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B=1,4.15/49-(4/5+2/3):2.1/5
B=3/7-22/15:2.1/5
B=3/7-11/15.1/5
B=3/7-11/75
B=148/525
78,25%=\(\frac{313}{400}\)
số vải hoa là :356,5:(313+400).313=156,5 (m)
số vải trắng là :356,5-156,5=200(m)
Đ/S
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{5}{6}\) -\(\frac{3}{4}\) + \(\frac{2}{3}\) -\(\frac{1}{2}\)
x . \(\frac{1}{2}\)- x.\(\frac{2}{3}\) + x.\(\frac{3}{4}\)- x. \(\frac{5}{6}\) = \(\frac{10}{12}\)-\(\frac{9}{12}\)+\(\frac{8}{12}\)-\(\frac{6}{12}\)
=>x.(1/2-2/3+3/4)=1/4
=>x.7/12=1/4
=>x=1/4:7/12
=>x=1/4.12/7
=>x=3/7
\(A=\frac{1}{2}\left(1^2+1^2\right)\left(x^2+y^2\right)\ge\frac{1}{2}\left(x+y\right)^2=\frac{1}{2}\)
Min A= 1/2 khi x = y =1/2
Vì x+y=1
=>y=1-x
Ta có: \(A=x^2+y^2=x^2+\left(1-x\right)^2=x^2+1\left(1-x\right)-x\left(1-x\right)=x^2+1-x-x+x^2\)
\(A=2x^2-2x+1=2.\left(x^2-x+\frac{1}{2}\right)\)
\(A=2.\left(x^2-\frac{1}{2}x-\frac{1}{2}x+\frac{1}{4}-\frac{1}{4}+\frac{1}{2}\right)=2\left[x\left(x-\frac{1}{2}\right)-\frac{1}{2}\left(x-\frac{1}{2}\right)+\frac{1}{4}\right]\)
\(A=2\left[\left(x-\frac{1}{2}\right)^2+\frac{1}{4}\right]=2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}\)
Vì \(2\left(x-\frac{1}{2}\right)^2>=0\) với mọi x
=>\(2\left(x-\frac{1}{2}\right)^2+\frac{1}{2}>=\frac{1}{2}\) với mọi x
Dấu "=" xảy ra <=>\(x=\frac{1}{2}\);mà x+y=1=>\(y=\frac{1}{2}\)
Khi đó GTNN của A=x2+y2 là 1/2 tại \(x=y=\frac{1}{2}\)
Câu 1 :
Đk: \(x\ge1\)
\(\sqrt{x-1}+\sqrt{2x-1}=5\\ \Leftrightarrow x-1+2\sqrt{\left(x-1\right)\left(2x-1\right)}+2x-1=25\\ \Leftrightarrow2\sqrt{2x^2-3x+1}=27-3x\\ \)
\(\Leftrightarrow\begin{cases}27-3x\ge0\\4\left(2x^2-3x+1\right)=9x^2-162x+729\end{cases}\) \(\Leftrightarrow\begin{cases}x\le9\\x^2-150x+725=0\end{cases}\)
\(\Leftrightarrow\begin{cases}x\le9\\x=145hoặcx=5\end{cases}\)
với x= 5 thoản mãn điều kiện, x=145 loại
Vậy \(S=\left\{5\right\}\)
\(\frac{1}{2}.x+\frac{3}{5}.\left(x-2\right)=3
\)
\(\frac{1}{2}x+\frac{3}{5}x-\frac{3}{5}.2=3\)
\(\left(\frac{1}{2}+\frac{3}{5}\right)x-\frac{6}{5}=3\)
\(\left(\frac{5}{10}+\frac{6}{10}\right).x=3+\frac{6}{5}\)
\(\frac{11}{10}.x=\frac{21}{5}\)
x =\(\frac{21}{5}:\frac{11}{10}\)
x=\(\frac{21}{5}.\frac{10}{11}\)
x-\(\frac{42}{11}\)
Bài 1:
a: \(\Leftrightarrow x\cdot\dfrac{3}{4}=-1\)
hay x=-4/3
b: =>x=4/8+3/7=1/2+3/7=7/14+6/14=13/14
Bài 3:
BCNN(16;32;5)=160
UCLN(16;32;5)=1
Bài 2 :
Ta có :
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}=\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\left(\frac{1}{8}-\frac{1}{16}\right)+\left(\frac{1}{16}-\frac{1}{32}\right)\)
\(=1-\frac{1}{32}\)
và \(B=\frac{2003}{2004}=1-\frac{1}{2004}\)
Vì \(\frac{1}{32}>\frac{1}{2004}\) nên A < B
a: \(=\dfrac{7}{5}\cdot\dfrac{15}{49}-\dfrac{12+10}{15}:\dfrac{11}{5}\)
\(=\dfrac{3}{7}-\dfrac{22}{15}\cdot\dfrac{5}{11}=\dfrac{3}{7}-\dfrac{2}{3}=\dfrac{9-14}{21}=\dfrac{-5}{21}\)
b: =>2,8x-32=-60
=>2,8x=-28
hay x=-10