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\(\frac{x+1}{125}+\frac{x+2}{124}+\frac{x+3}{123}+\frac{x+4}{122}+\frac{x+146}{5}=0\)
\(\left(\frac{x+1}{125}+1\right)+\left(\frac{x+2}{124}+1\right)+\left(\frac{x+3}{123}+1\right)+\left(\frac{x+4}{122}+1\right)+\left(\frac{x+146}{5}-4\right)=0\)
\(\frac{x+126}{125}+\frac{x+126}{124}+\frac{x+126}{123}+\frac{x+126}{122}+\frac{x+126}{5}=0\)
\(\left(x+126\right).\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)=0\)
vì \(\left(\frac{1}{125}+\frac{1}{124}+\frac{1}{123}+\frac{1}{122}+\frac{1}{5}\right)\ne0\)nên x + 126 = 0 \(\Rightarrow\)x = -126
\(a,\frac{2}{3}+\frac{7}{4}:x=\frac{5}{6}\)
\(\Leftrightarrow\frac{7}{4}:x=\frac{5}{6}-\frac{2}{3}\)
\(\Leftrightarrow\frac{7}{4}:x=\frac{1}{6}\)
\(\Leftrightarrow x=\frac{21}{2}\)
\(b,\left(x+\frac{5}{3}\right).\left(x-\frac{5}{4}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{5}{3}=0\\x-\frac{5}{4}=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\frac{5}{3}\\\frac{5}{4}\end{matrix}\right.\)
\(c,\left(x-1,2\right)^2=4\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1,2=2\\x-1,2=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3,2\\x=-0,8\end{matrix}\right.\)
\(d,\left(x+1\right)^3=-125\)
\(\Leftrightarrow\left(x+1\right)^3=\left(-5\right)^3\)
\(\Leftrightarrow x+1=-5\)
\(\Leftrightarrow x=-6\)
Vậy ...........................................................
cau (c) vo nghiem
cau(a) Ix(x-4)I=x
x>=0
Ix(x-4)I=x.Ix-4I=x
x =0 la nghiem
x khac 0 chia hai ve cho x
Ix-4I=1
x-4=+-1
x=3 hoac x=5
mik thấy câu c hơi vô lí
còn câu a = 5
câu b mik ko biết