b; \(\dfrac{2}{3}\) : \(\dfrac{1}{2}\) - \(\dfrac{1}{4.8}\)

=  \(\dfrac{2}{3}\) x \(\dfrac{2}{1}\)  - \(\dfrac{1}{32}\)

= \(\dfrac{4}{3}\) - \(\dfrac{1}{32}\)

= \(\dfrac{125}{96}\)

\(\left(-7+86,17:12,31\right)\left(\dfrac{2023}{2021}-\dfrac{1}{2022}\right)\)

\(=\left(-7+7\right)\cdot\left(\dfrac{2023}{2021}-\dfrac{1}{2022}\right)\)

=0

Lời giải:

a.

$1\frac{2}{5}x=(0,5)^2=0,25$

$1,4x=0,25$

$x=0,25:1,4=\frac{5}{28}$

b.

$2(2x+\frac{2}{3})-\frac{3}{4}=\frac{3}{12}:\frac{1}{2}$

$2(2x+\frac{2}{3})-\frac{3}{4}=\frac{1}{2}$

$2(2x+\frac{2}{3})=\frac{1}{2}+\frac{3}{4}=\frac{5}{4}$

$2x+\frac{2}{3}=\frac{5}{4}:2=\frac{5}{8}$

$2x=\frac{5}{8}-\frac{2}{3}=\frac{-1}{24}$

$x=\frac{-1}{24}:2=\frac{-1}{48}$

Bạn lưu ý lần sau gõ đề bằng công thức toán (nhấn vào biểu tượng $\sum$ góc trái khung soạn thảo) để mọi người đọc hiểu đề của bạn hơn nhé. 

I nằm giữa E và F

=>IE+IF=EF

=>IF+1=7

=>IF=6(cm)

M nằm giữa I và F

=>MI+MF=IF

=>\(\dfrac{1}{3}MF+MF=6\)

=>\(\dfrac{4}{3}MF=6\)

=>\(MF=6:\dfrac{4}{3}=4,5\left(cm\right)\)

Ta có: IM+MF=IF

=>IM+4,5=6

=>IM=1,5(cm)

Xét ΔOAB và ΔOCD có

\(\widehat{OAB}=\widehat{OCD}\)(AB//CD)

\(\widehat{AOB}=\widehat{COD}\)(hai góc đối đỉnh)

Do đó: ΔOAB~ΔOCD

=>\(\dfrac{S_{OAB}}{S_{OCD}}=\left(\dfrac{AB}{CD}\right)^2=\dfrac{1}{16}\)

=>\(S_{OCD}=16\cdot S_{OBA}\)

ta có: \(S_{OCD}-S_{OAB}=1995\)

=>\(16\cdot S_{OAB}-S_{OAB}=1995\)

=>\(15\cdot S_{OAB}=1995\)

=>\(S_{OAB}=1995:15=133\left(cm^2\right)\)

=>\(S_{OCD}=133+1995=2128\left(cm^2\right)\)

AB//CD

=>\(\dfrac{OA}{OC}=\dfrac{OB}{OD}=\dfrac{AB}{CD}=\dfrac{1}{4}\)

\(\dfrac{OA}{OC}=\dfrac{1}{4}\)

=>\(\dfrac{S_{BOA}}{S_{BOC}}=\dfrac{1}{4}\)

=>\(S_{BOC}=4\cdot S_{BOA}=4\cdot133=532\left(cm^2\right)\)

Vì OB/OD=1/4

nên \(\dfrac{S_{AOB}}{S_{AOD}}=\dfrac{1}{4}\)

=>\(S_{AOD}=532\left(cm^2\right)\)

\(S_{ABCD}=S_{ABO}+S_{BOC}+S_{COD}+S_{AOD}\)

\(=532+532+133+2128=3325\left(cm^2\right)\)

\(3\left(\dfrac{1}{2}x-1\right)=-\dfrac{3}{4}\)

=>\(\dfrac{1}{2}x-1=-\dfrac{3}{4}:3=-\dfrac{1}{4}\)
=>\(\dfrac{1}{2}x=-\dfrac{1}{4}+1=\dfrac{3}{4}\)

=>\(x=\dfrac{3}{4}\cdot2=\dfrac{3}{2}\)

a: \(3\cdot2,25-0,75=6,75-0,75=6\)

b: \(\left(-1,25\right)+3,5+1,25+36,5\)

\(=\left(-1,25+1,25\right)+\left(3,5+36,5\right)\)

=0+40

=40

mik camon^^

\(3x-x=20140+\left(-3\right)^2\\ \Rightarrow2x=20140+9\\ \Rightarrow2x=20149\\ \Rightarrow x=\dfrac{20149}{2}.\)

xin mng giúp tôi 

\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{2024\cdot2026}\right)\)

\(=\left(1+\dfrac{1}{\left(2-1\right)\left(2+1\right)}\right)\left(1+\dfrac{1}{\left(3-1\right)\left(3+1\right)}\right)\cdot...\cdot\left(1+\dfrac{1}{\left(2025-1\right)\left(2025+1\right)}\right)\)

\(=\left(1+\dfrac{1}{2^2-1}\right)\left(1+\dfrac{1}{3^2-1}\right)\cdot...\cdot\left(1+\dfrac{1}{2025^2-1}\right)\)

\(=\dfrac{2^2}{2^2-1}\cdot\dfrac{3^2}{3^2-1}\cdot...\cdot\dfrac{2025^2}{2025^2-1}\)

\(=\dfrac{2\cdot3\cdot...\cdot2025}{1\cdot2\cdot...\cdot2024}\cdot\dfrac{2\cdot3\cdot...\cdot2025}{3\cdot4\cdot...\cdot2026}\)

\(=\dfrac{2025}{1}\cdot\dfrac{2}{2026}=\dfrac{2025}{1013}\)

kO biết