tính
a) \(log_{\sqrt{2}}\sqrt{2};log_77\)
b) \(log_{10}1;log_91\)
c) \(3^{log_315};7^{log_7\sqrt{2}}\)
d) \(log_88^{-10};log_55^{\sqrt{3}}\)
Những câu hỏi liên quan
Tìm TXĐ:
a) y=\(\left(1-x\right)^{\dfrac{-1}{3}}\)
b) \(y=\sqrt{\log_{0,5}\dfrac{2x+1}{x+5}-2}\)
c) \(y=\log_{10}\sqrt{x^2-x-12}\)
d) \(y=\sqrt{\log_{10}x-1+\log_{10}x+1}\)
1.rút gọn A=3\(\log_4\sqrt{a}\)- \(\log_{\dfrac{1}{2}}a^2\)+ 2\(\log_{\sqrt{2}}a\)
2.bt \(\log_23=a\). tính \(\log_{12}36\) theo a
1.
\(A=3log_{2^2}\sqrt{a}-log_{2^{-1}}a^2+2log_{a^{\dfrac{1}{2}}}a\)
\(=3.\dfrac{1}{2}.\dfrac{1}{2}log_2a-\left(-1\right).2.log_2a+2.2.log_2a\)
\(=\dfrac{27}{4}log_2a\)
2.
\(log_{12}36=\dfrac{log_236}{log_212}=\dfrac{log_2\left(3^2.2^2\right)}{log_2\left(3.2^2\right)}=\dfrac{log_23^2+log_22^2}{log_23+log_22^2}\)
\(=\dfrac{2.log_23+2}{log_23+2}=\dfrac{2a+2}{a+2}\)
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28 tháng 7 2021 lúc 9:22
\(log_{\sqrt{3}}\left(\sqrt[5]{3}\right)=?\)
\(log_24.log_{\dfrac{1}{4}}2=?\)
m=? để \(log_{2\sqrt{2}+\sqrt{7}}\left(x-m+1\right)log_{2\sqrt{2}-\sqrt{7}}\left(mx-x^2\right)=0\)có nghiệm
ban solo voi minh khong
\((\log_{2} (4x))^2-\log_{\sqrt{}2} (2x)=5\)
\(\left[log_24x\right]^2-log_{\sqrt{2}}2x=5\)
=>\(\left[log_2\left(2\cdot2x\right)\right]^2-log_{2^{\dfrac{1}{2}}}2x=5\)
=>\(\left[1+log_22x\right]^2-1:\dfrac{1}{2}\cdot log_22x=5\)
=>\(\left(log_22x\right)^2+2\cdot log_22x+1-2\cdot log_22x=5\)
=>\(\left(log_22x\right)^2=4\)
=>\(\left[{}\begin{matrix}log_22x=2\\log_22x=-2\left(loại\right)\end{matrix}\right.\Leftrightarrow log_22x=2\)
=>\(2x=2^2=4\)
=>x=2
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Bất phương trình logarit
$$1) \sqrt{log_{1/2}^{2} \frac{2x}{4-x} - 4} \leq \sqrt{5}$$
$$2)log_{2}(x-1)^{2} > 2log_{2} (x^{3} +x +1)$$
$$3)\frac{1}{log_{2}(4x)^{2} +3 } + \frac{1}{log_{4} 16x^{3}-2} <-1$$
$$4)log_{2} (4^{x}+4) < log_{\frac{1}{2}} (2^{x+1} -2)$$
Bài 1: thực hiện phép tínha) sqrt{252}-sqrt{700}+sqrt{1008}-sqrt{448}Bài 2: Tínha) dfrac{sqrt{99999}}{sqrt{11111}} b) dfrac{sqrt[]{84^2-37^2}}{sqrt[]{47}} c) sqrt{dfrac{5left(38^2-17^2right)}{8left(47^2-19^2right)}} d) dfrac{sqrt{0,2.1,21.0,3}}{sqrt{7,5.3,2.0,64}}Bài 3: Tính (viết dưới dạng tích dưới dấu căn bậc hai)a) sqrt{27^2-23^2}b) sqrt{37^2-35^2}c)sqrt{65^2-63^2}d) sqrt{117^2-108^2}
Đọc tiếp
Bài 1: thực hiện phép tính
a) \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
Bài 2: Tính
a) \(\dfrac{\sqrt{99999}}{\sqrt{11111}}\)
b) \(\dfrac{\sqrt[]{84^2-37^2}}{\sqrt[]{47}}\)
c) \(\sqrt{\dfrac{5\left(38^2-17^2\right)}{8\left(47^2-19^2\right)}}\)
d) \(\dfrac{\sqrt{0,2.1,21.0,3}}{\sqrt{7,5.3,2.0,64}}\)
Bài 3: Tính (viết dưới dạng tích dưới dấu căn bậc hai)
a) \(\sqrt{27^2-23^2}\)
b) \(\sqrt{37^2-35^2}\)
c)\(\sqrt{65^2-63^2}\)
d) \(\sqrt{117^2-108^2}\)
Bài 1:
a: \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)
\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)
\(=8\sqrt{7}\)
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Bài 3:
a: \(\sqrt{27^2-23^2}=10\sqrt{2}\)
b: \(\sqrt{37^2-35^2}=12\)
c: \(\sqrt{65^2-63^2}=16\)
d: \(\sqrt{117^2-108^2}=45\)
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19 tháng 8 2021 lúc 10:51
Tính
a,\(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
b,\(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
a: Ta có: \(\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(=\dfrac{\sqrt{4-2\sqrt{3}}-\sqrt{4+2\sqrt{3}}}{\sqrt{2}}\)
\(=\dfrac{\sqrt{3}-1-\sqrt{3}-1}{\sqrt{2}}=-\sqrt{2}\)
b: Ta có: \(\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(=\dfrac{\left(\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\right)}{\sqrt{2}}\)
\(=\dfrac{\sqrt{5}+1+3-\sqrt{5}-2}{\sqrt{2}}=\sqrt{2}\)
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26 tháng 6 2023 lúc 6:54
tính
a.\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)
b. \(\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\)
\(a,\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\\ =\sqrt{\sqrt{5^2}+2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}-\sqrt{\sqrt{5^2}-2\sqrt{5}.\sqrt{3}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\\ =\left|\sqrt{5}+\sqrt{3}\right|-\left|\sqrt{5}-\sqrt{3}\right|\\ =\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\\ =2\sqrt{3}\)
\(b,\sqrt{5+2\sqrt{6}}+\sqrt{5-2\sqrt{6}}\\ =\sqrt{\sqrt{2^2}+2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}+\sqrt{\sqrt{2^2}-2.\sqrt{3}.\sqrt{2}+\sqrt{3^2}}\\ =\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}+\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}\\ =\left|\sqrt{2}+\sqrt{3}\right|+\left|\sqrt{2}-\sqrt{3}\right|\\ =\sqrt{2}+\sqrt{3}-\sqrt{2}+\sqrt{3}=2\sqrt{3}\)
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a) \(\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)
\(=\sqrt{5-2\cdot\sqrt{5\cdot3}+3}-\sqrt{5+2\cdot\sqrt{5\cdot3}+1}\)
\(=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}\)
\(=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)
\(=-2\sqrt{3}\)
b. \(\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}\)
\(=\sqrt{2+2\cdot\sqrt{2}\cdot\sqrt{3}+3}-\sqrt{3-2\cdot\sqrt{2}+2}\)
\(=\sqrt{\left(\sqrt{2}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\left(\sqrt{2}+\sqrt{3}\right)-\left(\sqrt{3}-\sqrt{2}\right)\)
\(=\sqrt{2}+\sqrt{3}-\sqrt{3}+\sqrt{2}\)
\(=2\sqrt{2}\)
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