\(A=\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\)
\(B=\frac{2}{5}.\frac{4}{7}.\frac{6}{9}....\frac{44}{47}.\frac{46}{49}\)
a) So sánh A và B
b) Chứng minh A<133
\(A=\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\)
\(B=\frac{2}{5}.\frac{4}{7}.\frac{6}{9}....\frac{44}{47}.\frac{46}{49}\)
a) So sánh A và B
b) Chứng minh A<133
Cho A=\(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{43}{46}.\frac{45}{48}\) và B=\(\frac{2}{5}.\frac{4}{7}.\frac{6}{9}.....\frac{44}{47}.\frac{46}{49}\).So sánh A và B
chứng minh rằng A =\(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}.....\frac{43}{46}.\frac{45}{48}\)<\(\frac{1}{133}\)
Cho A=\(\frac{1}{4}.\frac{3}{6}.\frac{5}{8}.....\frac{43}{46}.\frac{45}{48}\). CMR A <\(\frac{1}{133}\)
Cho : A = \(\frac{1}{4}\)\(\times\)\(\frac{3}{6}\)\(\times\)\(\frac{5}{8}\)\(\times\).......\(\times\)\(\frac{43}{46}\)
B=\(\frac{3}{5}\)\(\times\)\(\frac{4}{7}\)\(\times\)..........\(\times\)\(\frac{44}{47}\)
a) so sánh A với B
b) chứng minh A<\(\frac{1}{133}\)
Tim x, biet:
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
Ta có :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}+\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{49}+1\right)+\left(\frac{x+2}{48}+1\right)+\left(\frac{x+3}{47}+1\right)+\left(\frac{x+4}{46}+1\right)+\left(\frac{x+5}{45}+1\right)=-5+5\)
\(\Leftrightarrow\)\(\frac{x+50}{49}+\frac{x+50}{48}+\frac{x+50}{47}+\frac{x+50}{46}+\frac{x+50}{45}=0\)
\(\Leftrightarrow\)\(\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\right)=0\)
Vì \(\frac{1}{49}+\frac{1}{48}+\frac{1}{47}+\frac{1}{46}+\frac{1}{45}\ne0\)
Nên \(x+50=0\)
\(\Rightarrow\)\(x=-50\)
Vậy \(x=-50\)
Chúc bạn học tốt ~
Tìm x biết :
\(\frac{x+1}{49}+\frac{x+2}{48}+\frac{x+3}{47}\frac{x+4}{46}+\frac{x+5}{45}=-5\)
\(\frac{x+1}{49}+1+\frac{x+2}{48}+1+\frac{x+3}{47}+1+\frac{x+4}{46}+1+\frac{x+5}{45}+1=0\)
\(\Leftrightarrow\frac{x+50}{49}+\frac{x+50}{48}+...+\frac{x+50}{45}=0\)
\(\Leftrightarrow\left(x+50\right)\left(\frac{1}{49}+\frac{1}{48}+...+\frac{1}{45}\right)=0\)
Vì 1/49+1/48+...+1/45 khác 0
Nên x+50=0
do đó x=-50
Cho \(A=\frac{1}{4}.\frac{3}{6}.\frac{5}{8}....\frac{997}{100}\)
\(B=\frac{2}{5}.\frac{4}{7}.\frac{6}{9}...\frac{998}{1001}\)
So sánh A và B
bài 1: tính A:=\(\frac{1}{2}-\frac{2}{3}+\frac{3}{4}-\frac{4}{5}+\frac{5}{6}-\frac{6}{7}-\frac{5}{6}+\frac{4}{5}-\frac{3}{4}+\frac{2}{3}-\frac{2}{3}-\frac{1}{2}\)
Bài 2: Cho B=\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{49}-\frac{1}{50}\)
Chứng minh rằng: \(\frac{7}{12}< A< \frac{5}{6}\)