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4x-(7+5x)=15+(-29). b,x/11=x+4/33
4x-7-5x=-14. 33x=11(x+4)
4x-5x=14+7. 33x=11x+44
-x=21. 33x-11x=44
x=-21 22x=44
x=44:22=2
Câu 1:
Đặt: \(A=\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}+....+\frac{1}{100^2}\)
\(=\frac{1}{3.3}+\frac{1}{4.4}+\frac{1}{5.5}+\frac{1}{6.6}+....+\frac{1}{100.100}\)
\(A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+.....+\frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+....+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow A< \frac{49}{100}< \frac{50}{100}=\frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy:.............
Câu 2:
\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)...\left(\frac{1}{98}+1\right)\left(\frac{1}{99}+1\right)\)
\(=\left(\frac{1}{2}+\frac{2}{2}\right)\left(\frac{1}{3}+\frac{3}{3}\right)\left(\frac{1}{4}+\frac{4}{4}\right)...\left(\frac{1}{98}+\frac{98}{98}\right)\left(\frac{1}{99}+\frac{99}{99}\right)\)
\(=\frac{3}{2}.\frac{4}{3}.\frac{5}{4}....\frac{99}{98}.\frac{100}{99}\)
\(=\frac{3.4.5....99.100}{2.3.4...98.99}\)
\(=\frac{100}{2}=50\)
\(I=\frac{\frac{25}{17}-\frac{25}{27}-\frac{25}{37}-\frac{25}{47}}{\frac{45}{17}-\frac{45}{27}-\frac{45}{37}-\frac{45}{47}}\)
\(I=\frac{25.\left(\frac{1}{17}-\frac{1}{27}-\frac{1}{37}-\frac{1}{47}\right)}{45.\left(\frac{1}{17}-\frac{1}{27}-\frac{1}{37}-\frac{1}{47}\right)}\)
\(I=\frac{25}{45}=\frac{5}{9}\)
a ) \(\frac{2.9+6.7}{8.5+4.2}\)
= \(\frac{2.9+2.3.7}{2.2.2.5+4.2}\)
= \(\frac{9+3.7}{2.5+4.2}\)= \(\frac{30}{18}\)= \(\frac{5}{3}\)
a, 2.9+6.7/8.5+4.2
=2.3.3+2.3.7/2.4.5+2.4
=2.3.(3+7)/2.4.(5+1)
=30/24
=5/4
b,297-15/1188-60
=282/1128
=1/4
c, 2^5.3^7+2^5.3^2/2^6.3^10
=2^5.(3^7+3^2)/2.2^5.3^10
=3^9/2.3.3^9
=1/6
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}......\frac{889}{900}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{29\cdot31}{30.30}\)
\(=\frac{1.3.2.4.3.5.....29.31}{2.2.3.3.4.4....30.30}\)
\(=\frac{\left(1.2.3....29\right)\left(3.4.5...31\right)}{\left(2.3.4....30\right)\left(2.3.4.....30\right)}\)
\(=\frac{1.31}{30.2}=\frac{31}{60}\)