thoạt tiên bn đã vt sai đề :(chỗ in đậm)

196:{64-196:{64-[18+2(25-21)²]}}

=196:{64-196:{64-[18+2.4²]}}

=196:{64-196:{64-[18+32]}}

=196:{64-196:{64-50}}

=196:{64-196:14}}

=196:{64-14}

=196:50

=3.92

sai đề

d, ( 4¹⁷ + 64 ) : ( 4¹⁶ + 16 )

= ( 4¹⁷ + 4³ ) : ( 4¹⁶ + 4² ) 

= 4²⁰  : 4¹⁸ = 4² = 16 

Chúc bn hc tốt <3

Ai giúp toii trả lời câu hỏi với :v

\(a,248+14.36-3744:18=248+504-208\)

\(=8.31+8.63-8.26\)

\(=8\left(31+63-26\right)\)

\(=8.68=544\)

\(\left[347^0+\left(28-4^2:2^4\right)\right].2^5-9^3:9=\left[1+\left(28-1\right)\right].32-9^2\)

\(=28.32-81\)

\(=896-81=815\)

\(c,196.39+84.98+21.196-392=196.39+84.98+21.196-98.4\)

\(=196\left(39+21\right)+98\left(84-4\right)\)

\(=196.60+98.80\)

\(=20\left(196.3+98.4\right)\)

\(=20.980=19600\)

\(d,\left(4^{17}+64\right):\left(4^{16}+16\right)=\left(4^{17}+4^3\right):\left(4^{16}+4^2\right)\)

\(=\frac{4^3\left(4^{14}+1\right)}{4^2\left(4^{14}+1\right)}=4\)

\(\sqrt{64}-\sqrt{196}+\sqrt{25}\)

= 8 - 14 + 5

= -1

= 8 - 14 + 5

= -1

k mk nha

\(\sqrt{64-\sqrt{196}+\sqrt{25}}\)

\(=8-14+5\)

\(=-1\)

A : 196 x 56 + 196 x 41 + 4 x 196 - 196 x 1

= 196 x ( 56 + 41 + 4 - 1 )

= 196 x 100

= 19 600

B : 132 x 19 + 132 x 18 + 64 x 132 - 132 x 1

= 132 x ( 19 + 18 + 64 - 1 )

= 132 x 100

= 13 200

C : 145 x 69 + 145 x 32 - 145 x 1

= 145 x ( 69 + 32 - 1 )

= 145 x 100

= 14 500

A = 196 x 56 + 196 x 41 + 4 x 196 - 196

A = 196 x ( 56 + 41 + 4 - 1 )

A = 196 x 100 = 19600

B = 132 x 19 + 132 x 18 + 64 x 132 - 132

B = 132 x ( 19 + 18 + 64 - 1 )

B = 132 x 100 = 13200

C = 145 x 69 + 145 x 32 - 145

C = 145 x ( 69 + 32 - 1 )

C = 145 x 100 = 14500

a ) = 196 x 56 + 196 x 41 + 4 x 196 - 196 x 1

= 196 x ( 56 + 41 + 4 - 1 )

= 196 x 100

= 19600

b ) = 132 x 19 + 132 x 18 + 64 x 132 - 132 x 1 

= 132 x ( 19 + 18 + 64 - 1 )

= 132 x 100

= 13200

c ) = 145 x 69 + 145 x 32 - 145 x 1

= 145 x ( 69 + 32 - 1 )

= 145 x 100

= 14500

a,\( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\)

= \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+...+ \dfrac{1}{196} < \dfrac{1}{2^2-1}+ \dfrac{1}{4^2-1}+ \dfrac{1}{6^2-1}+...+ \dfrac{1}{14^2-1}\)

= \( \dfrac{1}{1.3}+ \dfrac{1}{3.5}+ \dfrac{1}{5.7}+...+ \dfrac{1}{13.15}\)

= \( \dfrac{1}{2}(1- \dfrac{1}{3}+ \dfrac{1}{3}- \dfrac{1}{5}+ \dfrac{1}{5}- \dfrac{1}{7}+ \dfrac{1}{7}-...- \dfrac{1}{13}+ \dfrac{1}{13}- \dfrac{1}{15})\)

= \( \dfrac{1}{2}(1- \dfrac{1}{15})< \dfrac{1}{2}\)

Vậy \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\) \(<\dfrac{1}{2} \)

b,A= \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)

\(=(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{40})+(\dfrac{1}{41}+...+1...\)
\(=(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40})+(40/...\)
\(20(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...\dfrac{1}{39.40})+40(\dfrac{1}{40}...\)
\(20(\dfrac{1}{20}-\dfrac{1}{40})+40(\dfrac{1}{40}-\dfrac{1}{60})>\dfrac{11}{15}\)
Lại có \(A<40(\dfrac{1}{20.21}+...\dfrac{1}{39.40})+60(\dfrac{1}{40.41}+...+...\)
\(=40(\dfrac{1}{20}-\dfrac{1}{40})+60(\dfrac{1}{40}-\dfrac{1}{60})<\dfrac{3}{2}\)

=> \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)

a: \(A=\dfrac{1-\dfrac{1}{\sqrt{49}}+\dfrac{1}{49}-\dfrac{1}{\left(7\sqrt{7}\right)^2}}{\dfrac{\sqrt{64}}{2}-\dfrac{4}{7}+\left(\dfrac{2}{7}\right)^2-\dfrac{4}{343}}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4-\dfrac{4}{7}+\dfrac{4}{49}-\dfrac{4}{343}}\)

\(=\dfrac{1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}}{4\left(1-\dfrac{1}{7}+\dfrac{1}{49}-\dfrac{1}{343}\right)}=\dfrac{1}{4}\)

b: \(M=1-\dfrac{5}{\sqrt{196}}-\dfrac{5}{\left(2\sqrt{21}\right)^2}-\dfrac{\sqrt{25}}{204}-\dfrac{\left(\sqrt{5}\right)^2}{374}\)

\(=1-\dfrac{5}{14}-\dfrac{5}{84}-\dfrac{5}{204}-\dfrac{5}{374}\)

\(=1-5\left(\dfrac{1}{14}+\dfrac{1}{84}+\dfrac{1}{204}+\dfrac{1}{374}\right)\)

\(=1-5\left(\dfrac{1}{2\cdot7}+\dfrac{1}{7\cdot12}+\dfrac{1}{12\cdot17}+\dfrac{1}{17\cdot22}\right)\)

\(=1-\left(\dfrac{5}{2\cdot7}+\dfrac{5}{7\cdot12}+\dfrac{5}{12\cdot17}+\dfrac{5}{17\cdot22}\right)\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{22}\right)\)

\(=1-\left(\dfrac{1}{2}-\dfrac{1}{22}\right)\)

\(=1-\dfrac{11-1}{22}=1-\dfrac{10}{22}=\dfrac{12}{22}=\dfrac{6}{11}\)

1/144 PHẢI ĐỔI THÀNH 1/100 MỚI ĐÚNG HƠN. BẠN XEM LẠI ĐỀ BÀI XEM

\(A< \frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+\frac{1}{11.13}+\frac{1}{13.15}.\)

\(2A< \frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+\frac{2}{11.13}+\frac{2}{13.15}\)

\(2A< \frac{3-1}{1.3}+\frac{5-3}{3.5}+\frac{7-5}{5.7}+\frac{9-7}{7.9}+\frac{11-9}{9.11}+\frac{13-11}{11.13}+\frac{15-13}{13.15}\)

\(2A< 1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}=1-\frac{1}{15}< 1\)

\(\Rightarrow2A< 1\Rightarrow A< \frac{1}{2}\)