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9) \(\dfrac{x}{4}=\dfrac{9}{x}\)
Theo định nghĩa về hai phân số bằng nhau, ta có:
\(4\cdot9=x^2\\ 36=x^2\Rightarrow\left[{}\begin{matrix}x=6\\x=-6\end{matrix}\right.\)
8)
\(x:\dfrac{5}{3}+\dfrac{1}{3}=-\dfrac{2}{5}\\ x:\dfrac{5}{3}=-\dfrac{2}{5}+\dfrac{1}{3}\\ x:\dfrac{5}{3}=-\dfrac{1}{15}\\ x=\dfrac{1}{15}\cdot\dfrac{5}{3}\\ x=\dfrac{1}{9}\)
7)
\(2x-16=40+x\\ 2x-x=40+16\\ x\left(2-1\right)=56\\ x=56\)
6)
\(1\dfrac{1}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}+x=\dfrac{3}{2}-7\\ \dfrac{3}{2}-\dfrac{3}{2}=-7-x\\ -7-x=0\\ x=-7-0\\ x=-7\)
5)
\(3\dfrac{1}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{7}{2}-\dfrac{1}{2}x=\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{7}{2}-\dfrac{2}{3}\\ \dfrac{1}{2}x=\dfrac{17}{6}\\ x=\dfrac{17}{6}:\dfrac{1}{2}\\ x=\dfrac{17}{3}\)
4)
\(x\cdot\left(x+1\right)=0\Rightarrow\left[{}\begin{matrix}x=0\\x+1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=-1\end{matrix}\right.\)
3)
\(\left(\dfrac{2x}{5}+2\right):\left(-4\right)=-1\dfrac{1}{2}\\ \left(\dfrac{2x}{5}+2\right):\left(-4\right)=-\dfrac{3}{2}\\ \dfrac{2x}{5}+2=-\dfrac{3}{2}\cdot\left(-4\right)\\ \dfrac{2x}{5}+2=6\\ \dfrac{2x}{5}=6-2\\ \dfrac{2x}{5}=4\\ 2x=4\cdot5\\ 2x=20\\ x=20:2\\ x=10\)
2)
\(\dfrac{1}{3}+\dfrac{1}{2}:x=-0,25\\ \dfrac{1}{3}+\dfrac{1}{2}:x=-\dfrac{1}{4}\\ \dfrac{1}{2}:x=-\dfrac{1}{4}-\dfrac{1}{3}\\ \dfrac{1}{2}:x=-\dfrac{7}{12}\\ x=\dfrac{1}{2}:-\dfrac{7}{12}\\ x=-\dfrac{6}{7}\)
1)
\(\dfrac{4}{3}+x=\dfrac{2}{15}\\ x=\dfrac{2}{15}-\dfrac{4}{3}x=-\dfrac{6}{5}\)
(x + \(\dfrac{1}{2}\) ) + (x + \(\dfrac{1}{8}\) ) + (x+\(\dfrac{1}{8}\))+(x+\(\dfrac{1}{16}\))=1
4x + \(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\)=1
4x +\(\dfrac{15}{16}\)=1
4x = 1 -\(\dfrac{15}{16}\)
4x=\(\dfrac{1}{16}\)
x=\(\dfrac{1}{16}\div4\)
x=\(\dfrac{1}{64}\)
a,Vì \(\dfrac{5}{6}\)=\(\dfrac{x}{24}\) nên ta có: 5.24:6= 20 \(\Rightarrow\)x =20
Mấy câu sau làm tương tự như vậy.
b,x =21
c,x =9
d,x = -5
\(\dfrac{3}{4}.\dfrac{8}{9}.\dfrac{15}{16}....\dfrac{624}{625}\)
\(=\dfrac{3.8.15...624}{4.9.16...625}\)
\(=\dfrac{\left(1.3\right)\left(2.4\right)\left(3.5\right)...\left(24.26\right)}{\left(2.2\right)\left(3.3\right)\left(4.4\right)...\left(25.25\right)}\)
\(=\dfrac{\left(1.2.3...24\right)\left(3.4.5...26\right)}{\left(2.3.4...25\right).\left(2.3.4...25\right)}\)
\(=\dfrac{1.26}{25.2}\)
\(=\dfrac{13}{25}\)
\(\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\cdot\cdot\dfrac{624}{625}\)
\(=\dfrac{1.3}{2.2}\cdot\dfrac{2.4}{3.3}\cdot\dfrac{3.5}{4.4}\cdot\cdot\cdot\dfrac{24.26}{25.25}\)
\(=\dfrac{1.2.3....24}{2.3.4....25}\cdot\dfrac{3.4.5....26}{2.3.4...25}\)
\(=\dfrac{1}{25}\cdot\dfrac{26}{2}=\dfrac{13}{25}\)
6. \(\dfrac{x}{4}=\dfrac{9}{x}\)
=>x2=4.9=36
=>x\(\in\)\(\left\{-6;6\right\}\)
\((\dfrac{2x}{5}+2):\left(-4\right)=-1\dfrac{1}{2}\)
(\(\dfrac{2x}{5}+2):\left(-4\right)=-\dfrac{3}{2}\)
\(\dfrac{2x}{5}=-\dfrac{3}{2}.\left(-4\right)\)
\(\dfrac{2x}{5}=6\)
\(\dfrac{2x}{5}=\dfrac{30}{5}\)
2x = 30
x = 30 : 2 = 15
a: \(\Leftrightarrow\dfrac{8}{5}+\dfrac{2}{5}\cdot x=\dfrac{16}{5}\)
=>2/5x=8/5
=>x=4
b: \(\Leftrightarrow\left(\dfrac{1}{24}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{26}+...+\dfrac{1}{39}-\dfrac{1}{40}\right)\cdot120+\dfrac{1}{3}x=-4\)
\(\Leftrightarrow x\cdot\dfrac{1}{3}+2=-4\)
=>1/3x=-6
=>x=-18
c: =>2|x-1/3|=0,24-4/5=-0,56<0
Bài 1: Tính ( hợp lý nếu có thể )
\(A=\dfrac{-3}{8}+\dfrac{12}{25}+\dfrac{5}{-8}+\dfrac{2}{-5}+\dfrac{13}{25}\)
\(=\left(\dfrac{-3}{8}+\dfrac{5}{-8}\right)+\left(\dfrac{12}{25}+\dfrac{13}{25}\right)+\dfrac{2}{-5}\)
\(=-1+1+\dfrac{2}{-5}\)
\(=0+\dfrac{2}{-5}\)
\(=\dfrac{2}{-5}\)
\(B=\dfrac{-3}{15}+\left(\dfrac{2}{3}+\dfrac{3}{15}\right)\)
\(=\left(\dfrac{-3}{15}+\dfrac{3}{15}\right)+\dfrac{2}{3}\)
\(=0+\dfrac{2}{3}\)
\(=\dfrac{2}{3}\)
\(C=\dfrac{-5}{21}+\left(\dfrac{-16}{21}+1\right)\)
\(=\left(\dfrac{-5}{21}+\dfrac{-16}{21}\right)+1\)
\(=-1+1\)
\(=0\)
\(D=\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)
\(=\left(\dfrac{5}{-12}+\dfrac{7}{12}\right)+\dfrac{-1}{6}\)
\(=\dfrac{1}{6}+\dfrac{-1}{6}\)
\(=0\)
Bài 2: Tìm x,biết:
a) \(x+\dfrac{2}{3}=\dfrac{4}{5}\)
\(x=\dfrac{4}{5}-\dfrac{2}{3}\)
\(x=\dfrac{2}{15}\)
Vậy \(x=\dfrac{2}{15}\)
b) \(x-\dfrac{2}{3}=\dfrac{7}{21}\)
\(\Rightarrow x-\dfrac{2}{3}=\dfrac{1}{3}\)
\(x=\dfrac{1}{3}+\dfrac{2}{3}\)
\(x=\dfrac{3}{3}=1\)
Vậy \(x=1\)
c) sai đề hay sao ấy bạn.bỏ dấu - ở x thì đúng đề.mk giải luôn nha!
\(x-\dfrac{3}{4}=\dfrac{-8}{11}\)
\(x=\dfrac{-8}{11}+\dfrac{3}{4}\)
\(x=\dfrac{1}{44}\)
Vậy \(x=\dfrac{1}{44}\)
d) \(\dfrac{11}{12}-\left(\dfrac{2}{5}+x\right)=\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{11}{12}-\dfrac{2}{3}\)
\(\dfrac{2}{5}+x=\dfrac{1}{4}\)
\(x=\dfrac{1}{4}-\dfrac{2}{5}\)
\(x=-\dfrac{3}{20}\)
Vậy \(x=-\dfrac{3}{20}\)
\(\dfrac{48}{25}\cdot\dfrac{27}{55}+2\dfrac{4}{9}\cdot\dfrac{14}{33}\)
\(=\dfrac{1296}{1375}+\dfrac{22}{9}\cdot\dfrac{14}{33}\\ =\dfrac{1296}{1375}+\dfrac{28}{27}\\ =\dfrac{34992}{37125}+\dfrac{38500}{37125}\\ =\dfrac{73492}{37125}\)
\(1\dfrac{19}{22}\cdot\left(\dfrac{47}{77}-\dfrac{16}{15}\right)\\ =\dfrac{41}{22}\cdot\dfrac{-527}{1155}\\ =\dfrac{-21607}{25410}\)
\(\left(3\dfrac{10}{99}+4\dfrac{11}{99}-5\dfrac{8}{299}\right)-\left(\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{6}\right)\\ =\left(\dfrac{307}{99}+\dfrac{37}{99}-\dfrac{1503}{299}\right)-0\\ =\dfrac{344}{99}-\dfrac{1053}{299}\\ =-\dfrac{107}{2277}\)
a: \(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{3}-\dfrac{1}{4}< x< \dfrac{1}{48}-\dfrac{1}{16}+\dfrac{1}{6}\)
\(\Leftrightarrow\dfrac{6}{12}-\dfrac{4}{12}-\dfrac{3}{12}< x< \dfrac{1}{48}-\dfrac{3}{48}+\dfrac{8}{48}\)
\(\Leftrightarrow\dfrac{-1}{12}< x< \dfrac{1}{8}\)
\(\Leftrightarrow-2< 24x< 3\)
=>x=0
b: \(\Leftrightarrow\dfrac{9-10}{12}< \dfrac{x}{12}< 1-\dfrac{8-3}{12}=\dfrac{7}{12}\)
=>-1<x<7
hay \(x\in\left\{0;1;2;3;4;5;6\right\}\)
\(\dfrac{4}{x}=\dfrac{x}{16}\\ \Leftrightarrow x.x=4.16\Leftrightarrow x^2=64\\ \Leftrightarrow\left[{}\begin{matrix}x^2=8^2\\x^2=\left(-8\right)^2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)
=> x\(\) . x = 4 . 16 => x2 = 64 => x2 = 82 => x = 8 vậy x = 8