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a) \(\left(\frac{2^2}{5}\right)+5\frac{1}{2}.\left(4,5-2,5\right)+\frac{2^3}{-4}\)
\(=\frac{4}{5}+\frac{11}{2}.2+\frac{-8}{4}\)
\(=\frac{4}{5}+11-2\)
\(=\frac{4}{5}+9\)
\(=\frac{49}{9}\)
b) \(\left(-2^3\right)+\frac{1}{2}:\frac{1}{8}-\sqrt{25}+\left|-64\right|\)
\(=-8+4-5+64\)
= 55
c) \(\frac{\sqrt{3^2+\sqrt{39}^2}}{\sqrt{91^2}-\sqrt{\left(-7\right)^2}}\)
\(=\frac{\sqrt{9+39}}{91-\sqrt{49}}\)
\(=\frac{\sqrt{48}}{91-7}\)
\(=\frac{4\sqrt{3}}{84}\)
\(=\frac{\sqrt{3}}{41}\)
d) Xem lại đề nhé em!
e) \(\sqrt{25}-3\sqrt{\frac{4}{9}}\)
\(=5-3.\frac{2}{3}\)
= 5 - 2
= 3
h) \(\left(-3^2\right).\frac{1}{3}-\sqrt{49}+\left(5^3\right):\sqrt{25}\)
\(=-9.\frac{1}{3}-7+125:5\)
\(=-3-7+25\)
= 15
\(3\frac{1}{2}-\frac{1}{2}.\left(-4,25-\frac{3}{4}\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-4,25-0,75\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.\left(-5\right)^2:\frac{5}{4}\)
\(=\frac{7}{2}-\frac{1}{2}.5.\frac{4}{5}\)
\(=\frac{7}{2}-2\)
\(=\frac{7}{2}-\frac{4}{2}\)
\(=\frac{3}{2}\)
\(\frac{3}{7}.1\frac{1}{2}+\frac{3}{7}.0,5-\frac{3}{7}.9\)
\(=\frac{3}{7}.\left(\frac{3}{2}+\frac{1}{2}-9\right)\)
\(=\frac{3}{7}.\left(2-9\right)\)
\(=\frac{3}{7}.\left(-7\right)\)
\(=-3\)
\(\frac{125^{2016}.8^{2017}}{50^{2017}.20^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^2\right)^{2017}.2^{2017}.\left(2^2\right)^{2018}.5^{2018}}=\frac{\left(5^3\right)^{2016}.\left(2^3\right)^{2017}}{\left(5^3\right)^{2017}.\left(2^3\right)^{2017}.2.5}=\frac{1}{5^4.2}=\frac{1}{1250}\)( tính nhẩm, ko chắc đúng )
1
a) \(3\frac{1}{2}-\frac{1}{2}\cdot\left(-4,25-\frac{3}{4}\right)^2\) : \(\frac{5}{4}\)
= \(3\cdot25:\frac{5}{4}\)
= \(3\cdot\left(25:\frac{5}{4}\right)\)
=\(3\cdot20\)
=60
b)=\(\frac{3}{7}\cdot\left(1\frac{1}{2}+0,5-9\right)\)
=\(\frac{3}{7}\cdot\left(-7\right)\)
=\(-3\)
c) =
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 \(x=\frac{1}{12}\)
1) \(\frac{1}{3}x-\frac{2}{5}=\frac{1}{3}\)
⇒ \(\frac{1}{3}x=\frac{1}{3}+\frac{2}{5}\)
⇒ \(\frac{1}{3}x=\frac{11}{15}\)
⇒ \(x=\frac{11}{15}:\frac{1}{3}\)
⇒ \(x=\frac{11}{5}\)
Vậy \(x=\frac{11}{5}.\)
2) \(2,5:7,5=x:\frac{3}{5}\)
⇒ \(\frac{5}{2}:\frac{15}{2}=x:\frac{3}{5}\)
⇒ \(\frac{1}{3}=x:\frac{3}{5}\)
⇒ \(x=\frac{1}{3}.\frac{3}{5}\)
⇒ \(x=\frac{1}{5}\)
Vậy \(x=\frac{1}{5}.\)
4) \(\left|x\right|+\left|x+2\right|=0\)
Có: \(\left\{{}\begin{matrix}\left|x\right|\ge0\\\left|x+2\right|\ge0\end{matrix}\right.\forall x.\)
⇒ \(\left|x\right|+\left|x+2\right|=0\)
⇒ \(\left\{{}\begin{matrix}x=0\\x+2=0\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=0-2\end{matrix}\right.\) ⇒ \(\left\{{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\)
Vô lí vì \(x\) không thể nhận cùng lúc 2 giá trị khác nhau.
⇒ \(x\in\varnothing\)
Vậy không tồn tại giá trị nào của \(x\) thỏa mãn yêu cầu đề bài.
10) \(5-\left|1-2x\right|=3\)
⇒ \(\left|1-2x\right|=5-3\)
⇒ \(\left|1-2x\right|=2\)
⇒ \(\left[{}\begin{matrix}1-2x=2\\1-2x=-2\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}2x=1-2=-1\\2x=1+2=3\end{matrix}\right.\) ⇒ \(\left[{}\begin{matrix}x=\left(-1\right):2\\x=3:2\end{matrix}\right.\)
⇒ \(\left[{}\begin{matrix}x=-\frac{1}{2}\\x=\frac{3}{2}\end{matrix}\right.\)
Vậy \(x\in\left\{-\frac{1}{2};\frac{3}{2}\right\}.\)
Chúc bạn học tốt!
9, \(13\frac{1}{3}:1\frac{1}{3}=26:\left(2x-1\right)\)
\(\frac{40}{3}:\frac{4}{3}=26:\left(2x-1\right)\)
\(10=26:\left(2x-1\right)\)
\(2x-1=26:10\)
\(2x-1=2,6\)
\(2x=2,6+1\)
\(2x=3,6\)
\(x=3,6:2\)
\(x=1,8\)
a) \(\sqrt{\frac{4}{81}}:\sqrt{\frac{25}{81}}-1\frac{2}{5}\)
\(=\frac{2}{9}:\frac{5}{9}-\frac{7}{5}\)
\(=\frac{2}{5}-\frac{7}{5}\)
\(=-1.\)
b) \(\sqrt{36}.\sqrt{\frac{25}{16}}+\frac{1}{4}\)
\(=6.\frac{5}{4}+\frac{1}{4}\)
\(=\frac{15}{2}+\frac{1}{4}\)
\(=\frac{31}{4}.\)
c) \(1\frac{1}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\frac{4}{7}:\left(-\frac{8}{9}\right)\)
\(=\frac{3}{2}+\left(-\frac{9}{14}\right)\)
\(=\frac{6}{7}.\)
d) \(1,17-0,4.\left(\frac{1}{2}\right)^2-\frac{1}{-5}\)
\(=\frac{117}{100}-\frac{2}{5}.\frac{1}{4}-\left(-\frac{1}{5}\right)\)
\(=\frac{117}{100}-\frac{1}{10}+\frac{1}{5}\)
\(=\frac{107}{100}+\frac{1}{5}\)
\(=\frac{127}{100}.\)
Chúc bạn học tốt!
a, \(\frac{4}{81}:\sqrt{\frac{25}{81}-1\frac{2}{5}}\)
\(\Rightarrow\frac{4}{81}:\frac{5}{9}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{81}.\frac{9}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{4}{9}.\frac{1}{5}-\frac{7}{5}\)
\(\Rightarrow\frac{-59}{45}\)
b,\(\sqrt{36}.\sqrt{\frac{25}{16}+\frac{1}{4}}\)
\(\Rightarrow6.\frac{5}{4}+\frac{1}{4}\)
\(\Rightarrow\frac{15}{2}+\frac{1}{4}\)
\(\Rightarrow\frac{31}{4}\)
c,\(1\frac{1}{2}+\frac{4}{7}:\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{4}{7}.\frac{-8}{9}\)
\(\Rightarrow\frac{3}{2}-\frac{9}{14}\)
\(\Rightarrow\frac{6}{7}\)
d, \(1,17-\left(0,4.\frac{1}{2}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\left(\frac{1}{5}\right)^2-\frac{1}{5}\)
\(\Rightarrow\frac{117}{100}-\frac{1}{25}-\frac{1}{5}\)
\(\Rightarrow\frac{93}{100}\)
\(A=\frac{99}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+..+\frac{1}{99.100}\right)\)
\(A=\frac{99}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{99}{100}-\frac{99}{100}\)
\(A=\frac{99-99}{100}=0\)
Bài 2
\(\left(3x+5\right).\left(2x-4\right)=0\)
\(TH1:3x+5=0\)
\(3x=-5\)
\(x=-\frac{5}{3}\)
\(TH2:2x-4=0\)
\(2x=4\)
\(x=2\)
\(\left(x^2-1\right).\left(x+3\right)=0\)
\(\Rightarrow x^2-1=0\)
\(x^2=1\)
\(\Rightarrow x=1\)
\(x+3=0\)
\(x=-3\)
\(5x^2-\frac{1}{2}x=0\)
\(\Rightarrow5x^2-\frac{x}{2}=0\)
\(\Rightarrow5x^2=\frac{5x^2}{1}=\frac{5x^2.2}{2}\)
\(10x^2-x=x.\left(10x-1\right)\)
\(\frac{x.\left(10x-1\right)}{2}=0\)
\(\frac{x.\left(10x-1\right)}{2}.2=0.2\)
\(10x-1=0\)
\(x=\frac{1}{10}=0.100\)
\(\Rightarrow\orbr{\begin{cases}x=\frac{1}{10}=0.100\\x=0\end{cases}}\)
\(\frac{x}{4}-\frac{1}{2}=\frac{3}{4}\)
\(\frac{x}{4}=\frac{3}{4}+\frac{1}{2}\)
\(\frac{x}{4}=\frac{5}{4}\)
\(\Rightarrow x=5\)
\(\frac{1}{8}+\frac{7}{8}:x=\frac{3}{4}\)
\(\frac{7}{8}:x=\frac{3}{4}-\frac{1}{8}\)
\(x=\frac{7}{8}:\frac{5}{8}\)
\(x=\frac{56}{40}=\frac{28}{20}=\frac{14}{10}=\frac{7}{5}\)
b) \(\left(x+\frac{1}{2}\right)^3:3=-\frac{1}{81}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{81}\right).3\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=-\frac{1}{27}\)
\(\Rightarrow\left(x+\frac{1}{2}\right)^3=\left(-\frac{1}{3}\right)^3\)
\(\Rightarrow x+\frac{1}{2}=-\frac{1}{3}\)
\(\Rightarrow x=\left(-\frac{1}{3}\right)-\frac{1}{2}\)
\(\Rightarrow x=-\frac{5}{6}\)
Vậy \(x=-\frac{5}{6}.\)
c) \(\frac{x-2}{2}=\frac{8}{x-2}\left(x\ne2\right).\)
\(\Rightarrow\left(x-2\right).\left(x-2\right)=8.2\)
\(\Rightarrow\left(x-2\right)^2=16\)
\(\Rightarrow\left(x-2\right)^2=\left(\pm4\right)^2\)
\(\Rightarrow x-2=\pm4.\)
\(\Rightarrow\left[{}\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=4+2\\x=\left(-4\right)+2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=6\left(TM\right)\\x=-2\left(TM\right)\end{matrix}\right.\)
Vậy \(x\in\left\{6;-2\right\}.\)
Chúc bạn học tốt!