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LG a
3√27−3√−8−3√125273−−83−1253
Phương pháp giải:
Tính từng căn bậc ba rồi thực hiện phép tính
Lời giải chi tiết:
3√27−3√−8−3√125=3√33−3√(−2)3−3√53273−−83−1253=333−(−2)33−533
=3−(−2)−5=3−(−2)−5
=3+2−5=0=3+2−5=0.
LG b
3√1353√5−3√54.3√4135353−543.43
Phương pháp giải:
Sử dụng các công thức:
3√a.b=3√a.3√ba.b3=a3.b3.
3√ab=3√a3√bab3=a3b3, với b≠0b≠0.
Lời giải chi tiết:
3√1353√5−3√54.3√4=3√27.53√5−3√54.4135353−543.43=27.5353−54.43
=3√5.3√273√5−3√216=53.27353−2163
=3√27−3√216=273−2163
=3√33−3√63=333−633
=3−6=−3=3−6=−3.
a) \(\sqrt{3}-2\sqrt{48}+3\sqrt{75}-4\sqrt{108}\)
= \(\sqrt{3}-8\sqrt{3}+15\sqrt{3}-24\sqrt{3}\)
= \(-16\sqrt{3}\)
b) \(\left(a.\sqrt{\dfrac{a}{b}}+2\sqrt{ab}+b.\sqrt{\dfrac{b}{a}}\right)\sqrt{\dfrac{a}{b}}\)
= \(\dfrac{a^2}{b}+2a+b\) = \(\dfrac{a^2+\left(2a+b\right)b}{b}\) = \(\dfrac{a^2+2ab+b^2}{b}\) = \(\dfrac{\left(a+b\right)^2}{b}\)
c) \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\) = \(3+2-5=0\)
d) \(3+\sqrt{18}+\sqrt{3}+\sqrt{8}\) = \(3+3\sqrt{2}+\sqrt{3}+2\sqrt{2}\)
= \(3+\sqrt{3}+5\sqrt{2}\)
a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)
b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)
c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)
d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
b: Ta có: \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}\cdot\sqrt[3]{4}\)
\(=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54\cdot4}\)
=3-6
=-3
Bài 1:
a: \(5\sqrt{8}-4\sqrt{27}-2\sqrt{75}+\sqrt{108}\)
\(=5\cdot2\sqrt{2}-4\cdot3\sqrt{3}-2\cdot5\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-12\sqrt{3}-10\sqrt{3}+6\sqrt{3}\)
\(=10\sqrt{2}-16\sqrt{3}\)
b: \(\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(1-\sqrt{6}\right)^2}\)
\(=\left|3-\sqrt{6}\right|+\left|1-\sqrt{6}\right|\)
\(=3-\sqrt{6}+\sqrt{6}-1\)
=3-1=2
c: \(\dfrac{5\sqrt{3}-3\sqrt{5}}{\sqrt{5}-\sqrt{3}}+\dfrac{1}{4+\sqrt{15}}\)
\(=\dfrac{\sqrt{15}\left(\sqrt{5}-\sqrt{3}\right)}{\sqrt{5}-\sqrt{3}}+\dfrac{1\left(4-\sqrt{15}\right)}{16-15}\)
\(=\sqrt{15}+4-\sqrt{15}=4\)
d: \(\dfrac{2\sqrt{3-\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)
\(=\dfrac{\sqrt{3-\sqrt{5}}\cdot\sqrt{2}\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)
\(=\dfrac{\sqrt{6-2\sqrt{5}}\cdot\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=\sqrt{\left(\sqrt{5}-1\right)^2}\cdot\dfrac{\left(3+\sqrt{5}\right)}{\sqrt{5}-1}-\dfrac{\sqrt{5}}{2}\)
\(=3+\sqrt{5}-\dfrac{\sqrt{5}}{2}=3+\dfrac{\sqrt{5}}{2}\)
Bài 2:
Vẽ đồ thị:
Phương trình hoành độ giao điểm là:
\(\dfrac{1}{2}x-4=-3x+3\)
=>\(\dfrac{1}{2}x+3x=3+4\)
=>\(\dfrac{7}{2}x=7\)
=>x=2
Thay x=2 vào y=-3x+3, ta được:
\(y=-3\cdot2+3=-3\)
Vậy: (d1) cắt (d2) tại A(2;-3)
Bài 1 :
a, ĐKXĐ : \(\dfrac{2x+1}{x^2+1}\ge0\)
Mà \(x^2+1\ge1>0\)
\(\Rightarrow2x+1\ge0\)
\(\Rightarrow x\ge-\dfrac{1}{2}\)
Vậy ...
b, Ta có : \(\sqrt[3]{-27}+\sqrt[3]{64}-\sqrt[3]{-\dfrac{128}{2}}\)
\(=-3+4-\left(-4\right)=-3+4+4=5\)
Bài 2 :
\(a,=2\sqrt{5}+6\sqrt{5}+5\sqrt{5}-12\sqrt{5}\)
\(=\sqrt{5}\left(2+6+5-12\right)=\sqrt{2}\)
\(b,=\sqrt{5}+\sqrt{5}+\left|\sqrt{5}-2\right|\)
\(=2\sqrt{5}+\sqrt{5}-2=3\sqrt{5}-2\)
\(c,=\dfrac{\left(5+\sqrt{5}\right)^2+\left(5-\sqrt{5}\right)^2}{\left(5-\sqrt{5}\right)\left(5+\sqrt{5}\right)}\)
\(=\dfrac{25+10\sqrt{5}+5+25-10\sqrt{5}+5}{25-5}\)
\(=3\)
a)\(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
\(=3+2-5\)
\(=0\)
b)\(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
\(=\sqrt[3]{\frac{153}{5}}-\sqrt[3]{54.4}\)
\(=\sqrt[3]{\frac{153}{5}}-6\)
Theo mình câu b như vậy
pham trung thanh câu b bn làm thiếu hay sao ý? Theo tôi, cả bài làm như thế này.
Giải:
a, \(\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{125}\)
\(=\sqrt[3]{27}-\sqrt[3]{-8}-\sqrt[3]{12}=3+2-5\)
\(=0\)
b, \(\frac{\sqrt[3]{153}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}\)
\(=\sqrt[3]{\frac{135}{5}}-\sqrt[3]{54.4}\)
\(=\sqrt[3]{27}-\sqrt[3]{216}\)
\(=3-6\)
\(=-3\)
Bài 68 :
a ) \(\sqrt[3]{27}-\sqrt[3]{8}-\sqrt[3]{125}=3-2-5=-4\)
b ) \(\dfrac{\sqrt[3]{135}}{\sqrt[3]{5}}-\sqrt[3]{54}.\sqrt[3]{4}=\sqrt[3]{\dfrac{135}{5}}-\sqrt[3]{54.4}=\sqrt[3]{27}-\sqrt[3]{216}=3-6=-3\)
Bài 69 :
a ) Ta có : \(\left\{{}\begin{matrix}3^3=27\\\left(\sqrt[3]{123}\right)^3=123\end{matrix}\right.\)
Vì 27 < 123 nên suy ra \(3< \sqrt[3]{123}\)
Vậy \(3< \sqrt[3]{123}\)