\(3.8\div\left(2x\right)=\frac{1}{4}\div2\frac{2}{3} \)
\(\left(0.25x\right)\div3=\frac{5}{6}\div0.125\)
\(0.001\div2.5=\left(0.75x\right)\div0.75\)
\(1\frac{1}{3}\div0.8=\frac{2}{3}\div\left(0.1x\right)\)
Tính:
a) \(25^{3\div}5^2\)
b) \(\left(\frac{3}{7}\right)^{21}\div\left(\frac{9}{49}\right)^6\)
c) \(3-\left(-\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2\div2\)
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) \(25^3:5^2=5^6:5^2=5^{6-2}=5^4\)
\(4\times\left(\frac{1}{4}\right)^2+25\times\left[\left(\frac{3}{4}\right)^3\div\left(\frac{5}{4}\right)^3\right]\div\left(\frac{3}{2}\right)^3\)
\(2^3+3\times\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2\div\frac{1}{2}\right]-8\)
\(4.\left(\frac{1}{4}\right)^2+25\left[\left(\frac{3}{4}\right)^3:\left(\frac{5}{4}\right)^3\right]:\left(\frac{3}{2}\right)^3=4.\frac{1}{16}+25\left(\frac{27}{64}.\frac{64}{125}\right).\frac{8}{27}\)
\(=\frac{1}{4}+25.\frac{27}{125}.\frac{8}{27}=\frac{1}{4}+\frac{8}{5}=\frac{37}{20}\)
\(2^3+3\left(\frac{1}{2}\right)^0-1+\left[\left(-2\right)^2:\frac{1}{2}\right]-8=8+3-1+4.2-8=10\)
Thực hiện phép tính (hợp lí nếu có thể):
\(1,\frac{-1}{3}-\frac{-3}{5}-\frac{1}{6}+\frac{1}{43}-\frac{-3}{7}+\frac{-1}{2}-\frac{1}{35}\\ \\ 2,\left(-\frac{1}{3}+\frac{7}{13}\right)-\left(\frac{-16}{24}+\frac{6}{26}+\frac{9}{13}\right)\)\(3,\frac{-7}{3}-\left[\frac{2}{5}-\left(\frac{1}{3}+\frac{-5}{25}\right)\right]\\ 4,\left(2\frac{1}{4}-3\frac{1}{5}\right)-\left[\frac{-3}{4}+\left(\frac{4}{5}-2019\right)\right]\)
GIẢI PHƯƠNG TRÌNH SAU
A) \(\frac{X^2+2X+1}{X^2+2X+2}+\frac{X^2+2X+2}{X^2+2X+3}=\frac{7}{6}\)
B) \(\frac{\left(X^2-3X-4\right)^4}{\left(X-3\right)^5\left(X+2\right)^3}+\frac{\left(X^2+4X+3\right)^6}{\left(X-3\right)^3\left(X+2\right)^5}=0\)
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\(=\left\{0;-2\right\}\)
\(\left(0,66-0,012\div0,2\right)\div\left(1-1\frac{4}{7}\times0,4\right)=2\frac{9}{13}\times x\div\left(3,125-5,6\div2\frac{2}{3}\right)\)
Cho mình xin cả cách làm nữa
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
\(\left(0,66-0,012:0,2\right):\left(1-1\frac{4}{4}\times0,4\right)=2\frac{9}{13}.x:\left(3,125-5,6:2\frac{2}{3}\right)\)
\(2\frac{9}{13}.x:\frac{41}{40}=\frac{3}{5}:\frac{13}{35}\)
\(2\frac{9}{13}.x:\frac{41}{40}=\frac{21}{13}\)
\(2\frac{9}{13}.x=\frac{861}{520}\)
\(x=\frac{123}{200}\)
Tìm x : \(\frac{2\left(x-1\right)\left(x-3\right)}{3}-\frac{4\left(2x-1\right)^2}{5}=\frac{\left(1+3x\right)^2}{2}-3x\left(1-x\right)\)
Tìm x : \(\frac{\left(x-3\right)^2}{2}-1\frac{1}{3}\left(x+2\right)^2-\frac{5}{4}\left(x-1\right)\left(x+1\right)=1\frac{1}{2}x\left(x-2\right)-x-4\)
A= \(-1,6\div\left(1+\frac{2}{3}\right)\)
B=\(1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right)\div2\frac{1}{5}\)
\(A=-1,6:\left(1+\frac{2}{3}\right)\)
\(A=-\frac{16}{10}:\frac{5}{3}\)
\(A=-\frac{16.3}{10.5}=-\frac{48}{50}=-\frac{24}{25}\)
\(B=1,4\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):2\frac{1}{5}\)
\(B=\frac{14}{10}\times\frac{15}{49}-\left(\frac{4}{5}+\frac{2}{3}\right):\frac{11}{5}\)
\(B=\frac{2.7.3.5}{2.5.7.7}-\left(\frac{12+10}{15}\right):\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}:\frac{11}{5}\)
\(B=\frac{3}{7}-\frac{22}{15}\times\frac{5}{11}=\frac{3}{7}-\frac{2.11.5}{3.5.11}\)
\(B=\frac{3}{7}-\frac{2}{3}=\frac{9-14}{21}=-\frac{5}{21}\)
Ủng hộ mk nha !!! ^_^
Tìm tập xác định
\(\left[\left(1+\frac{1}{x^2}\right)\div\left(1+2x+x^2\right)+\frac{2}{\left(x+1\right)^3}\times\left(1+\frac{1}{x}\right)\right]\div\frac{x-1}{x^3}\)
\(=\left[\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{2}{\left(x+1\right)^3}\times\frac{x+1}{x}\right]\div\frac{x-1}{x^3}\)
\(=\left(\frac{x^2+1}{x^2}\times\frac{1}{\left(x+1\right)^2}+\frac{1}{\left(x+1\right)^2}\times\frac{2}{x}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\left(\frac{x^2+1}{x^2}+\frac{2}{x}\right)\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x^3+2x^2+x}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x^2+2x+1\right)}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\left(\frac{1}{\left(x+1\right)^2}\times\frac{x\left(x+1\right)^2}{x^3}\right)\div\frac{x-1}{x^3}\)
\(=\frac{1}{x^2}\times\frac{x^3}{x-1}\)
\(=\frac{x}{x-1}\)