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a) \(\dfrac{1}{2}< \dfrac{x}{9}< \dfrac{y}{18}< \dfrac{2}{3}\)

\(\Leftrightarrow\dfrac{9}{18}< \dfrac{2x}{18}< \dfrac{y}{18}< \dfrac{12}{18}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=10\\y=11\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=5\\y=11\end{matrix}\right.\)

b) \(\dfrac{-1}{2}< \dfrac{x}{15}< \dfrac{y}{30}< \dfrac{-2}{5}\)

\(\Leftrightarrow\dfrac{-15}{30}< \dfrac{2x}{30}< \dfrac{y}{30}< \dfrac{-12}{30}\)

\(\Leftrightarrow\left\{{}\begin{matrix}2x=-14\\y=-13\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=-7\\y=-13\end{matrix}\right.\)

5 tháng 6 2023

\(3\dfrac{1}{2}+4\dfrac{5}{7}-5\dfrac{5}{14}\)

\(\dfrac{7}{2}+\dfrac{33}{7}-\dfrac{75}{14}\)

\(\dfrac{49}{14}+\dfrac{66}{14}-\dfrac{75}{14}\)

\(\dfrac{40}{14}=\dfrac{20}{7}\)

\(4\dfrac{1}{2}+\dfrac{1}{2}\div5\dfrac{1}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\div\dfrac{11}{2}\)

=\(\dfrac{9}{2}+\dfrac{1}{2}\times\dfrac{2}{11}\)

=\(\dfrac{9}{2}+\dfrac{1}{11}\)

=\(\dfrac{101}{22}\)

\(x\times3\dfrac{1}{3}=3\dfrac{1}{3}\div4\dfrac{1}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\div\dfrac{17}{4}\)

\(x\times\dfrac{10}{3}=\dfrac{10}{3}\times\dfrac{4}{17}\)

\(x\times\dfrac{10}{3}=\dfrac{40}{51}\)

\(x=\dfrac{40}{51}\div\dfrac{10}{3}\)

\(x=\dfrac{40}{51}\times\dfrac{3}{10}\)

\(x=\dfrac{120}{510}=\dfrac{12}{51}=\dfrac{4}{7}\)

\(5\dfrac{2}{3}\div x=3\dfrac{2}{3}-2\dfrac{1}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{11}{3}-\dfrac{5}{2}\)

\(\dfrac{17}{3}\div x=\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\div\dfrac{7}{6}\)

\(x=\dfrac{17}{3}\times\dfrac{6}{7}\)

\(x=\dfrac{102}{21}=\dfrac{34}{7}\)

7 tháng 5 2022

bài 1 :

\(a,\dfrac{2}{7}+\dfrac{1}{3}=\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

\(b,\dfrac{3}{5}-\dfrac{1}{3}=\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

\(c,\dfrac{13}{4}:5=\dfrac{13}{4}:\dfrac{5}{1}=\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

\(d,\dfrac{6}{23}x\dfrac{1}{18}=\dfrac{1}{69}\)

bài 2 :

\(a,x+\dfrac{1}{3}=\dfrac{5}{12}\)

   \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

  \(x=\dfrac{1}{12}\)

 

\(b,x:\dfrac{7}{4}=\dfrac{2}{5}\)

   \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

   \(x=\dfrac{7}{10}\)

bài 3 :

đổi : 2 dm 1cm = 21cm

chiều cao hình bình hành là;

       21 x\(\dfrac{3}{7}=\)9(cm)

diện tích hình bình hành là;

       21 x 9 =189 (cm2)

                 đáp số : 189 cm2

bài 4 :

\(\dfrac{2}{3}x\dfrac{2}{10}+\dfrac{2}{3}x\dfrac{5}{10}x\dfrac{3}{3}\)

\(\dfrac{2}{3}x\left(\dfrac{2}{10}+\dfrac{5}{10}\right)x\dfrac{2}{3}\)

=\(\dfrac{2}{3}x1x\dfrac{2}{3}\)

\(=\dfrac{2}{3}x\dfrac{2}{3}\)

=\(\dfrac{4}{9}\)

 

7 tháng 5 2022

Bài 1)

a) \(\dfrac{6}{21}+\dfrac{7}{21}=\dfrac{13}{21}\)

b) \(\dfrac{9}{15}-\dfrac{5}{15}=\dfrac{4}{15}\)

c) \(\dfrac{13}{4}x\dfrac{1}{5}=\dfrac{13}{20}\)

d) \(\dfrac{6}{414}=\dfrac{1}{69}\)

Bài 2)

a) \(x=\dfrac{5}{12}-\dfrac{1}{3}\)

\(x=\dfrac{1}{12}\)

b) \(x=\dfrac{2}{5}x\dfrac{7}{4}\)

\(x=\dfrac{7}{10}\)

Bài 3)

2dm 1cm = 21 cm

Chiều cao tấm bìa la

\(21x\dfrac{3}{7}=9\left(cm\right)\)

Diện tích tấm bìa là

\(21x9=189\left(cm2\right)\)

NV
27 tháng 12 2022

1.

Áp dụng BĐT Cauchy-Schwarz:

\(\dfrac{a}{2a+a+b+c}=\dfrac{a}{25}.\dfrac{\left(2+3\right)^2}{2a+a+b+c}\le\dfrac{a}{25}\left(\dfrac{2^2}{2a}+\dfrac{3^2}{a+b+c}\right)=\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{a}{a+b+c}\)

Tương tự:

\(\dfrac{b}{3b+a+c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{b}{a+b+c}\)

\(\dfrac{c}{a+b+3c}\le\dfrac{2}{25}+\dfrac{9}{25}.\dfrac{c}{a+b+c}\)

Cộng vế:

\(VT\le\dfrac{6}{25}+\dfrac{9}{25}.\dfrac{a+b+c}{a+b+c}=\dfrac{3}{5}\)

Dấu "=" xảy ra khi \(a=b=c\)

NV
27 tháng 12 2022

2.

Đặt \(\dfrac{x}{x-1}=a;\dfrac{y}{y-1}=b;\dfrac{z}{z-1}=c\)

Ta có: \(\dfrac{x}{x-1}=a\Rightarrow x=ax-a\Rightarrow a=x\left(a-1\right)\Rightarrow x=\dfrac{a}{a-1}\)

Tương tự ta có: \(y=\dfrac{b}{b-1}\) ; \(z=\dfrac{c}{c-1}\)

Biến đổi giả thiết:

\(xyz=1\Rightarrow\dfrac{abc}{\left(a-1\right)\left(b-1\right)\left(c-1\right)}=1\)

\(\Rightarrow abc=\left(a-1\right)\left(b-1\right)\left(c-1\right)\)

\(\Rightarrow ab+bc+ca=a+b+c-1\)

BĐT cần chứng minh trở thành:

\(a^2+b^2+c^2\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(ab+bc+ca\right)\ge1\)

\(\Leftrightarrow\left(a+b+c\right)^2-2\left(a+b+c-1\right)\ge1\)

\(\Leftrightarrow\left(a+b+c-1\right)^2\ge0\) (luôn đúng)

3 tháng 8 2023

a,a+1/4=2 3/4-1 1/2    

a+1/2=5/4

    a=5/4-1/2

     a=3/4

b,a-7/4=13/4-7/9

a-7/4=89/36

        a= 89/36+7/4

         a=152/36

c,3/2-a=17/6-1/6

3/2-a=8/3

       a= 3/2-8/3

       a= -7/6

16 tháng 6 2021

`A=(8 2/7-4 2/7)-3 4/9`

`=8+2/7-4-2/7-3-4/9`

`=4-3-4/9`

`=1-4/9=5/9`

`B=(10 2/9-6 2/9)+2 3/5`

`=10+2/9-6-2/9+2+3/5`

`=4+2+3/5`

`=6+3/5=33/5`

Bài 2:

`a)5 1/2*3 1/4`

`=11/2*13/4`

`=143/8`

`b)6 1/3:4 2/9`

`=19/3:38/9`

`=19/3*9/38=3/2`

`c)4 3/7*2`

`=31/7*2`

`=62/7`

Bài 1:

\(A=\left(8\dfrac{2}{7}-4\dfrac{2}{7}\right)-3\dfrac{4}{9}\) 

\(A=\left(\dfrac{58}{7}-\dfrac{30}{7}\right)-\dfrac{31}{9}\) 

\(A=4-\dfrac{31}{9}\) 

\(A=\dfrac{5}{9}\) 

 

\(B=\left(10\dfrac{2}{9}-6\dfrac{2}{9}\right)+2\dfrac{3}{5}\) 

\(B=\left(\dfrac{92}{9}-\dfrac{56}{9}\right)+\dfrac{13}{5}\) 

\(B=4+\dfrac{13}{5}\) 

\(B=\dfrac{33}{5}\)

7 tháng 9 2023

Bài 1:

a) \(3^7:3^5-\left(\dfrac{5}{17}\right)^0=3^{7-5}-1=3^2-1=9-1=8\)

b) \(\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{1}{2}+2\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{13}:\left(\dfrac{5}{2}\right)^3\)

\(=\left(\dfrac{5}{2}\right)^{10}\)

c) \(8.\left(\dfrac{1}{4}\right)^3+\left(\dfrac{2}{27}\right)^0-\dfrac{1}{8}\)

\(=8.\dfrac{1}{64}+1-\dfrac{1}{8}\)

\(=\dfrac{1}{8}+1-\dfrac{1}{8}\)

\(=1\)

Bài 2:

a) \(\dfrac{3^4.4^4}{6^4}=\dfrac{3^4.\left(2^2\right)^4}{\left(2.3\right)^4}=\dfrac{3^4.2^8}{2^4.3^4}=\dfrac{2^8}{2^4}=2^4=16\)

b) \(\dfrac{15^3}{10^3}=\dfrac{\left(3.5\right)^3}{ \left(2.5\right)^3}=\dfrac{3^3.5^3}{2^3.5^3}=3^3:2^3=\dfrac{27}{8}\)

c) \(\dfrac{4^2.12^5}{9^2.2^{10}}=\dfrac{\left(2^2\right)^2.\left[3.\left(2^2\right)\right]^5}{\left(3^2\right)^2.2^{10}}=\dfrac{2^4.3^5.2^{10}}{3^4.2^{10}}=2^4.3=16.3=48\)

d) \(\dfrac{6^2+5.2^2+4}{15}=\dfrac{\left(2.3\right)^2+5.2^2+2^2}{15}=\dfrac{2^2.3^2+5.2^2+2^2}{15}=\dfrac{2^2\left(3^2+5+1\right)}{15}=\dfrac{2^2.15}{15}=2^2=4\)

Bài 3:

a) \(\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.\left(-1\right)^5}{\left(\dfrac{2}{5}\right)^2.\left(\dfrac{-5}{12}\right)^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3.\left(\dfrac{-3}{4}\right)^2.-1}{\left[\dfrac{2}{5}.\left(\dfrac{-5}{12}\right)\right]^2}\)

\(=\dfrac{\left(\dfrac{2}{3}\right)^3. \left(\dfrac{-3}{4}\right)^2.-1}{\left(\dfrac{-1}{6}\right)^2}\)

\(=\left(\dfrac{2}{3}\right)^3.\left[\left(\dfrac{-3}{4}\right).-6\right]^2.-1\)

\(=\left(\dfrac{2}{3}\right)^3.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}\right)^2.\dfrac{2}{3}.\left(\dfrac{9}{2}\right)^2.-1\)

\(=\left(\dfrac{2}{3}.\dfrac{9}{2}\right)^2.\dfrac{2}{3}.-1\)

\(=9.\dfrac{2}{3}.-1\)

\(=6.-1=-6\)

b) \(\dfrac{6^6+6^3.3^3+3^6}{-73}=\dfrac{\left(2.3\right)^6+\left(2.3\right)^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^3.3^3+3^6}{-73}=\dfrac{2^6.3^6+2^3.3^6+3^6}{-73}=\dfrac{3^6\left(2^6+2^3+1\right)}{-73}=\dfrac{3^6.73}{-73}=\dfrac{3^6}{-1}=\left(-3\right)^6\)

\(#Wendy.Dang\)

7 tháng 9 2023

Lần sau bnn gửi từng bài thôi nha, chứ như vầy nhiều quá thì làm không nổi mất. đánh máy nãy giờ lú luôn gòi nè :))

23 tháng 12 2022

2.

\(P=\left(\dfrac{a+6}{3\left(a+3\right)}-\dfrac{1}{a+3}\right).\dfrac{27a}{a+2}=\left(\dfrac{a+3}{3\left(a+3\right)}\right).\dfrac{27a}{a+2}=\dfrac{27a}{3\left(a+2\right)}=\dfrac{9a}{a+2}\)

ĐKXĐ là :

\(a\ne0;-3;-2\)

Vs a = 1 ta có:

=> P=3

1.

\(M=\left(\dfrac{2a}{2a+b}-\dfrac{4a^2}{\left(2a+b\right)^2}\right):\left(\dfrac{2a}{\left(2a-b\right)\left(2a+b\right)}-\dfrac{1}{2a-b}\right)=\left(\dfrac{4a^2+2ab-4a^2}{\left(2a+b\right)^2}\right).\left(\dfrac{\left(2a+b\right)\left(2a-b\right)}{b}\right)=\dfrac{2a.\left(2a-b\right)}{\left(2a+b\right)}\)

Bài 1: B

Bài 2: A

5 tháng 4 2022

1: B

2: A