tính : 4\25 x 10\3
giải chi tiết hộ mình nhé
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a) 125 x 25 x 64
= ( 25 x 64 ) x 125
= 1600 x 125
= 200 000
b) 125 x 25 x 12 x 14
= ( 125 x 12 ) x ( 25 x 14 )
= 1500 x 350
= 525000
c) 1300 : 50 = 26
d) 700 : 25 = 28
a, = 125 x (25x64)=125 x 1600= 200 000
b, = (125 x 24) x ( 25 x 12) = 3000 x 300 = 900 000
c, 1300 : 50 = 26
d, 700 : 25 = 28.
1035 + 207 x 4 x 28 - 69 x 3 x 17
1035 + 207 x 112 - 207 x 17
1035 + 207 x(112-17)
207 x 5 + 207 x 95
207 x (5+95)
a)
`9/2-3`
`=9/2-6/2`
`=3/2`
b)
`8/5xx25/12`
`=10/3`
c)
`4:3/7`
`=4xx7/3`
`=28/3`
x x2 -x x 3 = 4
= ( 4 x 3 ) - ( 4 x 2 )
= 12 - 8
= 4
k mình nhé
a) \(x=-\dfrac{3}{5}\times\dfrac{9}{7}=-\dfrac{27}{35}\)
b) \(x\left(0,4-\dfrac{1}{5}\right)=\dfrac{3}{4}\)
\(x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
a, \(x=-3,5.\dfrac{9}{7}=-\dfrac{9}{2}\)
b, \(\dfrac{2}{5}x-\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow\dfrac{1}{5}x=\dfrac{3}{4}\Leftrightarrow x=\dfrac{3}{4}:\dfrac{1}{5}=\dfrac{15}{4}\)
1-1/3=2/3; 1-1/4=3/4; 1-1/5=4/5....; 1-1/99=98/99
=> A= (2.3.4.5....98):(3.4.5....99)=2/99
Đs: 2/99
\(\sqrt{13+\sqrt{48}}=\sqrt{13+\sqrt{4.12}}=\sqrt{13+2\sqrt{12}}=\sqrt{\left(\sqrt{12}+1\right)^2}\)
\(=\sqrt{12}+1=2\sqrt{3}+1\)
\(\Rightarrow\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}=\sqrt{\left(\sqrt{3}-1\right)^2}\)
\(=\sqrt{3}-1\)
\(\Rightarrow\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{3+\sqrt{3}-1}=\sqrt{2+\sqrt{3}}\)
\(\Rightarrow\sqrt{\dfrac{4+2\sqrt{3}}{2}}=\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}=\dfrac{\sqrt{3}+1}{\sqrt{2}}\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}==2.\dfrac{\sqrt{3}+1}{\sqrt{2}}=\sqrt{6}+\sqrt{2}\)
2) biến đổi khúc sau như câu 1:
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
1) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{5-\sqrt{13+\sqrt{4.12}}}=\sqrt{5-\sqrt{13+2\sqrt{12}}}\)
\(=\sqrt{5-\sqrt{\left(\sqrt{12}\right)^2+2.\sqrt{12}+1^2}}=\sqrt{5-\sqrt{\left(\sqrt{12}+1\right)^2}}=\sqrt{5-\left|\sqrt{4.3}+1\right|}\)
\(=\sqrt{5-\left(2\sqrt{3}+1\right)}=\sqrt{5-2\sqrt{3}-1}=\sqrt{4-2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2-2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}-1\right)^2}=\left|\sqrt{3}-1\right|=\sqrt{3}-1\)
\(\Rightarrow2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}=2\sqrt{3+\sqrt{3}-1}=2\sqrt{2+\sqrt{3}}\)
\(=2\sqrt{\dfrac{4+2\sqrt{3}}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}{2}}=2\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{2}}\)
\(=2.\dfrac{\left|\sqrt{3}+1\right|}{\sqrt{2}}=\sqrt{2}\left(\sqrt{3}+1\right)=\sqrt{6}+\sqrt{2}\)
2) Ta có: \(\sqrt{5-\sqrt{13+\sqrt{48}}}=\sqrt{3}-1\) (như trên)
\(\Rightarrow\sqrt{6+2\sqrt{5-\sqrt{13+\sqrt{48}}}}=\sqrt{6+2\left(\sqrt{3}-1\right)}=\sqrt{4+2\sqrt{3}}\)
\(=\sqrt{\left(\sqrt{3}\right)^2+2.\sqrt{3}.1+1^2}=\sqrt{\left(\sqrt{3}+1\right)^2}=\left|\sqrt{3}+1\right|=\sqrt{3}+1\)
a, \(-7|x+3|=-49\)
\(< =>|x+3|=-\frac{49}{-7}=7\)
\(< =>\orbr{\begin{cases}x+3=7\\x+3=-7\end{cases}}\)
\(< =>\orbr{\begin{cases}x=4\\x=-10\end{cases}}\)
b, \(10-2x=25-3x\)
\(< =>-2x+3x=25-10\)
\(< =>x=15\)
a, \(-7\left|x+3\right|=-49\Leftrightarrow\left|x+3\right|=7\)
TH1 : \(x+3=7\Leftrightarrow x=4\)
TH2 : \(x+3=-7\Leftrightarrow x=-10\)
b, \(10-2x=25-3x\Leftrightarrow x=15\)
4\25 x 10\3
= 4 x 10 / 25 x 3
= 40/75
= 40:5 / 75:5
= 8/15
4/25 x 10/3 = 4x10 / 25x3 = 40/75
rút gọn bạn tự làm nhé