Tìm x biết :
7 . x - ( 6.x+5 )=17
giải giúp mình với T_T
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a) \(\left(x+1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x^2-x+1\right)=7\)
\(\Leftrightarrow\left[\left(x+1\right)\left(x^2-x+1\right)\right]\left[\left(x-1\right)\left(x^2+x+1\right)\right]=7\)
\(\Leftrightarrow\left(x^3+1\right)\left(x^3-1\right)=7\)
\(\Leftrightarrow x^6-1=7\)
\(\Leftrightarrow x^6=8\Leftrightarrow x=1,414213562\) (số hơi lẻ)
b) \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)
\(\Leftrightarrow x^3-1-x\left(x^2-1\right)=5\)
\(\Leftrightarrow x^3-1-x^3+x=5\)
\(\Leftrightarrow-1+x=5\Leftrightarrow x=6\)
Vậy x = 6
\(a,\Leftrightarrow\left(5x+1\right)\left(x-4\right)-\left(x-4\right)=0\\ \Leftrightarrow\left(x-4\right)\left(5x+1-x\right)=0\\ \Leftrightarrow5x\left(x-4\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\end{matrix}\right.\\ b,\Leftrightarrow2x^2-10x-2x^2-3x=26\\ \Leftrightarrow-13x=26\\ \Leftrightarrow x=-2\\ c,\Leftrightarrow x^3+1-x^3+3x=15\\ \Leftrightarrow3x=14\\ \Leftrightarrow x=\dfrac{14}{3}\)
\(d,\Leftrightarrow x^3-5x+2x^2-10+5x-2x^2-17=0\\ \Leftrightarrow x^3-27=0\\ \Leftrightarrow x^3=27\\ \Leftrightarrow x=3\)
\(B\left(x\right)=x^5+3x^3+x=x\left(x^4+3x^2+1\right)=x\left(x^4+x^2+x^2+1+x^2\right)=x\left[x^2\left(x^2+1\right)+x^2+1+x^2\right]\)
\(=x\left[\left(x^2+1\right)\left(x^2+1\right)+x^2\right]=x\left[\left(x^2+1\right)^2+x^2\right]\)
Vì: \(x^2+1>0,x^2\ge0\)nên \(\left(x^2+1\right)^2+x^2>0\)
Vậy B(x) có nghiệm khi x=0
\(\frac{6}{8}=\frac{15}{x}\)
nhân chéo hai phân số với nhau ta được .
\(6\times x=15\times8\)
\(6\times x=120\)
\(x=120\div6\)
\(x=20\)
\(\Rightarrow x=20\)
\(\frac{6}{8}=\frac{15}{x}\)
=> 6x = 15.8
=> 6x = 120
=> x = 120 : 6
=> x = 20
a) 2 . x^3 - 2 = 52
=> 2 . x^3 = 52 + 2
=> 2 . x^3 = 54
=> x^3 = 54 :2
=> x^3 = 27
=> x =3
b) 2016 - 3 . ( x - 6 ) = 1446
\(\frac{x+2}{327}+\frac{x+3}{326}+\frac{x+4}{325}+\frac{x+5}{324}+\frac{x+349}{5}=0\)
=> \(\left(\frac{x+2}{327}+1\right)+\left(\frac{x+3}{326}+1\right)+\left(\frac{x+4}{325}+1\right)+\left(\frac{x+5}{324}+1\right)+\frac{x+349}{5}=0+1+1+1+1\)
=> \(\frac{x+2+327}{327}+\frac{x+3+326}{326}+\frac{x+4+325}{325}+\frac{x+5+324}{324}+\frac{x+329+20}{5}=4\)
=> \(\frac{x+329}{327}+\frac{x+329}{326}+\frac{x+329}{325}+\frac{x+329}{324}+\frac{x+329}{5}+\frac{20}{5}=4\)
=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)+4=4\)
=> \(\left(x+329\right)\left(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\right)=0\)
Ta có : \(\frac{1}{327}+\frac{1}{326}+\frac{1}{325}+\frac{1}{324}+\frac{1}{5}\ne0\)
=> \(x+329=0\)
=> \(x=-329\)
\(a)\frac{7}{3}-\frac{4}{3}\times x=\frac{5}{6}\)
\(-\frac{4}{3}\times x=\frac{3}{2}\)
\(x=\frac{-9}{8}\)
\(b)\frac{7}{4}-\frac{3}{4}\times x=\frac{5}{6}\)
\(-\frac{3}{4}\times x=\frac{11}{12}\)
\(x=-\frac{11}{9}\)
Bài 7:
a, \(x\) = \(\dfrac{1}{5}\) + \(\dfrac{2}{11}\)
\(x\) = \(\dfrac{11}{55}\) + \(\dfrac{10}{55}\)
\(x=\dfrac{21}{55}\)
b, \(\dfrac{x}{15}\) = \(\dfrac{3}{5}\) - \(\dfrac{2}{3}\)
\(\dfrac{x}{15}\) = \(\dfrac{9}{15}\) - \(\dfrac{10}{15}\)
\(\dfrac{x}{15}\) = \(\dfrac{1}{15}\)
\(x\) = 1
c, \(\dfrac{11}{8}\) + \(\dfrac{13}{6}\)= \(\dfrac{85}{x}\)
\(\dfrac{33}{24}\) + \(\dfrac{52}{24}\) = \(\dfrac{85}{x}\)
\(\dfrac{85}{24}\) = \(\dfrac{85}{x}\)
24 = \(x\)
7 . x - ( 6 . x + 5 ) = 17
7x - 6x - 5 = 17
x - 5 = 17
x = 22
hok tốt
..................
7x - (6x+5) = 17
7x - 6x - 5 = 17
x - 5 = 17
x = 22