a. \(678\times\left(276+49\right)-678\times\left(149+76\right)\)

\(=678\times\left[\left(276+49\right)-\left(149+76\right)\right]\)

\(=678\times\left(276+49-149-76\right)\)

\(=678\times\left(200-100\right)\)

\(=67800\)

b. \(23\times157+15\times46-28\times87\)

\(=3611+690-2436\)

\(=1865\)

c. \(12:\left\{390:\left[500-\left(125+35\times7\right)\right]\right\}\)

\(=12:\left\{390:\left[500-\left(125+245\right)\right]\right\}\)

\(=12:\left\{390:\left[500-370\right]\right\}\)

\(=12:\left\{390:130\right\}\)

\(=12:3\)

\(=4\)

a) 27 . 75 + 25 . 27 - 150 = 2025 + 675 - 150 = 2700 - 150 = 2550.

Lưu ý. Có thể dùng tính chất phân phối của phép nhân đối với phép cộng để nhẩm:

27 . 75 + 25 . 27 - 150 = 27 . (75 + 25) - 150 = 27 . 100 - 150 = 2700 - 150 = 2550.

b) 12 : {390 : [500 - (125 + 35 . 7)]} = 12 : {390 : [500 - (125 + 245)]}

= 12 : [390 : (500 - 370)] = 12 : (390 : 130) = 12 : 3 = 4.

biểu thức đâu

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

khó quá

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

mẹ ơi , nổ mắt con ròi @@

quá nổ ấy chứ lòi lun rùi @@@@@@@@@@@@@@@@@@@@@@@@

Bài 2 :

a) Phân thức A xác định \(\Leftrightarrow\hept{\begin{cases}x-2\ne0\\x+2\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne2\\x\ne-2\end{cases}}}\)

b) \(A=\left(\frac{1}{x-2}-\frac{1}{x+2}\right)\cdot\frac{x^2-4x+4}{4}\)

\(A=\left(\frac{x+2}{\left(x-2\right)\left(x+2\right)}-\frac{x-2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)

\(A=\left(\frac{x+2-x+2}{\left(x-2\right)\left(x+2\right)}\right)\cdot\frac{\left(x-2\right)^2}{4}\)

\(A=\frac{4}{\left(x-2\right)\left(x+2\right)}\cdot\frac{\left(x-2\right)^2}{4}\)

\(A=\frac{4\cdot\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)\cdot4}\)

\(A=\frac{x-2}{x+2}\)

c) Thay x = 4 ta có :

\(A=\frac{4-2}{4+2}=\frac{2}{6}=\frac{1}{3}\)

Vậy.........

\(4x^2y^3.\frac{2}{4}x^3y=4x^2y^3.\frac{1}{2}x^3y=2x^5y^4\)

\(\left(5x-2\right)\left(25x^2+10x+4\right)\)

\(=\left(5x-2\right)\left[\left(5x\right)^2+5x.2+2^2\right]\)

\(=\left(5x\right)^3-2^3\)

\(=125x^3-8\)

Bài 1

a,\(4x^2y^3.\frac{2}{4}x^3y=2x^5y^4\)