1) Tính giá trị biểu thức:
a. ( 102 + 112 + 122 ) : ( 132 + 142 )
b. 9! - 8! - 7! . 82
c.( 3 . 4 . 216 ) 2
11 . 213 . 411 - 169
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Ta có: 102+112+122 = 100 + 121 + 144 = 365
132+142 = 169 + 196 = 365
Vậy 102+112+122 = 132+142
(102 + 112 + 122) : (132 + 142)
= (100 + 121 + 144) :( 169 + 196)
= 365: 365
= 1
a) \(\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{4}{9}\times2=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{23}{9}\)
b) \(\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{2}{5}\times\dfrac{3}{2}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
Câu 4
\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times4\times3\times5\times4\times5}{5\times2\times4\times4\times5\times5}=\dfrac{3\times3}{5\times2}=\dfrac{9}{10}\)
Câu 3:
\(a.\dfrac{5}{3}+\dfrac{4}{9}:\dfrac{1}{2}=\dfrac{5}{3}+\dfrac{8}{9}=\dfrac{15}{9}+\dfrac{8}{9}=\dfrac{23}{9}\)
\(b.\dfrac{11}{10}-\dfrac{2}{5}:\dfrac{2}{3}=\dfrac{11}{10}-\dfrac{3}{5}=\dfrac{11}{10}-\dfrac{6}{10}=\dfrac{5}{10}=\dfrac{1}{2}\)
Câu 4:
\(\dfrac{12\times15\times20}{10\times16\times25}=\dfrac{3\times3\times1}{2\times1\times5}=\dfrac{9}{10}\)
`A=sqrt{8+2sqrt7}-sqrt{8-2sqrt7}`
`=sqrt{7+2sqrt7+1}-sqrt{7-2sqrt7+1}`
`=sqrt{(sqrt7+1)^2}-sqrt{(sqrt7-1)^2}`
`=sqrt7+1-sqrt7+1=2`
`B=sqrt{11-6sqrt2}+sqrt{6-4sqrt2}`
`=sqrt{9-2.3.sqrt2+2}+sqrt{4-2.2.sqrt2+2}`
`=sqrt{(3-sqrt2)^2}+sqrt{(2-sqrt2)^2}`
`=3-sqrt2+2-sqrt2=5-2sqrt2`
|(5/8-5/14)+(3/8-9/14)|:4/7
=|(5/8+3/8)+(-5/14-9/14)|:4/7
=|1+(-1)|:4/7
=0
a) 32 . 53 + 92 = 9 . 125 + 81
= 1 125 + 81 = 1 206
b) 83 : 42 - 52 = 512 : 16 - 25 = 32 - 25 = 7
c) 33 . 92 - 52.9 + 18 : 6 = 27 . 81 - 25 . 9 + 3
= 2 187 - 225 + 3 = 1 962 + 3 = 1 965
Bài 1:
a: x+1/2=5/6
nên x=5/6-1/2=1/3
b: x+1/4=3/4
nên x=3/4-1/4=2/4=1/2
c: x+3/10=1/2
nên x=1/2-3/10=5/10-3/10=1/5
d: x+1/4=3/8
nên x=3/8-1/4=3/8-2/8=1/8
\(b)\)\(9!-8!-7!.8^2\)
\(=\)\(8!\left(9-1\right)-7!.8^2\)
\(=\)\(7!.8.8-7!.8^2\)
\(=\)\(7!.8^2-7!.8^2\)
\(=\)\(0\)
\(c)\)\(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}\)
\(=\)\(\frac{\left(2^2\right)^2.\left(2^{16}\right)^2.3^2}{2^{13}.\left(2^2\right)^{11}.11-\left(2^4\right)^9}\)
\(=\)\(\frac{2^4.2^{32}.3^2}{2^{13}.2^{22}.11-2^{36}}\)
\(=\)\(\frac{2^{36}.3^2}{2^{35}.11-2^{36}}\)
\(=\)\(\frac{2^{36}.3^2}{2^{35}\left(11-2\right)}\)
\(=\)\(\frac{2.3^2}{9}\)
\(=\)\(\frac{2.3^2}{3^2}\)
\(=\)\(2\)