a: Ta có: \(\left(5x+1\right)^2-\left(5x-3\right)\left(5x+3\right)=30\)

\(\Leftrightarrow25x^2+10x+1-25x^2+9=30\)

\(\Leftrightarrow10x=20\)

hay x=2

b: Ta có: \(\left(x-1\right)\left(x^2+x+1\right)-x\left(x+2\right)\left(x-2\right)=5\)

\(\Leftrightarrow x^3-1-x^3+4x=5\)

\(\Leftrightarrow4x=6\)

hay \(x=\dfrac{3}{2}\)

a) x.31+(1+2+3+4+.....+30)=1240

x.31+465=1240

x.31=1240-465

x.31=775

x=775:31

x=25

b) (1+x).x:2=210

(1+x).x=210.2

(1+x).x=420

Vì 1+x và x là 2 số nhiên liên tiếp và 420=20.21

=> x=20

a)

x + (x + 1) + (x + 2) + ... + (x + 30) = 1240

x + x + 1 + x + 2 + ... + x + 30 = 1240

(x + x + ... + x) + (1 + 2 + ... + 30) = 1240

(x . [30 - 1 + 1 + 1]) + ([30 + 1] . [30 - 1 + 1] : 2) = 1240

31x + 465 = 1240

           31x = 1240 - 465

           31x = 775

\(a,x+\left(x+1\right)+\left(x+2\right)+...+\left(x+30\right)=1240\)

\(31x+1+2+3+...+30=1240\)

\(31x+465=1240\)

\(31x=775\)

\(x=25\)

\(b,1+2+3+...+x=210\)

Theo công thức tính tổng ta có:

\(\left(x+1\right)\cdot x:2=210\)

\(\left(x+1\right)\cdot x=420\)

Suy ra hai số liên tiếp nhân với nhau được 420 là: 20.21

\(\Rightarrow x=20\)

Bài 1:

a: Ta có: \(48751-\left(10425+y\right)=3828:12\)

\(\Leftrightarrow y+10425=48751-319=48432\)

hay y=38007

b: Ta có: \(\left(2367-y\right)-\left(2^{10}-7\right)=15^2-20\)

\(\Leftrightarrow2367-y=1222\)

hay y=1145

Bài 2: 

Ta có: \(8\cdot6+288:\left(x-3\right)^2=50\)

\(\Leftrightarrow288:\left(x-3\right)^2=2\)

\(\Leftrightarrow\left(x-3\right)^2=144\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=12\\x-3=-12\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=15\\x=-9\end{matrix}\right.\)

1) x = 25

2) x = 20

Ta có : x + x + 1 + x + 2 + ... + x + 30 = 1240

<=> (x + x + x + ... + x) + (1 + 2 + 3 + .. + 30) = 1240

<=> 31x + 30 = 1240

<=> 31x = 1240 - 30

<=> 31x = 1210

<=> x = 1210/31

a/ Tìm x

\(1+2+3+4+5+6+7+8+9+10+11+12+13+14+15+16+17+18+19+20=210\)

Câu \(b\) thì emdixaqua làm rồi nhé, chưa có câu a thì mình làm thôi

~~~HỌC TỐT NHA~~~

x(x+1)+(x+2)+(x+3)+..+( x+30)

31x+(1+2+3+.....+30)

31x+210=1240

x=31x=1030

=> x = 1030/31

1 + 2 + 3 +... + x = 210

\(\frac{x.\left(x+1\right)}{2}=210\)

x.(x+1)=210.2

x.(x+1)=420

x.(x+1)=20.21

=>x=20

a) x+(x+1)+(x+2)+…+(x+30)=1240

=>x+x+1+x+2+x+3+…+x+30=1240

=>x+x+x+…+x+1+2+3+…+30=1240

Từ 1->30 có: (30-1):1+1=30(số)

=>31.x+(30+1).30:2=1240

=>31.x+31.15=1240

=>31.x+465=1240

=>31.x=1240-465

=>31x=775

=>x=775:31

=>x=25

b) 1+2+3+…+x=210

=>x.(x+1):2=210

=>x.(x+1)=420

=>x.(x+1)=20.21=20.(20+1)

=>x=20

Bài 1:

a,x + ( x + 1) + (x + 2) + (x + 3) +....+ (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

b, 1+2+3+...+x=210

\(\frac{x.\left(x+1\right)}{2}=210\)

x(x+1)=210.2

x(x+1)=420

x(x+1)=20.21

=>x=20

BAi 1:

x + (x+1) + (x+2) + (x+3) + ...+ (x+30) = 1240

x + x + x +...+ x + 1 + 2 + 3 + 4 +..........+ 30 =1240

day so 1 + 2 + 3 + ... + 30 co so so hang la:

(30-1):1+1=30

Hoctot

a: Ta có: \(100-7\left(x-5\right)=58\)

\(\Leftrightarrow7\left(x-5\right)=42\)

\(\Leftrightarrow x-5=6\)

hay x=11

b: Ta có: \(12\left(x-1\right):3=4^3+2^3\)

\(\Leftrightarrow12\left(x-1\right)=216\)

\(\Leftrightarrow x-1=18\)

hay x=19