\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
Tính:
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(\dfrac{10^2+11^2+12^2}{13^2+14^2}=\dfrac{100+121+144}{169+196}\)
\(=\dfrac{365}{365}=1\)
Tính giá trị của biểu thức: \(D=\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
HELP ME!
D=(102+112+122):(132+142)
D =(100+121+144) / (169+196)
D =365 / 365
D = 1
sửa lại :
D=(102+112+122) : (32+142)
D =(100+121+144) / (169+196)
D =365 / 365
D = 1
sửa lại :
D=(102+112+122) : (132+142)
D =(100+121+144) / (169+196)
D =365 / 365
D = 1
Tính
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
(102 +112+122):(132 +142)
=(100+121+144):(169+196)
=365:365
=1
\(\text{Thực hiện các phép tính sau một cách hợp lý:}\)
\(a\)) \(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(b\)) \(1.2.3...9-1.2.3...8-1.2.3...7.8^2\)
\(c\)) \(\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(d\)) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(e\)) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
tính hợp lý:
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
( 100 + 121 + 144 ) : ( 169 + 196 )
= 365 : 365 = 1
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(=\left(100+121+144\right):\left(169+196\right)\)
\(=\left(221+144\right):365\)
\(=365:365\)
\(=1\)Vậy kết quả = 1
1. tính hợp lí
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
\(=\left(100+121+144\right):\left(169+196\right)\)
\(=365:365\)
\(=1\)
RÚT GỌN
\(a,\dfrac{121.75.130.169}{39.60.11.198}\)
\(b,\dfrac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}.16^9}\)
\(c,\dfrac{5.\left(2^2.3^2\right)^9.\left(2^2\right)^6-2.\left(2^2.3\right)^{14}.3^6}{5.2^{28}.3^{19}-7.2^{29}.3^{18}}\)
TÍNH HỢP LÍ
\(a,\left(10^2+11^2+12^2\right):\left(13^2+14^2\right)\)
TÍNH NHANH
\(a,13-12+11+10-9+8-7-6+-4+3+2-1\)
\(a,\dfrac{121.75.130.169}{39.60.11.198}=\dfrac{11.11.25.3.13.10.169}{13.3.6.10.11.11.18}=\dfrac{25.169}{6.18}\)
tìm x,biết:
a)\(\frac{2}{\left(x+2\right)\left(x+4\right)}+\frac{4}{\left(x+4\right)\left(x+8\right)}+\frac{6}{\left(x+8\right)\left(x+14\right)}=\frac{x}{\left(x+2\right)\left(x+14\right)}\)
b)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)
c)\(\left(x+2\right)^2=\frac{38}{25}+\frac{9}{10}-\frac{11}{15}+\frac{13}{21}-\frac{15}{28}+\frac{17}{36}-...+\frac{197}{4851}-\frac{199}{4950}\)
giúp tớ với,huhu
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}-\sqrt{12+2\sqrt{11}}}\right)\left(\sqrt{11}+\sqrt{3}\right)\)
\(\left(\sqrt{12+2\sqrt{14+2\sqrt{13}}}-\sqrt{12+2\sqrt{11}}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{\left(\sqrt{13+1}\right)^2}}-\sqrt{\left(\sqrt{11+1}\right)^2}\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{12+2\sqrt{13+2}}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{\left(\sqrt{13}+1\right)^2}-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)
\(=\left(\sqrt{13}+1-\sqrt{11}-1\right)\left(\sqrt{11}+\sqrt{13}\right)\)\(=\left(\sqrt{13}-\sqrt{11}\right)\left(\sqrt{11}+\sqrt{13}\right)=13-11=2\)
sao dấu= thứ 2 lại ra như vậy