34,8 + 5 : y = 35,6
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![](https://rs.olm.vn/images/avt/0.png?1311)
A = \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{23.24.25}\)
= \(\frac{1}{2}.\left(\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{23.24.25}\right)\)
= \(\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{23.24}-\frac{1}{24.25}\right)\)
= \(\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{24.25}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{600}\right)=\frac{1}{2}.\frac{299}{600}=\frac{299}{1200}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
A + B = a + b - 5 + ( -b - c + 1 )
= a + b - 5 - b - c + 1
= a - c - 4 (1)
C - D = b - c - 4 - ( b - a )
= b - c - 4 - b + a
= -c - 4 + a
= a - c - 4 (2)
Từ (1) và (2)
=> A + B = C - D
Vậy A + B = C - D
![](https://rs.olm.vn/images/avt/0.png?1311)
áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{c+a}{5}=\frac{b+c}{4}=\frac{a+b}{3}=\frac{c+b-b-c+a+b}{5-4+3}=\frac{2a}{4}=\frac{a}{4}\left(1\right)\)
Từ (1) có: \(\frac{b+c}{4}=\frac{a+b}{3}\Leftrightarrow3b+3c=4a+4b\Leftrightarrow b=3c-4a\left(2\right)\)
Thế 2 vào biểu thức M ta có: \(M=10a+3c-4a-7c+2017=6a-4c+2017\left(3\right)\)
Từ (1) có\(:\frac{c+a}{5}=\frac{a}{2}\Leftrightarrow2c+2a=5a\Leftrightarrow2c=3a\Leftrightarrow4c=6a\left(4\right)\)
Thế (4) vào (3) ta có: \(M=6a-6a+2017=2017\)
Vậy GT M = 2017
+ Ta có : \(\frac{a+b}{3}=\frac{b+c}{4}\Rightarrow4a+4b=3b+3c\)
\(\Rightarrow4a+b=3c\)
+ \(\frac{a+b}{3}=\frac{c+a}{5}\Rightarrow5a+5b=3c+3a\)
\(\Rightarrow2a+5b=3c\)
+ \(\frac{b+c}{4}=\frac{c+a}{5}\Rightarrow5b+5c=4c+4a\)
\(\Rightarrow5b+c=4a\)
+ Ta có : \(\hept{\begin{cases}4a+b=3c\\5b+3a=3c\end{cases}\Rightarrow4a+b=5b+2a}\)
\(\Rightarrow2a=4b\)
\(\Rightarrow a=2b\)
+ Ta có : \(4a+b=3c\)
\(\Rightarrow4.2b+b=3c\)
\(9b=3c\)
\(\Rightarrow3b=c\)
+ Ta có : \(M=10a+b-7c+2017\)
\(=10.2b+b-7.3b+2017\)
\(=20b+b-7.3b+2017\)
\(=21b-21b+2017\)
\(=0+2017=2017\)
Vậy M =2017
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cauchy - Schwarz ta có :
\(VT=\frac{x}{\sqrt[3]{yz}}+\frac{y}{\sqrt[3]{xz}}+\frac{z}{\sqrt[3]{xy}}=\frac{x^2}{\sqrt[3]{x^3yz}}+\frac{y^2}{\sqrt[3]{y^3xz}}+\frac{z^2}{\sqrt[3]{z^3xy}}\)
\(\ge\frac{\left(x+y+z\right)^2}{\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}}\left(1\right)\)
Áp dụng BĐT : AM - GM :
\(\sqrt[3]{x^3yz}\le\frac{x^2+xyz+1}{3};\sqrt[3]{y^3xz}\le\frac{y^2+xyz+1}{3};\sqrt[3]{z^3xy}\le\frac{z^2+xyz+1}{3}\)
\(\Rightarrow\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}\le\frac{x^2+y^2+z^2+3xyz+3}{3}=2+xyz\)
Theo BĐT AM - GM :
\(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\Leftrightarrow3\sqrt[3]{x^2y^2z^2}\le3\Leftrightarrow xyz\le1\)
Do đó : \(\sqrt[3]{x^3yz}+\sqrt[3]{y^3xz}+\sqrt[3]{z^3xy}\le3\left(2\right)\)
Tư (1) , (2) và sử dụng hệ quả :
\(x^2+y^2+z^2\ge xy+yz+zx:\)
\(\Rightarrow VT\ge\frac{\left(x+y+z\right)^2}{3}=\frac{x^2+y^2+z^2+2\left(xy+yz+xz\right)}{3}\ge\frac{3\left(xy+yz+xz\right)}{3}\)\(=xy+yz+xz\)
Ta có đpcm
Dấu " = " xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(A=\frac{1}{101^2}+\frac{1}{102^2}+......\frac{1}{105^2};\frac{1}{2^2.3.5^2.7}\)
\(A>\frac{1}{\left(101.101\right)}+\frac{1}{\left(101.102\right)}+\frac{1}{\left(102.103\right)}+......\frac{1}{\left(104.105\right)}\)
Ta thấy mỗi mẫu đều < thì => sẽ lớn hơn
\(A>\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+\frac{1}{102}-\frac{1}{103}+........\)
\(A>\frac{1}{100}-\frac{1}{105}=\frac{1}{2100}=\frac{1}{\left(2^2.3.5^2.7\right)}=B\)
=> gọi vế \(\frac{1}{\left(2^2.2.5^2.7\right)}\) là B
=> A>B
\(\text{Ta có :}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100.101}+\frac{1}{101.102}+.....+\frac{1}{105.106}\)
\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{101}+\frac{1}{101}-\frac{1}{102}+....+\frac{1}{105}-\frac{1}{106}\)\
\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{100}-\frac{1}{105}\)
\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2100}\)
\(\text{Mà :}\)\(\frac{1}{2100}=\frac{1}{2^2.3.5^2.7}\)
\(\text{Nên:}\)\(A=\frac{1}{101^2}+\frac{1}{102^2}+....+\frac{1}{105^2}< \)\(\frac{1}{2^2.3.5^2.7}\)
34,8 + 5 : y = 35,6
5 : y = 35,6 - 34,8
5 : y = 0,4
y = 5 : 0,4
y = 12,5
34,8 + 5 : y = 35,6
5 : y = 35,6 - 34,8
5 : y = 0,8
y = 5 : 0,8
y = 6,25