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a) Ta có: 1 + 5 + 6 = 12 ; 2 + 3 + 7 = 12
Vậy 1 + 5 + 6 = 2 + 3 + 7
b) Ta có:\(1^2\)\(+5^2\)\(+6^2\)\(=1+25+36=62\)
\(2^2\)\(+4^2\)\(+7^2\)\(=4+16+49=62\)
\(=>1^2\)\(+5^2\)\(+6^2\)\(=2^2\)\(+3^2\)\(+7^2\)
c) Ta có 1 + 6 +8 = 15; 2 + 4 + 9 = 15
Vậy 1 + 6 + 8 = 2 + 4 + 9
\(d,1^2\)\(+6^2\)\(+8^2\)\(=1+36+64=101\)
\(2^2\)\(+4^2\)\(+9^2\)\(=4+16+81=101\)
\(1^2\)\(+6^2\)\(+8^2\)\(=2^2\)\(+4^2\)\(+9^2\)
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a)
\(S=1-2+3-4+...+99-100\)
\(S=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)( Có 50 cặp số -1 )
\(S=\left(-1\right).50\)
\(S=\left(-50\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\dfrac{200-\left(3+\dfrac{2}{3}+\dfrac{2}{4}+\dfrac{2}{5}+...+\dfrac{2}{100}\right)}{\dfrac{1}{2}+\dfrac{2}{3}+\dfrac{3}{4}+...+\dfrac{99}{100}}\\ =\dfrac{200-2-\left(1+\dfrac{2}{3}+\dfrac{2}{4}+...+\dfrac{2}{100}\right)}{\left(1-\dfrac{1}{2}\right)+\left(1-\dfrac{1}{3}\right)+\left(1-\dfrac{1}{4}\right)+...+\left(1-\dfrac{99}{100}\right)}\\ =\dfrac{198-\left(\dfrac{2}{2}+\dfrac{2}{3}+\dfrac{2}{4}...+\dfrac{2}{100}\right)}{\left(1+1+1+...+1\right)-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot99-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}\\ =\dfrac{2\cdot\left[99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)\right]}{99-\left(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+...+\dfrac{1}{100}\right)}=2\left(đpcm\right)\)
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1. y = 7
2. (- 4)
3. - 21
4. 1
5. - 20
6. 30 cm
7. 3
8. - 100
9. Không biết giải
10. 9
câu 9 số nguyên đó là 0
Vì : 0+2=2 GTTĐ của nó là 2
0-2=-2 GTTĐ của nó là 2
Đáp án đúng là 0
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
\(x-\dfrac{20}{11\cdot13}-\dfrac{20}{13\cdot15}-...-\dfrac{20}{53\cdot55}=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}+...+\dfrac{2}{53\cdot55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\left(\dfrac{1}{11}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{15}+...+\dfrac{1}{53}-\dfrac{1}{55}\right)=\dfrac{3}{11}\)
\(\Leftrightarrow x-10\cdot\dfrac{4}{55}=\dfrac{3}{11}\)
hay x=1
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1) 167 + ( -252 ) + 52 + ( -67 )
= 167 + 52 + ( -252 ) + ( - 67 )
= 219 + - ( 252 + 67 )
= 219 + - ( 319 )
= -100
2) ( -215 ) + ( - 115 ) + ( - 80 )
= - ( 215 + 115 ) + ( - 80 )
= - 330 + ( - 80 )
= - ( 330 - 80 )
= - 250
3) 118 + 107 - ( 118 - 93 )
= 118 + ( 107 - 93 )
= 118 + 14
= 132
4) 1 + 5 + 9 + ..... + 97 + 101
= ( 101 : 1) - 4 = 96
= 101 x 96
= 9696
5) 38 - 138 + 250 - 350
= - 100 + - 100
= 0
6) - ( - 357 ) + ( - 27 ) + ( - 32 )
= - ( - 357 ) + - ( 27 + 32 )
= - ( - 357 ) + - 59
= - ( 357 + 59 )
= - 416
7) 40 + ( 139 - 172 + 99 ) - ( 139 + 199 - 172 )
= 40 + ( 100 - 100 )
= 40 + 0
= 40
8) ( -1 ) + 2 + ( - 3 ) + 4 + ( - 99 ) + 100
b)\(\frac{11}{4}:\left(2.y+\frac{5}{7}\right)=\frac{1}{2}\)
\(\left(2.y+\frac{5}{7}\right)=\frac{11}{4}:\frac{1}{2}\)
\(\left(2.y+\frac{5}{7}\right)=\frac{11}{2}\)
\(2y=\frac{67}{14}\)
\(y=\frac{67}{28}\)
Vậy \(y=\frac{67}{28}\)
\(\frac{11}{4}:\left(2y+\frac{5}{7}\right)=\frac{1}{2}\)
<=> \(\left(2y+\frac{5}{7}\right)=\frac{11}{4}:\frac{1}{2}=\frac{11}{4}.2=\frac{11}{2}\)
<=> \(2y=\frac{11}{2}-\frac{5}{7}=\frac{67}{14}\)
<=> y = 67/28
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