Bài 1 : Tìm y
y x 5 + 5 = 50
y : 5 - 5 = 90
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\(y.3\dfrac{7}{12}=6\dfrac{1}{4}\)
\(y.\dfrac{43}{12}=\dfrac{25}{4}\)
\(y=\dfrac{25}{4}:\dfrac{43}{12}\)
\(y=\dfrac{25.12}{4.43}\)
\(y=\dfrac{75}{43}\)
YYY +YY+ x + x + x + 1 = 1001
3x= 1000-yyy-yy
x= (1000 - yyy - yy) :3
YY + x + 5 = 113
x= 133-5 - yy
x= 128-yy
4725 + YYY + YY + x = 5463
x= 5463- 4725 - yyy - yy
x= 738 - yyy - yy
Bài 1:
\(101\cdot125+101\cdot25-101\cdot50\)
\(=101\cdot\left(125+25-50\right)\)
\(=101\cdot100\)
\(=10100\)
Bài 2:
\(76\cdot115+56\cdot24+59\cdot24\)
\(=76\cdot115+24\cdot\left(56+59\right)\)
\(=76\cdot115+24\cdot115\)
\(=115\cdot\left(76+24\right)\)
\(=115\cdot100\)
\(=11500\)
\(2x^2+30xy=5\left(x+5y\right)\sqrt{5xy}-50y^2\)\(\left(đk:x;y\ge0\right)\)
\(\Leftrightarrow2x^2+30xy-5\left(x+5y\right)\sqrt{5xy}+50y^2=0\left(1\right)\)
\(đặt:\sqrt{5xy}=b\ge0\Rightarrow5xy=b^2\Rightarrow10xy=2b^2\)
\(x+5y=a\ge0\Rightarrow x^2+10xy+25y^2=â^2\)
\(\Rightarrow2a^2=2x^2+20xy+50y^2\)
\(\Leftrightarrow\left(1\right)\Leftrightarrow2a^2+2b^2-5ab=0\Leftrightarrow\left(2a-b\right)\left(a-2b\right)=0\Leftrightarrow\left[{}\begin{matrix}b=2a\left(2\right)\\a=2b\left(3\right)\end{matrix}\right.\)
\(\left(2\right)\Rightarrow\sqrt{5xy}=2x+10y\Leftrightarrow4x^2+35xy+100y^2=0\left(4\right)\)
\(với:y=0\) \(ko\) \(là\) \(nghiệm\)
\(với:y\ne0\Rightarrow\left(4\right)\Leftrightarrow4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100=0\)\(\left(vô-lí\right)\)
\(do:4\left(\dfrac{x}{y}\right)^2+35\left(\dfrac{x}{y}\right)+100>0\)
\(\left(3\right)\Rightarrow x+5y=2\sqrt{5xy}\Leftrightarrow x^2+10xy+25y^2=20xy\Leftrightarrow x^2-10xy+25y^2=0\Leftrightarrow\left(x-5y\right)^2=0\Leftrightarrow x=5y\)
\(thay:x=5y\) \(vào:2x^2+y^2=51\Rightarrow2\left(5y\right)^2+y^2-51=0\Leftrightarrow51y^2-51=0\Leftrightarrow\left[{}\begin{matrix}y=1\left(tm\right)\Rightarrow x=5\left(tm\right)\\y=-1\left(loại\right)\end{matrix}\right.\)
x + ( 9 - 8 + 7 - 6 + 5 - 4 + 3 ) = 90,28
x + 6 = 90,28
x = 90,28 - 6
x = 84,28
Bài 1:
Theo đề ra ta có:
$a-2\vdots 3; a-3\vdots 5$
$a-2-2.3\vdots 3; a-3-5\vdots 5$
$\Rightarrow a-8\vdots 3; a-8\vdots 5$
$\Rightarrow a-8=BC(3,5)$
$\Rightarrow a-8\vdots 15$
$\Rightarrow a=15k+8$ với $k$ tự nhiên.
Mà $a$ chia 11 dư 6
$\Rightarrow a-6\vdots 11$
$\Rightarrow 15k+8-6\vdots 11$
$\Rightarrow 15k+2\vdots 11\Rightarrow 4k+2\vdots 11$
$\Rightarrow 4k+2-22\vdots 11\Rightarrow 4k-20\vdots 11$
$\Rightarrow 4(k-5)\vdots 11\Rightarrow k-5\vdots 11$
$\Rightarrow k=11m+5$
Vậy $a=15k+8=15(11m+5)+8=165m+83$ với $m$ tự nhiên.
Vì $a<500\Rightarrow 165m+83<500\Rightarrow m< 2,52$
$\Rightarrow m=0,1,2$
Nếu $m=0$ thì $a=165.0+83=83$
Nếu $m=1$ thì $a=165.1+83=248$
Nếu $m=2$ thì $a=165.2+83=413$
Bài 2:
$a=BC(60,85,90)$
$\Rightarrow a\vdots BCNN(60,85,90)$
$\Rightarrow a\vdots 3060$
Mà $a<1000$ nên $a=0$
y x 5 + 5 = 50
y = 50 - 5 : 5
y = 9