\(\sqrt[3]{64}+\sqrt[3]{-127}\)
K
Khách
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29 tháng 8 2021
26: \(x^2-\sqrt{x}+x-1\)
\(=\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)+\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x\sqrt{x}+x+\sqrt{x}+\sqrt{x}+1\right)\)
\(=\left(\sqrt{x}-1\right)\left(x\sqrt{x}+x+2\sqrt{x}+1\right)\)
30 tháng 8 2021
25: Ta có: \(-6x+7\sqrt{x}-2\)
\(=-6x+3\sqrt{x}+4\sqrt{x}-2\)
\(=-3\sqrt{x}\left(2\sqrt{x}-1\right)+2\left(2\sqrt{x}-1\right)\)
\(=\left(2\sqrt{x}-1\right)\left(2-3\sqrt{x}\right)\)
27: Ta có: \(2a-5\sqrt{ab}+3b\)
\(=2a-2\sqrt{ab}-3\sqrt{ab}+3b\)
\(=2\sqrt{a}\left(\sqrt{a}-\sqrt{b}\right)-3\sqrt{b}\left(\sqrt{a}-\sqrt{b}\right)\)
\(=\left(\sqrt{a}-\sqrt{b}\right)\left(2\sqrt{a}-3\sqrt{b}\right)\)
28: Ta có: \(\sqrt{ab}+2\sqrt{a}+3\sqrt{b}+6\)
\(=\sqrt{a}\left(\sqrt{b}+2\right)+3\left(\sqrt{b}+2\right)\)
\(=\left(\sqrt{b}+2\right)\left(\sqrt{a}+3\right)\)
30 tháng 6 2021
a) Ta có: \(-3\sqrt{16}\cdot\sqrt{90}\)
\(=-3\cdot4\cdot3\sqrt{10}\)
\(=-36\sqrt{10}\)
b) Ta có: \(3\sqrt{\dfrac{4}{3}}-3\sqrt{48}+5\sqrt{75}\)
\(=3\cdot\dfrac{2}{\sqrt{3}}-3\cdot4\sqrt{3}+5\cdot5\sqrt{3}\)
\(=2\sqrt{3}-12\sqrt{3}+25\sqrt{3}\)
\(=15\sqrt{3}\)
c) Ta có: \(4\sqrt[3]{27}-\sqrt[3]{64}-2\sqrt[3]{8}\)
\(=4\cdot3-4-2\cdot2\)
\(=12-4-4=4\)
30 tháng 8 2021
b: Ta có: \(\sqrt[3]{-0.008}-\dfrac{1}{5}\cdot\sqrt[3]{64}+5\cdot\sqrt[3]{\left(-5\right)^3}\)
\(=-\dfrac{1}{5}-\dfrac{1}{5}\cdot4+5\cdot\left(-5\right)\)
\(=-\dfrac{1}{5}-\dfrac{4}{5}-25\)
=-26
DM
1
22 tháng 7 2023
a: Sửa đề: căn 6+2căn 5-căn 5
\(a=\dfrac{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}{\sqrt{5}+1-\sqrt{5}}=\dfrac{2}{1}=2\)
b: \(a^3=2-\sqrt{3}+2+\sqrt{3}+3a\)
=>a^3-3a-4=0
=>a^3-3a=4
\(\dfrac{64}{\left(a^2-3\right)^3}-3a=\left(\dfrac{4}{a^2-3}\right)^3-3a\)
\(=\left(\dfrac{a^3-3a}{a^2-3}\right)^3-3a=a^3-3a\)
=4
13 tháng 8 2023
a: \(=\dfrac{5}{4}\cdot3=\dfrac{15}{4}\)
b: \(=\sqrt[5]{\dfrac{98}{64}\cdot343}=\sqrt[5]{\left(\dfrac{7}{2}\right)^5}=\dfrac{7}{2}\)
NC
1
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 6 2021
\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)
\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)
\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)
\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)
LN
9 tháng 7 2021
\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)
Sửa đề: \(\sqrt[3]{64}+\sqrt[3]{-27}\)
Ta có: \(\sqrt[3]{64}+\sqrt[3]{-27}\)
\(=4+\left(-3\right)=1\)