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\(2^{10}.2^{x+4}=64^5\)
\(\Leftrightarrow2^{x+14}=2^{30}\)
\(\Leftrightarrow x+14=30\)
\(\Leftrightarrow x=16\)
\(5^x+5^{x+3}=630\)
\(\Rightarrow5^x.1+5^x.125=630\)
\(\Rightarrow5^x.126=630\)
\(\Rightarrow5^x=5\)
\(\Rightarrow x=1\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+..............+\left(x+100\right)=7450\)
\(\Rightarrow\left(x+x+x+x+.........+x\right)+\left(1+2+3+..........+100\right)=7450\)
\(101x+5050=7450\)
Đến đây tự tính
\(a,\frac{6}{7}+\frac{5}{8}:5-\frac{3}{16}\cdot(-2)^2\)
\(=\frac{6}{7}+\frac{5}{8}:\frac{5}{1}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{5}{8}\cdot\frac{1}{5}-\frac{3}{16}\cdot4\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot4}{16}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3\cdot1}{4}\)
\(=\frac{6}{7}+\frac{1}{8}-\frac{3}{4}=\frac{48+7-42}{56}=\frac{13}{56}\)
\(b,\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-2}{3}+\frac{5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\left[\frac{-4+5}{6}\right]:\frac{2}{3}\)
\(=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}:\frac{2}{3}=\frac{2}{3}+\frac{1}{3}\cdot\frac{1}{6}\cdot\frac{3}{2}=\frac{2}{3}+\frac{1}{12}=\frac{8}{12}+\frac{1}{12}=\frac{9}{12}=\frac{3}{4}\)
c, Xem lại đề
d, \(\frac{-3}{5}+\left[\frac{-2}{5}-99\right]\)
\(=\frac{-3}{5}+\frac{-497}{5}=\frac{-500}{5}=-100\)
b, Tìm x
\(\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{1}{7}-\frac{1}{8}\right]\cdot56\)
\(\Rightarrow\left[\frac{2}{11}+\frac{1}{3}\right]\cdot x=\left[\frac{8}{56}-\frac{7}{56}\right]\cdot56\)
\(\Rightarrow\left[\frac{6}{33}+\frac{11}{33}\right]\cdot x=1\)
\(\Rightarrow\frac{17}{33}\cdot x=1\)
\(\Rightarrow x=1:\frac{17}{33}=1\cdot\frac{33}{17}=\frac{33}{17}\)
Đặt \(A=1+2+...+2^{97}+2^{98}+2^{99}\)\(\Rightarrow\)\(2^{100}-A=2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)\)
Ta có: \(2A=2+2^2...+2^{98}+2^{99}+2^{100}\)
Lấy \(2A-A\)theo vế, ta có:
\(2A-A=\left(2+2^2...+2^{98}+2^{99}+2^{100}\right)-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)\)
\(\Leftrightarrow2A-A=2+2^2...+2^{98}+2^{99}+2^{100}-1-2-...-2^{97}-2^{98}-2^{99}\)
\(\Leftrightarrow A=2^{100}-1\)
\(\Rightarrow2^{100}-A=2^{100}-2^{100}+1=1\)
Vậy \(2^{100}-\left(1+2+...+2^{97}+2^{98}+2^{99}\right)=1\)
Bài 3 :
a) \(1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right)\)\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[19+\left(-20\right)\right]\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\cdot10=-10\)
b) \(1-2+3-4+...+99-100=\left(1-2\right)+\left(3-4\right)+...+\left(99-100\right)\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)
\(=\left(-1\right)\cdot50=-50\)
c) \(2-4+6-8+...+46-48+50-52=\left(2-4\right)+\left(6-8\right)+...+\left(50-52\right)\)
\(=\left(-2\right)+\left(-2\right)+...+\left(-2\right)\)
\(=\left(-2\right)\cdot13=-26\)
d) \(-1+3-5+7-...-97+99\)\(=\left(-1+3\right)+\left(-5+7\right)+...+\left(-97+99\right)\)
\(=2+2+...+2\)
\(=2\cdot25=50\)
e) \(1+\left(-2\right)+3+\left(-4\right)+...+1999+\left(-2000\right)+2001\)\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[1999+\left(-2000\right)\right]+2001\)
\(=\left(-1\right)+\left(-1\right)+...+\left(-1\right)+2001\)
\(=\left(-1\right)\cdot1000+2001=-1000+2001=1001\)
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Bài 4 :
a) \(\left(2ab^2\right):\left(abc\right)=\left[2\cdot4\cdot\left(-6^2\right)\right]:\left[4\cdot\left(-6\right)\cdot12\right]\)
\(=\left[2\cdot4\cdot36\right]:\left[4\cdot\left(-6\right)\cdot12\right]\)
\(=\left[8\cdot36\right]:\left[-24\cdot12\right]\)
\(=288:\left(-288\right)=-1\)
b) \(\left[\left(-25\right)\cdot\left(-27\right)\cdot\left(-x\right)\right]:y=\left[\left(-25\right)\cdot\left(-27\right)\cdot4\right]:\left(-9\right)\)
\(=\left[675\cdot4\right]:\left(-9\right)=2700:\left(-9\right)=-300\)
c) \(\left(a^2-b^2\right):\left(a+b\right)\left(a-b\right)=\left(5^2-\left(-3^2\right)\right):\left(5+\left(-3\right)\right)\left(5-\left(-3\right)\right)\)
\(=\left(25-9\right):\left(5+\left(-3\right)\right)\left(5-\left(-3\right)\right)\)
\(=16:2\cdot8=8\cdot8=64\)
bài 1 :
a) S1=( 1 + 3 - 5 - 7 )+(9+11-13-15)+...+(393+395-397-399)
S1=(-8)+(-8)+...+(-8)
S1=(-8)*199
S1=-1592
b)S2=(1-2-3+4)+( 5 - 6 - 7 +8)+...+( 97 - 98 - 99 + 100)
S2=0+0+...+0
S2=0*100
S2=0
phần c và d tương tự nhé
BÀI 2
c)<=>2(x-1)+4 chia hết x-3
=>8 chia hết x-3
=>x-3\(\in\){-1,-2,-4,-8,1,2,4,8}
=>x\(\in\){2,1,-1,-5,4,5,7,11}
C = 2 + 22 + 23 + ... + 299 + 2100
a)
C = ( 2 + 22 + 23 + 24 + 25 ) + ( 26 + 27 + 28 + 29 + 210 ) + ... + ( 296 + 297 + 298 + 299 + 2100 )
= 2( 1 + 2 + 22 + 23 + 24 ) + 26( 1 + 2 + 22 + 23 + 24 ) + ... + 296( 1 + 2 + 22 + 23 + 24 )
= 2.31 + 26.31 + ... + 296.31
= 31( 2 + 26 + ... + 296 ) chia hết cho 31 ( đpcm )
b)
C = 2 + 22 + 23 + ... + 299 + 2100
=> 2C = 2( 2 + 22 + 23 + ... + 299 + 2100 )
= 22 + 23 + ... + 2100 + 2101
=> C = 2C - C
= 22 + 23 + ... + 2100 + 2101 - ( 2 + 22 + 23 + ... + 299 + 2100 )
= 22 + 23 + ... + 2100 + 2101 - 2 - 22 - 23 - ... - 299 - 2100
= 2101 - 2
22x-1 - 2 = C
<=> 22x-1 - 2 = 2101 - 2
<=> 22x-1 = 2101 - 2 + 2
<=> 22x-1 = 2101
<=> 2x - 1 = 101
<=> 2x = 102
<=> x = 51
C=2+22+23+...+299+2100.
= 2 . ( 2+22+23+24 ) + 26 . ( 2+22+23+24 ) + ....... + 296 . ( 2+22+23+24 )
= 2 . 31 + 26 . 31 + ....... + 296 . 31
=31 . ( 2 + 26 + ..... + 296 )
\(\Rightarrow\)C=2+22+23+...+299+2100 \(⋮\)31