Q=15X(23 x5-27:32X4)2 -5(12-22 x3)
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5
8 tháng 3 2023
\(a.=\left(\dfrac{4}{5}.\dfrac{5}{6}\right).\dfrac{2}{3}=\dfrac{4}{6}.\dfrac{2}{3}=\dfrac{4}{9}\)
\(b.\dfrac{4}{5}.\dfrac{3}{4}+\dfrac{5}{4}.\dfrac{3}{4}=\dfrac{3}{5}+\dfrac{15}{16}=\dfrac{123}{80}\)
\(c.\left(\dfrac{11}{23}+\dfrac{9}{23}\right)+\left(\dfrac{2}{23}+\dfrac{18}{23}\right)=\dfrac{20}{23}+\dfrac{20}{23}=\dfrac{40}{23}\)
\(d.\left(\dfrac{27}{12}-\dfrac{25}{36}\right)+\left(\dfrac{17}{6}-\dfrac{15}{6}\right)=\dfrac{14}{9}+\dfrac{1}{3}=\dfrac{17}{9}\)
22 tháng 8 2017
Đặt \(\frac{x_1-1}{5}=\frac{x_2-2}{4}=\frac{x_3-3}{3}=\frac{x_4-4}{2}=\frac{x_5-5}{1}=k\)
Áp dụng TC DTSBN ta có :
\(k=\frac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)
\(=\frac{x_1+x_2+x_3+x_4+x_5-15}{15}=\frac{30-15}{15}=1\)
\(\frac{x_1-1}{5}=1\Rightarrow x_1=6;\frac{x_2-2}{4}=1\Rightarrow x_2=6;\frac{x_3-3}{3}=1\Rightarrow x_3=6;\frac{x_4-4}{2}=1\Rightarrow x_4=6;\frac{x^5-5}{2}=1\Rightarrow x_5=6\)
Vậy \(x_1=x_2=x_3=x_4=x_5=6\)
PN
6 tháng 7 2017
a) \(5\dfrac{4}{23}.27\dfrac{3}{47}+4\dfrac{3}{47}.\left(-5\dfrac{4}{23}\right)\)
\(=5\dfrac{4}{23}.27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right).5\dfrac{4}{23}\)
\(=5\dfrac{4}{23}.\left[27\dfrac{3}{47}+\left(-4\dfrac{3}{47}\right)\right]\)
\(=5\dfrac{4}{23}.\left(27\dfrac{3}{47}-4\dfrac{3}{27}\right)\)
\(=5\dfrac{4}{23}.23\)
\(=\dfrac{119}{23}.23\)
\(=\dfrac{119}{23}\)
b) \(4.\left(\dfrac{-1}{2}\right)^3+\dfrac{3}{2}\)
\(=4.\dfrac{-1}{6}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{3}{2}\)
\(=\dfrac{-2}{3}+\dfrac{3}{2}\)
\(=\dfrac{-4}{6}+\dfrac{9}{6}\)
\(=\dfrac{5}{6}\)
c) \(\left(\dfrac{1999}{2011}-\dfrac{2011}{1999}\right)-\left(\dfrac{-12}{1999}-\dfrac{12}{2011}\right)\)
\(=\dfrac{1999}{2011}-\dfrac{2011}{1999}-\dfrac{-12}{1999}+\dfrac{12}{2011}\)
\(=\left(\dfrac{1999}{2011}+\dfrac{12}{2011}\right)-\left(\dfrac{2011}{1999}+\dfrac{-12}{1999}\right)\)
\(=\dfrac{2011}{2011}-\dfrac{1999}{1999}\)
\(=1-1\)
\(=0\)
d) \(\left(\dfrac{-5}{11}+\dfrac{7}{22}-\dfrac{-4}{33}-\dfrac{5}{44}\right):\left(\dfrac{381}{22}-39\dfrac{7}{22}\right)\)
(đợi đã, mình chưa tìm được hướng làm...)
11 tháng 12 2017
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x_1-1}{5}=\dfrac{x_2-2}{4}=\dfrac{x_3-3}{3}=\dfrac{x_4-4}{2}=\dfrac{x_5-5}{1}\)
\(=\dfrac{\left(x_1-1\right)+\left(x_2-2\right)+\left(x_3-3\right)+\left(x_4-4\right)+\left(x_5-5\right)}{5+4+3+2+1}\)
\(=\dfrac{\left(x_1+x_2+x_3+x_4+x_5\right)-\left(1+2+3+4+5\right)}{15}\)
\(=\dfrac{30-15}{15}=1\)
\(\Rightarrow x_1=x_2=x_3=x_4=x_5=6\)
Vậy...
11 tháng 12 2017
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{x1-1}{5}\)=\(\dfrac{x2-2}{4}\)\(\dfrac{x3-3}{3}\)=\(\dfrac{x4-4}{2}\)=\(\dfrac{x5-5}{1}\)=\(\dfrac{x1-1+x2-2+x3-3+x4-4+x5-5}{5+4+3+2+1}\)=\(\dfrac{x1+x2+x3+x4+x5-\left(1+2+3+4+5\right)}{15}\)=\(\dfrac{30-15}{15}\)=\(\dfrac{15}{15}\)=1
\(\dfrac{x1-1}{5}\)=1 => x1-1=5 => x1 =6
\(\dfrac{x2-2}{4}\)=1 => x2-2=4 => x2 =6
\(\dfrac{x3-3}{3}\)=1 => x3-3=3 => x3 =6
\(\dfrac{x4-4}{2}\)=1 => x4-4=2 => x4 =6
\(\dfrac{x5-5}{1}\)=1 => x5-5=1 => x5 = 6
Vậy x1=x2=x3=x4=x5 =6
29 tháng 1 2023
\(\dfrac{22}{5}\times\dfrac{6}{121}\times\dfrac{11}{4}\times\dfrac{3}{5}\times\dfrac{1}{3}\times\dfrac{5}{4}\)
\(=\left(\dfrac{22}{5}\times\dfrac{5}{4}\right)\times\left(\dfrac{6}{121}\times\dfrac{11}{4}\right)\times\left(\dfrac{3}{5}\times\dfrac{1}{3}\right)\)
\(=\dfrac{11}{2}\times\dfrac{3}{22}\times\dfrac{1}{5}\)
\(=\dfrac{3}{20}\)
TD
29 tháng 1 2023
\(=\dfrac{22\times6\times11\times3\times1\times5}{5\times121\times4\times5\times3\times4}=\dfrac{11\times2\times6\times11\times1}{11\times11\times4\times5\times4}=\dfrac{2\times6\times1}{4\times5\times4}=\dfrac{18}{100}=\dfrac{9}{50}\)
5 tháng 4 2020
Vì P(x) có hệ số bậc cao nhất là 1
Nên P(x) có thể được viết dưới dạng: \(P\left(x\right)=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)
Và \(P\left(-1\right)=\left(-1\right)^5-5\left(-1\right)^3+4\left(-1\right)+1=1\)
\(P\left(\frac{1}{2}\right)=\frac{77}{32}\)
Ta có: \(Q\left(x\right)=2x^2+x-1=2x^2+2x-x-1=2x\left(x+1\right)-\left(x+1\right)=\left(x+1\right)\left(2x-1\right)\)
=> \(Q\left(x_1\right).\text{}\text{}Q\left(x_2\right).\text{}\text{}Q\left(x_3\right).\text{}\text{}Q\left(x_4\right).\text{}\text{}Q\left(x_5\right)\text{}\text{}\)
\(=\left(x_1+1\right)\left(2x_1-1\right)\left(x_2+1\right)\left(2x_2-1\right)\left(x_3+1\right)\left(2x_3-1\right)\left(x_4+1\right)\left(2x_4-1\right)\left(x_5+1\right)\left(2x_5-1\right)\)
\(=32\left(-1-x_1\right)\left(\frac{1}{2}-x_1\right)\left(-1-x_2\right)\left(\frac{1}{2}-x_2\right)\left(-1-x_3\right)\left(\frac{1}{2}-x_3\right)\left(-1-x_4\right)\left(\frac{1}{2}-x_4\right)\left(-1-x_5\right)\left(\frac{1}{2}-x_5\right)\)\(=32.P\left(-1\right).P\left(\frac{1}{2}\right)=32.1.\frac{77}{32}=77\)
7 tháng 4 2020
\(p\left(x\right)=x^5-5x^3+4x+1=\left(x-x_1\right)\left(x-x_2\right)\left(x-x_3\right)\left(x-x_4\right)\left(x-x_5\right)\)
\(Q\left(x\right)=2\left(\frac{1}{2}-x\right)\left(-1-x\right)\)
Do đó \(Q\left(x_1\right)\cdot Q\left(x_2\right)\cdot Q\left(x_3\right)\cdot Q\left(x_4\right)\cdot Q\left(x_5\right)\)
\(=2^5\left[\left(\frac{1}{2}-x_1\right)\left(\frac{1}{2}-x_2\right)\left(\frac{1}{2}-x_3\right)\left(\frac{1}{2}-x_4\right)\left(\frac{1}{2}-x_5\right)\right]\)
\(=\left(-1-x_1\right)\left(-1-x_2\right)\left(-1-x_3\right)\left(-1-x_4\right)\left(-1-x_5\right)\)
\(=32P\left(\frac{1}{2}\right)\cdot\left[P\left(-1\right)\right]\)
\(=32\cdot\left(\frac{1}{32}-\frac{5}{8}+\frac{4}{2}+1\right)\left(-1+5-4+1\right)\)
\(=4300\)
*Mình không chắc*