1.tính
a)162 b)32 2 c) 256 : 32 d) 532 : 59
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a: \(A=\left(1-\sqrt{7}\right)\cdot\left(1+\sqrt{7}\right)=1-7=-6\)
b: \(B=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)
c: \(C=4\sqrt{2}-5\sqrt{2}+3\sqrt{2}=2\sqrt{2}\)
Bài 1: Thực hiện các phép tính sau:
\(a)\)Chưa rỏ đề
\(b)\)\(5025\div5-25\div5\)
\(=\)\(1005-5\)
\(=\)\(1000\)
\(c)\)\(218-180\div2\div9\)
\(=\)\(218-10\)
\(=\)\(208\)
\(d)\)\(\left(328-8\right)\div32\)
\(=\)\(320\div32\)
\(=\)\(10\)
Bài 1:
a) ( Tôi không nhìn rõ đầu bài )
b) 5025 : 5 - 25 : 5
= ( 5025 - 25 ) : 5
= 5000 : 5
= 1000
c) 218 - 180 : 2 : 9
= 218 - 180 : ( 2 . 9 )
= 218 - 180 : 18
= 218 - 10
= 208
d) ( 328 - 8 ) : 32
= 320 : 32
= 10
a) Ta có: \(19\cdot64+36\cdot19\)
\(=19\cdot\left(64+36\right)\)
\(=19\cdot100=1900\)
b) Ta có: \(150-\left[10^2-\left(14-11\right)^2\cdot2700^0\right]\)
\(=150-\left[100-3^2\cdot1\right]\)
\(=150-91\)
\(=69\)
c) Ta có: \(22^3-\left(1^{10}+8\right):3^2\)
\(=22^3-\left(1+8\right):3^2\)
\(=22^3-9:9\)
\(=22^3-1\)
\(=10647\)
d) Ta có: \(59-\left[90-\left(17-8\right)^2\right]\cdot1^{4514}\)
\(=59-\left[90-9^2\right]\cdot1\)
\(=59-\left(90-81\right)\)
\(=59-9=50\)
e) Ta có: \(7^2-36:3^2\)
\(=7^2-36:9\)
\(=49-4=45\)
a: =2/5+3/5=1
b: =1/3+2/3=1
c: =7/8+5/8=12/8=3/2
d: =2/7+3/7=5/7
\(a,\dfrac{2}{5}+\dfrac{3}{5}=\dfrac{5}{5}=1\\ b,\dfrac{1}{3}+\dfrac{2}{3}=\dfrac{3}{3}=1\\ c,\dfrac{7}{8}+\dfrac{5}{8}=\dfrac{12}{8}=\dfrac{3}{2}\\ d,\dfrac{1}{7}+\dfrac{3}{7}=\dfrac{4}{7}\)
a: \(\dfrac{-35}{-7}=\dfrac{35}{7}=5\)
b: \(\dfrac{42}{-21}=-\dfrac{42}{21}=-2\)
c: \(\dfrac{55}{-5}=-\dfrac{55}{5}=-11\)
d: \(\dfrac{46}{-23}=-\dfrac{46}{23}=-2\)
e: \(-30:\left(-2\right)=\dfrac{30}{2}=15\)
f: \(23\cdot\left(-4\right)=-23\cdot4=-92\)
g: \(15\cdot\left(-3\right)\cdot0=15\cdot0=0\)
h: \(-32\cdot14=-32\cdot10-32\cdot4=-320-128=-448\)
B)A*2=(1/2+1/4+....+1/256)*2
=1+1/2+1/4+....+1/128)
A*2-A=(1+1/2+1/4+...+1/128)-(1/2+1/4+...+1/256)
=1-1/256
=255/256
a) Đặt A = \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\)
\(\Rightarrow\frac{1}{3}\times A=\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\)
Lấy \(A-\frac{1}{3}\times A\)theo vế ta có :
\(A-\frac{1}{3}\times A=\left(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}\right)-\left(\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}+\frac{5}{486}\right)\)
\(\Rightarrow\frac{2}{3}\times A=\frac{5}{2}-\frac{5}{486}\)
\(\Rightarrow\frac{2}{3}\times A=\frac{605}{243}\)
\(\Rightarrow A=\frac{605}{243}:\frac{2}{3}\)
\(\Rightarrow A=\frac{605}{162}\)
Vậy \(\frac{5}{2}+\frac{5}{6}+\frac{5}{18}+\frac{5}{54}+\frac{5}{162}=\frac{605}{162}\)
b) Đặt B = \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}+\frac{1}{256}\)
=> \(\frac{1}{2}\times B=\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}+\frac{1}{512}\)
Lấy B trừ \(\frac{1}{2}\times B\)theo vế ta có :
\(B-\frac{1}{2}\times B=\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...++\frac{1}{128}+\frac{1}{256}\right)-\left(\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{512}\right)\)
\(\Rightarrow\frac{1}{2}\times B=\frac{1}{2}-\frac{1}{512}\)
\(\Rightarrow\frac{1}{2}\times B=\frac{255}{512}\)
\(\Rightarrow B=\frac{255}{512}:\frac{1}{2}\)
\(\Rightarrow B=\frac{255}{256}\)
Vậy \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...+\frac{1}{256}=\frac{255}{256}\)
a) Ta có : 2 x : 2 2 = 2 5 nên x = 7.
b) Ta có: 3 x : 3 2 = 3 5 nên x = 7.
c) Ta có : 4 4 : 4 x = 4 2 nên x = 2.
d) Ta có : 5 x : 5 2 = 5 2 nên x = 4,
e) Ta có: 5 x + 1 : 5 = 5 4 nên x = 4.
f) Ta có : 4 2 x - 1 : 4 = 4 2 nên x = 2
\(a,16^2=\left(2^4\right)^2=2^8\)
\(b,32^2=\left(2^5\right)^2=2^{10}\)
\(c,2^{56}:32=2^{56}:2^5=2^{51}\)
\(d,5^{32}:5^9=5^{23}\)