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DP
21 tháng 7 2017
a)
\(\frac{x-3}{x+5}=\frac{5}{7}\)
\(\Leftrightarrow\left(x-3\right).7=5.\left(x+5\right)\)
\(\Leftrightarrow7x-21=25+5x\)
\(\Leftrightarrow7x-5x=25+21\)
\(\Leftrightarrow2x=46\)
\(\Leftrightarrow x=23\)
b)
\(\frac{7}{x-1}=\frac{x+1}{9}\)
\(\Leftrightarrow\left(x+1\right)\left(x-1\right)=7.9\)
\(\Leftrightarrow x^2-1=63\)
\(\Leftrightarrow x^2=64\)
\(\Leftrightarrow x=8\)
Mẫy bài còn lại làm tương tự
28 tháng 6 2019
\(c,\frac{x-1}{x+2}=\frac{x-2}{x+3}\)
\(\Leftrightarrow(x-1)(x+3)=(x-2)(x+2)\)
\(\Leftrightarrow x^2+2x-3=x^2-4\)
\(\Leftrightarrow x^2+2x-3-x^2=-4\)
\(\Leftrightarrow x^2-x^2+2x-3=-4\)
\(\Leftrightarrow2x-3=-4\Leftrightarrow2x=-1\Leftrightarrow x=-\frac{1}{2}\)
NN
24 tháng 4 2020
1.a) Sửa lại đề: \(\frac{11}{17}\)ở mẫu chuyển thành \(\frac{11}{7}\)
\(\frac{0,75+0,6-\frac{3}{7}-\frac{3}{13}}{2,75+2,2-\frac{11}{7}-\frac{11}{13}}=\frac{\frac{3}{4}+\frac{3}{5}-\frac{3}{7}-\frac{3}{13}}{\frac{11}{4}+\frac{11}{5}-\frac{11}{7}-\frac{11}{13}}\)\(=\frac{3\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}{11\left(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\right)}=\frac{3}{11}\)
( vì \(\frac{1}{4}+\frac{1}{5}-\frac{1}{7}-\frac{1}{13}\ne0\))
2.a) \(\frac{3}{5}+\frac{3}{2}.x=\frac{-5}{7}\)\(\Leftrightarrow\frac{3}{2}.x=\frac{-5}{7}-\frac{3}{5}\)
\(\Leftrightarrow\frac{3}{2}.x=\frac{-46}{35}\)\(\Leftrightarrow x=\frac{-46}{35}:\frac{3}{2}\)\(\Leftrightarrow x=\frac{-92}{105}\)
Vậy \(x=\frac{-92}{105}\)
b) \(\left(4x-\frac{1}{3}\right).\left(\frac{3}{2}x+\frac{5}{6}\right)=0\)\(\Leftrightarrow\orbr{\begin{cases}4x-\frac{1}{3}=0\\\frac{3}{2}x+\frac{5}{6}=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}4x=\frac{1}{3}\\\frac{3}{2}x=\frac{-5}{6}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=\frac{1}{12}\\x=\frac{-5}{9}\end{cases}}\)
Vậy \(x=\frac{-5}{9}\)hoặc \(x=\frac{1}{12}\)
21 tháng 6 2020
a) Ta có: \(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}\)
\(=\frac{15}{40}-\frac{8}{40}+\frac{3}{40}\)
\(=\frac{10}{40}=\frac{1}{4}\)
b) Ta có: \(\frac{21}{4}\cdot\frac{3}{8}+\frac{43}{4}\cdot\frac{3}{8}-4\cdot\frac{1}{2}\)
\(=\frac{3}{8}\left(\frac{21}{4}+\frac{43}{4}\right)-2\)
\(=\frac{3}{8}\cdot16-2\)
\(=6-2=4\)
c) Ta có: \(\frac{-5}{9}+\frac{7}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{8}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=\frac{-2}{11}\)
d) Ta có: \(125\%\cdot\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1.5\right)+2016^0\)
\(=\frac{5}{4}\cdot\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}\cdot3+1\)
\(=\frac{15}{16}+\frac{16}{16}=\frac{31}{16}\)
23 tháng 5 2020
Bài 1:
\(A=\left(x^3.x^3.x^2\right).\left(y.y^4\right).\left(\frac{2}{5}.\frac{-5}{4}\right)\)
\(A=x^8.y^5.\left(-\frac{1}{2}\right)\)
\(B=\left(x^5.x.x^2\right).\left(y^4.y^2.y\right).\left(\frac{-3}{4}.\frac{-8}{9}\right)\)
\(B=x^8.y^7.\frac{2}{3}\)
Bài 2:
\(A=\left(15.x^2.y^3-12.x^2.y^3\right)+\left(11x^3.y^2-8.x^3.y^2\right)+\left(7x^2-12x^2\right)\)
\(A=3.x^2.y^3+2.x^3.y^2-5x^2\)
B tương tự nhé, đáp án là (theo mình)
\(B=\frac{5}{2}.x^5.y+\frac{7}{3}.x.y^4-\frac{1}{4}.x^2.y^3\)
16 tháng 9 2017
a) \(x:\left(\frac{3}{4}\right)^3=\left(\frac{3}{4}\right)^2\)
\(x=\left(\frac{3}{4}\right)^2.\left(\frac{3}{4}\right)^3\)
\(x=\left(\frac{3}{4}\right)^5\)
\(x=\frac{243}{1024}\)
vay \(x=\frac{243}{1024}\)
b) \(\left(\frac{2}{5}\right)^5.x=\left(\frac{2}{5}\right)^8\)
\(x=\left(\frac{2}{5}\right)^8:\left(\frac{2}{5}\right)^5\)
\(x=\left(\frac{2}{5}\right)^3\)
\(x=\frac{8}{125}\)
vay \(x=\frac{8}{125}\)
4) \(\left(0,36\right)^8=\left(0,6^2\right)^8=\left(0,6\right)^{16}\)
\(\left(0,216\right)^4=\left(0,6^3\right)^4=\left(0,6\right)^{12}\)
5) a) \(\left(3,5\right)^3=42,875\)
b) \(\left(-\frac{4}{11}\right)^2=\frac{16}{121}\)
c) \(\left(0,5\right)^4.6^4=3^4=81\)
d) \(\left(-\frac{1}{3}\right)^5:\left(\frac{1}{6}\right)^5=\left(-2\right)^5=-32\)
TP
10 tháng 9 2020
giúp mik vs, mik bik các pạn giờ này đang ngủ rùi nhưng giúp mik lần này thui.yêu các pạn nhìu
12 tháng 9 2020
\(5\frac{1}{2}+\left(-3\right)=\frac{11}{2}+\frac{-3}{1}\)\(=\frac{11}{2}+\frac{-6}{2}=\frac{5}{2}\)\(;\)
\(4\frac{9}{11}+\left(-2\frac{1}{11}\right)=\frac{53}{11}+\frac{-23}{11}\)\(=\frac{30}{11}\)\(;\)
\(2\frac{1}{2}+\left(-6\right)=\frac{5}{2}+\frac{-6}{1}\)\(=\frac{5}{2}+\frac{-12}{2}=\frac{-7}{2}\)\(;\)
\(\left(-\frac{4}{5}\right)+\frac{1}{2}=\frac{-4}{5}+\frac{1}{2}\)\(=\frac{-8}{10}+\frac{5}{10}=\frac{-3}{10}\)\(;\)
\(4,3-\left(-1,2\right)=4,3+1,2=5,5\)\(=\frac{55}{10}=\frac{11}{2}\)\(;\)
\(0-\left(-0,4\right)=0+0,4=0,4\)\(=\frac{4}{10}=\frac{2}{5}\)\(;\)
\(\frac{-2}{3}-\frac{-1}{3}=\frac{-2}{3}+\frac{1}{3}=\frac{-1}{3}\)\(;\)
\(\frac{-1}{2}-\frac{-1}{6}=\frac{-1}{2}+\frac{1}{6}\)\(=\frac{-3}{6}+\frac{1}{6}=\frac{-2}{6}=\frac{-1}{3}\)\(;\)
\(x+\frac{1}{3}=\frac{3}{4}\) \(;\) \(x-\frac{2}{5}=\frac{5}{7}\) \(;\)
\(x=\frac{3}{4}-\frac{1}{3}\) \(x=\frac{5}{7}+\frac{2}{5}\)
\(x=\frac{5}{12}\) \(x=\frac{39}{35}\)
\(-x-\frac{2}{3}=-\frac{6}{7}\) \(;\) \(\frac{4}{7}-x=\frac{1}{3}\)
\(\frac{6}{7}-\frac{2}{3}=x\) \(\frac{4}{7}-\frac{1}{3}=x\)
\(\frac{4}{21}=x\) \(\Leftrightarrow\)\(x=\frac{4}{21}\) \(\frac{5}{21}=x\)\(\Leftrightarrow\)\(x=\frac{5}{12}\)
\(-x-\frac{3}{4}=-\frac{8}{11}\Rightarrow-x=-\frac{8}{11}+\frac{3}{4}\Leftrightarrow-x=\frac{1}{44}\Leftrightarrow x=-\frac{1}{44}\)
Vậy \(11x=11.\left(-\frac{1}{44}\right)=-\frac{11}{44}\)