rút gọn biểu thức
\(D=\left(\sqrt{a}\right)^7.\left(\sqrt[3]{a}\right).\left(\sqrt[4]{a}\right)^7\) (a>0)
\(D=a^{\sqrt{2}-1}.\left(a^2\right)^{\sqrt{2}}.\left(a^3\right)^{1-\sqrt{2}}\)
Những câu hỏi liên quan
Rút gọn biểu thức:
\(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(B=3\sqrt{\left(1,5\right)^2}-4\sqrt{\left(3-\sqrt{2}\right)^2}\)
\(A=\left|2-\sqrt{7}\right|+7-2\sqrt{7}+1\)
\(=\sqrt{7}-2+8-2\sqrt{7}\) \(=6-\sqrt{7}\)
\(B=3\cdot1,5-4\cdot\left|3-\sqrt{2}\right|\) \(=4,5-4\left(3-\sqrt{2}\right)\)
\(=4,5-12+4\sqrt{2}\) \(=4\sqrt{2}-7,5\)
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Ta có: \(A=\sqrt{\left(2-\sqrt{7}\right)^2}+\left(\sqrt{7}-1\right)^2\)
\(=\sqrt{7}-2+8-2\sqrt{7}\)
\(=6-\sqrt{7}\)
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rút gọn biểu thứcGdfrac{sqrt[3]{a}.a^{dfrac{2}{3}}}{left(a^{4-2sqrt{3}}right)^{4+2sqrt{3}}}Gdfrac{a^{sqrt{7}+1}.a^{2-sqrt{7}}}{left(a^{sqrt{2}-2}right)^{sqrt{2}+2}}Hdfrac{a^2.left(a^{-2}.b^3right).b^{-1}}{left(a^{-1}.bright)^3.a^{-5}.b^{-2}}Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2} Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2} Hdfrac{b^3.a^{-4}.left(ab^2right)^3}{left(a^2right)^{-2}.left(ab^3right)^2.b^2}
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rút gọn biểu thức
\(G=\dfrac{\sqrt[3]{a}.a^{\dfrac{2}{3}}}{\left(a^{4-2\sqrt{3}}\right)^{4+2\sqrt{3}}}\)
\(G=\dfrac{a^{\sqrt{7}+1}.a^{2-\sqrt{7}}}{\left(a^{\sqrt{2}-2}\right)^{\sqrt{2}+2}}\)
\(H=\dfrac{a^2.\left(a^{-2}.b^3\right).b^{-1}}{\left(a^{-1}.b\right)^3.a^{-5}.b^{-2}}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
\(H=\dfrac{b^3.a^{-4}.\left(ab^2\right)^3}{\left(a^2\right)^{-2}.\left(ab^3\right)^2.b^2}\)
Rút gọn các biểu thức sau :a) Aleft(0,04right)^{-1,5}-left(0,125right)^{frac{-2}{3}}b) Bleft(6^{frac{-2}{7}}right)^{-7}-left[left(left(0,2right)^{0,75}right)^{-4}right]c) Cfrac{a^{sqrt{5}+3}.a^{sqrt{5}left(sqrt{5}-1right)}}{left(a^{2sqrt{2}-1}right)^{2sqrt{2}+1}}d) Dleft(a^{frac{1}{2}}-b^{frac{1}{2}}right)^2:left(b-2bsqrt{frac{b}{a}}+frac{b^2}{a}right)left(a,b0right)
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Rút gọn các biểu thức sau :
a) \(A=\left(0,04\right)^{-1,5}-\left(0,125\right)^{\frac{-2}{3}}\)
b) \(B=\left(6^{\frac{-2}{7}}\right)^{-7}-\left[\left(\left(0,2\right)^{0,75}\right)^{-4}\right]\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\left(a,b>0\right)\)
a) \(A=\left[\left(\frac{1}{5}\right)^2\right]^{\frac{-3}{2}}-\left[2^{-3}\right]^{\frac{-2}{3}}=5^3-2^2=121\)
b) \(B=6^2+\left[\left(\frac{1}{5}\right)^{\frac{3}{4}}\right]^{-4}=6^2+5^3=161\)
c) \(C=\frac{a^{\sqrt{5}+3}.a^{\sqrt{5}\left(\sqrt{5}-1\right)}}{\left(a^{2\sqrt{2}-1}\right)^{2\sqrt{2}+1}}=\frac{a^{\sqrt{5}+3}.a^{5-\sqrt{5}}}{a^{\left(2\sqrt{2}\right)^2-1^2}}\)
\(=\frac{a^{\sqrt{5}+3+5-\sqrt{5}}}{a^{8-1}}=\frac{a^8}{a^7}=a\)
d) \(D=\left(a^{\frac{1}{2}}-b^{\frac{1}{2}}\right)^2:\left(b-2b\sqrt{\frac{b}{a}}+\frac{b^2}{a}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left[1-2\sqrt{\frac{b}{a}}+\left(\sqrt{\frac{b}{a}}\right)^2\right]\)
\(=\left(\sqrt{a}-\sqrt{b}\right)^2:b\left(1-\sqrt{b}a\right)^2\)
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Rút gọn các biểu thức:
\(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right):\dfrac{x-4}{3\sqrt{x}}\)
\(B=\left(\dfrac{\sqrt{a}}{\sqrt{a}-2}+\dfrac{1}{\sqrt{a}+2}+\dfrac{6-7\sqrt{a}}{a-4}\right).\left(\sqrt{a}+2\right)\)
a: Ta có: \(A=\left(\dfrac{1}{\sqrt{x}+2}+\dfrac{1}{\sqrt{x}-2}\right)\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{\sqrt{x}-2+\sqrt{x}+2}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{x-4}{3\sqrt{x}}\)
\(=\dfrac{2}{3}\)
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rút gọn biểu thức
\(E=\left(a^{12}b^3\right):\left(a^4b^7\right)\)
\(E=\left(a.b^3\right)^4:a^6\)
\(F=\dfrac{\left(a^{\sqrt{3}-1}\right)^{\sqrt{3}+1}}{a^{\sqrt{5}-3}.a^{4-\sqrt{5}}}\) (a>0)
\(E=a^{12-4}.b^{3-7}=\dfrac{a^8}{b^4}\)
\(E=a^{4-6}.b^{3.4}=\dfrac{b^{12}}{a^2}\)
\(F=\dfrac{a^{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}}{a^{\left(\sqrt{5}-3\right)+\left(4-\sqrt{5}\right)}}=\dfrac{a^2}{a^1}=a\)
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Thực hiện phép tính (rút gọn biểu thức) a)sqrt{left(3+sqrt{2}right)^2}-sqrt{left(3-2sqrt{2}right)^2}b) sqrt{left(sqrt{7}-2sqrt{2}right)^2}-sqrt{left(sqrt{7}+2sqrt{2}right)^2}c)sqrt{left(3+sqrt{5}right)^2}+sqrt{left(3-sqrt{5}right)^2}d) sqrt{left(2-sqrt{3}right)^2}-sqrt{left(2+sqrt{3}right)^2}
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Thực hiện phép tính (rút gọn biểu thức)
a)\(\sqrt{\left(3+\sqrt{2}\right)^2}\)-\(\sqrt{\left(3-2\sqrt{2}\right)^2}\)
b) \(\sqrt{\left(\sqrt{7}-2\sqrt{2}\right)^2}\)-\(\sqrt{\left(\sqrt{7}+2\sqrt{2}\right)^2}\)
c)\(\sqrt{\left(3+\sqrt{5}\right)^2}\)+\(\sqrt{\left(3-\sqrt{5}\right)^2}\)
d) \(\sqrt{\left(2-\sqrt{3}\right)^2}\)-\(\sqrt{\left(2+\sqrt{3}\right)^2}\)
Lời giải:
a. $=|3+\sqrt{2}|-|3-2\sqrt{2}|=(3+\sqrt{2})-(3-2\sqrt{2})$
$=3\sqrt{2}$
b. $=|\sqrt{7}-2\sqrt{2}|-|\sqrt{7}+2\sqrt{2}|$
$=(2\sqrt{2}-\sqrt{7})-(\sqrt{7}+2\sqrt{2})$
$=-2\sqrt{7}$
c.
$=|3+\sqrt{5}|+|3-\sqrt{5}|=(3+\sqrt{5})+(3-\sqrt{5})=6$
d.
$=|2-\sqrt{3}|-|2+\sqrt{3}|=(2-\sqrt{3})-(2+\sqrt{3})=-2\sqrt{3}$
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Rút gọn biểu thức :
\(\frac{\sqrt{7-4\sqrt{3}}}{\sqrt{2-\sqrt{3}}}.\sqrt{2+\sqrt{3}}\)
\(\left[\left(a-b\right)\sqrt{\frac{a+b}{a-b}}+a-b\right]\left(a-b\right)\left(\sqrt{\frac{a+b}{a-b}}-1\right)\)với a>b>0
Chứng minh rằng :
\(\left(4+\sqrt{15}\right)\left(\sqrt{10}-\sqrt{6}\right)\sqrt{4-\sqrt{15}}=2\)
Bài 1:
a: \(=\sqrt{\dfrac{7-4\sqrt{3}}{2-\sqrt{3}}}\cdot\sqrt{2+\sqrt{3}}\)
\(=\sqrt{2-\sqrt{3}}\cdot\sqrt{2+\sqrt{3}}=1\)
Bài 2:
\(VT=\left(4+\sqrt{15}\right)\cdot\left(\sqrt{5}-\sqrt{3}\right)\cdot\sqrt{8-2\sqrt{15}}\)
\(=\left(4+\sqrt{15}\right)\left(8-2\sqrt{15}\right)\)
\(=32-8\sqrt{15}+8\sqrt{15}-30=2\)
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Rút gọn các biểu thức sau:a) left(sqrt{8}-3sqrt{2}+sqrt{10}right)left(sqrt{2}-3sqrt{0,4}right) b) left(sqrt{28}-2sqrt{14}+sqrt{7}right).sqrt{7}+7sqrt{8}c) 2sqrt{left(sqrt{2}-3right)^2}+sqrt{2left(-3right)^2}-5sqrt{left(-1right)^4} d) left(frac{sqrt{14}-sqrt{7}}{1-sqrt{2}}+frac{sqrt{15}-sqrt{5}}{1-sqrt{3}}right):frac{1}{sqrt{7}-sqrt{5}}
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Rút gọn các biểu thức sau:
a) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\left(\sqrt{2}-3\sqrt{0,4}\right)\) b) \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right).\sqrt{7}+7\sqrt{8}\)
c) \(2\sqrt{\left(\sqrt{2}-3\right)^2}+\sqrt{2\left(-3\right)^2}-5\sqrt{\left(-1\right)^4}\) d) \(\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)
Rút gọn các biểu thức sau:Aleft(dfrac{1}{sqrt{x}-3}+dfrac{1}{sqrt{x}+3}right)left(1-dfrac{3}{sqrt{x}}right)Bleft(dfrac{sqrt{x}}{sqrt{x}-2}+dfrac{1}{sqrt{x}+2}+dfrac{6-7sqrt{x}}{x-4}right)left(sqrt{x}+2right)Cleft(dfrac{sqrt{a}}{sqrt{a}-1}-dfrac{sqrt{a}}{a-sqrt{1}}right):dfrac{sqrt{a}+1}{a-1}Dleft(dfrac{x-2}{x+2sqrt{x}}+dfrac{1}{sqrt{x}+2}right).dfrac{sqrt{x}+1}{sqrt{x}-1}Eleft(1+dfrac{x+sqrt{x}}{sqrt{x}+1}right)left(1+dfrac{x-sqrt{x}}{1-sqrt{x}}right)giúp mình với ạ!mình đang cần gấp
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Rút gọn các biểu thức sau:
\(A=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{1}{\sqrt{x}+3}\right)\left(1-\dfrac{3}{\sqrt{x}}\right)\)
\(B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-2}+\dfrac{1}{\sqrt{x}+2}+\dfrac{6-7\sqrt{x}}{x-4}\right)\left(\sqrt{x}+2\right)\)
\(C=\left(\dfrac{\sqrt{a}}{\sqrt{a}-1}-\dfrac{\sqrt{a}}{a-\sqrt{1}}\right):\dfrac{\sqrt{a}+1}{a-1}\)
\(D=\left(\dfrac{x-2}{x+2\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\right).\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
\(E=\left(1+\dfrac{x+\sqrt{x}}{\sqrt{x}+1}\right)\left(1+\dfrac{x-\sqrt{x}}{1-\sqrt{x}}\right)\)
giúp mình với ạ!mình đang cần gấp
1. ĐKXĐ: $x>0; x\neq 9$
\(A=\frac{\sqrt{x}+3+\sqrt{x}-3}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2\sqrt{x}}{(\sqrt{x}-3)(\sqrt{x}+3)}.\frac{\sqrt{x}-3}{\sqrt{x}}=\frac{2}{\sqrt{x}+3}\)
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2. ĐKXĐ: $x\geq 0; x\neq 4$
\(B=\left[\frac{\sqrt{x}(\sqrt{x}+2)+\sqrt{x}-2}{(\sqrt{x}-2)(\sqrt{x}+2)}+\frac{6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}\right](\sqrt{x}+2)\)
\(=\frac{x+3\sqrt{x}-2+6-7\sqrt{x}}{(\sqrt{x}-2)(\sqrt{x}+2)}.(\sqrt{x}+2)=\frac{x-4\sqrt{x}+4}{\sqrt{x}-2}=\frac{(\sqrt{x}-2)^2}{\sqrt{x}-2}=\sqrt{x}-2\)
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3. ĐKXĐ: $a\geq 0; a\neq 1$
\(C=\frac{\sqrt{a}(\sqrt{a}+1)-\sqrt{a}}{(\sqrt{a}+1)(\sqrt{a}-1)}:\frac{\sqrt{a}+1}{(\sqrt{a}-1)(\sqrt{a}+1)}\)
\(\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}:\frac{1}{\sqrt{a}-1}=\frac{a}{(\sqrt{a}-1)(\sqrt{a}+1)}.(\sqrt{a}-1)=\frac{a}{\sqrt{a}+1}\)
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