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b/ x + (x + 1) + (x + 2) + .... + (x + 30) = 1240
x + (x + x + ....+ x) + (1 + 2 + ... + 30) = 1240
x + 30x + 465 = 1240
31x = 1240 - 465
31x = 775
=> x = 775 : 31 = 25
c/ Sửa đề (x + 4) + (x + 6) + (x + 8) + .... + (x + 26) = 210
(x + x + x + .... + x) + (4 + 6 + 8 + .... + 26) = 210
12x + 180 = 210
12x = 210 - 180 = 30
=> x = 30 : 12 = 2,5
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a) \(x+\left(x+1\right)\left(x+2\right)+...+\left(x+30\right)=1240\)
\(\Rightarrow\left(x+x+...+x\right)\left(1+2+...+30\right)=1240\)
\(\Rightarrow31x+465=1240\)
\(\Rightarrow31x=775\)
\(\Rightarrow x=25\)
Vậy x = 25
b) \(2\left|x-1\right|^2-3=5\)
\(\Rightarrow2\left|x-1\right|^2=8\)
\(\Rightarrow\left|x-1\right|^2=4\)
\(\Rightarrow x-1=\pm2\)
Nếu x - 1 = 2=> x = 3
Nếu x - 1 = -2 => x = -1
Vậy x = {3;-1}
c)Vì \(x;y\in N\)
\(\Rightarrow\left(2x+1\right)\left(y-5\right)=12=1.12=12.1=2.6=6.2=3.4=4.3\)
Đến đây thì dễ rồi, lập bảng tìm ra
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\(x+(x+1)+(x+2)+(x+3)+....+(x+30)=1240\)
\(=x+\left(30x+1+2+3+...+30\right)=1240\)
\(=31x+1+2+3+...+30\)(có 30 số hạng) \(=1240\)
\(=31x+\left(30+1\right).30\div2=1240\)
\(=31x+465=1240\)
\(\Rightarrow31x=1240-465\)
\(\Rightarrow31x=775\)
\(\Rightarrow x=775\div31\)
\(\Rightarrow x=25\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\text{x+(x+1)+(x+2)+(x+3)+.............+(x+30)=1240}\)
\(=x+x+1+x+2+x+3+...+x+30=1240\)
\(=31x+1+2+3+...+30=1240\)
\(=31x+465=1240\)
\(\Rightarrow31x=1240-465=775\)
\(\Rightarrow x=775:31=25\)
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=>(x+x+x+..+x)+(1+2+3+..+30)=1240
=>x.31+465=1240
=>x.31=1240-465
=>x.31=775
=>x=775:31
=>x=25
vậy x=25
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25 nha
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a) (x+10)(2y-5) = 143
=> (x+10);(2y-5) thuộc Ư(143)={-1,-143,1,143}
\(\orbr{\begin{cases}x+10=-143\\2y-5=-1\end{cases}}\Rightarrow\orbr{\begin{cases}x=-153\\y=2\end{cases}}\)
\(\orbr{\begin{cases}x+10=-1\\2y-5=-143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-11\\y=-69\end{cases}}\)
\(\orbr{\begin{cases}x+10=1\\2y-5=143\end{cases}}\Rightarrow\orbr{\begin{cases}x=-9\\y=74\end{cases}}\)
\(\orbr{\begin{cases}x+10=143\\2y-5=1\end{cases}}\Rightarrow\orbr{\begin{cases}x=133\\y=3\end{cases}}\)
Vậy ta có các cặp x,y thõa mãn : (-153,2);(-11,-69);(-9,74);(113,3)
b) x+(x+1)+(x+2)+..+(x+30)=1240
=> (x+x+x+...+x)+(1+2+3+...+30)=1240
=> 31x+465=1240
31x = 1240-465
31x = 775
x = 775 : 31
x= 25
c) 1+2+3+...+x=210
\(\frac{\left(x-1\right)}{1}+1=x\)
=> \(\frac{\left(x+1\right).x}{2}=210\)
(x+1)x = 210:2
(x+1)x = 105
chắc ko có x thõa mãn
d) 2+4+6+...+2x=210
=> 2(1+2+3+...+x)=210
1+2+3+..+x= 210:2 = 105
\(\frac{\left(x-1\right)}{1}+1\) = x
\(\frac{\left(x+1\right).x}{2}=105\)
(x+1)x = 105:2
(x+1)x = 52,5
ko có x thõa mãn đề bài
a, x + 10 và 2y - 5 thuộc Ư(143) = {1;-1;143;-143}
x + 10 | 1 | -1 | 143 | -143 |
2y - 5 | 143 | -143 | 1 | -1 |
x | -9 | -11 | 133 | -153 |
y | 74 | -69 | 3 | 2 |
b, x+(x+1)+(x+2)+........+(x+30) = 1240
=> x+x+1+x+2+...+x+30=1240
=> 31x+(1+2+...+30) = 1240
=> 31x + 465 = 1240
=> 31x = 775
=> x = 25
c, 1+2+...+x=210
=> \(\frac{x\left(x+1\right)}{2}=210\)
=> x(x+1) = 420
Mà 420 = 20.21
=> x = 20
d, 2+4+...+2x = 210
=> 2(1+2+...+x) = 210
=> \(\frac{2x\left(x+1\right)}{2}=210\)
=> x(x + 1) = 210
Mà 210 = 14.15
=> x = 14
e, 1+3+5+...+(2x-1) = 225
=> \(\frac{\left[\left(2x-1\right)+1\right].x}{2}=225\)
=> \(\frac{2x^2}{2}=225\)
=> x2 = \(\left(\pm15\right)^2\)
=> x = 15 hoặc x = -15
\(2^x+2^{x+3}=288\)
\(\Rightarrow2^x+2^x.2^3=288\)
\(\Rightarrow2^x\left(1+2^3\right)=288\)
\(\Rightarrow2^x.9=288\)
\(\Rightarrow2^x=288:9\)
\(\Rightarrow2^x=32\)
\(\Rightarrow2^x=2^5\)
\(\Rightarrow x=5\)
Vậy \(x=5\)
\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+30\right)=1240\)
\(\Rightarrow x+x+1+x+2+x+3+...+x+30=1240\)
\(\Rightarrow x+x+x+...+x+1+2+3+...+30=1240\)
Từ \(1\rightarrow30\)có: \(\left(30-1\right):1+1=30\)( số )
\(\Rightarrow31.x+\left(30+1\right).30:2=1240\)
\(\Rightarrow31.x+31.15=1240\)
\(\Rightarrow31.x+465=1240\)
\(\Rightarrow31.x=1240-465\)
\(\Rightarrow31.x=775\)
\(\Rightarrow x=775:31\)
\(\Rightarrow x=25\)
Chúc bạn học tốt !!!
x+(x+1)+(x+2)+....+(x+30)=1240
31 . x + (1 + 2 + 3 + 4 +...+ 29 + 30) = 1240
31 . x + 31.15 = 1240
31 . x = 1240 - 31.15
31 . x = 775
x = 775 : 31
x = 25
Vậy x=25