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a: \(A=\dfrac{3}{5}\cdot xy\cdot\dfrac{4}{25}x^2y^4z^2=\dfrac{12}{125}x^3y^5z^2\)
b: Hệ số là 12/125
Bậc là 10
\(A=\dfrac{x^2-2x-3-x^2+x-1+4x+4}{\left(x+1\right)\left(x^2-x+1\right)}\)
\(=\dfrac{3x}{\left(x+1\right)\left(x^2+x+1\right)}\)
\(A=\left(5a-5\right)^2+10\left(a-3\right)\left(1+a\right).3a\)
\(A=25a^2-50a+25+30a\left(a-3+a^2-3a\right)\)
\(A=25a^2-50a+25+30a^2-90a+30a^3-90a^2\)
\(A=30a^3-35a^2-140a+25\)
Ta có: \(A=\left(5a-5\right)^2+10\left(a-3\right)\left(a+1\right)\cdot3a\)
\(=25a^2-50a+25+30a\left(a^2-2a-3\right)\)
\(=25a^2-50a+25+30a^3-60a^2-90a\)
\(=30a^3-35a^2-140a+25\)
A = 3:5xy {-2:5\(xy^2z\)}\(^2\)
A= \(\dfrac{3}{5}\)xy (\(-\dfrac{2}{5}\))\(^2\).\(x^2\). \(\left(y^2\right)^2\). \(z^2\)
A= \(\dfrac{3}{5}\)xy. \(\dfrac{4}{25}x^2y^4z^2\)
A= \(\left(\dfrac{3}{5}.\dfrac{4}{25}\right).\left(x.x^2\right).\left(y.y^4\right).z^2\)
A= \(\dfrac{12}{125}x^3y^5z^2\)
Hệ số: \(\dfrac{12}{125}\)
Bậc của đơn thức A là 10
mik chỉ biết giải A thôi B mik không biết làm
Bài 1:
\(\dfrac{x^2-3}{x+\sqrt{3}}=\dfrac{\left(x+\sqrt{3}\right)\left(x-\sqrt{3}\right)}{x+\sqrt{3}}=x-\sqrt{3}\)
Bài 2:
a) Ta có: \(A=\sqrt{16x+16}-\sqrt{9x+9}+\sqrt{4x+4}+\sqrt{x+1}\)
\(=4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}+\sqrt{x+1}\)
\(=4\sqrt{x+1}\)
b) Để A=16 thì \(\sqrt{x+1}=4\)
\(\Leftrightarrow x+1=16\)
hay x=15
\(log_{a^3}b.log_ba=\dfrac{1}{3}.log_ab.log_ba=\dfrac{1}{3}\)
\(log_{a^{10}}b^5.log_{b^3}a^9=\dfrac{1}{10}.5.log_ab.\dfrac{1}{3}.9.log_ba=\dfrac{3}{2}\)
\(log_{a^{107}}b^{101}.log_{b^{303}}a^{428}=\dfrac{1}{107}.101.log_ab.\dfrac{1}{303}.428.log_ba=\dfrac{4}{3}.log_ab.log_ba=\dfrac{4}{3}\)
a: \(log_{a^3}b\cdot log_ba=\dfrac{1}{3}\cdot log_ab\cdot log_ba=\dfrac{1}{3}\)
b: \(log_{a^{10}}b^5\cdot log_{b^3}a^9\)
\(=\dfrac{1}{10}\cdot log_ab^5\cdot\dfrac{1}{3}\cdot log_ba^9\)
\(=\dfrac{1}{30}\cdot5\cdot log_ab\cdot9\cdot log_ba=\dfrac{45}{30}=\dfrac{3}{2}\)
c: \(log_{a^{107}}b^{101}\cdot log_{b^{303}}a^{428}\)
\(=\dfrac{1}{107}\cdot log_ab^{101}\cdot\dfrac{1}{303}\cdot log_ba^{428}\)
\(=\dfrac{1}{107}\cdot101\cdot log_ab\cdot\dfrac{1}{303}\cdot428\cdot log_ba\)
\(=4\cdot\dfrac{1}{3}=\dfrac{4}{3}\)
Giải:
a,\(\dfrac{2^5.3^2}{2^3.3}=2^2.3=12\)
b, \(\dfrac{5^3.6}{5.2^2}=\dfrac{5^3.2.3}{5.2^2}=\dfrac{5^2.3}{2}=37,5\)
c, \(\dfrac{8.5^2-8.4^2}{2^2.3^2}=\dfrac{8.\left(5^2-4^2\right)}{2^2.3^2}=\dfrac{8.9}{4.9}=2\)
d, \(\dfrac{7^4.2^2+7^4.3}{49.26}=\dfrac{7^4.\left(2^2+3\right)}{7^2.26}=\dfrac{7^4.7}{7^2.26}=\dfrac{343}{26}\)
Have a nice day!
`a) (2^5 .3^2)/(2^3 .3)`
`= (2^2 .3)/1`
`= 4.3`
`=12`
`b) (5^3 .6)/(5.2^2)`
`= (5^3 .2.3)/(5.2^2)`
`= (5^2 .3)/2`
`= 75/2`
`c) (8.5^2-8.4^2)/(2^2 .3^2)`
`= (8.(5^2-4^2))/(4.9)`
`= (8.9)/(4.9)`
`= (2.4.9)/(4.9)`
`=2`
`d) (7^4 .2^2+7^4 .3)/(49.26)`
`= (7^4.(2^2+3))/(7^2 .26)`
`= (7^4 .7)/(7^2 .26)`
`= (7^5)/(7^2. 26)`
`= (7^3)/26`
`= 343/26`
đây mà là toán lớp 5????????
Toán lớp 6 chứ
'Lớp 5 
còn chưa học số mũ mà'
bn bị lag à
a) \(\left|a\right|+a=a+a=2a\)
b) \(\left|a\right|-a=a-a=0\)
c) \(\left|a\right|.a=a.a=a^2\)
d) \(\left|a\right|:a=a:a=1\)( Với \(a\ne0\) )
e) \(3\left(x-1\right)-2\left|x+3\right|=3x-3-2.\left(x+3\right)\)
\(=3x-3-2x-6=x-9\)
g) \(2\left|x-3\right|-\left|4x-1\right|=2.\left(x-3\right)-\left(4x-1\right)\)
\(=2x-6-4x+1=-2x-5\)
a,\(|a|+a=a+a=2a\)
b,\(|a|-a=a-a=0\)
c,\(|a|.a=a.a=a^2\)
d,\(|a|:a=a:a=1\)
e,\(3\left(x-1\right)-2|x+3|=3x-3-2\left(x+3\right)\)
\(=3x-3-2x-6=x-9\)
g,\(2|x-3|-|4x-1|=2\left(x-3\right)-\left(4x-1\right)\)
\(=2x-6-4x+1=4x-5\)