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b) áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\frac{a^2}{2^2}=\frac{b^2}{3^2}=\frac{2c^2}{2\cdot4^2}=\frac{a^2-b^2+2c^2}{2^2-3^2+2\cdot4^2}=\frac{108}{27}=4\)
\(\frac{a^2}{2^2}=4\Rightarrow a^2=4\cdot2^2=16\Rightarrow a=\sqrt{16}=4\)
\(\frac{b^2}{3^2}=4\Rightarrow b^2=4\cdot3^2=36\Rightarrow b=\sqrt{36}=6\)
\(\frac{2c^2}{2\cdot4^2}=4\Rightarrow2c^2=4\cdot2\cdot4^2=128\Rightarrow c^2=128:2=64\Rightarrow c=\sqrt{64}=8\)
vậy a = 4
b = 6
c = 8
a)
a:b:c = 2:4:5
=> a/2 = b/4 =c/5
áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\frac{2a}{2\cdot2}=\frac{b}{4}=\frac{c}{5}=\frac{2a-b+c}{2\cdot2-4+5}=\frac{7}{5}\)
\(\frac{2a}{2\cdot2}=\frac{7}{5}\Rightarrow2a=\frac{7\cdot2\cdot2}{5}=\frac{28}{5}\Rightarrow a=\frac{28}{5}:2=\frac{14}{5}=2,8\)
\(\frac{b}{4}=\frac{7}{5}\Rightarrow b=\frac{7\cdot4}{5}=\frac{28}{5}=5,6\)
\(\frac{c}{5}=\frac{7}{5}\Rightarrow c=\frac{7\cdot5}{5}=7\)
vậy a = 2,8
b = 5,6
c = 7
1) a) Ta có: \(\frac{x}{-15}=\frac{-60}{x}\) \(\Rightarrow x^2=\left(-15\right).\left(-60\right)=900\)
\(\Rightarrow x=30\)
b) \(\frac{-2}{x}=\frac{-x}{\frac{8}{25}}\) \(\Rightarrow x.\left(-x\right)=\left(-2\right).\frac{8}{25}\)
\(\Rightarrow x.\left(-x\right)=\frac{-16}{25}\)
\(\Rightarrow x.\left(-x\right)=\left(\frac{-4}{5}\right).\frac{4}{5}\)
Vậy \(x=\frac{4}{5}\)
2) a) \(3,8: \left(2x\right)=\frac{1}{4}:2\frac{2}{3}\)
\(\Rightarrow3,8: \left(2x\right)=\frac{3}{32}\)
\(\Rightarrow2x=\frac{3}{32}:3,8=\frac{15}{608}\)
\(x=\frac{15}{608}:2=\frac{15}{1216}\)
Vậy \(x=\frac{15}{1216}\)
b) \(\left(0,25x\right):3=\frac{5}{6}:0,125\)
\(\Rightarrow\left(0,25x\right):3=\frac{20}{3}\)
\(\Rightarrow0,25x=\frac{20}{3}.3=20\)
\(\Rightarrow x=20:0,25=80\)
Vậy x = 80
c) \(0,01:2,5=\left(0,75x\right):0,75\)
\(\Rightarrow\frac{1}{250}=\left(0,75x\right):0,75\)
\(\Leftrightarrow0,75x=\frac{1}{250}.0,75=\frac{3}{1000}\)
\(\Rightarrow x=\frac{3}{1000}:0,75=\frac{1}{250}\)
Vậy \(x=\frac{1}{250}\)
d) \(1\frac{1}{3}:0,8=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow\frac{5}{3}=\frac{2}{3}:\left(0,1x\right)\)
\(\Rightarrow0,1x=\frac{5}{3}.\frac{2}{3}=\frac{10}{9}\)
\(\Rightarrow x=\frac{10}{9}:0,1=\frac{100}{9}\)
Vậy \(x=\frac{100}{9}\)
a) \(\frac{x}{-15}=\frac{-60}{x}\Leftrightarrow x.x=-15.\left(-60\right)\Leftrightarrow x^2=900\Leftrightarrow x^2=\orbr{\begin{cases}30^2\\\left(-30\right)^2\end{cases}}\Leftrightarrow x=\orbr{\begin{cases}30\\-30\end{cases}}\)
Vì \(a:b:c=\frac{2}{5}:\frac{3}{4}:\frac{1}{6}\)
\(\Rightarrow\frac{a}{\frac{2}{5}}=\frac{b}{\frac{3}{4}}=\frac{c}{\frac{1}{6}}\Rightarrow\frac{a}{\frac{2}{5}.60}=\frac{b}{\frac{3}{4}.60}=\frac{c}{\frac{1}{6}.60}\Leftrightarrow\frac{a}{24}=\frac{b}{45}=\frac{c}{10}\)
Theo t)c của dãy tỉ số bằng nhau ta có :
\(\frac{a^2}{576}=\frac{b^2}{2025}=\frac{c^2}{100}=\frac{a^2+b^2+c^2}{576+2025+100}=\frac{24309}{2701}=9\)
\(\Rightarrow a^2=9.576=5184\Rightarrow a=72\left(a>0\right)\)
\(b^2=9.2025=18225\Rightarrow b=135\left(b>0\right)\)
\(c^2=9.100=900\Rightarrow c=30\left(c>0\right)\)
\(\Rightarrow A=a+b+c=72+135+30=237\)
Bài 3:
a) Ta có: \(1.25\cdot\left(-3\frac{3}{8}\right)\)
\(=\frac{5}{4}\cdot\frac{-27}{8}\)
\(=\frac{-135}{32}\)
b) Ta có: \(\frac{-9}{34}\cdot\frac{17}{4}\)
\(=\frac{-9}{4}\cdot\frac{17}{34}\)
\(=-\frac{9}{4}\cdot\frac{1}{2}\)
\(=-\frac{9}{8}\)
c) Ta có: \(-\frac{20}{41}\cdot\frac{-4}{5}\)
\(=\frac{20}{41}\cdot\frac{4}{5}\)
\(=\frac{16}{41}\)
d) Ta có: \(\frac{-6}{7}\cdot\frac{21}{2}\)
\(=-\frac{6}{2}\cdot\frac{21}{7}\)
\(=-3\cdot3=-9\)
Bài 4:
a) Ta có: \(-\frac{5}{2}\cdot\frac{3}{4}\)
\(=-\frac{5\cdot3}{2\cdot4}=\frac{-15}{8}\)
b) Ta có: \(4\frac{1}{5}:\left(-2\frac{4}{5}\right)\)
\(=-\frac{21}{5}:\frac{14}{5}\)
\(=-\frac{21}{5}\cdot\frac{5}{14}\)
\(=-\frac{21}{14}=-\frac{3}{2}\)
c) Ta có: \(1.8:\left(-\frac{3}{4}\right)\)
\(=\frac{9}{5}:\frac{-3}{4}\)
\(=\frac{9}{5}\cdot\frac{4}{-3}\)
\(=-\frac{12}{5}\)
d) Ta có: \(\frac{17}{15}:\frac{4}{3}\)
\(=\frac{17}{15}\cdot\frac{3}{4}\)
\(=\frac{17}{20}\)
a) Vì \(\left(2x-3\right)^2\ge0\Rightarrow\left(2x-3\right)^2+5\ge5\Rightarrow C=\frac{-4}{\left(2x-3\right)^2+5}\ge-\frac{4}{5}\)
<=>\(C_{min}=-\frac{4}{5}\Leftrightarrow\left(2x-3\right)^2=0\Leftrightarrow2x-3=0\Leftrightarrow2x=3\Leftrightarrow x=\frac{3}{2}\)
Vậy C đạt GTNN là -4/5 tại x=3/2
b) \(\frac{a+b}{a-b}=\frac{c+a}{c-a}\Leftrightarrow\left(a+b\right)\left(c-a\right)=\left(a-b\right)\left(c+a\right)\)
\(\Leftrightarrow ac+bc-a^2-ab=ac-bc+a^2-ab\)
\(\Leftrightarrow bc-a^2=-bc+a^2\)
\(\Leftrightarrow2bc=2a^2\)
\(\Leftrightarrow bc=a^2\) (đpcm)
\(a.=\frac{16}{21}+\left(\frac{5}{21}-\frac{5}{21}\right)+\left(\frac{4}{23}-\frac{4}{23}\right).\)
\(=\frac{16}{21}+0+0=\frac{16}{21}\)
\(b.=\left(\frac{3}{4}\right)^{3-2}=\left(\frac{3}{4}\right)^1=\frac{3}{4}\)
\(c.=\frac{1}{3}.\left(\frac{-4}{5}+\frac{-6}{5}\right)\)
\(=\frac{1}{3}.\left(-2\right)=\frac{-2}{3}\)
Nhớ k cho mình nhé! Thank you!!!