0,( 17) : 2,(3) = x : 0,(3)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a) x÷0,(7)=0,(32):2,(4)
\(x:\frac{7}{9}=\frac{32}{99}:\frac{22}{9}\)
\(x:\frac{7}{9}=\frac{16}{121}\)
\(x=\frac{16}{121}.\frac{7}{9}\)
\(x=\frac{112}{1089}\)
b)0,(17):2,(3)=x:0,(3)
\(\frac{17}{99}:\frac{7}{3}=x:\frac{1}{3}\)
\(\frac{17}{231}=x:\frac{1}{3}\)
x=\(\frac{17}{231}.\frac{1}{3}\)
\(x=\frac{17}{693}\)
a) X : \(\frac{7}{9}=\frac{32}{99}:\left(2+\frac{4}{9}\right)\) => X : \(\frac{7}{9}=\frac{32}{99}:\frac{22}{9}\)=> X : \(\frac{7}{9}=\frac{64}{81}\) => X = \(\frac{64}{81}.\frac{7}{9}=\frac{64}{63}\)
b) \(\frac{17}{99}:\left(2+\frac{3}{9}\right)=X:\frac{3}{9}\)=> \(\frac{17}{99}:\frac{7}{3}=X:\frac{1}{3}\)=> \(\frac{17}{231}=X:\frac{1}{3}\)=> X = \(\frac{17}{231}.\frac{1}{3}=\frac{17}{693}\)
Vậy...
1) Ta có: \(\left(x-2\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-1\end{matrix}\right.\)
Vậy \(x=2\) hoặc \(x=-1\)
2) Ta có: \(\left(3-x\right)x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}3-x=0\\x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=0\end{matrix}\right.\)
Vậy \(x=3\) hoặc \(x=0\)
3) Ta có: \(2x-17=-\left(3x-18\right)\)
\(\Leftrightarrow2x-17=18-3x\)
\(\Leftrightarrow2x+3x=18+17\)
\(\Leftrightarrow5x=35\Leftrightarrow x=\dfrac{35}{5}=7\)
Vậy \(x=7\)
1/ x(x+17)=0
⇒ \(\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
2/ (x+1112)(x-3)=0
⇒\(\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
3/ (-x+25)(3-x)=0
⇒\(\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
4/ x(12+x)(7-x)=0
⇒ \(\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
5/ (x-15)(x+2)(-x-3)=0
⇒\(\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
\(x\left(x+17\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x+17=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-17\end{matrix}\right.\)
Vậy \(x\in\left\{0;-17\right\}\)
\(\left(x+1112\right)\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x+1112=0\\x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-1112\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{-1112;3\right\}\)
\(\left(-x+25\right)\left(3-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}-x+25=0\\3-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=25\\x=3\end{matrix}\right.\)
Vậy \(x\in\left\{25;3\right\}\)
\(x\left(12+x\right)\left(7-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\12+x=0\\7-x=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-12\\x=7\end{matrix}\right.\)
Vậy \(x\in\left\{0;-12;7\right\}\)
\(\left(x-15\right)\left(x+2\right)\left(-x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-15=0\\x+2=0\\-x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=15\\x=-2\\x=-3\end{matrix}\right.\)
Vậy \(x\in\left\{15;-2;-3\right\}\)
a)
(2x-4)(x-22)=0
<=>2x-4=0 hoặc x-22=0
<=>x=2 hoặc x=22
b
(x-17)(x^2-16)=0
<=>x-17 =0 hoặc x^2-16=0
<=>x=17 hoặc x=4 hoặc x=-4
c
(x^2+3)(x+8)=0
Vì x^2+3>0
=>x+8=0
<=>x=-8
| x-3 | = 4x
☛ x-3 = 4x
⇒ -3 = 4x -x
⇒ -3 = 3x
☛ x = ( -3):3
x = -1
HOẶC x-3= -(4x)
x-3 = (-4)(-x)
=> -3=(-4)(-x)-x
=> -3 = -3 . x
x = 1
vay x=1 hoac -1
1) Do x ∈ Z và 0 < x < 3
⇒ x ∈ {1; 2}
2) Do x ∈ Z và 0 < x ≤ 3
⇒ x ∈ {1; 2; 3}
3) Do x ∈ Z và -1 < x ≤ 4
⇒ x ∈ {0; 1; 2; 3; 4}
5x(2x-1)+12x3=41
5x(2x-1)+36 =41
5x(2x-1) =41-36
5x(2x-1) =5
2x-1 =5:5
2x-1 =1
2x =1+1
2x =2
x =1
Ta có: \(x:\left(0,3\right)\)= \(\frac{17}{231}\)
\(\Rightarrow\)\(x=\frac{17}{77}\)
Mình lộn dấu ngoặc nha bạn sử chỗ \(0,\left(3\right)\)nha ^.^