Tìm x thuộc N, biết
a, (7.x-11)3=25.52+200
b, (2.x-15)5=(2.x-15)3
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b: =>x(8-7)=-33
=>x=-33
c: =>-12x+60+21-7x=5
=>-19x=-76
hay x=4
d: =>-2x-2-x+5+2x=0
=>3-x=0
hay x=3
Bài 1:
a) Ta có: \(\dfrac{17}{6}-x\left(x-\dfrac{7}{6}\right)=\dfrac{7}{4}\)
\(\Leftrightarrow\dfrac{17}{6}-x^2+\dfrac{7}{6}x-\dfrac{7}{4}=0\)
\(\Leftrightarrow-x^2+\dfrac{7}{6}x+\dfrac{13}{12}=0\)
\(\Leftrightarrow-12x^2+14x+13=0\)
\(\Delta=14^2-4\cdot\left(-12\right)\cdot13=196+624=820\)
Vì Δ>0 nên phương trình có hai nghiệm phân biệt là:
\(\left\{{}\begin{matrix}x_1=\dfrac{14-2\sqrt{205}}{-24}=\dfrac{-7+\sqrt{205}}{12}\\x_2=\dfrac{14+2\sqrt{2015}}{-24}=\dfrac{-7-\sqrt{205}}{12}\end{matrix}\right.\)
b) Ta có: \(\dfrac{3}{35}-\left(\dfrac{3}{5}-x\right)=\dfrac{2}{7}\)
\(\Leftrightarrow\dfrac{3}{5}-x=\dfrac{3}{35}-\dfrac{10}{35}=\dfrac{-7}{35}=\dfrac{-1}{5}\)
hay \(x=\dfrac{3}{5}-\dfrac{-1}{5}=\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{4}{5}\)
\(a,-\dfrac{13}{20}+x=\dfrac{-11}{15}\\ \Rightarrow x=\dfrac{-11}{15}+\dfrac{13}{20}\\ \Rightarrow x=-\dfrac{1}{12}\\ b,\left(x-3,5\right):3\dfrac{1}{2}-2,5=-1\dfrac{3}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}-\dfrac{5}{2}=\dfrac{-7}{4}\\ \Rightarrow\left(x-\dfrac{7}{2}\right):\dfrac{7}{2}=\dfrac{3}{4}\\ \Rightarrow x-\dfrac{7}{2}=\dfrac{21}{8}\\ \Rightarrow x=\dfrac{49}{8}\)
1:
a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)
b:
Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\)
\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)
2:
a: x=2,4-0,4=2
b: =>2x=-1,5+0,8=-0,7
=>x=-0,35
c: =>x-16=-15
=>x=1
a) \(\left(7x-11\right)^3=2^5.5^2+200\)
\(\Rightarrow\left(7x-11\right)^3=32.25+200\)
\(\Rightarrow\left(7x-11\right)^3=800+200\)
\(\Rightarrow\left(7x-11\right)^3=1000\)
\(\Rightarrow\left(7x-11\right)^3=10^3\)
\(\Rightarrow7x-11=10\)
\(\Rightarrow7x=10+11\)
\(\Rightarrow7x=21\)
\(\Rightarrow x=21:7\)
\(\Rightarrow x=3\)
Vậy x = 3
b) \(\left(2x-15\right)^5=\left(2x-15\right)^3\)
\(\Rightarrow\left(2x-15\right)^5-\left(2x-15\right)^3=0\)
\(\Rightarrow\left(2x-15\right)^3\left[\left(2x-15\right)^2-1\right]=0\)
\(\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}\left(2x-15\right)^3=0\\\left(2x-15\right)^2=1\end{cases}}\)
TH 1 : \(\left(2x-15\right)^3=0\Rightarrow2x-15=0\Rightarrow2x=15\Rightarrow x=\frac{15}{2}\)
TH 2 : \(\left(2x-15\right)^2=1\Rightarrow\orbr{\begin{cases}2x-15=1\\2x-15=-1\end{cases}}\Rightarrow\orbr{\begin{cases}2x=16\\2x=14\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=7\end{cases}}\)
Vậy \(x\in\left\{\frac{15}{2};8;7\right\}\)
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