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\(\dfrac{2\times x-4,36}{0,125}=0,25\times42,9-11,7\times0,25+0,25\times0,8\)
\(\dfrac{2\times x-4,36}{0,125}=0,25\times\left(42,9-11,7+0,8\right)\)
\(\dfrac{2\times x-4,36}{0,125}=0,25\times32=8\)
\(2\times x-4,36=8\times0,125=1\)
\(2\times x=1+4,36=5,36\)
\(x=5,36\div2\)
\(x=2,68\)
\(\frac{2x-4.36}{0.125}=0.25\cdot42.9-11.7\cdot0.25+0.25\cdot0.8\)
\(\frac{2x-4.36}{0.125}=0.25\left(42.9-11.7+0.8\right)\)
\(\frac{2x-4.36}{0.125}=0.25\cdot32\)
\(\frac{2x-4.36}{0.125}=8\)
2x - 4.36 = 8 * 0.125 = 1
2x = 1 + 4.36 = 5.36
x = 5.36 : 2 = 2.68
a)\(6:\left(\frac{\frac{3}{5}x+3}{20}+9\right)=\frac{3}{5}\)
\(\frac{\frac{3}{5}x+3}{20}+9=6:\frac{3}{5}\)
\(\frac{\frac{3}{5}x+3}{20}+9=6.\frac{5}{3}\)
\(\frac{\frac{3}{5}x+3}{20}+9=10\)
\(\frac{\frac{3}{5}x+3}{20}=10-9\)
\(\left(\frac{3}{5}x+3\right):20=1\)
\(\frac{3}{5}x+3=1.20\)
\(\frac{3}{5}x+3=20\)
\(\frac{3}{5}x=20-3\)
\(\frac{3}{5}x=17\)
\(x=17:\frac{3}{5}\)
\(x=17.\frac{5}{3}\)
\(x=\frac{85}{3}\)
b)l x+3 l = l -9 l
l x+3l=9
\(\orbr{\begin{cases}x+3=9\\x+3=-9\end{cases}\Rightarrow\orbr{\begin{cases}x=9-3\\x=-9-3\end{cases}\Rightarrow}\orbr{\begin{cases}x=6\\x=-12\end{cases}}}\)
Giải:
Theo bài ra ta có:
\(\frac{-5}{6}+\frac{8}{3}+\frac{29}{-6}\le x\le\frac{-1}{2}+2+\frac{5}{12}\)
\(\Rightarrow-3\le x\le\frac{23}{12}\)
\(\Rightarrow x\varepsilon\left\{-2;-1;0;1\right\}\)
\(\frac{-5}{6}+\frac{16}{6}+-\frac{29}{6}\le x\le\frac{-6}{12}+\frac{24}{12}+\frac{5}{12}\)
=>-3\(\le\) x\(\le\) 23/12
=> x thuộc{-2-1;0;1}
ta có : \(\frac{y}{-78}=\frac{-y}{78};\frac{7}{-6}=\frac{-7}{6}\)
\(\frac{-y}{78}=\frac{-7}{6}\)
=> ( -y ) . 6 = ( -7 ) . 78
=> -y = \(\frac{\left(-7\right).78}{6}=-91\)
=> y = 91
\(\frac{x}{102}=\frac{-91}{78}\)
=> x . 78 = ( -91 ) . 102
=> x = \(\frac{\left(-91\right).102}{78}=-119\)
vậy x = -119 ; y = 91
Theo đề ta có: \(x=\frac{102.7}{-6}=\frac{714}{-6}=-119\)
\(y=\frac{-78.7}{-6}=\frac{-546}{-6}=91\)
Vậy \(x=-119;y=91\)
a) \(\frac{3}{4}.x+40\%=\frac{-1}{4}\)
\(\frac{3}{4}.x+\frac{2}{5}=\frac{-1}{4}\)
\(\frac{3}{4}.x=\frac{-13}{20}\)
\(x=\frac{-13}{15}\)
Vậy \(x=\frac{-13}{15}\)
c) \(|x-1|=2^3+\left(-5\right)\)
\(|x-1|=3\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)
Vậy \(x\in\left\{-2;4\right\}\)
Ta có: \(-\frac{8}{15}< \frac{x}{40}\le-\frac{7}{15}\)
\(\Leftrightarrow-\frac{8.8}{15.8}< \frac{3x}{40.3}\le\frac{-7.8}{15.8}\)
\(\Leftrightarrow-\frac{64}{120}< \frac{3x}{120}\le-\frac{56}{120}\)
=> 3x từ -56 đến -63
=> 3x = -57 ; -60 ; -63
=> x = -19 ; -20 ; -21
\(\frac{-15}{25}\)= \(\frac{-3}{5}\)
\(\frac{a}{b}\)= \(\frac{-3}{5}\)Suy ra 5a= -3b (*)
\(b-a=32\)
\(a=b-32\)(**)
\(5.\left(b-32\right)\)=\(-3b\)
\(5b-160=-3b\)
\(5b+3b=160\)
\(b=160:8=20\)
a) \(\frac{x}{3}-\frac{10}{21}=-\frac{1}{7}\)
\(\Rightarrow\frac{x}{3}=-\frac{1}{7}+\frac{10}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{7}{21}\)
\(\Rightarrow\frac{x}{3}=\frac{1}{3}\)
\(\Rightarrow x=1\)
\(x-25\%=\frac{1}{2}\)
\(\Rightarrow x-\frac{1}{4}=\frac{1}{2}\)
\(\Rightarrow x=\frac{1}{2}+\frac{1}{4}\)
\(\Rightarrow x=\frac{3}{4}\)
c) \(-\frac{5}{6}+\frac{8}{3}+-\frac{29}{6}\le x\le-\frac{1}{2}+2+\frac{5}{2}\)
\(\Rightarrow-3\le x\le4\)
\(\Rightarrow x\in\left\{-3;-2;-1;0;1;2;3;4\right\}\)
Ta có:
0,25 . 42,9 - 11,7 . 0,25 + 0,25 . 0,8
= 0,25 ( 42,9 - 11,7 + 0,8 )
= 0,25 . 32
= 8
\(\frac{2.x-4,36}{0,125}=8\)
\(\Rightarrow\left(2.x-4,36\right):0,125=8\)
\(\Rightarrow2x-4,36=8.0,125=1\)
\(2x=1+4,36=5,36\)
\(x=5,36:2=2,68\)
\(\frac{2x-4,36}{0,125}=0,25.42,9-11,7.0,25+0,25.0,8\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.\left(42,9-11,7+0,8\right)\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=0,25.32\)
\(\Leftrightarrow\frac{2x-4,36}{0,125}=8\)
\(\Leftrightarrow2x-4,36=1\)
\(\Leftrightarrow2x=5,36\)
\(\Leftrightarrow x=5,36:2=2,68\)