√25 + 10√3
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a: =25(15+45*3)
=25*150
=3750
b: \(=-10\left(25+75-50\right)=-10\cdot50=-500\)
c: =>3^x-2=27
=>x-2=3
=>x=5
d: =>2x-5=-4
=>2x=1
=>x=1/2
e: =>2(x-1)^2=32
=>(x-1)^2=16
=>x-1=4 hoặc x-1=-4
=>x=-3 hoặc x=5
f: =>25(x+3)=75
=>x+3=3
=>x=0
\(=\dfrac{25}{9}.\dfrac{3}{10}+\dfrac{25}{9}.\dfrac{7}{10}+\dfrac{25}{3}=\dfrac{25}{9}\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\dfrac{25}{3}=\dfrac{25}{9}+\dfrac{25}{3}=\dfrac{100}{9}\)
\(A=\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}=1+\dfrac{2}{25^{10}-1}\left(1\right)\)
\(B=\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\left(2\right)\)
Từ \(\left(1\right)+\left(2\right)\Leftrightarrow A< B\)
Ta có : \(\dfrac{25^{10}+1}{25^{10}-1}=\dfrac{25^{10}-1+2}{25^{10}-1}=\dfrac{25^{10}-1}{25^{10}-1}+\dfrac{2}{25^{10}-1}\)\(=1+\dfrac{2}{25^{10}-1}\)
Ta có : \(\dfrac{25^{10}-1}{25^{10}-3}=\dfrac{25^{10}-3+2}{25^{10}-3}=\dfrac{25^{10}-3}{25^{10}-3}+\dfrac{2}{25^{10}-3}=1+\dfrac{2}{25^{10}-3}\)
Vì \(25^{10}-1>25^{10}-3\)
\(\Rightarrow\dfrac{2}{25^{10}-1}< \dfrac{2}{25^{10}-3}\)
\(\Rightarrow1+\dfrac{2}{25^{10}-1}< 1+\dfrac{2}{25^{10}-3}\)
\(\Rightarrow A< B\)
Vậy \(A< B\)
1/2×(2×x-3)+105/2=-137/2
1/2×(2×x-3)=-137/2-105/2
1/2×(2×x-3)=-242/2
2×x-3=-242/2:1/2
2×x-3=-242
2.x=(-242)+3
2.x=239
x=239:2
x=239/2
A,-12×25-25×75-25×13
=[(-12)-75-13].25
=(-100).25
-2500
B,-50/7×49/10-35/2×-10/7+ -25/3× -9/5
=(-35)-(-25)+(-13)
=-23
C,-354+265-156+125
=[(-354)-156]+(265+125)
=(-510)+277
=-233
D,-74+40-50+16-35
=(-74)+40+(-50)+16+(-35)
=[(-74)+(-35)]+[40+(-50)+16]
=(-109)+26
=-83
E,-2/3×4/5+-4/5×4/3-0,125
=-2/3.4/5+4/5.(-4/3)-0,125
=4/5.[-2/3+(-4/3)]-0,125
=4/5.(-2)-0,125
=-8/5-0,125
=(-1,6)+(-0,125)
=-1,725
-23/25.10/13+-23/25.3/13+-27/25
= -23/25(10/13+3/13)+-27/25
= -23/25.1+-27/25
= -23/25+-27/25
= -2
-23/25.10/13+-23/25.3/13+27/25
= -23/25.(10/13+3/10)+ -27/25
=23/25.1+-27/25
=23/25+ -27/25
=-2
\(\sqrt{25\left(x-3\right)}-10\sqrt{\dfrac{x-3}{25}}-1=3+\sqrt{x-3}\left(đk:x\ge3\right)\)
\(\Leftrightarrow5\sqrt{x-3}-10.\dfrac{1}{5}\sqrt{x-3}-1=3+\sqrt{x-3}\)
\(\Leftrightarrow2\sqrt{x-3}=4\Leftrightarrow\sqrt{x-3}=2\)
\(\Leftrightarrow x-3=4\Leftrightarrow x=7\)
Ta có: \(\sqrt{25\left(x-3\right)}-10\sqrt{\dfrac{x-3}{25}}-1=3+\sqrt{x-3}\)
\(\Leftrightarrow5\sqrt{x-3}-2\sqrt{x-3}-\sqrt{x-3}=4\)
\(\Leftrightarrow2\sqrt{x-3}=4\)
\(\Leftrightarrow x-3=4\)
hay x=7
Nhầm, mình sửa lại nha
( 103 - 102 + 252 ) : 53
= ( 1000 - 100 + 625 ) : 125
= ( 900 + 625 ) : 125
= 1525 : 125
= 12,2
Ta có:
Tập hợp A:
\(A=\left\{1;5;9;13;17;21;25\right\}\)
Tập hợp B:
\(B=\left\{0;1;3;5;10;13\right\}\)
Mà: \(A\cap B\)
\(\Rightarrow A\cap B=\left\{1;5;13\right\}\)
⇒ Chọn B
`@` `\text {Ans}`
`\downarrow`
`\sqrt {25} + 10 \sqrt {3}`
`= 5 + 10 \sqrt {3}`
\(\sqrt{25}+10\sqrt{3}=5+10\sqrt{3}\)