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a) \(\left(\dfrac{1}{16}\right)^{-\dfrac{3}{4}}+810000^{0.25}-\left(7\dfrac{19}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{4.\left(-\dfrac{3}{4}\right)}+\left(30\right)^{4.0,25}-\left(\dfrac{243}{32}\right)^{\dfrac{1}{5}}\)
\(=\left(\dfrac{1}{2}\right)^{-3}+30-\left(\dfrac{3}{2}\right)^{5.\dfrac{1}{5}}\)
\(=2^3+30-\dfrac{3}{2}\)
\(=36,5\)
b) \(=\left(0,1\right)^{3.\left(-\dfrac{1}{3}\right)}-2^{-2}.2^{6.\dfrac{2}{3}}-\left[\left(2\right)^3\right]^{-\dfrac{4}{3}}\)
\(=0,1^{-1}-2^2-2^{-4}\)
\(=10-4-\dfrac{1}{16}\)
\(=\dfrac{95}{16}\)
Lời giải:
Giả sử \(\log _{3}a=\log_4b=\log_{12}c=\log_{13}(a+b+c)=t\)
\(\Rightarrow 13^t=3^t+4^t+12^t\)
\(\Rightarrow \left ( \frac{3}{13} \right )^t+\left ( \frac{4}{13} \right )^t+\left ( \frac{12}{13} \right )^t=1\)
Xét vế trái , đạo hàm ta thấy hàm luôn nghịch biến nên phương trình có duy nhất một nghiệm \(t=2\)
Khi đó \(\log_{abc}144=\log_{144^t}144=\frac{1}{t}=\frac{1}{2}\)
Đáp án B
cho em hỏi tại sao lại có 3^t +4^t +12^t=13^t. Với lại em không hiểu chỗ tại sao hàm số nghịch biến. Và tại sao từ \(\log_{abc}144=\log144_{144^t}=\dfrac{1}{t}\)
a) =
=
b) =
=
=
. ( Với điều kiện b # 1)
c) \(\dfrac{a^{\dfrac{1}{3}}b^{-\dfrac{1}{3}-}a^{-\dfrac{1}{3}}b^{\dfrac{1}{3}}}{\sqrt[3]{a^2}-\sqrt[3]{b^2}}\)= =
=
( với điều kiện a#b).
d) \(\dfrac{a^{\dfrac{1}{3}}\sqrt{b}+b^{\dfrac{1}{3}}\sqrt{a}}{\sqrt[6]{a}+\sqrt[6]{b}}\) = =
=
=
a.
\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)
b.
\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)
c.
\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)
d.
\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)
e.
\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)
g.
\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)
Giải:
Gọi tọa độ điểm \(H=(a,b,c)\)
Ta có
\(\overrightarrow{AH}=(a,b,c-1)\perp \overrightarrow{BC}=(3,3,-1)\Rightarrow 3a+3b-(c-1)=0(1)\)
\(H\in BC\Rightarrow \) tồn tại \(k\in\mathbb{R}\) sao cho \(\overrightarrow {BH}=k\overrightarrow {BC}\)
\(\Leftrightarrow (a+1,b+2,c)=k(3,3,-1)\Rightarrow \frac{a+1}{3}=\frac{b+2}{3}=\frac{c}{-1}=k\)
\(\Rightarrow a=3k-1,b=3k-2,c=-k\)
Thay vào \((1)\Rightarrow 19k-8=0\rightarrow k=\frac{8}{19}\)
\(\Rightarrow (a,b,c)=\left(\frac{5}{19},\frac{-14}{19},\frac{-8}{19}\right)\)
Đáp án A.
a) \(\left(\dfrac{1}{2}\right)^n\le10^{-9}\)\(\Leftrightarrow2^{-n}\le10^{-9}\)\(\Leftrightarrow-n\le log^{10^{-9}}_2\)\(\Leftrightarrow-n\le-9log^{10}_2\)\(\Leftrightarrow n\ge9log^{10}_2\)\(\Leftrightarrow n\ge30\).
Vậy \(n=30\).
b) \(3-\left(\dfrac{7}{5}\right)^n\le0\)
\(\Leftrightarrow-\left(\dfrac{7}{5}\right)^n\le-3\)
\(\Leftrightarrow\left(\dfrac{7}{5}\right)^n\ge3\)\(\Leftrightarrow n\ge log^3_{\dfrac{7}{5}}\)
\(\Rightarrow\)\(n\in\left\{4;5;6;7;...\right\}\Rightarrow n=4\)
c) \(1-\left(\dfrac{4}{5}\right)^n\ge0,97\)
\(\Leftrightarrow-\left(\dfrac{4}{5}\right)^n\ge-0,3\)
\(\Leftrightarrow\left(\dfrac{4}{5}\right)^n\le0,3\)\(\Leftrightarrow n\ge log^{0,3}_{\dfrac{4}{5}}\)
\(\Rightarrow n\in\left\{6;7;8;9...\right\}\Rightarrow n=6\)
d)\(\left(1+\dfrac{5}{100}\right)^n\ge2\)
\(\Leftrightarrow1,05^n\ge2\)
\(\Rightarrow n\in\left\{15;16;17;18;...\right\}\Rightarrow n=15\)
\(\Leftrightarrow\left(\dfrac{3}{4}\right)^x.\left(\dfrac{4}{3}\right)^{\dfrac{4}{x}}=\dfrac{9}{16}\)
\(\Rightarrow\left(\dfrac{3}{4}\right)^x.\left(\dfrac{3}{4}\right)^{-\dfrac{4}{x}}=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow\left(\dfrac{3}{4}\right)^{x-\dfrac{4}{x}}=\left(\dfrac{3}{4}\right)^2\)
\(\Rightarrow x-\dfrac{4}{x}=2\)
\(\Rightarrow x^2-2x-4=0\)
Viet: \(x_1+x_2=2\)
\(\left(\dfrac{-2}{5}+\dfrac{3}{7}\right)-\left(\dfrac{4}{9}+\dfrac{12}{20}-\dfrac{13}{35}\right)+\dfrac{7}{35}\)
\(=-\dfrac{2}{5}+\dfrac{3}{7}-\dfrac{4}{9}-\dfrac{3}{5}+\dfrac{13}{35}+\dfrac{7}{35}\\ =\left(-\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{13}{35}+\dfrac{7}{35}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+\left(\dfrac{4}{7}+\dfrac{3}{7}\right)-\dfrac{4}{9}\\ =-1+1-\dfrac{4}{9}\\ =-\dfrac{4}{9}\)
a)
. 
= 
= 
= 
= 
= 9.
b)
: 
= 
= 
= 
= 
= 
= 8.
c)
+ 
= 
+ 
= 
+ 
= 
+ 
= 
+ 
= 40.
d)
- 
= 
- 
= 
- 
= 
- 
= 121.
a) \(9^{\dfrac{2}{5}}.27^{\dfrac{2}{5}}=\left(9.27\right)^{\dfrac{2}{5}}=\left(3^2.3^3\right)^{\dfrac{2}{5}}=3^{5.\dfrac{2}{5}}=3^2=9\)
b) \(=\left(\dfrac{144}{9}\right)^{\dfrac{3}{4}}=\left(\dfrac{12}{3}\right)^{2.\dfrac{3}{4}}=4^{\dfrac{3}{2}}=2^{2.\dfrac{3}{2}}=2^3=8\)
c) \(=\left(\dfrac{1}{2}\right)^{4.\left(-0,75\right)}+\left(\dfrac{1}{4}\right)^{-\dfrac{5}{2}}\)
\(=\left(\dfrac{1}{2}\right)^{-3}+\left(\dfrac{1}{2}\right)^{-5}\)
\(=2^3+2^5=40\)
d) \(=\left(0,2\right)^{2.\left(-1.5\right)}-\left(0,5\right)^{3.\dfrac{-2}{3}}\)
\(=\left(\dfrac{1}{5}\right)^{-3}-\left(\dfrac{1}{2}\right)^{-2}\)
\(=5^3-2^2=121\)