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2/5 + 11/15 = 17/15
7/8 - 7/9 = 7/72
11/13 x 26/31 = 22/31
16/24 : 4 = 1/6
30 : 6/5 = 25
9/16 : 4 = 9/64
1/2 : 1/3 x 8/15 = 3/2 x 8/15 = 4/5
![](https://rs.olm.vn/images/avt/0.png?1311)
y \(\times\) \(\dfrac{16}{64}\) + y \(\times\) \(\dfrac{25}{100}\) + y \(\times\) \(\dfrac{1}{4}\) + y \(\times\) \(\dfrac{15}{60}\) - \(\dfrac{13}{15}\) = \(\dfrac{17}{15}\)
y \(\times\) ( \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\)+ \(\dfrac{1}{4}\)) - \(\dfrac{13}{15}\) = \(\dfrac{17}{15}\)
y = \(\dfrac{17}{15}\) + \(\dfrac{13}{15}\)
y = \(\dfrac{30}{15}\)
y = 2
Cái này bằng rủ em chơi oẳn tù tì ,em ra lá còn anh ra hôn má em
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![](https://rs.olm.vn/images/avt/0.png?1311)
\(46\cdot99\)
\(=46\cdot\left(100-1\right)\)
\(=46\cdot100-46\)
\(=4600-46\)
\(=4554\)
___________
\(34\cdot11\)
\(=34\cdot\left(10+1\right)\)
\(=34\cdot10+34\)
\(=340+34\)
\(=374\)
___________
\(25\cdot12\)
\(=5\cdot5\cdot12\)
\(=5\cdot60\)
\(=300\)
__________
\(15\cdot4\)
\(=15\cdot2\cdot2\)
\(=30\cdot2\)
\(=60\)
____________
\(45\cdot6\)
\(=45\cdot2\cdot3\)
\(=90\cdot3\)
\(=2700\)
________
\(13\cdot99\)
\(=13\cdot\left(100-1\right)\)
\(=13\cdot100-13\)
\(=1300-13\)
\(=1287\)
______________
\(16\cdot19\)
\(=16\cdot\left(20-1\right)\)
\(=16\cdot20-16\)
\(=320-16\)
\(=304\)
_____________
\(35\cdot98\)
\(=35\cdot\left(100-2\right)\)
\(=35\cdot100-35\cdot2\)
\(=3500-70\)
\(=3430\)
____________
\(125\cdot16\)
\(=125\cdot8\cdot2\)
\(=1000\cdot2\)
\(=2000\)
___________
\(47\cdot101\)
\(=47\cdot\left(100+1\right)\)
\(=47\cdot100+47\)
\(=4700+47\)
\(=4747\)
46 x 99 =4554 , 34 x 11 = 374 , 25 x 12 =- 300 , 15 x 4 = 60 , 45 x 6 = 270 , 13 x 99 = 1287 , 16 x 19 = 304
35 x 98 = 3430 , 125 x 16 = 2000 , 47 x 101 = 4747
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a) \(\frac{3}{16}+\frac{4}{15}+\frac{5}{16}+\frac{1}{15}\)
\(=\left(\frac{3}{16}+\frac{5}{16}\right)+\left(\frac{4}{15}+\frac{1}{15}\right)\)
\(=\frac{1}{2}+\frac{1}{3}\)
\(=\frac{5}{6}\)
b) \(\frac{6}{7}\times\frac{8}{15}\times\frac{7}{6}\times\frac{15}{16}\)
\(=\left(\frac{6}{7}\times\frac{7}{6}\right)\times\left(\frac{8}{15}\times\frac{15}{16}\right)\)
\(=1\times\frac{1}{2}=\frac{1}{2}\)
c) \(\frac{19}{20}\times\frac{13}{21}+\frac{9}{20}\times\frac{8}{21}\)
\(=\frac{19\times13}{20\times21}+\frac{9\times8}{20\times21}\)
\(=\frac{247}{420}+\frac{72}{420}\)
\(=\frac{319}{420}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left\{{}\begin{matrix}26⋮x\\x\ge13\end{matrix}\right.\Rightarrow x\in\left\{13;26\right\}\)
\(\left\{{}\begin{matrix}16⋮x\\x< 8\end{matrix}\right.\Rightarrow x\in\left\{1;2;4\right\}\)
\(\left\{{}\begin{matrix}18⋮x\\0< x< 40\end{matrix}\right.\Rightarrow x\in\left\{1;2;3;6;9;18\right\}\)
\(\left\{{}\begin{matrix}x⋮15\\30< x< 40\end{matrix}\right.\Rightarrow x\in\varnothing\)
\(\left\{{}\begin{matrix}x⋮12\\22\le5x\le50\end{matrix}\right.\Rightarrow x\in\varnothing\)
\(\left\{{}\begin{matrix}x⋮4\\16\le x\le36\end{matrix}\right.\Rightarrow x\in\left\{16;20;24;28;32;36\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: x=3/4-1/2=3/4-2/4=1/4
2: x-1/5=2/11
=>x=2/11+1/5=21/55
3: x-5/6=16/42-8/56
=>x-5/6=8/21-4/28=5/21
=>x=5/21+5/6=15/14
4: x/5=5/6-19/30
=>x/5=25/30-19/30=6/30=1/5
=>x=1
5: =>|x|=1/3+1/4=7/12
=>x=7/12 hoặc x=-7/12
6: x=-1/2+3/4
=>x=3/4-1/2=1/4
11: x-(-6/12)=9/48
=>x+1/2=3/16
=>x=3/16-1/2=-5/16
1)x= 1/4
2)x= 2/11+ 1/5
x= 21/55
3)x - 5/6 = 5/21
x = 5/21+5/6
x = 15/14
4)x/5 = 5/6 + -19/30
x:5 = 1/5
x = 1/5.5
x = 1
5) |x| - 1/4 = 6/18
|x| = 6/18 - 1/4
|x| =7/12
⇒x= 7/12 hoặc -7/12
6)x = -1/2 +3/4
x= 1/4
7) x/15 = 3/5 + -2/3
x:15 = -1/15
x = -1/15. 15
x = -1
8)11/8 + 13/6 = 85/x
85/24 = 85/x
⇒ x = 24
9) x - 7/8 = 13/12
x = 13/12 + 7/8
x = 47/24
10)x - -6/15 = 4/27
x = 4/27 + (-6/15)
x = -34/135
11) -(-6/12)+x = 9/48
x= 9/48 - 6/12
x = -5/16
12) x - 4/6 = 5/25 + -7/15
x -4/6 = -4/15
x = -4/15 + 4/6
x = 2/5
\(\left(x+13\right)^4+\left(x+15\right)^4=16\)
Đặt \(x+14=a\), phương trình trở thành:
\(\left(a-1\right)^4+\left(a+1\right)^4=16\)
\(\Leftrightarrow a^4-4a^3+6a^2-4a+1\)\(+a^4+4a^3+6a^2+4a+1=16\)
\(\Leftrightarrow2a^4+12a^2+2=16\).
\(\Leftrightarrow a^4+6a^2+1=8\)
\(\Leftrightarrow a^4+6a^2-7=0\)
\(\Leftrightarrow\left(a^2-1\right)\left(a^2+7\right)=0\)
\(\Leftrightarrow\left(a-1\right)\left(a+1\right)\left(a^2+1\right)=0\)
\(\Leftrightarrow\left(x+13\right)\left(x+15\right)\left[\left(x+14\right)^2+7\right]=0\)
Vì \(\left(x+14\right)^2+7\ge7>0\forall x\)nên:
\(\left(x+13\right)\left(x+15\right)=0:\left[\left(x+14\right)^2+7\right]\)
\(\Leftrightarrow\left(x+13\right)\left(x+15\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x+13=0\\x+15=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-13\\x=-15\end{cases}}\)
Vậy phương trình có tập nghiệm: \(S=\left\{-15;-13\right\}\)