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a,ĐK : x \(\ne\)3/7
\(\frac{24}{7x-3}=-\frac{4}{25}\Leftrightarrow600=-28x+12\Leftrightarrow-28x=588\Leftrightarrow x=-21\)
b, ĐK : x;y \(\ne\)6
Xét : \(\frac{4}{x-6}=-\frac{12}{18}\Leftrightarrow72=-12x+72\Leftrightarrow x=0\)
Xét : \(\frac{y}{24}=-\frac{12}{18}\Leftrightarrow18y=-288\Leftrightarrow y=-16\)
\(\frac{24}{7.x-3}=-\frac{4}{25}\)
24.25=7.x-3.-4
600=7.x-3.-4
7.x-3.-4=600
7.x-3=600:-4
7.x-3=-150
7.x=-150+3
7.x=-147
x=-147:7
x=-21
vậy x=-21
a, \(\frac{x+3}{y+4}=\frac{3}{4}\)
\(< =>\frac{x+3}{3}=\frac{y+4}{4}< =>\frac{x}{3}+1=\frac{y}{4}+1\)
\(< =>\frac{x}{3}=\frac{y}{4}\)
Theo tinh chat cua day ti so bang nhau ta co
\(\frac{x}{3}=\frac{y}{4}=\frac{x+y}{3+4}=\frac{21}{7}=3\)
\(=>\hept{\begin{cases}x=3.3=9\\y=4.3=12\end{cases}}\)
a)
\(\frac{x+3}{y+4}=\frac{3}{4}\)
\(\Leftrightarrow\frac{x+3}{3}=\frac{y+4}{4}\)
Áp dụng tính chất của dãy tỉ số bằng nhau :
\(\frac{x+3}{3}=\frac{y+4}{4}=\frac{x+y+3+4}{3+4}=\frac{28}{7}=4\)
Do đó
\(\frac{x+3}{3}=4\Rightarrow x+3=12\Rightarrow x=9\)
\(\frac{y+4}{4}=4=>y+4=16\Rightarrow y=12\)
\(\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)
\(\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)
\(\frac{3}{x}=\frac{-1}{2}\)
\(\Rightarrow3.2=\left(-1\right).x\)
\(\Rightarrow6=\left(-1\right).x\)
\(\Rightarrow x=6:\left(-1\right)\)
\(\Rightarrow x=-6\)
\(\frac{x}{2}-\frac{2}{y}=\frac{1}{2}\)
\(\Rightarrow\frac{x}{2}-\frac{1}{2}=\frac{2}{y}\)
\(\Rightarrow\frac{x-1}{2}=\frac{2}{y}\)
\(\Rightarrow\hept{\begin{cases}x-1=2\\2=y\end{cases}\Rightarrow}\hept{\begin{cases}x=3\\y=2\end{cases}}\)
\(b,\frac{3}{x}+\frac{4}{3}=\frac{5}{6}\)
\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{4}{3}\)
\(\Rightarrow\frac{3}{x}=\frac{5}{6}-\frac{8}{6}\)
\(\Rightarrow\frac{3}{x}=\frac{-3}{6}\)
\(\Rightarrow x\cdot(-3)=18\Rightarrow x=-6\)
1)
A = \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+..+\frac{2}{99.101}\)
A = \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{99}-\frac{1}{101}\)
A = \(\frac{1}{1}-\frac{1}{101}\)
A = \(\frac{100}{101}\)
Vậy A = \(\frac{100}{101}\)
B = \(\frac{5}{1.3}+\frac{5}{3.5}+...+\frac{5}{99.101}\)
B = \(\frac{5}{2}\left(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}\left(\frac{1}{1}-\frac{1}{101}\right)\)
B = \(\frac{5}{2}.\frac{100}{101}\)
B = \(\frac{250}{101}\)
Vậy B = \(\frac{250}{101}\)
2)
Gọi ƯCLN ( 2n + 1 ; 3n + 2 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+1⋮d\\3n+2⋮d\end{cases}\Rightarrow\hept{\begin{cases}3\left(2n+1\right)⋮d\\2\left(3n+2\right)⋮d\end{cases}}}\)
\(\Rightarrow\hept{\begin{cases}6n+3⋮d\\6n+4⋮d\end{cases}\Rightarrow\left(6n+4\right)-\left(6n+3\right)⋮d\Rightarrow1⋮d}\)
\(\Rightarrow d=1\)
Vậy \(\frac{2n+1}{3n+2}\)là p/s tối giản
Gọi ƯCLN ( 2n+3 ; 4n+4 ) = d ( d \(\in\)N* )
\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\4n+4⋮d\end{cases}\Rightarrow\hept{\begin{cases}2n+3⋮d\\\left(4n+4\right):2⋮d\end{cases}}}\)\(\Rightarrow\hept{\begin{cases}2n+3⋮d\\2n+2⋮d\end{cases}\Rightarrow\left(2n+3\right)-\left(2n+2\right)⋮d}\)
\(\Rightarrow1⋮d\Rightarrow d=1\)
Vậy ...
\(\frac{5}{11}.\frac{3}{7}+\frac{5}{7}.\frac{4}{11}-\frac{3}{11}\)
\(=\frac{3}{11}.\frac{5}{3}.\frac{3}{7}+\frac{5}{7}.\frac{4}{3}.\frac{3}{11}-\frac{3}{11}\)
\(=\frac{3}{11}.\left(\frac{5}{7}+\frac{20}{21}-1\right)\)
\(=\frac{3}{11}.\left(\frac{15+20}{21}-1\right)\)
\(=\frac{3}{11}.\left(\frac{35}{21}-1\right)\)
\(=\frac{3}{11}.\frac{14}{21}\)
\(=\frac{14}{77}\)
5/11 x 3/7 + 5/7 x 4/11 - 3/11
= 5/11 x 3/7 + 5/11 x 4/7 - 3/11
= 5/11 x ( 3/7 + 4/7 ) - 3/11
= 5/11 x 1 - 3/11
= 5/11 - 3/11
= 2/11
Tk nha !!
2/9.-11/5+4/9.-5/11+1/3.-5/11
-22/45+-20/99+-5/33
-38/55+-5/33
-139/165
Ta có : \(\frac{x}{3}=\frac{y}{8};x+y=-22\)
Áp dụng tính cất của dãy tỉ số bằng nhau , ta có :
\(\frac{x}{3}=\frac{y}{8}=\frac{x+y}{3+8}=\frac{-22}{11}=-2\)
\(\Rightarrow\frac{x}{3}=-2\Rightarrow x=-6\)
\(\Rightarrow\frac{y}{8}=-2\Rightarrow y=-16\)
Vậy x = -6 và y = -16
Áp dụng tính chất của dãy tỉ số bằng nhau ta có :
\(\frac{x}{3}=\frac{y}{8}=\frac{x+y}{3+8}=-\frac{22}{11}=-2\)
Ta có :
\(\frac{x}{3}=-2\Rightarrow x=-6\)
\(\frac{y}{8}=-2\Rightarrow y=-16\)
Vậy x = -6 ; y = -16
\(-\dfrac{4}{11}=\dfrac{x}{22}\Rightarrow x=\dfrac{\left(-4\right)\cdot22}{11}=-8\\ -\dfrac{4}{11}=\dfrac{40}{y}\Rightarrow y=\dfrac{40\cdot11}{-4}=-110\\ -\dfrac{4}{11}=-\dfrac{8}{22}=\dfrac{40}{-110}\)
\(\frac{-4}{11}=\frac{x}{22}=\frac{40}{y}\)
\(+)\) Với \(\frac{-4}{11}=\frac{x}{22}\) ta có:
\(\left(-4\right).22=11.x\)
\(-88=11.x\)
\(x=\left(-88\right):11\)
\(x=-8\)
\(+)\) Với \(\frac{-4}{11}=\frac{40}{y}\) ta có:
\(\left(-4\right).y=11.40\)
\(\left(-4\right).y=440\)
\(y=440:\left(-4\right)\)
\(y=-110\)
Vậy \(x=-8\) và \(y=-110\)