Đặt A = 1.2 + 2.3 + 3.4 + ... + 98.99

A = 1/3 . ( 1.2.3 + 2.3.3 + 3.4.3 + ... + 98.99.3)

A = 1/3 . [ 1.2.(3-0) + 2.3.(4-1) + 3.4.(5-2) + ... + 98.99.(100-97)

A = 1/3 . ( 1.2.3 - 0.1.2 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 98.99.100 - 97.98.99)

A = 1/3 . [(1.2.3 + 2.3.4 + 3.4.5 + ... + 98.99.100) - ( 0.1.2 + 1.2.3 + 2.3.4 + ... + 97.98.99)]

A = 1/3 . ( 98.99.100 - 0.1.2)

A = 1/3 .98.99.100

A = 323400

Ta có: 323400x/26950 = 12/6/7 : 3/2

12x = 14 × 2/3

12x = 28/3

x = 28/3 : 12

x = 28/3 × 1/12 = 7/9

Vậy x = 7/9

`1)(x+2)(x+3)(x-7)(x-8)=144`
`<=>[(x+2)(x-7)][(x+3)(x-8)]=144`
`<=>(x^2-5x-14)(x^2-5x-24)=144`
`<=>(x^2-5x-19)^2-25=144`
`<=>(x^2-5x-19)^2-169=0`
`<=>(x^2-5x-6)(x^2-5x-32)=0`
`+)x^2-5x-6=0`
`<=>` $\left[ \begin{array}{l}x=6\\x=-1\end{array} \right.$
`+)x^2-5x-32=0`
`<=>` $\left[ \begin{array}{l}x=\dfrac{5+3\sqrt{17}}{2}\\x=\dfrac{5-3\sqrt{17}}{2}\end{array} \right.$
Vậy `S={-1,6,\frac{5+3\sqrt{17}}{2},\frac{5-3\sqrt{17}}{2}}`

1: Ta có: \(\left(x+2\right)\left(x+3\right)\left(x-7\right)\left(x-8\right)=144\)

\(\Leftrightarrow\left(x^2-7x+2x-14\right)\left(x^2-8x+3x-24\right)=144\)

\(\Leftrightarrow\left(x^2-5x-14\right)\left(x^2-5x-24\right)-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+336-144=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-38\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)^2-6\left(x^2-5x\right)-32\left(x^2-5x\right)+192=0\)

\(\Leftrightarrow\left(x^2-5x\right)\left(x^2-5x-6\right)-32\left(x^2-5x-6\right)=0\)

\(\Leftrightarrow\left(x^2-5x-6\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left(x-6\right)\left(x+1\right)\left(x^2-5x-32\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-6=0\\x+1=0\\x^2-5x-32=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-1\\x=\dfrac{5-3\sqrt{17}}{2}\\x=\dfrac{5+3\sqrt{17}}{2}\end{matrix}\right.\)

Vậy: \(S=\left\{6;-1;\dfrac{5-3\sqrt{17}}{2};\dfrac{5+3\sqrt{17}}{2}\right\}\)

Tính tổng dãy dấu ngoặc trước 

Đặt \(S=1\cdot2+2\cdot3+3\cdot4+...+98\cdot99\)

\(3S=1\cdot2\cdot3+2\cdot3\cdot(4-1)+...+98\cdot99\cdot(100-97)\)

\(3S=1\cdot2\cdot3+2\cdot3\cdot4-1\cdot3\cdot4+...+98\cdot99\cdot100-97\cdot98\cdot99\)

\(3S=98\cdot99\cdot100\Rightarrow S=\frac{1}{3}\cdot98\cdot99\cdot100\)

Thay vào đề bài,ta có :

\(\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}:\frac{-3}{2}\)

\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=12\frac{6}{7}\cdot\frac{2}{-3}\)

\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{90}{7}\cdot\frac{2}{-3}\)

\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-30}{7}\cdot\frac{2}{-1}\)

\(\Leftrightarrow\frac{\frac{1}{3}\cdot98\cdot99\cdot100\cdot x}{26950}=\frac{-60}{-7}=\frac{60}{7}\)

\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=\frac{60}{7}\cdot26950\)

\(\Leftrightarrow\frac{1}{3}\cdot98\cdot99\cdot100\cdot x=231000\)

\(\Leftrightarrow323400\cdot x=231000\)

\(\Leftrightarrow x=231000:323400=\frac{5}{7}\)

Tử thần sai từ dòng:

\(\frac{\frac{1}{3}.98.99.100.x}{26950}=\frac{30}{7}.\frac{2}{-1}\Leftrightarrow12x=-\frac{60}{7}\Leftrightarrow x=\frac{-5}{7}\)

Bài 2:

a)|x| < 3

x\(\in\){-2;-1;0;1;2}

b)|x - 4 | < 3

x\(\in\){ 6 ; 5 ; 4 ; 3 ; 2 }

c) | x + 10 | < 2

x\(\in\){ -2 ; -10 }

Bài 1:

A = 1 + 2 - 3 + 4 + 5 - 6 +...+98 - 99

A = (1 + 4 + 7 +...+97) + [(2-3)+(5-6)+...+(98-99)]

A = 1617 + [(-1)+(-1)+...+(-1)]

A = 1617 + (-49)

A = +(1617-49) = A = 1568

B = - 2 - 4 + 6 - 8 + 10 + 12 - .... + 60

B =  

2) 

a) \(x\in\left\{2;1;0;-1;-2\right\}\)

b) \(x\in\left\{6;-6;5;-5;4\right\}\)

c) \(x\in\left\{-9;-11;-10\right\}\)

3)

\(\left(a;b\right)\in\left\{\left(0;1\right);\left(0;-1\right);\left(1;0\right);\left(-1;0\right)\right\}\)

1) \(1+\left(-2\right)+3+\left(-4\right)+...+19+\left(-20\right).\)

\(=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+...+\left[19+\left(-20\right)\right].\)

\(=-1+\left(-1\right)+...+\left(-1\right)\) (10 số hạng -1).

\(=-1.10=-10.\)

Vậy..........

2); 3); 4): làm tương tự 1).

5) \(1+2-3-4+...+97+98-99-100.\)

\(=1+\left(2-3-4+5\right)+\left(6-7-8+9\right)...+\left(94-95-96+97\right)+\left(98-99-100\right).\)

\(=1+0+0+...+0+\left(-101\right).\)

\(=1+\left(-101\right).\)

\(=-100.\)

Vậy..........

Câu 5 mình viết sai đầu bài nhé

Cau đúng là1+2-3-4+.......+97+98-99+100