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a) \(\frac{1}{9}\cdot27^n=3^n\)
\(\frac{1}{9}=\frac{3^n}{27^n}\)
\(\frac{1}{9}=\frac{3^n}{3^{3n}}\)
\(\frac{1}{9}=\frac{1}{3^{2n}}\)
=> 32n = 9 = 32
=> 2n = 2
=> n = 1
a,\(A=x^2-2x+\frac{1}{x-1}\)
\(A=x^2-2x+1-\frac{x-2}{x-1}\)
\(A=\left(x-1\right)^2+\frac{-\left(x-2\right)}{x-1}\ge\frac{-\left(x-2\right)}{x-1}\)
Do \(x-2>x-1\Rightarrow-\left(x-2\right)< x-1\)
Mà \(\frac{-\left(x-2\right)}{x-1}\ge-1\)
Vậy Min A = -1 <=> x = 1
Câu 1:
\(4\sqrt[4]{\left(a+1\right)\left(b+4\right)\left(c-2\right)\left(d-3\right)}\le a+1+b+4+c-2+d-3=a+b+c+d\)
Dấu = xảy ra khi a = -1; b = -4; c = 2; d= 3
\(\frac{a^2}{b^5}+\frac{1}{a^2b}\ge\frac{2}{b^3}\)\(\Leftrightarrow\)\(\frac{a^2}{b^5}\ge\frac{2}{b^3}-\frac{1}{a^2b}\)
\(\frac{2}{a^3}+\frac{1}{b^3}\ge\frac{3}{a^2b}\)\(\Leftrightarrow\)\(\frac{1}{a^2b}\le\frac{2}{3a^3}+\frac{1}{3b^3}\)
\(\Rightarrow\)\(\Sigma\frac{a^2}{b^5}\ge\Sigma\left(\frac{5}{3b^3}-\frac{2}{3a^3}\right)=\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}+\frac{1}{d^3}\)
a) 3/4x16/9-7/5:(-21/20)
=4/3-(-4/3)
=8/3
b)=7/3-1/3x[-3/2+(2/3+2)]
=7/3-1/3x[-3/2+8/3]
=7/3-1/3x7/6
=7/3-7/18
=35/18
c)=(20+37/4):9/4
=117/4:9/4
=13
d)=6-14/5x25/8-8/5:1/4
=6-35/4-32/5
=-11/4-32/5
=-183/20
Bài 1
\(a,\frac{3}{5}+\left(-\frac{1}{4}\right)=\frac{7}{20}\)
\(b,\left(-\frac{5}{18}\right)\cdot\left(-\frac{9}{10}\right)=\frac{1}{4}\)
\(c,4\frac{3}{5}:\frac{2}{5}=\frac{23}{5}\cdot\frac{5}{2}=\frac{23}{2}\)
Bài 2
\(a,\frac{12}{x}=\frac{3}{4}\Rightarrow3x=12\cdot4\)
\(\Rightarrow3x=48\)
\(\Rightarrow x=16\)
\(b,x:\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^2\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)^2\cdot\left(-\frac{1}{3}\right)^3=\left(-\frac{1}{3}\right)^5\)
\(\Rightarrow x=-\frac{1}{243}\)
\(c,-\frac{11}{12}\cdot x+0,25=\frac{5}{6}\)
\(\Rightarrow-\frac{11}{12}x=\frac{5}{6}-\frac{1}{4}=\frac{7}{12}\)
\(\Rightarrow x=\frac{7}{12}:\left(-\frac{11}{12}\right)\)
\(\Rightarrow x=-\frac{7}{11}\)
\(d,\left(x-1\right)^5=-32\)
\(\left(x-1\right)^5=-2^5\)
\(x-1=-2\)
\(x=-2+1=-1\)
Bài 3
\(\left|m\right|=-3\Rightarrow m\in\varnothing\)
Bài 3
Gọi 3 cạnh của tam giác lần lượt là a;b;c ( a,b,c>0)
Ta có
\(a+b+c=13,2\)
\(\frac{a}{3};\frac{b}{4};\frac{c}{5}\)
Ap dụng tính chất DTSBN ta có
\(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=\frac{a+b+c}{3+4+5}=\frac{13,2}{12}=\frac{11}{10}\)
\(\hept{\begin{cases}\frac{a}{3}=\frac{11}{10}\\\frac{b}{4}=\frac{11}{10}\\\frac{c}{5}=\frac{11}{10}\end{cases}}\Rightarrow\hept{\begin{cases}a=\frac{33}{10}\\b=\frac{44}{10}=\frac{22}{5}\\c=\frac{55}{10}=\frac{11}{2}\end{cases}}\)
Vậy 3 cạnh của tam giác lần lượt là \(\frac{33}{10};\frac{22}{5};\frac{11}{2}\)
a)\(\frac{3}{5}+\left(-\frac{1}{4}\right)\)
\(=\frac{3}{5}-\frac{1}{4}\)
\(=\frac{12}{20}-\frac{5}{20}=\frac{7}{20}\)
b)\(\left(-\frac{5}{18}\right)\left(-\frac{9}{10}\right)\)
\(=\frac{\left(-5\right)\left(-9\right)}{18.10}\)
\(=\frac{\left(-1\right)\left(-1\right)}{2.2}=\frac{1}{4}\)
c)\(4\frac{3}{5}:\frac{2}{5}\)
\(=\frac{23}{5}:\frac{2}{5}\)
\(=\frac{23}{5}.\frac{5}{2}\)
\(=\frac{23.1}{1.2}=\frac{23}{2}\)
1/
a)\(\frac{12}{x}=\frac{3}{4}\)
\(\Rightarrow x.3=12.4\)
\(\Rightarrow x.3=48\)
\(\Rightarrow x=48:3=16\)
b)\(x:\left(\frac{-1}{3}\right)^3=\left(\frac{-1}{3}\right)^2\)
\(x=\left(\frac{-1}{3}\right)^2.\left(\frac{-1}{3}\right)^3\)
\(x=\frac{\left(-1\right)^2}{3^2}.\frac{\left(-1\right)^3}{3^3}\)
\(x=\frac{1}{9}.\frac{-1}{27}=-\frac{1}{243}\)
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