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\(\frac{x-1}{9}=\frac{8}{3}\)
\(\left(x-1\right)\cdot3=8\cdot9\)
\(\left(x-1\right)\cdot3=72\)
\(x-1=\frac{72}{3}\)
\(x-1=24\)
\(x=24+1\)
\(x=25\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(x\cdot\left(x+1\right)=18\cdot4\)
\(x\left(x+1\right)=72\)
\(x\left(x+1\right)=8\cdot9\)
\(x=8\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\frac{x}{-4}=\frac{-9}{x}\)
\(x\cdot x=\left(-9\right)\cdot\left(-4\right)\)
\(x^2=36\)
\(x^2=\left(-6\right)^2\)hoặc \(x^2=6^2\)
\(x=-6\) hoặc\(x=6\)
a)x-1/9=24/9 => x-1=24 =>x=23
b)x(x+1)=18*4 =>x=8
c)-x:4=-9:x =>-1.x2=-1.36 =>x=6
k mik nha!
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
Bài 1:
a)\(\frac{3}{7}.\frac{-5}{9}+\frac{4}{9}.\frac{3}{7}-\frac{3}{7}.\frac{8}{9}\) b,\(1\frac{13}{15}.0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(=\frac{3}{7}.(\frac{-5}{9}+\frac{4}{9}-\frac{8}{9})\) \(=\frac{28}{15}.\frac{3}{4}-\left(\frac{11}{20}+\frac{5}{20}\right):\frac{7}{5}\)
\(=\frac{3}{7}.\frac{-9}{9}\) \(=\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(=\frac{-3}{7}\) \(=\frac{7}{5}-\frac{4}{7}\)
\(=\frac{29}{35}\)
Bài 2:
a)\(\frac{-3}{4}x-\frac{4}{10}=\frac{1}{5}\) b,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
\(a,\frac{1}{2}+\frac{2}{3}x=\frac{4}{5}\)
=> \(\frac{2}{3}x=\frac{4}{5}-\frac{1}{2}=\frac{3}{10}\)
=> \(x=\frac{3}{10}:\frac{2}{3}=\frac{9}{20}\)
Vậy \(x\in\left\{\frac{9}{20}\right\}\)
\(b,x+\frac{1}{4}=\frac{4}{3}\)
=> \(x=\frac{4}{3}-\frac{1}{4}=\frac{13}{12}\)
Vậy \(x\in\left\{\frac{13}{12}\right\}\)
\(c,\frac{3}{5}x-\frac{1}{2}=-\frac{1}{7}\)
=> \(\frac{3}{5}x=-\frac{1}{7}+\frac{1}{2}=\frac{5}{14}\)
=> \(x=\frac{5}{14}:\frac{3}{5}=\frac{25}{42}\)
Vậy \(x\in\left\{\frac{25}{42}\right\}\)
\(d,\left|x+5\right|-6=9\)
=> \(\left|x+5\right|=9+6=15\)
=> \(\left[{}\begin{matrix}x+5=15\\x+5=-15\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=15-5=10\\x=-15-5=-20\end{matrix}\right.\)
Vậy \(x\in\left\{10;-20\right\}\)
\(e,\left|x-\frac{4}{5}\right|=\frac{3}{4}\)
=> \(\left[{}\begin{matrix}x-\frac{4}{5}=\frac{3}{4}\\x-\frac{4}{5}=-\frac{3}{4}\end{matrix}\right.\)
=> \(\left[{}\begin{matrix}x=\frac{3}{4}+\frac{4}{5}=\frac{31}{20}\\x=-\frac{3}{4}+\frac{4}{5}=\frac{1}{20}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{31}{20};\frac{1}{20}\right\}\)
\(f,\frac{1}{2}-\left|x\right|=\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{2}-\frac{1}{3}\)
=> \(\left|x\right|=\frac{1}{6}\)
=> \(\left[{}\begin{matrix}x=\frac{1}{6}\\x=-\frac{1}{6}\end{matrix}\right.\)
Vậy \(x\in\left\{\frac{1}{6};-\frac{1}{6}\right\}\)
\(g,x^2=16\)
=> \(\left|x\right|=\sqrt{16}=4\)
=> \(\left[{}\begin{matrix}x=4\\x=-4\end{matrix}\right.\)
vậy \(x\in\left\{4;-4\right\}\)
\(h,\left(x-\frac{1}{2}\right)^3=\frac{1}{27}\)
=> \(x-\frac{1}{2}=\sqrt[3]{\frac{1}{27}}=\frac{1}{3}\)
=> \(x=\frac{1}{3}+\frac{1}{2}=\frac{5}{6}\)
Vậy \(x\in\left\{\frac{5}{6}\right\}\)
\(i,3^3.x=3^6\)
\(x=3^6:3^3=3^3=27\)
Vậy \(x\in\left\{27\right\}\)
\(J,\frac{1,35}{0,2}=\frac{1,25}{x}\)
=> \(x=\frac{1,25.0,2}{1,35}=\frac{5}{27}\)
Vậy \(x\in\left\{\frac{5}{27}\right\}\)
\(k,1\frac{2}{3}:x=6:0,3\)
=> \(\frac{5}{3}:x=20\)
=> \(x=\frac{5}{3}:20=\frac{1}{12}\)
Vậy \(x\in\left\{\frac{1}{12}\right\}\)
\(a,\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x-1=24\)
\(\Rightarrow x=25\)
\(b,-\frac{x}{4}=-\frac{9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(c,\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x^2+x=72\)
\(\Leftrightarrow x\left(x+1\right)=72..\)
ấn nhầm: lm tiếp nhé!
\(x\left(x+1\right)=72\)
\(\text{Mà x thuộc Z nên }x\left(x+1\right)=8\left(8+1\right)\)
\(\Leftrightarrow x=8\)
a, \(\frac{x-1}{9}=\frac{8}{3}\)
\(\Rightarrow\left(x-1\right).3=8.9\)
\(\Rightarrow\left(x-1\right).3=72\)
\(\Rightarrow x-1=72:3\)
\(\Rightarrow x-1=24\)
\(\Rightarrow x=24+1\)
\(\Rightarrow x=25\)
b, \(\frac{-x}{4}=\frac{-9}{x}\)
\(\Rightarrow-x.x=-9.4\)
\(\Rightarrow-\left(x^2\right)=-36\)
\(\Rightarrow x^2=36\)
\(\Rightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
c, \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x\left(x+1\right)=4.18\)
\(\Rightarrow x.x+x.1=72\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x^2+x-72=0\)
\(\Rightarrow x^2+x-8^2+8=0\)
\(\Rightarrow x=8\)
a,\(\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}.\left(\frac{3}{4}-\frac{6}{5}\right)=\frac{5}{2-2x};Đkxđ:x\ne1\)
\(\Rightarrow\frac{1}{x-1}+\frac{-2}{3}\left(\frac{-9}{20}\right)=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}+\frac{3}{10}=\frac{5}{2-2x}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{2-2x}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}-\frac{5}{-2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow\frac{1}{x-1}+\frac{5}{2\left(x-1\right)}=\frac{3}{10}\)
\(\Rightarrow\frac{7}{2\left(x-1\right)}=\frac{-3}{10}\)
\(\Rightarrow70=-6\left(x-1\right)\)
\(\Rightarrow6x=6-70\)
\(\Rightarrow6x=-64\)
\(\Rightarrow x=\frac{-32}{3}x\ne1\)
mik ko chép lại đề, mik làm luôn:
a) x - \(\frac{31}{36}=\frac{-13}{38}\)
x = \(\frac{-13}{18}+\frac{31}{36}\)
\(x=\frac{5}{36}\)
b)\(2-x-\frac{3}{7}=\frac{9}{-21}\)
\(\frac{11}{7}-x=\frac{3}{7}\)
x = \(\frac{11}{7}-\frac{3}{7}\)
x = 8/7
c) x + 3/11 = 23/44
x = 23/44 - 3/11
x = 1/4
d) \(\frac{1}{12}-x=\frac{-11}{9}\)
x = \(\frac{1}{12}+\frac{11}{9}\)
x = 47/36
e) \(x-\frac{2}{3}=\frac{-17}{3}\)
x= -17/3 + 2/3
x = -5
f) \(x-\frac{1}{2}=\frac{11}{4}.\frac{3}{11}\)
x - 1/2 = 3/4
x = 3/4 + 1/2
x = 5/4
g) \(2x+\frac{3}{8}=\frac{-21}{32}.\frac{4}{7}\)
2x + 3/8 = -3 / 8
2x = -3/8 - 3/8
2x = -9/8
x = -9/8.1/2
x = -9/16
h) x - \(\frac{x}{3}=\frac{3}{57}.\frac{19}{12}\)
x - \(\frac{x}{3}=\frac{1}{12}\)
x = \(\frac{1}{12}+\frac{x}{3}\)
x = \(\frac{1+4x}{12}\)
=> 12x = 1+4x
12x - 4x = 1
8x = 1
x = 1/8
giup j zay bn
thiếu đề bạn ah