Chương I - Căn bậc hai. Căn bậc ba
Các câu hỏi tương tự
Cho a,b,c >0 thỏa mãn ab+bc+ca=3abc
Tìm GTNN của \(Q=\frac{a^2}{c\cdot\left(c^2+a^2\right)}+\frac{b^2}{a\cdot\left(a^2+b^2\right)}+\frac{c^2}{b\cdot\left(b^2+c^2\right)}\)
\(A=\dfrac{1}{\sqrt{8}+\sqrt{7}}+\sqrt{175}+2\sqrt{2}\\ B=\left(5+2\sqrt{6}\right)\cdot\left(49-20\sqrt{6}\right)\cdot\sqrt{5-2\sqrt{6}}\)
\(C=\dfrac{1}{2}\left(\sqrt{6}+\sqrt{5}\right)^2-\dfrac{1}{4}\sqrt{120}-\sqrt{\dfrac{15}{2}}\)
\(D=\left(2\sqrt{6}-4\sqrt{3}+5\sqrt{2}+\dfrac{1}{4}\sqrt{8}\right)\cdot3\sqrt{6}\)
Tính: a. \(\left(3\sqrt{2}+\sqrt{6}\right)\cdot\left(6-3\sqrt{3}\right)\)
b. \(\left(4+\sqrt{15}\right)\cdot\left(\sqrt{10}-\sqrt{6}\right)\cdot\sqrt{4-\sqrt{15}}\)
c. \(\left(3-\sqrt{5}\right)\cdot\left(10-\sqrt{2}\right)\cdot\sqrt{3+\sqrt{5}}\)
Rút gọn biểu thức:
1) Pfrac{x^2-sqrt{x}}{x+sqrt{x}+1}-frac{2x+sqrt{x}}{sqrt{x}}+frac{2cdotleft(x-1right)}{sqrt{x}-1}
2) Pleft(frac{sqrt{x}-2}{sqrt{x}-1}-frac{sqrt{x}+2}{x+2sqrt{x}+1}right)cdotfrac{left(1-xright)^2}{2}
3) Bleft(frac{1-asqrt{a}}{1-sqrt{a}}+sqrt{a}right)cdotleft(frac{1+asqrt{a}}{1+sqrt{a}}-sqrt{a}right)
4) Kleft(frac{sqrt{a}}{sqrt{a}-1}-frac{1}{a-sqrt{a}}right)divleft(frac{1}{sqrt{a}+1}-frac{2}{a-1}right)
Đọc tiếp
Rút gọn biểu thức:
1) \(P=\frac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\frac{2x+\sqrt{x}}{\sqrt{x}}+\frac{2\cdot\left(x-1\right)}{\sqrt{x}-1}\)
2) \(P=\left(\frac{\sqrt{x}-2}{\sqrt{x}-1}-\frac{\sqrt{x}+2}{x+2\sqrt{x}+1}\right)\cdot\frac{\left(1-x\right)^2}{2}\)
3) \(B=\left(\frac{1-a\sqrt{a}}{1-\sqrt{a}}+\sqrt{a}\right)\cdot\left(\frac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right)\)
4) \(K=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right)\div\left(\frac{1}{\sqrt{a}+1}-\frac{2}{a-1}\right)\)
1) Rút gọn các đa thức:
a) dfrac{1}{m.n^2}cdotsqrt{dfrac{m^2.n^4}{5}} với m 0;nne0
b) sqrt{dfrac{m^4}{9-12m+4m^2}} với mle1,5
c) dfrac{a-1}{sqrt{a}-1}:sqrt{dfrac{left(a-1right)^4}{a-2sqrt{a}+1}} với 0 a 1
d) dfrac{a-b}{sqrt{a+b}}:sqrt{dfrac{left(a-bright)^2}{aleft(a+bright)}} với ab0
2) Chứng minh rằng:
dfrac{a-b}{b^2}:sqrt{dfrac{a^2-2ab+b^2}{a^2.b^2}}left{{}begin{matrix}aleft(ab0right)-aleft(0 a bright)end{matrix}right.
Đọc tiếp
1) Rút gọn các đa thức:
a) \(\dfrac{1}{m.n^2}\cdot\sqrt{\dfrac{m^2.n^4}{5}}\) với \(m< 0;n\ne0\)
b) \(\sqrt{\dfrac{m^4}{9-12m+4m^2}}\) với \(m\le1,5\)
c) \(\dfrac{a-1}{\sqrt{a}-1}:\sqrt{\dfrac{\left(a-1\right)^4}{a-2\sqrt{a}+1}}\) với \(0< a< 1\)
d) \(\dfrac{a-b}{\sqrt{a+b}}:\sqrt{\dfrac{\left(a-b\right)^2}{a\left(a+b\right)}}\) với \(a>b>0\)
2) Chứng minh rằng:
\(\dfrac{a-b}{b^2}:\sqrt{\dfrac{a^2-2ab+b^2}{a^2.b^2}}=\left\{{}\begin{matrix}a\left(a>b>0\right)\\-a\left(0< a< b\right)\end{matrix}\right.\)
B = \(\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}-\dfrac{1}{\sqrt{5}+\sqrt{2}}+1\right)\cdot\dfrac{1}{\left(\sqrt{2}+1\right)^2}\)
Cho a,b,c >0 thỏa mãn: ab+ bc+ca=1. Rút gọn biểu thức:
A= \(a\sqrt{\dfrac{\left(b^2+1\right)\left(c^2+1\right)}{a^2+1}}+b\sqrt{\dfrac{\left(a^2+1\right)\left(c^2+1\right)}{b^2+1}}+c\sqrt{\dfrac{\left(b^2+1\right)\left(a^2+1\right)}{c^2+1}}\)
cho ba số dương a,b,c .Chứng minh rằng \(\dfrac{1}{a^2\left(b+c\right)}+\dfrac{1}{b^2\left(a+c\right)}+\dfrac{1}{c^2\left(b+a\right)}\ge\dfrac{3}{2}\)
B1 : Tính
a) sqrt{1,2.270} ; sqrt{55.77.35}
b) (sqrt{3}-sqrt{2})^2 ; (3sqrt{2}-1)left(3sqrt{2}+1right) ; left(sqrt{6}+2right)left(sqrt{3}-2right)
c) left(sqrt{dfrac{3}{2}}-sqrt{dfrac{2}{3}}right) ; left(sqrt{dfrac{8}{3}}-sqrt{24}+sqrt{dfrac{50}{3}}right).sqrt{6}
B2 : Thực hành phép tính :
a) sqrt{dfrac{1}{8}}.sqrt{2}.sqrt{125}.sqrt{dfrac{1}{5}} ; sqrt{left(sqrt{2}-1right)}.sqrt{left(sqrt{2}+1right)}
b) sqrt{left(sqrt{2}-3righ...
Đọc tiếp
B1 : Tính
a) \(\sqrt{1,2.270}\) ; \(\sqrt{55.77.35}\)
b) (\(\sqrt{3}\)-\(\sqrt{2}\))\(^2\) ; (\(3\sqrt{2}-1\))\(\left(3\sqrt{2}+1\right)\) ; \(\left(\sqrt{6}+2\right)\left(\sqrt{3}-2\right)\)
c) \(\left(\sqrt{\dfrac{3}{2}}-\sqrt{\dfrac{2}{3}}\right)\) ; \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}\)
B2 : Thực hành phép tính :
a) \(\sqrt{\dfrac{1}{8}}.\sqrt{2}.\sqrt{125}.\sqrt{\dfrac{1}{5}}\) ; \(\sqrt{\left(\sqrt{2}-1\right)}.\sqrt{\left(\sqrt{2}+1\right)}\)
b) \(\sqrt{\left(\sqrt{2}-3\right)^2}\cdot\sqrt{11\cdot6\sqrt{2}}\) ; \(\sqrt{\left(\sqrt{3}-3\right)^2}\cdot\sqrt{\dfrac{1}{3-\sqrt{3}}}\)