Độn dài đường chéo lớn :

\(2\dfrac{3}{5}x2=\dfrac{13}{5}x2=\dfrac{26}{5}\left(dm\right)\)

Diện tích tờ giấy :

\(\dfrac{1}{2}x\dfrac{13}{5}x\dfrac{26}{5}=\dfrac{169}{25}\left(dm^2\right)\)

Bài giải

Đổi: \(2\dfrac{3}{5}dm=\dfrac{13}{5}dm\) 

Độ dài đường chéo lớn là:

\(\dfrac{13}{5}\times2=\dfrac{26}{5}\left(dm\right)\)

Diện tích tờ giấy là:

\(\dfrac{13}{5}\times\dfrac{26}{5}:2=\dfrac{169}{25}\left(dm^2\right)\)

Đ/s: \(\dfrac{169}{25}dm^2\)

\(\left(x-12\right)\times4=18\\ x-12=18:4\\ x-12=4,5\\ x=4,5+12\\ x=16,5\)

=>x-12=18:4=9/2

=>x=9/2+12=33/2

\(...\dfrac{9}{x}+\dfrac{3}{x}=6\Rightarrow\dfrac{1}{x}\left(9+3\right)=6\Rightarrow\dfrac{12}{x}=6\Rightarrow x=12:6=2\)

\(9:x+3:x=6\\ 9\times\dfrac{1}{x}+3\times\dfrac{1}{x}=6\\ \left(9+3\right)\times\dfrac{1}{x}=6\\ 12\times\dfrac{1}{x}=6\\ \dfrac{1}{x}=6:12\\ \dfrac{1}{x}=\dfrac{1}{2}\\ x=2\)

Bài 7: Tính

a) \(4\dfrac{2}{5}\times8\dfrac{3}{4}-2\dfrac{3}{4}\)

\(=\dfrac{22}{5}\times\dfrac{35}{4}-\dfrac{11}{4}\)

\(=\dfrac{77}{2}-\dfrac{11}{4}\)

\(=\dfrac{143}{4}\)

b) \(2\dfrac{2}{3}+1\dfrac{2}{5}-\dfrac{2}{15}\)

\(=\dfrac{8}{3}+\dfrac{7}{5}-\dfrac{2}{15}\)

\(=\dfrac{47}{15}-\dfrac{2}{15}\)

\(=\dfrac{45}{15}=3\)

c) \(3\dfrac{1}{3}-2\dfrac{2}{3}+1\dfrac{5}{6}\)

\(=\dfrac{10}{3}-\dfrac{8}{3}+\dfrac{11}{6}\)

\(=\dfrac{2}{3}+\dfrac{11}{6}\)

\(=\dfrac{15}{6}\)

Hiển nhiên \(P=4^{2010}+2^{2014}⋮2\). Ta chỉ cần chứng minh \(P⋮5\) là xong.

Trước hết ta chứng minh \(A=4^{2n}-1⋮5\), với mọi \(n\inℕ\)     (*)

 Với \(n=0\) thì \(A=0⋮5\). Với \(n=1\) thì \(A=15⋮5\).

 Giả sử (*) đúng đến \(n=k\). Với \(n=k+1\), ta có:

 \(A=4^{2\left(k+1\right)}-1\) \(=16.4^{2k}-1\) \(=16\left(4^{2k}-1\right)+15⋮5\), vậy (*) được chứng minh. Do đó \(4^{2010}-1⋮5\)              (1)

 Bây giờ ta sẽ chứng minh \(B=2^{4n+2}+1⋮5\) với mọi \(n\inℕ\).     (**)

 Với \(n=0\) thì \(B=5⋮5\). Với \(n=1\) thì \(B=65⋮5\).

 Giả sử (**) đúng đến \(n=k\). Với \(n=k+1\)  thì

 \(B=2^{4\left(k+1\right)+2}+1\) \(=16.2^{4k+2}+1\) \(=16\left(2^{4k+2}+1\right)-15⋮5\)

 Vậy (**) được chứng minh. Do đó \(2^{2014}+1⋮5\)         (2)

 Từ (1) và (2), suy ra \(P=4^{2010}+2^{2014}=\left(4^{2010}-1\right)+\left(2^{2014}+1\right)⋮5\)

 Như vậy \(2|P,5|P\Rightarrow10|P\) (đpcm)

` a/` 

` 3 1/5 : 2 1/3 : y = 12/7 `

` 48/35 : y = 12/7 `

` y = 48/35 : 12/7 `

` y = 48/35 xx 7/12 `

` y = 4/5`

Vậy ` y = 4/5` 

`b/`

` 3 : y xx 3 1/2 = 2/3 xx 3/4 `

` 3 : y xx 7/2 =  1/2`

` 3 : y = 1/2 : 7/2 `

` 3 : y = 1/2 xx 2/7 `

` 3 : y = 1/7 `

`      y = 3 : 1/7 `

`      y = 3 xx 7`

`      y = 21 `

 Vậy ` y = 21 `

`c/` 

` 3 2/3 - y + 1 3/4 = 2`

` 11/3 - y + 7/4 = 2 `

`   11/3 -y  = 2 - 7/4`

`    11/3 - y = 1/4 `

`        y = 11/3 - 1/4 `

`        y = 41/12  `

Vậy ` y = 41/12`

\(A=\left(2x-1\right)^4+3\)

mà \(\left(2x-1\right)^4\ge0,\forall x\)

\(\Rightarrow A=\left(2x-1\right)^4+3\ge0+3=3\)

\(\Rightarrow GTNN\left(A\right)=3\left(x=\dfrac{1}{2}\right)\)

\(B=-\left(8x-\dfrac{4}{5}\right)^6+1\)

mà \(-\left(8x-\dfrac{4}{5}\right)^6\le0,\forall x\)

\(\Rightarrow B=-\left(8x-\dfrac{4}{5}\right)^6+1\le0+1=1\)

\(\Rightarrow GTLN\left(B\right)=1\left(x=\dfrac{1}{10}\right)\)

S \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+...+\dfrac{1}{9.10}\)

\(S=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)

\(S=1-\dfrac{1}{10}\)

\(S=\dfrac{9}{10}\)

N= \(\dfrac{1}{9.11}+\dfrac{1}{11.13}+\dfrac{1}{13.15}+\dfrac{1}{15.16}+...+\dfrac{1}{43.47}\)

N= \(\dfrac{1}{2}.\left(\dfrac{2}{9.11}+\dfrac{2}{11.13}+\dfrac{2}{13.15}+...+\dfrac{2}{43.45}\right)\)

N= \(\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{43}-\dfrac{1}{45}\right)\)

N= \(\dfrac{1}{2}.\left(\dfrac{1}{9}-\dfrac{1}{45}\right)\)

N=\(\dfrac{1}{2}\) . \(\dfrac{4}{45}\)

N= \(\dfrac{2}{45}\)

\(\dfrac{3}{25}=\dfrac{3\times4}{25\times4}=\dfrac{12}{100}\\ \dfrac{7}{8}=\dfrac{7\times125}{8\times125}=\dfrac{875}{1000}\\ \dfrac{9}{45}=\dfrac{9:9}{45:9}=\dfrac{1}{5}=\dfrac{1\times2}{5\times2}=\dfrac{2}{10}\)

a) 12/100

b) 875/1000

c) 18/100

a,A,C,D

b,C

a) Câu A;D

b) Câu C