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a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
a)\(\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
\(=\left(100+121+144\right)\div\left(169+196\right)\)
\(=365\div365\)
\(=1\)
b) \(1.2.3...9-1.2.3...8-1.2.3...8^2\)
\(=1.2.3...8\left(9-1-8\right)\)
\(=1.2.3...8.0\)
\(=0\)
d) \(1152-\left(374+1152\right)+\left(-65+374\right)\)
\(=1152-374-1152-65+374\)
\(=\left(1152-1152\right)-65+\left(374-374\right)\)
\(=0-65+0\)
\(=-65\)
e) \(13-12+11+10-9+8-7-6+5-4+3+2-1\)
\(=13-\left(12-11\right)+\left(10-9\right)+\left(8-7\right)-\left(6-5\right)-\left(4-3\right)\)\(+\left(2-1\right)\)
\(=13-1+1+1-1-1+1\)
\(=13+0+0+0\)
\(=13\)
a) Đặt \(A=\left(10^2+11^2+12^2\right)\div\left(13^2+14^2\right)\)
- Ta có: \(A=\left(100+121+144\right)\div\left(169+196\right)\)
\(\Leftrightarrow A=365\div365=1\)
Vậy \(A=1\)
b) Đặt \(B=1.2.3.....9-1.2.3.....8-1.2.3.....8^2\)
- Ta có: \(B=1.2.3.....8.\left(9-1\right)-1.2.3.....8^2\)
\(\Leftrightarrow B=1.2.3.....8.8-1.2.3.....8.8=0\)
Vậy \(B=0\)
c) Đặt \(C=\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
- Ta có: \(C=\frac{3^2.4^2.2^{32}}{11.2^{13}.2^{22}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^4.2^{32}}{11.2^{35}-2^{36}}\)
\(\Leftrightarrow C=\frac{3^2.2^{36}}{2^{35}.\left(11-2\right)}\)
\(\Leftrightarrow C=\frac{9.2^{36}}{2^{35}.9}\)
\(\Leftrightarrow C=2\)
Vậy \(C=2\)
d) Đặt \(D=1152-\left(374+1152\right)+\left(-65+374\right)\)
- Ta có: \(D=1152-374-1152-65+374\)
\(\Leftrightarrow D=\left(1152-1152\right)+\left(374-374\right)-65\)
\(\Leftrightarrow D=-65\)
Vậy \(D=-65\)
Bài 1:
A,
(100+121+144)÷(169+196)
=365:365=1
B
,1.2.3...7.8.(9-1-8)
=1.2.3...7.8.0=0
C,
3^2.2^4.2^32/11.2^13.2^22-2^26
=3^2.2^36/11.2^35-2^36
=3^2.2^36/2^35-(11-2)
=9.2/9=2
D,
=1152-374-1152+(-65)+374
=(1152-1152)+(-374+374)+(-65)
=0+0+(-65)=-65
E,
=13-(12-11-10+9)+(8-7-6+5)-(4-3-2+1)
=13-0+0-0=13
Bài 2:
A,(19x+50)÷14=25-16
(19x+50)÷14=9
19x+50=126
19x=76
x=4
B,
31x+(1+2+3+...+30)=1240
31x+465=1240
31x=775
×=25
a: \(\dfrac{10^2+11^2+12^2}{13^2+14^2}=\dfrac{100+121+144}{169+196}=\dfrac{365}{365}=1\)
c: \(=\dfrac{3^2\cdot2^{36}}{11\cdot2^{13}\cdot2^{22}-2^{36}}=\dfrac{3^2\cdot2^{36}}{2^{35}\cdot\left(11-2\right)}=2\)