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![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{x-1}{9}=\frac{8}{3}\)
\(\left(x-1\right)\cdot3=8\cdot9\)
\(\left(x-1\right)\cdot3=72\)
\(x-1=\frac{72}{3}\)
\(x-1=24\)
\(x=24+1\)
\(x=25\)
\(\frac{x}{4}=\frac{18}{x+1}\)
\(x\cdot\left(x+1\right)=18\cdot4\)
\(x\left(x+1\right)=72\)
\(x\left(x+1\right)=8\cdot9\)
\(x=8\)
\(\frac{-x}{4}=\frac{-9}{x}\)
\(\frac{x}{-4}=\frac{-9}{x}\)
\(x\cdot x=\left(-9\right)\cdot\left(-4\right)\)
\(x^2=36\)
\(x^2=\left(-6\right)^2\)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!
![](https://rs.olm.vn/images/avt/0.png?1311)
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
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\frac{x-1}{9}=\frac{8}{3}\)
\(\Leftrightarrow x-1=24\)
\(\Rightarrow x=25\)
\(b,-\frac{x}{4}=-\frac{9}{x}\)
\(\Leftrightarrow x^2=36\)
\(\Leftrightarrow\orbr{\begin{cases}x=6\\x=-6\end{cases}}\)
\(c,\frac{x}{4}=\frac{18}{x+1}\)
\(\Leftrightarrow x^2+x=72\)
\(\Leftrightarrow x\left(x+1\right)=72..\)
ấn nhầm: lm tiếp nhé!
\(x\left(x+1\right)=72\)
\(\text{Mà x thuộc Z nên }x\left(x+1\right)=8\left(8+1\right)\)
\(\Leftrightarrow x=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(bai1:a,\frac{3}{7}\cdot\frac{-5}{9}+\frac{4}{9}\cdot\frac{3}{7}-\frac{3}{7}\cdot\frac{8}{9}\)
\(< =>\frac{-15}{63}+\frac{12}{63}-\frac{24}{63}\)
\(< =>\frac{-15+12-24}{63}\)
\(< =>\frac{-3}{7}\)
\(b,1\frac{13}{15}\cdot0,75-\left(\frac{11}{20}+25\%\right):\frac{7}{5}\)
\(< =>\frac{28}{15}\cdot\frac{3}{4}-\left(\frac{11}{20}+\frac{1}{4}\right):\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{5}:\frac{7}{5}\)
\(< =>\frac{7}{5}-\frac{4}{7}\)
\(< =>\frac{29}{35}\)
\(bai2:\)
\(a,\frac{-3}{4}\cdot x-\frac{4}{10}=\frac{1}{5}\)
\(< =>\frac{-3}{4}\cdot x=\frac{1}{5}+\frac{4}{10}\)
\(< =>\frac{-3}{4}\cdot x=\frac{3}{5}\)
\(< =>x=\frac{3}{5}:\frac{-3}{4}\)
\(< =>x=\frac{-4}{5}\)
\(b,3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(< =>3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(< =>\left[3\left(x-\frac{1}{3}\right)\right]=\frac{1}{12}< =>x-\frac{1}{3}=\frac{1}{12}:3=\frac{1}{36}=>x=\frac{1}{36}+\frac{1}{3}=>x=\frac{13}{36}\)
\(< =>\left[\frac{1}{3}\cdot x\right]=\frac{1}{12}< =>x=\frac{1}{12}:\frac{1}{3}=>x=\frac{1}{4}\)
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,\(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{19}:\frac{12}{19}\)
\(\frac{-3}{4}x\) \(=\frac{1}{5}+\frac{4}{10}\) \(3\left(x-\frac{1}{3}\right)+\frac{1}{3}x=\frac{1}{12}\)
\(\frac{-3}{4}x\) \(=\frac{3}{5}\) \(\left(x.3-\frac{1}{3}.3\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{3}{5}:\frac{-3}{4}\) \(\left(x.3-1\right)+\frac{1}{3}x=\frac{1}{12}\)
\(x\) \(=\frac{4}{-5}\) \(x.\left(3+\frac{1}{3}\right)-1=\frac{1}{12}\)
\(x.\left(3+\frac{1}{3}\right)=\frac{1}{12}+1\)
\(x.\frac{10}{3}=\frac{13}{12}\)
\(x=\frac{13}{12}:\frac{10}{3}\)
\(x=\frac{13}{40}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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, \(\frac{x}{4}=\frac{18}{x+1}\)
\(\Rightarrow x\left(x+1\right)=4.18\)
\(\Rightarrow x.x+x.1=72\)
\(\Rightarrow x^2+x=72\)
\(\Rightarrow x^2+x-72=0\)
\(\Rightarrow x^2+x-8^2+8=0\)
\(\Rightarrow x=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a)x+30\%x=-1,31\)
\(\Leftrightarrow x+\frac{3x}{10}=-1,31\)
\(\Leftrightarrow10x+3x=-13,1\)
\(\Leftrightarrow13x=-13,1\Leftrightarrow x=-\frac{131}{130}\)
\(b)\left(x-\frac{1}{2}\right):\frac{1}{3}+\frac{5}{7}=9\frac{5}{7}\)
\(\Leftrightarrow\frac{2x-1}{2}.3+\frac{5}{7}=\frac{68}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=\frac{63}{7}\)
\(\Leftrightarrow\frac{6x-3}{2}=9\)
\(\Leftrightarrow6x-3=18\)
\(\Leftrightarrow x=\frac{7}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=\frac{1}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}.\frac{5.5}{4.6}.\frac{6.6}{5.7}=\frac{6}{2.7}=\frac{3}{7}\)
B) \(=\frac{70}{11}+\frac{1}{9}-\frac{37}{11}-\frac{1}{9}=\left(\frac{70}{11}-\frac{37}{11}\right)+\left(\frac{1}{9}-\frac{1}{9}\right)=\frac{33}{11}+0=3\)
BÀI 2:
A) \(\Leftrightarrow\frac{7}{2}x-\frac{x}{2}+\frac{2x}{2}=\frac{7}{2}.\frac{5}{6}\)
\(\Leftrightarrow\frac{7x-x+2x}{2}=\frac{35}{12}\)
\(\Leftrightarrow\frac{8x}{2}=\frac{35}{12}\)
\(\Leftrightarrow8x.12=35.2\Leftrightarrow96x=70\Leftrightarrow x=\frac{70}{96}=\frac{35}{48}\)
b) \(\left(x-\frac{3}{1.2}\right)+\left(x-\frac{3}{2.3}\right)+...+\left(x-\frac{3}{99.100}\right)=1\)
\(x-\frac{3}{1.2}+x-\frac{3}{2.3}+....x+\frac{3}{99.100}=1\)
\(\Leftrightarrow\left(x+x+x+...+x\right)-3\left(\frac{1}{1.2}+\frac{1}{1.3}+....+\frac{1}{99.100}\right)=1\)
ngoặc 1 có 99 số hạng x
\(\Leftrightarrow99x-3\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{99}-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3\left(1-\frac{1}{100}\right)=1\)
\(\Leftrightarrow99x-3.\frac{99}{100}=1\)
\(\Leftrightarrow99x=1+\frac{3.99}{100}\)
\(\Leftrightarrow99x=\frac{397}{100}\)
\(\Leftrightarrow x=\frac{397}{100.99}=\frac{397}{9900}\)
giup j zay bn
thiếu đề bạn ah