bài 1: Tính giá trị biểu thức:
1) H= [0,(32) . 1,(5) - 0,(25)] . \(\dfrac{11}{83}\)
2) A= \(\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
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M
22 tháng 2 2020
Trl:
Đặt giá trị biểu thức là A , ta có :
\(A=\frac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{\frac{5}{2}+\frac{5}{3}-\frac{5}{6}}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}-\frac{1}{6}}{5\left(\frac{1}{2}+\frac{1}{3}-\frac{1}{6}\right)}=\frac{1}{5}\)
Nguồn : mạng
10 tháng 12 2022
\(G=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{6}}{\dfrac{5}{2}+\dfrac{5}{3}-\dfrac{5}{6}}=\dfrac{1}{5}\)
\(H=\left[\dfrac{32}{99}\cdot\dfrac{14}{9}-\dfrac{25}{99}\right]\cdot\dfrac{11}{83}\)
\(=\dfrac{223}{891}\cdot\dfrac{11}{83}=\dfrac{223}{6723}\)
9 tháng 6 2021
a, ĐKXĐ: x≠±2
A=\(\left(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}\right)\left(x-2+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{x}{x^2-4}-\dfrac{2x+4}{x^2-4}+\dfrac{x-2}{x^2-4}\right)\left(\dfrac{x^2+2x}{x+2}-\dfrac{2x+4}{x+2}+\dfrac{10-x^2}{x+2}\right)\)
A=\(\left(\dfrac{-6}{x^2-4}\right)\left(\dfrac{6}{x+2}\right)\)
A=\(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\)
b, |x|=\(\dfrac{1}{2}\)
TH1z: x≥0 ⇔ x=\(\dfrac{1}{2}\) (TMĐKXĐ)
TH2: x<0 ⇔ x=\(\dfrac{-1}{2}\) (TMĐXĐ)
Thay \(\dfrac{1}{2}\), \(\dfrac{-1}{2}\) vào A ta có:
\(\dfrac{-36}{\left(\dfrac{1}{2}-2\right)\left(\dfrac{1}{2}+2\right)^2}\)=\(\dfrac{96}{25}\)
\(\dfrac{-36}{\left(\dfrac{-1}{2}-2\right)\left(\dfrac{-1}{2}+2\right)^2}\)=\(\dfrac{32}{5}\)
c, A<0 ⇔ \(\dfrac{-36}{\left(x-2\right)\left(x+2\right)^2}\) ⇔ (x-2)(x+2)2 < 0
⇔ {x-2>0 ⇔ {x>2
[ [
{x+2<0 {x<2
⇔ {x-2<0 ⇔ {x<2
[ [
{x+2>0 {x>2
⇔ x<2
Vậy x<2 (trừ -2)
22 tháng 5 2019
\(\frac{0,5+\frac{1}{3}-\frac{1}{6}}{2,5+\frac{5}{3}-\frac{5}{6}}\)
= \(\frac{\frac{3}{6}+\frac{2}{6}-\frac{1}{6}}{\frac{15}{6}+\frac{10}{6}-\frac{5}{6}}\)
= \(\frac{\frac{4}{6}}{\frac{20}{6}}\)
= \(\frac{4}{6}:\frac{20}{6}\)
= \(\frac{2}{3}.\frac{3}{10}\)
= \(\frac{1}{5}\)
18 tháng 9 2018
Đặt GTBT là A, ta có:
A=0,5+0,(3)−0,1(6)2,5+1,(6)−0,8(3)
A=12 +13 −16 52 +53 −56
A=12 +13 −16 5(12 +13 −16 ) =15
18 tháng 9 2018
bạn tự điền mấy cái dấu gach p/s nha
_____________________
chúc bạn học tốt
9 tháng 6 2021
a, ĐKXĐ: x≠±3
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{x^2+6x+9}{x^2-9}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x+3}.\dfrac{\left(x+3\right)^2}{\left(x+3\right)\left(x-3\right)}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{3-x}{x-3}+\dfrac{x}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{9-x^2}{x^2-9}+\dfrac{x^2-3x}{x^2-9}\right):\dfrac{3x^2}{x+3}\)
A=\(\left(\dfrac{-3}{x+3}\right):\dfrac{3x^2}{x+3}\)
A=\(\dfrac{-1}{x^2}\)
b, Thay x=\(-\dfrac{1}{2}\) (TMĐKXĐ) vào A ta có:
\(\dfrac{-1}{\left(-\dfrac{1}{2}\right)^2}\)=-4
c, A<0 ⇔ \(\dfrac{-1}{x^2}< 0\) ⇔ x2>0 (Đúng với mọi x)
Vậy để A<0 thì x đúng với mọi giá trị (trừ ±3)
\(H=\left[0,\left(32\right).1,\left(5\right)-0,\left(25\right)\right].\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{32}{99}.\dfrac{14}{9}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{25}{99}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\left(\dfrac{448}{891}-\dfrac{225}{891}\right).\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{448-225}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{223}{891}.\dfrac{11}{83}\)
\(\Leftrightarrow H=\dfrac{2453}{73953}\)
\(\Leftrightarrow H=\dfrac{223}{6723}\)
2) \(A=\dfrac{0,5+0,\left(3\right)-0,1\left(6\right)}{2,5+1,\left(6\right)-0,8\left(3\right)}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3}{6}+\dfrac{2}{6}-\dfrac{1}{6}}{\dfrac{15}{6}+\dfrac{10}{6}-\dfrac{5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{3+2-1}{6}}{\dfrac{15+10-5}{6}}\)
\(\Leftrightarrow A=\dfrac{\dfrac{4}{6}}{\dfrac{20}{6}}\)
\(\Leftrightarrow A=\dfrac{4}{6}.\dfrac{6}{20}\)
\(\Leftrightarrow A=\dfrac{24}{120}\)
\(\Leftrightarrow A=\dfrac{1}{5}\)