(12x15-x)x1/4=120 x 1/4
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a) x : ( 12 x 15 ) = 8/3 x 4 x 15
x : 180 = 160
x = 160 : 180
x = 8/9
Vậy x = 8/9
b) ( 1/2 x 2/3 x 3/4 x 4/5 ) : x = 1 x 2 x 3 x 4/ 5 x 6 x 7 x 8
1/5 : x = 1/70
x = 1/5 : 1/70
x = 14
Vậy x = 14
(x + 5 ) x \(\frac{19}{13}=57\)
x + 5 = \(57\div\frac{19}{13}=39\)
x = 39 -5 = 34
b) 12 . 15 - x = 120 \(\cdot\frac{1}{4}\div\frac{1}{4}=120\)
180 - x = 120
x = 180 - 120 = 60
\(x\)là dấu nhân hả bạn? Nếu vậy thì mk làm cho nhé
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot....\cdot\left(1-\frac{1}{20}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot.......\cdot\frac{17}{18}\cdot\frac{18}{19}\cdot\frac{19}{20}=\frac{1}{20}\)
Vậy \(A=\frac{1}{20}\)
\(B=1\frac{1}{2}\cdot1\frac{1}{3}\cdot1\frac{1}{4}\cdot........\cdot1\frac{1}{2005}\cdot1\frac{1}{2006}\cdot1\frac{1}{2007}\)
\(B=\frac{3}{2}\cdot\frac{4}{3}\cdot\frac{5}{4}\cdot......\cdot\frac{2006}{2005}\cdot\frac{2007}{2006}\cdot\frac{2008}{2007}=\frac{2008}{2}=1004\)
Vậy \(B=1004\)
DẤU CHẤM LÀ DẤU NHÂN
a,
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{19}{20}=\frac{1}{20}\)
b, \(1\frac{1}{2}.1\frac{1}{3}....1\frac{1}{2017}=\frac{3}{2}.\frac{4}{3}....\frac{2018}{2017}=\frac{2018}{2}=1009\)
\(\dfrac{1}{120}\cdot120+x:\dfrac{1}{3}=-4\)
\(\Leftrightarrow1+x\cdot3=-4\)
\(\Leftrightarrow3x=-5\)
\(\Leftrightarrow x=-\dfrac{5}{3}\)
\(\dfrac{1}{120}.120+x:\dfrac{1}{3}=-4\)
\(1+x:\dfrac{1}{3}=-4\)
\(x:\dfrac{1}{3}=-4-1\)
\(x:\dfrac{1}{3}=-5\)
\(x=-5.\dfrac{1}{3}\)
\(x=\dfrac{-5}{3}\)
\(x^4-1-2\left(m+1\right)x^2+2\left(m+1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2+1\right)-2\left(m+1\right)\left(x^2-1\right)=0\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-2m-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^2=1\\x^2=2m+1\end{matrix}\right.\)
Pt có 4 nghiệm pb khi: \(\left\{{}\begin{matrix}2m+1>0\\2m+1\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m>-\dfrac{1}{2}\\m\ne0\end{matrix}\right.\)
Do \(x=\pm1< 3\) nên để \(x_1< x_2< x_3< x_4< 3\) thì:
\(\sqrt{2m+1}< 3\Leftrightarrow m< 4\) \(\Rightarrow\left\{{}\begin{matrix}-\dfrac{1}{2}< m< 4\\m\ne0\end{matrix}\right.\)
b. \(\left\{{}\begin{matrix}x_1-x_3=x_3-x_2\\x_1-x_3=x_2-x_1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_1=-x_2\\x_1-x_3=-x_1-x_1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x_2=-x_1\\x_3=3x_1\end{matrix}\right.\)
Do vai trò \(x_1;x_2\) như nhau, giả sử \(x_1< 0\) \(\Rightarrow x_1;x_3\) là 2 nghiệm âm
TH1: \(\left\{{}\begin{matrix}x_1=-1\\x_2=1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x_3=-\sqrt{2m+1}\\x_3=3x_1\end{matrix}\right.\) \(\Rightarrow-\sqrt{2m+1}=-3\Rightarrow m=4\)
TH2: \(x_1=-\sqrt{2m+1}\Rightarrow\left\{{}\begin{matrix}x_3=-1\\x_3=3x_1\end{matrix}\right.\) \(\Rightarrow-1=-3\sqrt{2m+1}\) \(\Rightarrow m=-\dfrac{4}{9}\)
thầy cho em hỏi nếu bài này đặt \(x^2=t^{ }\left(t\ge0\right)\)
thì giải pt ẩn t có 2 nghiệm phân biệt dương
\(=>\left\{{}\begin{matrix}\Delta>0\\S>0\\P>0\end{matrix}\right.\) em giải ra thì m>0 =)))
1) ( y-25):4 - 120 = 0 2) 120 - ( x+25 )x4 = 0
( y-25 ):4 = 0+120 (x+25) x4 =120-0
( y-25):4 =120 (x+25) x4 =120
( y-25)=120x4 (x+25)=120:4
y-25=480 x+25=30
y=480+25 x=30-25
y=505 x=5
k mk nha
\(\frac{1}{1}\cdot\frac{1}{2}+\frac{1}{2}\cdot\frac{1}{3}+\frac{1}{3}\cdot\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{5}+\frac{1}{5}\cdot\frac{1}{6}=\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}=\)
\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}=\)
\(\frac{1}{1}-\frac{1}{6}=\frac{5}{6}\)
k nha gõ mỏi tay lắm
\(\left(12\cdot15-x\right)\cdot\dfrac{1}{4}=120\cdot\dfrac{1}{4}\)
\(\Rightarrow\left(180-x\right)\cdot\dfrac{1}{4}=30\)
\(\Rightarrow180-x=30:\dfrac{1}{4}\)
\(\Rightarrow180-x=120\)
\(\Rightarrow x=180-120\)
\(\Rightarrow x=60\)