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Xét \(\frac{-5}{x}=\frac{-18}{72}\)
\(\Rightarrow\)(-5) . 72 = x . (-18)
\(\Rightarrow\)-360 = x . (-18)
\(\Rightarrow\)x = (-360) : (-18) = 20
Xét \(\frac{y}{16}=\frac{-18}{72}\)
\(\Rightarrow\)72y = 16 . (-18)
\(\Rightarrow\)72y = -288
\(\Rightarrow\)y = (-288) : 72 = -4
Vậy x = 20 ; y = -4
Ta có :
\(\frac{-5}{x}=\frac{-18}{72}\Rightarrow\frac{-5}{x}=\frac{-1}{4}\)
\(\Rightarrow-5.4=x.-1\)
\(\Rightarrow-20=-x\)
\(\Rightarrow x=20\)
Thay x = 20 vào \(\frac{-5}{x}\)ta được :
\(\frac{-5}{20}=\frac{y}{16}\)\(\Rightarrow\frac{-1}{4}=\frac{y}{16}\Rightarrow-1.16=y.4\)
\(\Rightarrow-16=y.4\)
\(\Rightarrow y=-4\)
Vậy x = 20 , y = -4
Tk mk nha !!!
\(\dfrac{-5}{x}=\dfrac{y}{16}=\dfrac{-18}{72}\)
=>\(\dfrac{-5}{x}=\dfrac{-18}{72}=>x=\dfrac{72.-5}{-18}=20\)
\(\dfrac{y}{16}=\dfrac{-18}{72}=>y=\dfrac{16.\left(-18\right)}{72}=-4\)
chúc bạn học tốt ^ ^
mk nghĩ đây là toán 8.
\(Pt\Leftrightarrow\left(x-2010\right)\left(\frac{1}{12}+\frac{1}{20}+\frac{1}{30}+....+\frac{1}{72}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right)\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+....+\frac{1}{8.9}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right)\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+....-\frac{1}{9}\right)=\frac{16}{9}\Leftrightarrow\left(x-2010\right).\frac{2}{9}=\frac{16}{9}\Leftrightarrow x-2010=8\Leftrightarrow x=2018.\text{ Vậy: x=2018}\)
a/ \(\dfrac{-5}{x}\) = \(\dfrac{-18}{72}\)
\(\Rightarrow\dfrac{5}{x}=\dfrac{18}{72}=\dfrac{1}{4}\)
\(\Rightarrow x=5:\dfrac{1}{4}=20\)
\(\dfrac{y}{16}=\dfrac{-18}{72}\) \(=\dfrac{-1}{4}\)
\(\Leftrightarrow\) 4y = - 16
\(\Leftrightarrow\) y = -4
Vậy x = 20 ; y = -4
b/ \(\dfrac{1}{2}=\dfrac{x-1}{8}=\dfrac{10}{y-5}\)
\(\Rightarrow\dfrac{x-1}{8}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{x-1}{8}=\dfrac{4}{8}\)
\(\Rightarrow x-1=4\)
\(\Rightarrow\) \(x=4+1\)
\(x=5\)
\(\dfrac{10}{y-5}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{10}{y-5}=\dfrac{10}{20}\)
\(\Rightarrow\) \(y-5=20\)
\(\Rightarrow y=20+5\)
\(y=25\)
Vậy x = 5 ; y = 25
a)
\(-\dfrac{5}{x}=-\dfrac{18}{72}\\ \Leftrightarrow\dfrac{5}{x}=\dfrac{18}{72}=\dfrac{1}{4}\\ \Leftrightarrow x=20\)
\(\dfrac{y}{16}=\dfrac{-18}{72}=-\dfrac{1}{4}\\ \Leftrightarrow4y=-16\\ \Leftrightarrow y=-4\)
b)
\(\dfrac{x-1}{8}=\dfrac{1}{2}\Rightarrow\dfrac{x-1}{8}=\dfrac{4}{8}\Rightarrow x-1=4\Rightarrow x=5\)
\(\dfrac{10}{y-5}=\dfrac{1}{2}\Leftrightarrow y-5=20\Leftrightarrow y=25\)
-18.16=72.y ( nhân chéo )
-288=72.y
y=-4
-5.16=-4(thay vào y) .x
-80=-4.x
x=20
ta có : \(\frac{y}{16}=-\frac{18}{72}\Rightarrow-18.16=y.72\Rightarrow-288=y.72\Rightarrow y=-288:72=-4\)
\(\Rightarrow-\frac{5}{x}=-\frac{4}{16}\Rightarrow x.\left(-4\right)=-5.16\Rightarrow80:\left(-4\right)=x=>x=-20\)
vậy.....
a) x \(\in\) B(BC(35; 63; 105)) = 315
=> x = 325k (k \(\in\) N*) . Mà 315 < x < 632 nên 315 < 325k < 632
hay 1 < k < 3. Do đó k = 2
b) x - 2 \(\in\) Ư(ƯC(8; 32; 48)) = 8
Vì 0 < x < 100 nên -2 < x - 2 < 98
Do đó x - 2 \(\in\) {-2; -1; 1; 2; 4; 8}
\(\Leftrightarrow\) x \(\in\) {0; 1; 3; 4; 6; 10}
a) \(\overline{3x}+\overline{5x}-\overline{1x}=72\)
\(\Leftrightarrow\left(30+x\right)+\left(50+x\right)-\left(10+x\right)=72\)
\(\Leftrightarrow30+x+50+x-10-x=72\)
\(\Leftrightarrow x+70=72\)
\(\Leftrightarrow x=2\)
Vậy \(x=2\)
b) \(x.\left(x+1\right)=132\)
\(\Leftrightarrow x^2+x=132\)
\(\Leftrightarrow x^2+x-132=0\)
\(\Leftrightarrow x^2+12x-11x-132=0\)
\(\Leftrightarrow x.\left(x+12\right)-11.\left(x+12\right)=0\)
\(\Leftrightarrow\left(x-11\right).\left(x+12\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-11=0\\x+12=0\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=11\\x=-12\end{cases}}\)
Vậy \(x=11\)hoặc \(x=-12\)
x= (-72):63*(-16)=\(\dfrac{128}{7}\)
Ta có: (-72):63=-16:x
\(\Leftrightarrow\dfrac{-16}{x}=\dfrac{-72}{63}\)
\(\Leftrightarrow\dfrac{-16}{x}=\dfrac{-8}{7}\)
\(\Leftrightarrow x=\dfrac{-16\cdot7}{-8}=\dfrac{16\cdot7}{8}=14\)
Vậy: x=14