tim x biet: (x-3)*(x-5)+1=0 giup minh voi. Arigato gozaimasu!
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9
9 tháng 6 2016
( x + 1 ) + ( x + 2 ) + ( x + 3 ) +... + ( x + 100 ) = 5750
( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 100 ) = 5750
( x . 100 ) + ( 1 . 100 ) . 100 : 2 = 5750
( x . 100 ) + 5050 = 5750
x . 100 = 5750 - 5050
x . 100 = 700
x = 700 : 100
x = 7
Vậy x = 7
NH
0
25 tháng 7 2020
\(\frac{5}{3}-\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{7}{6}\)
=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)
=> \(\frac{1}{3}:\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{2}\)
=> \(\left(1-x\cdot\frac{1}{3}\right)=\frac{1}{3}:\frac{1}{2}=\frac{1}{3}\cdot2=\frac{2}{3}\)
=> \(1-\frac{x}{3}=\frac{2}{3}\)
=> \(\frac{x}{3}=1-\frac{2}{3}=\frac{1}{3}\)
=> x = 1
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)
=> \(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}=\frac{3}{4}\)
=> \(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}=\frac{9}{4}\)
=> \(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)
=> \(x:\frac{1}{2}=\frac{3}{4}\)
=> \(x=\frac{3}{4}\cdot\frac{1}{2}=\frac{3}{8}\)
25 tháng 7 2020
\(\frac{5}{3}-\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{7}{6}\)
\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{5}{3}-\frac{7}{6}\)
\(\frac{1}{3}:\left(1-x\times\frac{1}{3}\right)=\frac{1}{2}\)
\(1-x\times\frac{1}{3}=\frac{1}{3}:\frac{1}{2}\)
\(1-x\times\frac{1}{3}=\frac{2}{3}\)
\(x\times\frac{1}{3}=1-\frac{2}{3}\)
\(x\times\frac{1}{3}=\frac{1}{3}\)
\(x=\frac{1}{3}:\frac{1}{3}\)
\(x=1\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)-\frac{1}{4}=\frac{1}{2}\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{1}{2}+\frac{1}{4}\)
\(3-\left(x:\frac{1}{2}+\frac{3}{2}\right)=\frac{3}{4}\)
\(x:\frac{1}{2}+\frac{3}{2}=3-\frac{3}{4}\)
\(x:\frac{1}{2}+\frac{3}{2}=\frac{9}{4}\)
\(x:\frac{1}{2}=\frac{9}{4}-\frac{3}{2}\)
\(x:\frac{1}{2}=\frac{3}{4}\)
\(x=\frac{3}{4}\times\frac{1}{2}\)
\(x=\frac{3}{8}\)
NH
1
\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Leftrightarrow x^2-5x-3x+15+1=0\)
\(\Leftrightarrow x^2-8x+16=0\)
\(\Leftrightarrow\left(x-4\right)^2=0\)
\(\Leftrightarrow x-4=0\)
\(\Leftrightarrow x=4\)
Vậy \(x=4\)
\(\left(x-3\right)\left(x-5\right)+1=0\)
\(\Rightarrow x^2-3x-5x+15+1=0\)
\(\Rightarrow x^2-8x+16=0\)
\(\Rightarrow x^2-2x.4+4^2=0\)
\(\Rightarrow\left(x-4\right)^2=0\)
\(\Rightarrow x-4=0\)
\(\Rightarrow x=4\)
Vậy \(x=4\)