Bài 1 : Tính
a) \(\left(5\sqrt{18}-3\sqrt{18}+4\sqrt{2}\right):\sqrt{2}\)
b) \(\left(\sqrt{\dfrac{a^2}{d}}+\sqrt{\dfrac{b^2}{d}}-\sqrt{d}\right):\sqrt{d}\) Với a,b là các số hữu tỉ dương , d là số nguyên tố dương
Tính:
\(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
a) Ta có: \(A=\sqrt{20}-2\sqrt{45}+3\sqrt{18}+\sqrt{72}\)
\(=2\sqrt{5}-6\sqrt{5}+9\sqrt{2}+6\sqrt{2}\)
\(=-4\sqrt{5}+15\sqrt{2}\)
b) Ta có: \(B=4\sqrt{\left(\sqrt{3}-1\right)^2}+2\sqrt{12}+4\sqrt{\dfrac{1}{2}}\)
\(=4\left(\sqrt{3}-1\right)+2\cdot2\sqrt{3}+\dfrac{4}{\sqrt{2}}\)
\(=4\sqrt{3}-4+4\sqrt{3}+2\sqrt{2}\)
\(=8\sqrt{3}+2\sqrt{2}-4\)
c) Ta có: \(C=\left(3+\dfrac{3-\sqrt{3}}{\sqrt{3}-1}\right)\left(3-\dfrac{3+\sqrt{3}}{1+\sqrt{3}}\right)\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3
=6
d) Ta có: \(D=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}\)
\(=2-\sqrt{3}+2+\sqrt{3}\)
=4
Cho a, b, x là những số dương. Đơn giản các biểu thức sau :
a) \(A=\left[\dfrac{2a+\left(ab\right)^{\dfrac{1}{2}}}{3a}\right]^{-1}\left[\dfrac{a^{\dfrac{3}{2}}-b^{\dfrac{3}{2}}}{a-\left(ab\right)^{\dfrac{1}{2}}}-\dfrac{a-b}{\sqrt{a}+\sqrt{b}}\right]\)
b) \(B=\left(\dfrac{\sqrt{a}-\sqrt{x}}{\sqrt{a+x}}-\dfrac{\sqrt{a+x}}{\sqrt{a}+\sqrt{x}}\right)^{-2}-\left(\dfrac{\sqrt{a}-\sqrt{x}}{\sqrt{a+x}}-\dfrac{\sqrt{a+x}}{\sqrt{a}-\sqrt{x}}\right)^{-2}\)
c) \(C=\sqrt{16^{\dfrac{1}{\log_74}}+81^{\dfrac{1}{\log_69}}+15}\)
d) \(D=49^{1-\log_72}+5^{-\log_54}\)
* Rút gọn biểu thức
a. \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}\)
b. \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
c. \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}\)
d. \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
Cho a,b,c,d,A,B,C,D là các số nguyên dương và \(\dfrac{a}{A}=\dfrac{b}{B}=\dfrac{c}{C}=\dfrac{d}{D}\)
CMR \(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}\)
Giúp mình với các cao nhân
Theo tính chất dãy tỉ số bằng nhau, đặt:
\(\dfrac{a}{A}=\dfrac{b}{B}=\dfrac{c}{C}=\dfrac{d}{D}=\dfrac{a+b+c+d}{A+B+C+D}=k>0\)
\(\Rightarrow a=kA;b=kB;c=kC;d=kD;a+b+c+d=k\left(A+B+C+D\right)\)
Do đó:
\(\sqrt{aA}+\sqrt{bB}+\sqrt{cC}+\sqrt{dD}=\sqrt{kA^2}+\sqrt{kB^2}+\sqrt{kC^2}+\sqrt{kD^2}\)
\(=\sqrt{k}\left(A+B+C+D\right)\) (1)
\(\sqrt{\left(a+b+c+d\right)\left(A+B+C+D\right)}=\sqrt{k\left(A+B+C+D\right)^2}=\sqrt{k}\left(A+B+C+D\right)\) (2)
Từ (1);(2) suy ra điều phải c/m
* Rút gọn biểu thức
a. \(\left(2\sqrt{125}-3\sqrt{5}-\sqrt{180}\right):\left(-\sqrt{5}\right)+\sqrt{8}\)
b. \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}\)
c. \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
d.\(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\left(\dfrac{1}{\sqrt{5}-\sqrt{2}}\right)\)
a) \(\dfrac{2\sqrt{125}-3\sqrt{5}-\sqrt{180}}{-\sqrt{5}}+\sqrt{8}=\dfrac{2\sqrt{25.5}-3\sqrt{5}-\sqrt{36.5}}{-\sqrt{5}}+\sqrt{8}\)
\(=\dfrac{10\sqrt{5}-3\sqrt{5}-6\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=\dfrac{\sqrt{5}}{-\sqrt{5}}+2\sqrt{2}=2\sqrt{2}-1\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}\)
\(=\sqrt{3}-\sqrt{2}+3\sqrt{2}=2\sqrt{2}+\sqrt{3}\)
c) \(\sqrt{48}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}=\sqrt{16.3}-2\sqrt{9.\dfrac{1}{3}}+\dfrac{\sqrt{3}\left(1-\sqrt{3}\right)}{\sqrt{3}}\)
\(=4\sqrt{3}-2\sqrt{3}+1-\sqrt{3}=1+\sqrt{3}\)
d) \(\left(\dfrac{\sqrt{6}-\sqrt{2}}{1-\sqrt{3}}-\dfrac{5}{\sqrt{5}}\right):\dfrac{1}{\sqrt{5}-\sqrt{2}}=\left(\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}-\sqrt{5}\right).\left(\sqrt{5}-\sqrt{2}\right)\)
\(=\left(-\sqrt{2}-\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)=-\left(\sqrt{5}+\sqrt{2}\right)\left(\sqrt{5}-\sqrt{2}\right)=-3\)
Tính:
\(A=\left(\sqrt{72}-3\sqrt{24}+5\sqrt{8}\right)\sqrt{2}+4\sqrt{27}\)
\(B=\dfrac{1}{\sqrt{2}-1}+\dfrac{14}{3+\sqrt{2}}\)
\(C=\dfrac{5+3\sqrt{5}}{\sqrt{5}}+\dfrac{3\sqrt{3}}{\sqrt{3}+1}-\left(\sqrt{5}+3\right)\)
\(D=\sqrt{\left(1-\sqrt{2}\right)^2}-3\sqrt{18}+4\sqrt{\dfrac{1}{2}}\)
* Thực hiện phép tính.
a.\(2\sqrt{18}-9\sqrt{50}+3\sqrt{8}\)
b.\(\left(\sqrt{7}-\sqrt{3}\right)^2+7\sqrt{84}\)
c.\(\left(\dfrac{6-2\sqrt{2}}{3-\sqrt{2}}-\dfrac{5}{\sqrt{5}}\right).\dfrac{1}{2-\sqrt{5}}\)
d.\(\sqrt{\left(2-\sqrt{5}\right)^2-\sqrt{5}}\)
a) \(\text{2}\sqrt{\text{18}}-9\sqrt{50}+3\sqrt{8}\)
= \(\text{6}\sqrt{\text{2}}-45\sqrt{2}+6\sqrt{2}\)
= \(-33\sqrt{2}\)
b) = \(7-2.\sqrt{7}.\sqrt{3}+3+7.2\sqrt{21}\)
= \(10-2\sqrt{21}+14\sqrt{21}\)
= \(10+12\sqrt{21}\)
Tính:
\(a.\) \(A=\sqrt{12}-2\sqrt{48}+\dfrac{7}{5}\sqrt{75}\)
\(b.\) \(B=\sqrt{14-6\sqrt{5}}+\sqrt{\left(2-\sqrt{5}\right)^2}\)
\(c.\) \(C=\left(\sqrt{6}-\sqrt{2}\right)\sqrt{2+\sqrt{3}}\)
\(d.\) \(D=\dfrac{5+\sqrt{5}}{\sqrt{5}+2}+\dfrac{\sqrt{5}-5}{\sqrt{5}}-\dfrac{11}{2\sqrt{5}+3}\)
a)A=\(2\sqrt{3}-8\sqrt{3}+7\sqrt{3}=\sqrt{3}\)
b)B\(=\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(2-\sqrt{5}\right)^2}=3-\sqrt{5}+\sqrt{5}-2=1\)
d)\(=\dfrac{\left(5+\sqrt{5}\right)\left(\sqrt{5}-2\right)}{1}+1-\sqrt{5}-\dfrac{11\left(2\sqrt{5}-3\right)}{11}=5\sqrt{5}+5-10-2\sqrt{5}+1-\sqrt{5}-2\sqrt{5}+3=-1\)
Cho a và b là các số dương. Đơn giản các biểu thức sau :
a) \(\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}\)
b) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\)
c) \(\left(\sqrt[3]{a}+\sqrt[3]{b}\right)\left(a^{\dfrac{2}{3}}+b^{\dfrac{2}{3}}-\sqrt[3]{ab}\right)\)
d) \(\left(a^{\dfrac{1}{3}}+b^{\dfrac{1}{3}}\right):\left(2+\sqrt[3]{\dfrac{a}{b}}+\sqrt[3]{\dfrac{b}{a}}\right)\)
a)
\(A=\dfrac{a^{\dfrac{4}{3}}\left(a^{-\dfrac{1}{3}}+a^{\dfrac{2}{3}}\right)}{a^{\dfrac{1}{4}}\left(a^{\dfrac{3}{4}}+a^{-\dfrac{1}{4}}\right)}=\dfrac{a^{\left(\dfrac{4}{3}-\dfrac{1}{3}\right)+}a^{\left(\dfrac{4}{3}+\dfrac{2}{3}\right)}}{a^{\left(\dfrac{1}{4}+\dfrac{3}{4}\right)}+a^{\left(\dfrac{1}{4}-\dfrac{1}{4}\right)}}=\dfrac{a+a^2}{a+1}=\dfrac{a\left(a+1\right)}{a+1}\)
\(a>0\Rightarrow a+1\ne0\) \(\Rightarrow A=a\)