\(M=\left(x-2\right)\left(x^2+2x+4\right)-\left(x+2\right)\left(x^2+2x+4\right)\)

\(=x^3-8-\left(x^3+2x^2+4x+2x^2+4x+8\right)\)

\(=x^3-8-x^3-4x^2-8x-8=-4x^2-8x-16\)

\(C=2\cdot7+3\cdot8+...+41\cdot46+42\cdot47\)

\(=2\left(2+5\right)+3\left(3+5\right)+...+41\left(41+5\right)+42\left(42+5\right)\)

\(=5\left(2+3+...+41+42\right)+\left(2^2+3^2+...+42^2\right)\)

\(=5\left[\left(42-2+1\right)\cdot\dfrac{\left(42+2\right)}{2}\right]+\left(1^2+2^2+...+42^2-1\right)\)

\(=5\cdot41\cdot\dfrac{44}{2}+\left[\dfrac{42\left(42+1\right)\left(2\cdot42+1\right)}{6}-1\right]\)

\(=5\cdot41\cdot22+7\cdot43\cdot85-1=30094\)

1:

2:

`2(x-5) - 17 = 25`

`=> 2(x - 5) = 25 + 17`

`=> 2(x-5) = 42`

`=> x - 5 = =  42 : 2`

`=> x -5 = 21`

`=> x = 21 + 5`

`=> x = 26`

Vậy `x = 26`

Rồi , ở dưới ý , câu hỏi nãy ý .

a: Đặt \(A=\dfrac{x^2+y^2+2xy}{x+y}\)

\(=\dfrac{\left(x+y\right)^2}{x+y}=x+y\)

Khi x=5 và y=-1/2 thì \(A=5-\dfrac{1}{2}=\dfrac{9}{2}\)

a: Chu vi hình vuông có cạnh là a là \(C=4a\)

b: Diện tích hình chữ nhật là: \(S=5\cdot x=5x\left(cm^2\right)\)

c: Chu vi hình chữ nhật là \(C=2\left(x+y\right)=2x+2y\left(cm\right)\)

sửa \(\dfrac{2}{1.4}+\dfrac{2}{4.7}+...+\dfrac{2}{2022.2023}\)

\(=\dfrac{2}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2022}-\dfrac{1}{1023}\right)=\dfrac{2}{3}.\left(1-\dfrac{1}{2023}\right)=\dfrac{2.2022}{3.2023}=\dfrac{1348}{2023}\)

\(\left(x-2\right)^2+\left(x-1\right)\left(x+5\right)\)

\(=x^2-4x+4+x^2+5x-x-5\)

\(=2x^2-1\)