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Lời giải:
$\frac{4}{9}\times \frac{-13}{-17}+\frac{-2}{17}\times \frac{4}{9}+\frac{2}{9}\times \frac{4}{17}$
$=\frac{4}{9}\times (\frac{13}{17}+\frac{-2}{17})+\frac{2}{9}\times \frac{4}{17}$
$=\frac{4}{9}\times \frac{11}{17}+\frac{4}{9}\times \frac{2}{17}$
$=\frac{4}{9}\times (\frac{11}{17}+\frac{2}{17})$
$=\frac{4}{9}\times \frac{13}{17}=\frac{52}{153}$
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: \(B=\left(2^{17}+17^2\right)\cdot\left(9^{15}-15^9\right)\cdot\left(16-16\right)=0\)
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)
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 = \(\pm2\)
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
Dễ mà,e cứ chia 2 TH là đc
Vd:<0 thì chia ra x+2>0 hoac x<0 và nguoc lai roi tìm x
1/ Suy ra: -35 + 7x - 2x + 20 = 15
5x - 15 = 15
5x = 30
x = 6
2/ Suy ra: 4x - 20 - 3x - 21 = -19
x - 41 = -19
x = 22
3/ Suy ra: 8 - 4x + 3x - 15 = 14
-7 -x = 14
x = -7 -14
x = -21
4/ Suy ra: 7x - 63 - 30 + 5x = -6 + 11x
7x + 5x - 11x = -6 +63 + 30
x = 87
5/ Suy ra: -21x +35 + 14x - 28 = 28
-21x + 14x = 28 +28 - 35
-7x = 21
x= -3
6/ Suy ra: 20 - 5x + 7x - 14 = -2
2x = -8
x = -4
7/ Suy ra: 4x +15 -5x = 7
-x = -8
x = 8
8/ Suy ra: 5x - 35 +30 -10x = 20
-5x - 5 = 20
-5x = 25
x = -5
9/ Suy ra: 35 - 7x + 5x - 10 = 15
-7x + 5x = 15 +10 - 35
-2x = -10
x = 5
- \(x\cdot2\frac{4}{9}=4\frac{2}{5}-1\frac{4}{7}\)
\(x\cdot\frac{22}{9}=\frac{22}{5}-\frac{11}{7}\)
\(x\cdot\frac{22}{9}=\frac{99}{35}\)
\(x=\frac{99}{35}:\frac{22}{9}\)
\(x=\frac{81}{70}\)
Vậy \(x=\frac{81}{70}\).
- \(x:10\frac{6}{5}=4\frac{6}{5}+9\frac{7}{5}\)
\(x:\frac{56}{5}=\frac{26}{5}+\frac{52}{5}\)
\(x:\frac{56}{5}=\frac{78}{5}\)
\(x=\frac{78}{5}\cdot\frac{56}{5}\)
\(x=\frac{4368}{25}\)
Vậy \(x=\frac{4368}{25}\).
- \(x:3\frac{10}{3}=9\frac{1}{9}-5\frac{1}{9}\)
\(x:\frac{19}{3}=4\)
\(x=4\cdot\frac{19}{3}\)
\(x=\frac{76}{3}\)
Vậy \(x=\frac{76}{3}\).
\(\dfrac{x+2}{-4}=\dfrac{-9}{x+2}\)(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