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a, \(-\frac{2}{3}\sqrt{ab}=-\sqrt{\frac{4ab}{9}}\)
b, \(a\sqrt{\frac{3}{a}}=\sqrt{\frac{3a^2}{a}}=\sqrt{3a}\)
c, \(a\sqrt{7}=\sqrt{7a^2}\)
d, \(b\sqrt{3}=\sqrt{3b^2}\)
e, \(ab\sqrt{\frac{a}{b}}=\sqrt{\frac{a^3b^2}{b}}=\sqrt{a^3b}\)
f, \(ab\sqrt{\frac{1}{a}+\frac{1}{b}}=\sqrt{\frac{a^2b^2}{a}+\frac{a^2b^2}{b}}=\sqrt{ab^2+a^2b}\)
d: \(D=\dfrac{2}{x^2-y^2}\cdot\sqrt{\dfrac{9\left(x^2+2xy+y^2\right)}{4}}\)
\(=\dfrac{2}{\left(x-y\right)\left(x+y\right)}\cdot\dfrac{3\left(x+y\right)}{2}\)
\(=\dfrac{3}{x-y}\)
em xin lỗi nhưng em chưa đủ tuổi để làm bài này xin cáo từ
xin lỗi quản lý olm ạ
a) Ta có:
f(−2)=23.(−2)=−43;f(−1)=23.(−1)=−23;f(0)=23.0=0;f(12)=23.12=13;f(1)=23.1=23;f(2)=23.2=43;f(3)=23.3=2.f(−2)=23.(−2)=−43;f(−1)=23.(−1)=−23;f(0)=23.0=0;f(12)=23.12=13;f(1)=23.1=23;f(2)=23.2=43;f(3)=23.3=2.
b) Ta có:
g(−2)=23.(−2)+3=53;g(−1)=23.(−1)+3=73;g(0)=23.0+3=3;g(12)=23.12+3=103;g(1)=23.1+3=113;g(2)=23.2+3=133;g(3)=23.3+3=5.g(−2)=23.(−2)+3=53;g(−1)=23.(−1)+3=73;g(0)=23.0+3=3;g(12)=23.12+3=103;g(1)=23.1+3=113;g(2)=23.2+3=133;g(3)=23.3+3=5.
c) Khi biến xx lấy cùng một giá trị thì giá trị của hàm số y=f(x)y=f(x) luôn nhỏ hơn giá trị tương ứng của hàm số y=g(x)y=g(x) là 3 đơn vị.
a) \(E=2\sqrt{40\sqrt{12}}+3\sqrt{5\sqrt{48}}-2\sqrt{\sqrt{75}}-4\sqrt{15\sqrt{27}}.\)
\(=8\sqrt{5\sqrt{3}}+6\sqrt{5\sqrt{3}}-2\sqrt{5\sqrt{3}-12\sqrt{5\sqrt{3}}}\)
\(=0\)
b) \(F=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}+\frac{1}{\sqrt{3}}\sqrt{\frac{5}{12}-\frac{1}{\sqrt{6}}}.\)
Vì \(=\frac{5}{12}-\frac{1}{\sqrt{6}}=\frac{5-2\sqrt{6}}{12}=\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}\)
\(\frac{1}{\sqrt{3}}+\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{3}+\frac{\sqrt{2}}{6}=\frac{2\sqrt{3}+\sqrt{2}}{6}\)
Nên \(F=\frac{2\sqrt{3}+\sqrt{2}}{6}+\frac{1}{\sqrt{3}}\sqrt{\frac{\left(\sqrt{3}-\sqrt{2}\right)^2}{12}}=\frac{2\sqrt{3}+\sqrt{2}+\sqrt{3}-\sqrt{2}}{6}=\frac{3\sqrt{3}}{6}=\frac{\sqrt{3}}{2}\)
9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)
10: \(A=\dfrac{\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)
11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)
12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)
\(=\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)\)
=9-3=6
13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)
\(=\sqrt{5}-2-3+\sqrt{5}=2\sqrt{5}-5\)
a) \(A=2\sqrt{8}-3\sqrt{32}+\sqrt{50}\)
\(A=2\sqrt{4.2}-3\sqrt{16.2}+\sqrt{25.2}\)
\(A=2.2\sqrt{2}-3.4\sqrt{2}+5\sqrt{2}\)
\(A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}\)
\(A=\left(4-12+5\right)\sqrt{2}\)
\(A=-3\sqrt{2}\)
b) \(B=\sqrt{12}+4\sqrt{27}-3\sqrt{48}\)
\(B=\sqrt{4.3}+4\sqrt{9.3}-3\sqrt{16.3}\)
\(B=2\sqrt{3}+4.3\sqrt{3}-3.4\sqrt{3}\)
\(B=2\sqrt{3}\)
c) \(C=\sqrt{20a}+4\sqrt{45a}-2\sqrt{125a}\left(a\ge0\right)\)
\(C=\sqrt{4.5a}+4\sqrt{9.5a}-2\sqrt{25.5a}\)
\(C=2\sqrt{5a}+4.3\sqrt{5a}-2.5\sqrt{5a}\)
\(C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}\)
\(C=\left(2+12-10\right)\sqrt{5a}\)
\(C=4\sqrt{5a}\)
a) ta có \(2\sqrt{8}=2\sqrt{4.2}=4\sqrt{2},3\sqrt{32}=3\sqrt{16.2}=12\sqrt{2},\sqrt{50}=\sqrt{25.2}=5\sqrt{2}\) \(\Rightarrow A=4\sqrt{2}-12\sqrt{2}+5\sqrt{2}=-3\sqrt{2}\) b) ta có \(\sqrt{12}=\sqrt{4.3}=2\sqrt{3},4\sqrt{27}=4\sqrt{9.3}=12\sqrt{3},3\sqrt{48}=3\sqrt{16.3}=12\sqrt{3}\Rightarrow B=2\sqrt{3}+12\sqrt{3}-12\sqrt{3}=26\sqrt{3}\)c) ta có \(\sqrt{20a}=\sqrt{4.5a}=2\sqrt{5a},4\sqrt{45a}=4\sqrt{9.5a}=12\sqrt{5a},2\sqrt{125a}=2\sqrt{25.5a}=10\sqrt{5a}\Rightarrow C=2\sqrt{5a}+12\sqrt{5a}-10\sqrt{5a}=4\sqrt{5a}\)
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) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)
30,001x3=3(0,1x)3=0,1x;
\sqrt[3]{-125 a^{12}}=\sqrt[3]{\left(-5 a^{4}\right)^{3}}=-5 a^{4};3−125a12=3(−5a4)3=−5a4;
\sqrt[3]{27 x^{6}}=\sqrt[3]{\left(3 x^{2}\right)^{3}}=3 x^{2};327x6=3(3x2)3=3x2;
\sqrt[3]{-0,343 a^{3}}=\sqrt[3]{(-0,7 a)^{3}}=-0,7 a;3−0,343a3=3(−0,7a)3=−0,7a;
Ta rút gọn các biểu thức như sau:
\(\sqrt[3]{0,001x^3}=\sqrt[3]{\left(0,1x\right)^3}=0,1x.\)
\(\sqrt[3]{-125a^{12}}=\sqrt[3]{\left(-5a^4\right)^3}=-5a^4\)
\(\sqrt[3]{27x^6}=\sqrt[3]{\left(3x^2\right)^3}=3x^2\)
\(\sqrt[3]{-0,343a^3}=\sqrt[3]{\left(-0,7a\right)^3}=-0,7a\)