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G = 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210
2.G = 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211
2G - G = (22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210 + 211) - (21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210)
G = 22 + 23 + 24 +25 + 26 + 27 + 28 + 29 + 210 + 211 - 21 -22 -23 -24 - 25 - 26 - 27 - 28 - 29 - 210
G = (22 -22) +(23 - 23) + (24 - 24) + (25 -25) + (26 - 26) +(27 - 27) +(28 -28) + (29 - 29) + (210 - 210) + (211 - 21)
G = 211 - 2
G = 2048 - 2 (đpcm)
b,
G = 21 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29 + 210
D = 2.(1+ 2 + 22 + 23 + 24 + 25 + 26 + 27 + 28 + 29)
Vì 2 ⋮ 2 nên D = 2.(1+2+22+23+24+25+26+27+28+29)⋮2 (đpcm)
Lời giải:
\(P=1+2+22+23+24+25+26+27\)
\(=(22+23)+24+(25+2)+(26+1)+27\)
\(=45+24+27+27+27=3.15+3.8+3.27\)
\(=3(15+8+27)\vdots 3\)
1/
Tổng A là tổng các số hạng cách đều nhau 4 đơn vị.
Số số hạng: $(101-1):4+1=26$
$A=(101+1)\times 26:2=1326$
2/
$B=(1+2+2^2)+(2^3+2^4+2^5)+(2^6+2^7+2^8)+(2^9+2^{10}+2^{11})$
$=(1+2+2^2)+2^3(1+2+2^2)+2^6(1+2+2^2)+2^9(1+2+2^2)$
$=(1+2+2^2)(1+2^3+2^6+2^9)$
$=7(1+2^3+2^6+2^9)\vdots 7$
`Answer:`
a. \(\frac{17}{2}-\left|2x-\frac{3}{4}\right|=-\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{17}{2}+\frac{7}{4}\)
\(\Leftrightarrow\left|2x-\frac{3}{4}\right|=\frac{41}{4}\)
\(\Leftrightarrow\orbr{\begin{cases}2x-\frac{3}{4}=\frac{41}{4}\\2x-\frac{3}{4}=-\frac{41}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=\frac{41}{4}+\frac{3}{4}\\2x=-\frac{41}{4}+\frac{3}{4}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}2x=11\\2x=-\frac{19}{2}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=11:2\\x=-\frac{19}{2}:2\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{11}{2}\\x=-\frac{19}{4}\end{cases}}\)
b. \(\left(x+\frac{1}{5}\right)^2+\frac{17}{25}=\frac{26}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{26}{25}-\frac{17}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)^2=\frac{9}{25}\)
\(\Leftrightarrow\left(x+\frac{1}{5}\right)=\left(\frac{3}{5}\right)^2\)
\(\Leftrightarrow\orbr{\begin{cases}x+\frac{1}{5}=\frac{3}{5}\\x+\frac{1}{5}=-\frac{3}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{3}{5}-\frac{1}{5}\\x=-\frac{3}{5}-\frac{1}{5}\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=\frac{2}{5}\\x=-\frac{4}{5}\end{cases}}\)
c. \(-1\frac{5}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow-\frac{32}{27}-\left(3x-\frac{7}{9}\right)^3=-\frac{24}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{32}{27}-\left(-\frac{24}{27}\right)\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=-\frac{8}{27}\)
\(\Leftrightarrow\left(3x-\frac{7}{9}\right)^3=\left(-\frac{2}{3}\right)^3\)
\(\Leftrightarrow3x-\frac{7}{9}=-\frac{2}{3}\)
\(\Leftrightarrow3x=-\frac{2}{3}+\frac{7}{9}\)
\(\Leftrightarrow3x=\frac{1}{9}\)
\(\Leftrightarrow x=\frac{1}{9}:3\)
\(\Leftrightarrow x=\frac{1}{27}\)
\(A=2\left(1+2\right)+...+2^7\left(1+2\right)=3\left(2+...+2^7\right)⋮3\)
Tìm x biết
a) (x-1/2)^2=4
b) 10/1/2-(x+1/3)^2=1/1/2
c) (x-1/5)^2+17/25=26/25
d) 1/5/27+(3x-7/9)^3=24/27
a) (x - 1/2)2 = 4
<=> (x - 1/2)2 = 22
<=> x - 1/2 = -2; 2
<=> x - 1/2 = 2 hoặc x - 1/2 = -2
x = 2 + 1/2 x = -2 + 1/2
x = 5/2 x = -3/2
=> x = 5/2 hoặc x = -3/2
b) 10/1/2 - (x + 1/3)2 = 1/1/2
<=> -(x + 1/3)2 = 1/1/2 - 10/1/2
<=> -(x + 1/3)2 = 1/2 - 5
<=> -(x + 1/3)2 = -5.2 + 1/2
<=> -(x + 1/3)2 = -9/2
<=> (x + 1/3)2 = 9/2
<=> x + 1/3 = \(\sqrt{\frac{9}{2}}\) hoặc x + 1/3 = \(-\sqrt{\frac{9}{2}}\)
x = \(\frac{3\sqrt{2}}{2}\) - 1/3 x = \(-\frac{3\sqrt{2}}{2}\) -1/3
=> x = \(\frac{3\sqrt{2}}{2}\) - 1/3 hoặc x = \(-\frac{3\sqrt{2}}{2}\) -1/3
c) (x - 1/5)2 + 17/25 = 26/25
<=> (x - 1/5)2 = 26/25 - 17/25
<=> (x - 1/5)2 = (3/5)2
<=> x - 1/5 = -3/5; 3/5
<=> x - 1/5 = 3/5 hoặc x - 1/5 = -3/5
x = 3/5 + 1/5 x = -3/5 + 1/5
x = 4/5 x = -2/5
=> x = 4/5 hoặc x = -2/5
A=27.(45+55)/[(18-2):2+1].(18+2):2
A=27.100/9.10
A=2700/90
A=30
B=135.1420+45.3.780/[(27-3):3+1].(3+27):2
B=135.(1420+780)/9.15
B=135.2200/135
B=2200
C=24347.78-78.4.6000/[(26-4):1+1].(26+4):2
C=78.(24347-24000)/23.15
C=78.347/23.15
C=9022/115
chúc bạn học tốt nha
a, Ta có :
\(2^4+2^5+2^6+2^7\)
\(=2^4\left(1+2+2^2+2^3\right)\)
\(=2^4\cdot15\)
\(=2^4\cdot3\cdot5⋮3\) (đpcm)
b, \(6^{19}+6^{20}+6^{21}+6^{22}\)
\(=6^{19}\left(1+6+6^2+6^3\right)\)
\(=6^{19}\cdot159\)
\(=6^{19}\cdot37\cdot7⋮7\) (đpcm)
Dòng 2 từ dưới lên là 259 nhé mình đánh nhầm