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a)\(1,5:2,16=15:216=5:72\)
b)\(4\dfrac{2}{7}:\dfrac{3}{5}=\dfrac{30}{7}:\dfrac{3}{5}=\dfrac{30}{7}.\dfrac{5}{3}=\dfrac{50}{7}=50:7\)
c)\(\dfrac{\dfrac{2}{9}}{0,31}=\dfrac{2}{9}:\dfrac{31}{100}=\dfrac{2}{9}.\dfrac{100}{31}=\dfrac{31}{450}=31:450\)
1. Tìm x thuộc N:
\(\left(x-3\right)^6=\left(x-3\right)^7\)
\(\Leftrightarrow\left(x-3\right)^6-\left(x-3\right)^7=0\)
\(\Leftrightarrow\left(x-3\right)^6.\text{[}1-\left(x-3\right)\text{]}=0\)
\(\Leftrightarrow\left(x-3\right)^6.\left(4-x\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\4-x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\)(thỏa mãn \(x\in N\))
2.
Ta có: 6x=4y=3z
\(\Rightarrow\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}=\dfrac{2x}{4}=\dfrac{3y}{9}=\dfrac{5z}{20}\)
\(=\dfrac{2x+3y-5z}{4+9-20}=\dfrac{-21}{-7}=3\)
\(\Rightarrow\left\{{}\begin{matrix}x=3.2=6\\y=3.3=9\\z=3.4=12\end{matrix}\right.\)
\(\frac{2}{7}\)A = \(\frac{3}{10}\)B x \(\frac{4}{9}\)= \(\frac{2}{15}\)B => \(\frac{A}{B}\)= \(\frac{2}{15}\)\(=\frac{7}{15}\)
\(\frac{2}{7}\)
Mấy bài dễ tự làm nhé:D
1)
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Leftrightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{a}{a+b}=\dfrac{bk}{bk+b}=\dfrac{bk}{b\left(k+1\right)}=\dfrac{k}{k+1}\\\dfrac{c}{c+d}=\dfrac{dk}{dk+d}=\dfrac{dk}{d\left(k+1\right)}=\dfrac{k}{k+1}\end{matrix}\right.\)
Ta có điều phải chứng minh
\(\left\{{}\begin{matrix}\dfrac{a}{a-b}=\dfrac{bk}{bk-b}=\dfrac{bk}{b\left(k-1\right)}=\dfrac{k}{k-1}\\\dfrac{c}{c-d}=\dfrac{dk}{dk-d}=\dfrac{dk}{d\left(k-1\right)}=\dfrac{k}{k-1}\end{matrix}\right.\)
Ta có điều phải chứng minh
1) Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{2010}=\dfrac{2010}{a}=\dfrac{a+b+c+2010}{b+c+2010+a}=1\)
\(\dfrac{2010}{a}=1\Rightarrow a=2010\);
\(\dfrac{c}{2010}=1\Rightarrow c=2010\);
\(\dfrac{b}{c}=1\Rightarrow\dfrac{b}{2010}=1\Rightarrow b=2010\).
Vậy (a, b, c) = (2010; 2010; 2010)
3)
a) \(A=\sqrt{x+24}+\dfrac{4}{7}\)
Có: \(\sqrt{x+24}\ge0\forall x\in R\)
\(\Rightarrow\sqrt{x+24}+\dfrac{4}{7}\ge\dfrac{4}{7}\forall x\in R\)
\(\Rightarrow A\ge\dfrac{4}{7}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+24}=0\Rightarrow x+24=0\Rightarrow x=-24\)
Vậy GTNN của \(A=\dfrac{4}{7}\Leftrightarrow x=-24\)
b) \(B=\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\)
Có: \(\sqrt{2x+\dfrac{4}{13}}\ge0\forall x\in R\)
\(\Rightarrow\sqrt{2x+\dfrac{4}{13}}-\dfrac{13}{191}\ge-\dfrac{13}{191}\forall x\in R\)
\(\Rightarrow B\ge-\dfrac{13}{191}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{2x+\dfrac{4}{13}}=0\)
\(\Rightarrow2x+\dfrac{4}{13}=0\)
\(\Rightarrow2x=-\dfrac{4}{13}\)
\(\Rightarrow x=-\dfrac{2}{13}\)
Vậy GTNN của \(B=-\dfrac{13}{191}\Leftrightarrow x=-\dfrac{2}{13}\)
4)
a) \(A=-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\)
Có: \(\sqrt{x+\dfrac{5}{41}}\ge0\forall x\in R\)
\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}\le0\forall x\in R\)
\(\Rightarrow-\sqrt{x+\dfrac{5}{41}}+\dfrac{7}{12}\le\dfrac{7}{12}\forall x\in R\)
\(\Rightarrow A\le\dfrac{7}{12}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x+\dfrac{5}{41}}=0\)
\(\Rightarrow x+\dfrac{5}{41}=0\)
\(\Rightarrow x=-\dfrac{5}{41}\)
Vậy GTLN của \(A=\dfrac{7}{12}\Leftrightarrow x=-\dfrac{5}{41}\)
b) \(B=\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\)
Có: \(\sqrt{x-\dfrac{2}{3}}\ge0\forall x\in R\)
\(\Rightarrow-\sqrt{x-\dfrac{2}{3}}\le0\forall x\in R\)
\(\Rightarrow\dfrac{-5}{13}-\sqrt{x-\dfrac{2}{3}}\le\dfrac{-5}{13}\forall x\in R\)
\(\Rightarrow B\le\dfrac{-5}{13}\forall x\in R\)
Đẳng thức xảy ra \(\Leftrightarrow\sqrt{x-\dfrac{2}{3}}=0\)
\(\Rightarrow x-\dfrac{2}{3}=0\)
\(\Rightarrow x=\dfrac{2}{3}\)
Vậy GTLN của \(B=\dfrac{-5}{13}\Leftrightarrow x=\dfrac{2}{3}\)
Theo bài ra ta có:
\(\dfrac{2}{7}A=\dfrac{9}{4}.\dfrac{3}{10}B\\ \Rightarrow\dfrac{2}{7}A=\dfrac{27}{40}B\\ \Rightarrow\dfrac{A}{B}=\dfrac{\dfrac{27}{40}}{\dfrac{2}{7}}=\dfrac{189}{80}\)
\(\dfrac{2}{7}\)A=\(\dfrac{3}{10}\).\(\dfrac{4}{9}\)
⇒\(\dfrac{2}{7}\)A=\(\dfrac{2}{15}\)B
⇒\(\dfrac{A}{\dfrac{2}{15}}=\dfrac{B}{\dfrac{2}{7}}\)
\(\dfrac{A}{B}=\dfrac{2}{\dfrac{15}{\dfrac{2}{7}}}\)
⇒\(\dfrac{A}{B}=\dfrac{7}{15}\)