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a) Ta có: \(\dfrac{36}{45}=\dfrac{4}{5}\)
BCNN (4;5)=20
Mà BCNN (a;b)=300
\(\Rightarrow\)300:20=15
\(\Rightarrow\)\(\dfrac{a}{b}=\dfrac{4\cdot15}{5\cdot15}=\dfrac{60}{75}\)
Vậy phân số \(\dfrac{a}{b}\) cần tìm là \(\dfrac{60}{75}\).
b) Ta có: \(\dfrac{a}{b}=\dfrac{21}{35}=\dfrac{3}{5}\)
ƯCLN (a;b)=30
\(\Rightarrow\)\(\dfrac{a}{b}=\dfrac{3\cdot30}{5\cdot30}=\dfrac{90}{15}\)
Vậy phân số \(\dfrac{a}{b}\) cần tìm là \(\dfrac{90}{15}\).
Câu 2:
Đặt a/b=c/d=k
=>a=bk; c=dk
\(\dfrac{a+b}{b}=\dfrac{bk+b}{b}=k+1\)
\(\dfrac{c+d}{d}=\dfrac{dk+d}{d}=k+1\)
Do đó: \(\dfrac{a+b}{b}=\dfrac{c+d}{d}\)
a)
\(A=\dfrac{3}{4}.\dfrac{8}{9}...\dfrac{9999}{10000}\)
\(=\dfrac{1.3}{2.2}.\dfrac{2.4}{3.3}...\dfrac{99.101}{100.100}\)
\(=\dfrac{1.2...99}{2.3...100}.\dfrac{3.4...101}{2.3...100}\)
\(=\dfrac{1}{100}.\dfrac{101}{2}\)
\(=\dfrac{101}{200}\)
Câu 2:
\(\Leftrightarrow x\left(\dfrac{1}{6}+\dfrac{1}{10}+\dfrac{1}{15}+...+\dfrac{1}{78}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow2x\left(\dfrac{1}{12}+\dfrac{1}{20}+...+\dfrac{1}{156}\right)=\dfrac{220}{39}\)
\(\Leftrightarrow x\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{12}-\dfrac{1}{13}\right)=\dfrac{110}{39}\)
\(\Leftrightarrow x\cdot\dfrac{10}{39}=\dfrac{110}{39}\)
=>x=11
Bài 1:
a)\(\left(5x+1\right)^2=\dfrac{36}{49}\)
\(\Leftrightarrow\left(5x+1\right)^2=\left(\dfrac{6}{7}\right)^2=\left(-\dfrac{6}{7}\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}5x+1=\dfrac{6}{7}\\5x+1=-\dfrac{6}{7}\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}5x=-\dfrac{1}{7}\\5x=-\dfrac{13}{7}\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{1}{35}\\x=-\dfrac{13}{35}\end{matrix}\right.\)
Bài 2:
a)\(x^2+\left(y-\dfrac{1}{10}\right)^4=0\)
Dễ thấy: \(\left\{{}\begin{matrix}x^2\ge0\\\left(y-\dfrac{1}{10}\right)^4\ge0\end{matrix}\right.\)
\(\Rightarrow x^2+\left(y-\dfrac{1}{10}\right)^4\ge0\)
Xảy ra khi \(\left\{{}\begin{matrix}x^2=0\\\left(y-\dfrac{1}{10}\right)^4=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=\dfrac{1}{10}\end{matrix}\right.\)
b)\(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\le0\)
Dễ thấy: \(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}\ge0\\\left(y^2-\dfrac{1}{4}\right)^{40}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\ge0\)
Mà \(\left(\dfrac{1}{2}x-5\right)^{20}+\left(y^2-\dfrac{1}{4}\right)^{40}\le0\)
Xảy ra khi \(\left\{{}\begin{matrix}\left(\dfrac{1}{2}x-5\right)^{20}=0\\\left(y^2-\dfrac{1}{4}\right)^{40}=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=\pm\dfrac{1}{2}\end{matrix}\right.\)
Mấy bài này bạn tự làm đi, chuyển vế tìm x gần giống cấp I mà.
b)\(\dfrac{-3}{5}.x=\dfrac{1}{4}+0,75\)
=>\(\dfrac{-3}{5}.x=1\)
=>\(x=1:\dfrac{-3}{5}\)
=>\(x=\dfrac{-5}{3}\)
Vậy \(x=\dfrac{-5}{3}\)
\(A=\left(-\dfrac{43}{51}\right)\left(-\dfrac{19}{80}\right)\)
=>A>0(1)
\(B=\left(-\dfrac{7}{13}\right)\left(-\dfrac{4}{65}\right)\left(-\dfrac{8}{21}\right)\)
=>B<0(2)
C\(=-\dfrac{5}{10}.\left(-\dfrac{4}{10}\right).....\left(\dfrac{4}{10}\right)\left(\dfrac{5}{10}\right)=0\)
=>C=0(3)
Từ 1;2;3 =>A>C>B
\(A=\dfrac{-43}{51}.\dfrac{-19}{80}\Leftrightarrow A>0\left(1\right)\)
\(B=\left(\dfrac{-7}{13}\right).\left(-\dfrac{4}{65}\right).\left(\dfrac{-8}{31}\right)\Leftrightarrow B< 0\left(2\right)\)
\(C=\dfrac{-5}{10}.\dfrac{-4}{10}...........\dfrac{3}{10}.\dfrac{4}{10}.\dfrac{5}{10}\Leftrightarrow C=0\left(3\right)\)
Từ \(\left(1\right)+\left(2\right)+\left(3\right)\Leftrightarrow A>C>B\)
Bài 1:
a: \(\Leftrightarrow\left|x+\dfrac{4}{15}\right|=-2.15+3.75=\dfrac{8}{5}\)
=>x+4/15=8/5 hoặc x+4/15=-8/5
=>x=4/3 hoặc x=-28/15
b: \(\Leftrightarrow\left[{}\begin{matrix}\dfrac{5}{3}x=-\dfrac{1}{6}\\\dfrac{5}{3}x=\dfrac{1}{6}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{6}:\dfrac{5}{3}=\dfrac{-3}{30}=\dfrac{-1}{10}\\x=\dfrac{1}{10}\end{matrix}\right.\)
c: \(\Leftrightarrow\left|x-1\right|-1=1\)
=>|x-1|=2
=>x-1=2 hoặc x-1=-2
=>x=3 hoặc x=-1
Bài 2:
b: \(\Leftrightarrow\left\{{}\begin{matrix}x-y=0\\y+\dfrac{9}{25}=0\end{matrix}\right.\Leftrightarrow x=y=-\dfrac{9}{25}\)
Bài 3:
a: \(A=\left|x+\dfrac{15}{19}\right|-1>=-1\)
Dấu '=' xảy ra khi x=-15/19
b: \(\left|x-\dfrac{4}{7}\right|+\dfrac{1}{2}>=\dfrac{1}{2}\)
Dấu '=' xảy ra khi x=4/7
Lời giải:
$\frac{a}{b}=\frac{42}{66}=\frac{7}{11}$ rồi đó bạn, cần gì tìm $\frac{a}{b}$ nữa
nhưng mà ước chung lớn nhất của a và b bằng 6 mà