\sqrt{\dfrac{25}{\left(a-3\right)^2b^4}}=\left(a-3\right)b^3.\sqrt{\left[\dfrac{5}{\left(a-3\right).b^2}\right]^2}\\ =\left(a-3\right)b^3.\left|\dfrac{5}{\left(a-3\right)b^2}\right|=5b\)
\(\sqrt{2x-5}=3\\ \Rightarrow2x-5=3^2\\ \Leftrightarrow2x=9+5=14\\ Vậy:x=\dfrac{14}{2}=7\\ \Rightarrow S=\left\{7\right\}\)
a) \(\sqrt{36\left(x-5\right)^2}=6\left|x-5\right|\)
\(=6\left(x-5\right)\) (khi \(x\ge5\))
hoặc \(=6\left(5-x\right)\) (khi \(x< 5\))
b) \(\sqrt{\dfrac{1}{4}\left(1-x\right)^2}=\dfrac{1}{2}\left|1-x\right|\)
\(=\dfrac{1}{2}\left(1-x\right)\) (khi \(x\le1\))
hoặc \(=\dfrac{1}{2}\left(x-1\right)\) (khi \(x>1\))
c) \(\sqrt{x^2\left(2x-4\right)^2}=\left|x\right|\left|2x-4\right|\)
\(=x\left(2x-4\right)\) (khi \(x\ge2\))
hoặc \(=x\left(4-2x\right)\) (khi \(0\le x< 2\))
hoặc \(=-x\left(4-2x\right)\) (khi \(x< 0\))
\(=2\left|3-\sqrt{2}\right|+\sqrt{18}-5.1=6-2\sqrt{2}+3\sqrt{2}-5\)
\(=1+\sqrt{2}\)
a) Ta có: \(\left(\sqrt{7}-\sqrt{2}\right)\cdot\sqrt{9+2\sqrt{14}}\)
\(=\left(\sqrt{7}-\sqrt{2}\right)\cdot\left(\sqrt{7}+\sqrt{2}\right)\)
=7-2
=5
d) Ta có: \(\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}-\dfrac{6\sqrt{2}-4}{3-\sqrt{2}}\)
\(=2\sqrt{2}-\sqrt{7}+5\sqrt{7}-\dfrac{2\sqrt{2}\left(3-\sqrt{2}\right)}{3-\sqrt{2}}\)
\(=2\sqrt{2}+4\sqrt{7}-2\sqrt{2}\)
\(=4\sqrt{7}\)
Đặt \(B=4^{1993}+4^{1992}+.......+4^2+1\)
\(\Rightarrow4B=4^{1994}+4^{1993}+....+4^3+4\)
\(\Rightarrow3B=4^{1994}-1\)
Mà: \(A=75B+25=25\left(3B+1\right)=25\left(4^{1994}-1+1\right)=25.4^{1994}\)
\(A=75\left(4^{1993}+5^{1992}+...+4^2+5\right)+25=75B+25\)
Xét \(B=4^{1993}+4^{1992}+...+4^2+5=4^{1993}+4^{1992}+...+4^2+4+1\)
\(\Rightarrow4B=4^{1994}+4^{1993}+...+4^2+4\)
\(\Rightarrow4B+1-4^{1994}=4^{1993}+4^{1992}+...+4^2+4+1=B\)
\(\Rightarrow3B=4^{1994}-1\Rightarrow B=\dfrac{4^{1994}-1}{3}\)
Vậy \(A=75.\dfrac{\left(4^{1994}-1\right)}{3}+25=25.4^{1994}-25+25\)
\(\Rightarrow A=25.4^{1994}\)