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Lời giải:
$\frac{4}{9}\times \frac{-13}{-17}+\frac{-2}{17}\times \frac{4}{9}+\frac{2}{9}\times \frac{4}{17}$
$=\frac{4}{9}\times (\frac{13}{17}+\frac{-2}{17})+\frac{2}{9}\times \frac{4}{17}$
$=\frac{4}{9}\times \frac{11}{17}+\frac{4}{9}\times \frac{2}{17}$
$=\frac{4}{9}\times (\frac{11}{17}+\frac{2}{17})$
$=\frac{4}{9}\times \frac{13}{17}=\frac{52}{153}$
a: =91/105+60/105-101/105
=50/105=10/21
c: \(\dfrac{3}{4}\cdot\dfrac{5}{2}\cdot\dfrac{7}{6}=\dfrac{3}{6}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{1}{2}\cdot\dfrac{7}{2}\cdot\dfrac{5}{4}=\dfrac{35}{16}\)
d: =2-2/9
=18/9-2/9
=16/9
e: =24/36-9/36+8/36
=23/36
g: =5/2+1/2
=3
Theo bài ra ta có:
\(9.x^4-x^4+2.x^4=2^5+2^9:2^2\)
\(=x^4\left(9-1+2\right)=2^5+2^7\)
\(\Rightarrow10x^4=160\)
\(\Rightarrow x^4=160\div10=16\)
\(\Rightarrow x=\pm2\)
Vậy giá trị của x = \(\pm2\)
Bài 1:
\(\Leftrightarrow2^n\cdot\dfrac{9}{2}=9\cdot5^n\)
\(\Leftrightarrow2^n=2\cdot5^n\)
\(\Leftrightarrow2^{n-1}=5^n\)
Bài 2:
a: \(A=\dfrac{2^8+3^8}{2^8}=1+\dfrac{3^8}{2^8}\)
b: \(B=\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(16-16\right)=0\)
B= {[ 5 * (2^2)^15 * (3^2)^9 ] - [ (2^2) * 3^20 * (2^3)^9 ]} / {[ 5 * (2^9) * (2^19)*(3^19) ] - [ 7 * (2^29) * (3^3)^6 ]}
B= {[ 5 * (2^30) * (3^18) ] - [ (3^20) * (2^29) ]} / {[ 5 * (2^28) * (3^19) ] - [ 7 * (2^29) * (3^18) ]}
B= {[ (2^29) * (3^18) ] * [(5 * 2) - 3^2 ]} / {[ (2^28) * (3^18) ] - [(5 * 3) - (7 * 2)] }
B= [ (2^29) * (3^18) ] / [ (2^28) * (3^18) ]
B= [ (2^1) * (2^28) * (3^18) ] / [ (2^28) * (3^18) ]
B = 2
dấu * là dấu nhân
B= {[ 5 * (2^2)^15 * (3^2)^9 ] - [ (2^2) * 3^20 * (2^3)^9 ]} / {[ 5 * (2^9) * (2^19)*(3^19) ] - [ 7 * (2^29) * (3^3)^6 ]}
B= {[ 5 * (2^30) * (3^18) ] - [ (3^20) * (2^29) ]} / {[ 5 * (2^28) * (3^19) ] - [ 7 * (2^29) * (3^18) ]}
B= {[ (2^29) * (3^18) ] * [(5 * 2) - 3^2 ]} / {[ (2^28) * (3^18) ] - [(5 * 3) - (7 * 2)] }
B= [ (2^29) * (3^18) ] / [ (2^28) * (3^18) ]
B= [ (2^1) * (2^28) * (3^18) ] / [ (2^28) * (3^18) ]
B = 2
dấu * là dấu nhân
1) |x + 2| = 4
\(\Leftrightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=2\\x=-6\end{cases}}\)
2) 3 – |2x + 1| = (-5)
\(\Leftrightarrow\left|2x+1\right|=8\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}\)
3) 12 + |3 – x| = 9
\(\Leftrightarrow\left|3-x\right|=-3\)(vô lí)
=>\(x=\varnothing\)
1) I x+2 I=4
\(\Rightarrow\orbr{\begin{cases}x+2=4\\x+2=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=2\\x=-6\end{cases}}}\)
2) \(3-|2x+1|=-5\)
\(\Leftrightarrow|2x+1|=8\)
\(\Leftrightarrow\orbr{\begin{cases}2x+1=8\\2x+1=-8\end{cases}\Leftrightarrow\orbr{\begin{cases}2x=7\\2x=-9\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=\frac{7}{2}\\x=\frac{-9}{2}\end{cases}}}\)
3) \(12+|3-x|=9\)
\(\Leftrightarrow|3-x|=-3\)(vô lí vì I 3-x I \(\ge\)0)
a: x*3/4=1/5
=>x=1/5:3/4=1/5*4/3=4/15
b: x*3/7=2/5
=>x=2/5:3/7=2/5*7/3=14/15
c: 1/3+2/9=2/12x
=>1/6x=3/9+2/9=5/9
=>x=5/9*6=30/9=10/3
d: 4/15*x-2/3=1/5
=>4/15*x=2/3+1/5=10/15+3/15=13/15
=>4x=13
=>x=13/4
e: x:1/7=2/3
=>x=2/3*1/7=2/21
f: 1/9:x=7/3
=>x=1/9:7/3=1/9*3/7=3/63=1/21
j: 1/4+5/12=8/3:x
=>8/3:x=3/12+5/12=8/12=2/3
=>x=4
h: =>7/4:x=1/5+1/2=7/10
=>x=7/4:7/10=10/4=5/2
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\(\dfrac{x+2}{-4}=\dfrac{-9}{x+2}\)(ĐKXĐ: x<>-2)
=>\(\left(x+2\right)^2=\left(-9\right)\cdot\left(-4\right)=36\)
=>\(\left[{}\begin{matrix}x+2=6\\x+2=-6\end{matrix}\right.\)
=>\(\left[{}\begin{matrix}x=4\left(nhận\right)\\x=-8\left(nhận\right)\end{matrix}\right.\)
sos