Tim x biet : (x+1) (x+6)-x2=20
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- a, Do \(\left[3x-15\right]^7=0\)
=> \(3x-15=0\)
=> \(3x=0+15\)
=> \(3x=15\)
=> \(x=15:3\)
=> \(x=5\)
\(\left(3x-5\right)^7=0\)
\(\Rightarrow3x-5=0\)
\(\Rightarrow3x=5\)
\(\Rightarrow x=\frac{5}{3}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{x}=\frac{1}{42}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-......+\frac{1}{5}-\frac{1}{6}+\frac{1}{x}=\frac{1}{42}\)
\(1-\frac{1}{6}+\frac{1}{x}=\frac{1}{42}\)
\(\frac{5}{6}+\frac{1}{x}=\frac{1}{42}\)
\(\frac{1}{x}=\frac{1}{42}-\frac{5}{6}\)
\(\frac{1}{x}=\frac{-34}{42}\)
x=\(\frac{-21}{17}\)
Ta có:\(6^x+99=20\times y\)
\(\Rightarrow6^x=20\times y-99\).Vì \(20\times y-99\) là số lẻ nên \(6^x\) là số lẻ
\(\Rightarrow6^x=1\Rightarrow x=0\)
\(\Rightarrow20\times y=1+99=100\Rightarrow y=5\)
Vậy x=0;y=5 thỏa mãn
(-25)-(5.x-3)=4
5.x-3 = -25 - 4
5.x-3=-29
5.x=-26
x=-26/5
x=-5,2
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\Leftrightarrow\frac{4}{20}-x=\frac{21}{128}.\frac{32}{9}\)
\(\Leftrightarrow\frac{1}{5}-x=\frac{7}{12}\)
\(\Leftrightarrow x=\frac{1}{5}-\frac{7}{12}\)
\(\Leftrightarrow x=-\frac{23}{60}\)
Vậy \(x=\frac{-23}{60}\)
Bài làm
\(\left(\frac{3}{20}+\frac{1}{20}-x\right):\frac{32}{9}=\frac{21}{128}\)
\(\left(\frac{3}{20}+\frac{1}{20}-x\right)=\frac{21}{128}.\frac{32}{9}\)
\(\frac{4}{20}-x=\frac{7}{4}.\frac{1}{3}\)
\(\frac{4}{20}-x=\frac{7}{12}\)
\(x=\frac{4}{20}-\frac{7}{12}\)
\(x=\frac{12}{60}-\frac{35}{60}\)
\(x=-\frac{23}{60}\)
Vậy \(x=-\frac{23}{60}\)
\(x=-\frac{37}{15}\)
Ta có: (x+1).(x+6)-x2=20
=> x.(x+6)+x+6-x2=20
=> x2+6.x+x+6-x2=20
=> (x2-x2)+(6.x+x)=20-6
=> 7.x=14
=> x=14:7
=> x=2
Vậy x=2
l-i-k-e cho mình nha bạn.