Question

[Nguyen Dieu Thao Ly]

1) Thực hiện phép tính

a) \(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}\)

2) Tìm x , biết

a) \(\frac{x}{7}\)+ \(\left(-\frac{3}{7}\right)^2\)= \(\frac{2}{7}:\frac{4}{3}\)

b) ( x-1 ) \(^{x+6}\) = ( x-1 ) \(^{x+4}\)

Answer 1

[Nguyễn Huy Tú]

Bài 1:
\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}=\frac{5}{7}\)

Bài 2:
a) \(\frac{x}{7}+\left(\frac{-3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)

\(\Rightarrow\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)

\(\Rightarrow\frac{x}{7}=\frac{3}{98}\)

\(\Rightarrow98x=21\)

\(\Rightarrow x=\frac{3}{14}\)

Vậy \(x=\frac{3}{14}\)

b) \(\left(x-1\right)^{x+6}=\left(x-1\right)^{x+4}\)

\(\Rightarrow\left(x-1\right)^{x+6}-\left(x-1\right)^{x+4}=0\)

\(\Rightarrow\left(x-1\right)^{x+4}.\left[\left(x-1\right)^2-1\right]=0\)

\(\Rightarrow\left(x-1\right)^{x+1}=0\) hoặc \(\left(x-1\right)^2-1=0\)

+) \(\left(x-1\right)^{x+1}=0\Rightarrow x-1=0\Rightarrow x=1\)

+) \(\left(x-1\right)^2-1=0\)

\(\Rightarrow\left(x-1\right)^2=1\)

\(\Rightarrow\left(x-1\right)=\pm1\)

+ \(x-1=1\Rightarrow x=2\)

+ \(x-1=-1\Rightarrow x=0\)

Vậy \(x\in\left\{0;2;1\right\}\)

Answer 2

[Hải Ninh]

1)

\(\frac{\frac{5}{131}+\frac{5}{141}-\frac{5}{191}-\frac{5}{4011}}{\frac{7}{131}+\frac{7}{141}+\frac{7}{-191}-\frac{7}{4011}}\)

\(=\frac{5\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}{7\left(\frac{1}{131}+\frac{1}{141}-\frac{1}{191}-\frac{1}{4011}\right)}\)

\(=\frac{5}{7}\)

2) \(\frac{x}{7}+\left(-\frac{3}{7}\right)^2=\frac{2}{7}:\frac{4}{3}\)

\(=\frac{x}{7}+\frac{9}{49}=\frac{3}{14}\)

\(=\frac{x}{7}=\frac{3}{14}-\frac{9}{49}=\frac{3}{98}\)

\(\Rightarrow98x=21\)

\(\Rightarrow x=\frac{3}{14}\)