a) \(25^3:5^2=5^6:5^2=5^4=625\)

b) \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}=\left(\frac{3}{7}\right)^9\)

c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2=3-1+\frac{1}{8}=\frac{17}{8}\)

a) \(25^3:5^2=5^6:5^2=5^{6-2}=5^4\)

\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)

\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)

\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)

a) ta có :x²+2x+2=(x+1)²+1>0,với mọi x

x²+2x+3=(x+1)²+2>0,với mọi x

ĐKXĐ:x\(\in\)R.Đặt x²+2x+2=a (a>0),ta có:\(\dfrac{a-1}{a}+\dfrac{a}{a+1}=\dfrac{7}{6}\)

<=>\(\dfrac{6\left(a-1\right)\left(a+1\right)}{6a\left(a+1\right)}+\dfrac{6a^2}{6a\left(a+1\right)}=\dfrac{7a\left(a+1\right)}{6a\left(a+1\right)}\)

=>6(a²-1)+6a²=7a²+7a<=>6a²-6+6a²=7a²+7a<=>12a²-7a²-7a-6=0

<=>5a²-7a-6=0<=>(a-2)(5a+3)=0<=>a-2=0(vì a>0,nên 5a+3>0)

<=>a=2=>x²+2x+2=2<=>x(x+2)=0<=>\(|^{x=0}_{x+2=0< =>x=-2}\)

Vậy tặp nghiệm của PT là S\(=\left\{0;-2\right\}\)

bn đổi ngược hai vế cho nhau là ra 1 bài toán bình thường thôi

Bài nỳ tuy rất dài nhưng cũng dễ

Chí cần cậu cuyển VT sang VP rồi tìm x bình thường

Chúc cậu học tốt

\(\left(0,66-0,012:0,2\right):\left(1-1\frac{4}{4}\times0,4\right)=2\frac{9}{13}.x:\left(3,125-5,6:2\frac{2}{3}\right)\)

\(2\frac{9}{13}.x:\frac{41}{40}=\frac{3}{5}:\frac{13}{35}\)

\(2\frac{9}{13}.x:\frac{41}{40}=\frac{21}{13}\)

\(2\frac{9}{13}.x=\frac{861}{520}\)

\(x=\frac{123}{200}\)

                               \(A=-1,6:\left(1+\frac{2}{3}\right)\)

                              \(A=-\frac{16}{10}:\frac{5}{3}\)

                             \(A=-\frac{16.3}{10.5}=-\frac{48}{50}=-\frac{24}{25}\)

                           \(B=1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)

                          \(B=\frac{14}{10}\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)

                         \(B=\frac{2.7.3.5}{2.5.7.7}-\left(\frac{12+10}{15}\right):\frac{11}{5}\)

                         \(B=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}\)

                        \(B=\frac{3}{7}-\frac{22}{15}\times\frac{5}{11}=\frac{3}{7}-\frac{2.11.5}{3.5.11}\)

                       \(B=\frac{3}{7}-\frac{2}{3}=\frac{9-14}{21}=-\frac{5}{21}\)

                       Ủng hộ mk nha !!! ^_^

\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)

\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)

\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)

\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)

\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)

\(=\frac{x}{x-1}\)