b: \(\left[\left(6+\dfrac{3}{5}-3-\dfrac{3}{14}\right)\cdot\dfrac{2}{5}\right]:\left(21-1.25\right)=x:\left(5+\dfrac{5}{6}\right)\)

\(\Leftrightarrow x:\dfrac{35}{6}=\dfrac{237}{175}:\dfrac{79}{4}\)

\(\Leftrightarrow x:\dfrac{35}{6}=\dfrac{12}{175}\)

\(\Leftrightarrow x=\dfrac{2}{5}\)

1) 2,75 - 5/6 × 2/5 = 2,75 - (5/6) × (2/5) = 2,75 - 1/3 = 2,75 - 0,33 = 2,42

2) 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 0,75) - 3/5 = 1,25 - (5/6 - 3/4) - 3/5 = 1,25 - (5/6 - 9/12) - 3/5 = 1,25 - (10/12 - 9/12) - 3/5 = 1,25 - 1/12 - 3/5 = 1,25 - 0,08 - 0,6 = 1,25 - 0,68 = 0,57

3) 4/9 × 0,75 + 8/5 + 3,125 = (4/9) × 0,75 + 8/5 + 3,125 = 0,44 + 8/5 + 3,125 = 0,44 + 1,6 + 3,125 = 0,44 + 4,725 = 5,165

4) 1,125 - 4/7 - 0,12 = 1,125 - (4/7) - 0,12 = 1,125 - 0,57 - 0,12 = 0,435 - 0,12 = 0,315

5) (1/3 + 0,4) × 3,5 + (1/6 + 0,75) × 6/5

a) y : 4/11 x 3/5 = 27/4

y : 4/11 = 27/4 : 3/5

y : 4/11 = 45/4

y = 45/4 x 4/11

y = 45/11

b) y + 7/2 - 5/9 = 14/3

y + 7/2 = 14/3 + 5/9

y + 7/2 = 47/9

y = 47/9 - 7/2

y = 31/18

a) y : 4/11 x 3/5 = 27/4

y : 4/11 = 27/4 : 3/5

y : 4/11 = 45/4

y = 45/4 x 4/11

y = 45/11
b) y + 7/2 - 5/9 = 14/3

y + 7/2 = 14/3 + 5/9

y + 7/2 = 47/9

y = 47/9 - 7/2

y = 31/18

\(\)a) \(y\) : \(\dfrac{4}{11}\) x \(\dfrac{3}{5}\) \(=\) \(\dfrac{27}{4}\)                   b) \(y\) \(+\) \(\dfrac{7}{2}\) \(-\) \(\dfrac{5}{9}\) \(=\) \(\dfrac{14}{3}\)

    \(y\) : \(\dfrac{4}{11}\)        \(=\) \(\dfrac{27}{4}:\dfrac{3}{5}\)                 \(y\) \(+\) \(\dfrac{7}{2}\)          \(=\) \(\dfrac{14}{3}\) \(+\) \(\dfrac{5}{9}\) 

    \(y\) : \(\dfrac{4}{11}\)        \(=\) \(\dfrac{45}{4}\)                       \(y\) \(+\) \(\dfrac{7}{2}\)           \(=\) \(\dfrac{47}{9}\)

    \(y\)                \(=\) \(\dfrac{45}{4}\) x \(\dfrac{4}{11}\)              \(y\)                    \(=\) \(\dfrac{47}{9}-\dfrac{7}{2}\)

    \(y\)                \(=\) \(\dfrac{45}{11}\)                      \(y\)                     \(=\) \(\dfrac{31}{18}\)

a) Ta có: \(\dfrac{x}{y}=\dfrac{10}{9}\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}\)

               \(\dfrac{y}{z}=\dfrac{3}{4}\Rightarrow\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\dfrac{y}{9}=\dfrac{z}{12}\)

\(\Rightarrow\dfrac{x}{10}=\dfrac{y}{9}=\dfrac{z}{12}=\dfrac{x-y+z}{10-9+12}=\dfrac{78}{13}=6\)

\(\Rightarrow\left\{{}\begin{matrix}x=6.10=60\\y=6.9=54\\z=6.12=72\end{matrix}\right.\)

b)Ta có:  \(\dfrac{x}{y}=\dfrac{9}{7}\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}\)

               \(\dfrac{y}{z}=\dfrac{7}{3}\Rightarrow\dfrac{y}{7}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x}{9}=\dfrac{y}{7}=\dfrac{z}{3}=\dfrac{x-y+z}{9-7+3}=-\dfrac{15}{5}=-3\)

\(\Rightarrow\left\{{}\begin{matrix}x=-3.9=-27\\y=-3.7=-21\\z=-3.3=-9\end{matrix}\right.\)

c) \(\dfrac{x}{3}=\dfrac{y}{4}=\dfrac{z}{3}\)

\(\Rightarrow\dfrac{x^2}{9}=\dfrac{y^2}{16}=\dfrac{z^2}{9}=\dfrac{x^2+y^2+z^2}{9+16+9}=\dfrac{200}{34}=\dfrac{100}{17}\)

\(\Rightarrow\left\{{}\begin{matrix}x^2=\dfrac{900}{17}\\y^2=\dfrac{1600}{17}\\z^2=\dfrac{900}{17}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=\pm\dfrac{30\sqrt{17}}{17}\\y=\pm\dfrac{40\sqrt{17}}{17}\\z=\pm\dfrac{30\sqrt{17}}{17}\end{matrix}\right.\)

Vậy\(\left(x;y;z\right)\in\left\{\left(\dfrac{30\sqrt{17}}{17};\dfrac{40\sqrt{17}}{17};\dfrac{30\sqrt{17}}{17}\right),\left(-\dfrac{30\sqrt{17}}{17};-\dfrac{40\sqrt{17}}{17};-\dfrac{30\sqrt{17}}{17}\right)\right\}\)

a = 2,45 x ( 46 + 54 ) + 8 x (0,5 + 0,75 )

   = 2,45 x 100 + 8 x 1,25

   = 245 + 10

   = 255

Bài làm:

a) Đáp án là 255

b) Đáp án là 75,25

c) Đáp án là Hình như là đề có 1 lỗi là:

 17,75+16,25+14,75+13,25+...+4,25+2,75+1,25

Chỗ mk gạch chân ấy .

(ý kiến riêng, sai mong M.n bỏ qua nhé )!

A.

Xét \(1+\frac{1}{3}-1\frac{1}{3}=1\frac{1}{3}-1\frac{1}{3}=0\)

=> A = 0 ( vì số nào nhân vs 0 cx bằng 0 )

4A:

a: \(A=a\left(b+3\right)-b\left(b+3\right)\)

\(=\left(b+3\right)\left(a-b\right)\)

\(=2000\cdot6=12000\)

b: \(B=b^2-8b-c\left(8-b\right)\)

\(=b\left(b-8\right)+c\left(b-8\right)\)

\(=\left(b-8\right)\left(b+c\right)\)

\(=100\cdot100=10000\)

a) \(A=a\left(b+3\right)-b\left(3+b\right)\)

\(=a\left(b+3\right)-b\left(b+3\right)\)

\(=\left(a+b\right)\left(b+3\right)\)

Thay a=2003 và b=1997 ta có:

\(A=\left(2003+1997\right)\left(1997+3\right)\)

\(=4000.2000\)

\(=8000000\)

\(4,\\ A=\left(b+3\right)\left(a-b\right)=\left(1997+3\right)\left(2003-1997\right)\\ A=2000\cdot6=12000\\ B=\left(b-8\right)\left(b+c\right)=\left(108-8\right)\left(108-8\right)\\ B=100\cdot100=10000\\ C=\left(x+y\right)\left(xy-2\right)=7\cdot10=70\\ D=\left(x+2y\right)\left(x^5-x^3y+x^2y^2\right)=\left(10-10\right)\left(x^5-x^3y+x^2y^2\right)=0\)