Exer 1: Given two natural numbers whose sum are 78293. The bigger number where 5 is the units digit and 2 is hundred digit. If we clean these digits then we obtain a number which equals the smaller number. Find two natural numbers.
Exer 2: Prove that: If x, y \(\in\) N and x + 2y divisible by 5 then 3x - 4y divisibles by 5.
Exer 3: Given that 2x + 5y \(⋮\) 7. Prove that 4x + 3y \(⋮\) 7.
Exer 1:
Solution:
Suppose that, the unknown number is: \(\overline{x215}\) (where x \(\in\) N).
When we clean three digits then the smaller number is \(\overline{x}\).
We have: \(\overline{x215}\) + \(\overline{x}\) = 78293
\(\Rightarrow\) 1000. \(\overline{x}\) + 215 + \(\overline{x}\) = 78293
1001. \(\overline{x}\) = 78078
x = 78
Thus, we found two natural number: 78215 and 78.
Exer 2:
Solution:
We have: x + 2y \(⋮\) 5
\(\Rightarrow\) 2x + 4y \(⋮\) 5
(2x + 4y) + (3x - 4y) = 5x \(⋮\) 5
\(\Rightarrow\) 2x + 4y \(⋮\) 5
Deduce 3x - 4y \(⋮\) 5.
Exer 3:
Solution:
We have: 2x + 5y \(⋮\) 7
4x + 10y \(⋮\) 7
(4x + 10y) - (4x + 3y) = 7y \(⋮\) 7
\(\Rightarrow\) 4x + 10y \(⋮\) 7
Deduce 4x + 3y \(⋮\) 7.
Given 3 number such that : x + y + z =3
Find the minimum P = xy + yz + zx
Hihi kb vs Linh nah m. n :3
<3 Girl 2k5 - FA
We have:
\(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2yx-2zx\ge0\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yx-zx\ge0\)
\(\Leftrightarrow\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\Leftrightarrow3\ge xy+yz+zx\)
"=" happen only and only x=y=z=1
find a 2 - digit natural number such that it is equal to two times product of two digits
In the figure below, x = Câu 2 Find the measure of angle J if \widehat{K}=29^o K =29 o and angle J and K are supplementary. Answer: \widehat{J}= J = ^o o . Câu 3 In the figure below, x = Câu 4 In the figure below, x = Câu 5 Solve: |2x-3|-4=3∣2x−3∣−4=3, where xx is a negative number. Answer: x=x= Câu 6 Calculate: 9+|-6|:1^2 =9+∣−6∣:1 2 = Câu 7 Compare: \dfrac{36}{27} 27 36 \dfrac{4}{3} 3 4 Câu 8 Find the negative number yy such that |y+\dfrac{7}{2}|-\dfrac{1}{2}=4∣y+ 2 7 ∣− 2 1 =4. Answer: y=y= Câu 9 Find xx such that \dfrac{32}{2^x}=2 2 x 32 =2. Answer: x=x= Câu 10 Find nn such that \dfrac{(-4)^n}{64}=-16 64 (−4) n =−16. Answer: n=n= Cần gấp nhé
Find all natural numbers having two digit , knowing that twice the units digit 1 more than the ten digit , and if we write that two digits in reverse order, we get a new number which is 27 less than the old one
Tell the numbers to find are \(\overline{ab}\)
\(\Rightarrow\left\{{}\begin{matrix}2b-a=1\\\overline{ab}-\overline{ba}=27\end{matrix}\right.\)
=> 10a + b - 10b - a = 27
=> 9a - 9b = 27
=> a-b = 3
=> a = b + 3
=> 2b - 1 = b+3
=> 2b - b = 3 + 1
=> b = 4
=> a = \(2\cdot4-1=7\)
So he number to find is 74
Let x, y, z be positive real numbers such that xy + yz + zx + xyz = 4 . Prove that :
\(3\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\ge\left(x+2\right)\left(y+2\right)\left(z+2\right)\)
Đặt \(x=\frac{2a}{b+c};y=\frac{2b}{c+a};z=\frac{2c}{a+b}\) Thì bài toán thành chứng minh
\(3\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
Áp dụng holder ta có:
\(\left(\sqrt{\frac{a+b}{2c}}+\sqrt{\frac{b+c}{2a}}+\sqrt{\frac{c+a}{2b}}\right)^2\left(2c\left(a+b\right)^2+2a\left(b+c\right)^2+2b\left(c+a\right)^2\right)\)
\(\ge\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]^3=8\left(a+b+c\right)^3\)
\(\Rightarrow VT\ge3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\)
Từ đây ta cần chứng minh:
\(3.\frac{8\left(a+b+c\right)^3}{2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2}\ge\frac{8\left(a+b+c\right)^3}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
\(\Leftrightarrow2a\left(b+c\right)^2+2b\left(c+a\right)^2+2c\left(a+b\right)^2\le3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(\Leftrightarrow a\left(b-c\right)^2+b\left(c-a\right)^2+c\left(a-b\right)^2\ge0\)( đúng )
Vậy có ĐPCM
Given that A=5^2+1 0n+601 where n is a natural number . Find the minmum sum of the digits of A.
TOÁN SONG NGỮ NHA! GIÚP MÌNH ĐI
THANKS
cho x,y,z là số thực dương thỏa mãn xy+yz+xz=xyz
cmr \(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}+\dfrac{xz}{y^3\left(1+x\right)\left(1+z\right)}\ge\dfrac{1}{16}\)
Gọi cái thiệt gớm đó là P
Ta có:
\(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64y}+\dfrac{1+y}{64x}\ge3\sqrt[3]{\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}.\dfrac{1+x}{64y}.\dfrac{1+y}{64x}}=\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{64x}-\dfrac{1}{64y}-\dfrac{1}{32}\left(1\right)\)
Tương tự ta cũng có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{64y}-\dfrac{1}{64z}-\dfrac{1}{32}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{64z}-\dfrac{1}{64x}-\dfrac{1}{32}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta được
\(P\ge\dfrac{3}{16}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
Đặt cái ban đầu là P
Ta có: \(xy+yz+zx=xyz\)
\(\Leftrightarrow\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\)
Ta lại có:
\(\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}+\dfrac{1+x}{64x}+\dfrac{1+y}{64y}\ge\dfrac{3}{16z}\)
\(\Leftrightarrow\dfrac{xy}{z^3\left(1+x\right)\left(1+y\right)}\ge\dfrac{3}{16z}-\dfrac{1}{32}-\dfrac{1}{64x}-\dfrac{1}{64y}\left(1\right)\)
Tương tự ta có:
\(\left\{{}\begin{matrix}\dfrac{yz}{x^3\left(1+y\right)\left(1+z\right)}\ge\dfrac{3}{16x}-\dfrac{1}{32}-\dfrac{1}{64y}-\dfrac{1}{64z}\left(2\right)\\\dfrac{zx}{y^3\left(1+z\right)\left(1+x\right)}\ge\dfrac{3}{16y}-\dfrac{1}{32}-\dfrac{1}{64z}-\dfrac{1}{64x}\left(3\right)\end{matrix}\right.\)
Từ (1), (2), (3) ta có:
\(P\ge\dfrac{3}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{1}{32}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)-\dfrac{3}{32}\)
\(=\dfrac{3}{16}-\dfrac{1}{32}-\dfrac{3}{32}=\dfrac{1}{16}\)
Dấu = xảy ra khi \(x=y=z=3\)
1.Giải hệ pt
1)\(\hept{\begin{cases}\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\\xy+yz+zx=3\\\frac{1}{1+x+xy}+\frac{1}{1+y+yz}+\frac{1}{1+z+zx}=x\end{cases}}\)
2)\(\hept{\begin{cases}xy+yz+zx=3\\\left(x+y\right)\left(y+z\right)=\sqrt{3}z\left(1+y^2\right)\\\left(y+z\right)\left(z+x\right)=\sqrt{3}x\left(1+z^2\right)\end{cases}}\)
3)\(\hept{\begin{cases}xy+yz+zx=3\\1+x^2\left(y+z\right)+xyz=4y\\1+y^2\left(z+x\right)+xyz=4z\end{cases}}\)
Hinh câu 1
