= 51/20

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= 3/14

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= 4/7 - 2/7 

= 2/7

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= 17/45

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= 2/3 

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= 2 x 1/4 

= 1/2

\(a,\dfrac{7}{12}+\dfrac{3}{4}\times\dfrac{2}{9}=\dfrac{7}{12}+\dfrac{1}{6}=\dfrac{7}{12}+\dfrac{2}{12}=\dfrac{9}{12}=\dfrac{3}{4}\)

\(b,\dfrac{8}{9}-\dfrac{4}{15}:\dfrac{2}{5}=\dfrac{8}{9}-\dfrac{4}{15}\times\dfrac{5}{2}=\dfrac{8}{9}-\dfrac{2}{3}=\dfrac{8}{9}-\dfrac{6}{9}=\dfrac{2}{9}\)

đùa thì kệ con mẹ mày xàm loz

Đáp án a là 0,75 

Đáp án b: là 0.86222222222

*\(\frac{\left(\frac{3}{10}-\frac{4}{15}-\frac{7}{20}\right).\frac{5}{19}}{\left[\frac{1}{14}+\frac{1}{7}-\left(-\frac{3}{35}\right)\right].\frac{4}{3}}=\frac{\left(\frac{18}{60}-\frac{16}{60}-\frac{21}{60}\right).\frac{5}{19}}{\left(\frac{5}{70}+\frac{10}{70}+\frac{6}{70}\right).\frac{4}{3}}=\frac{\frac{-19}{60}.\frac{5}{19}}{\frac{21}{70}.\frac{4}{3}}=\frac{\frac{-1}{12}}{\frac{14}{35}}=-\frac{1}{12}.\frac{35}{14}=\frac{-35}{168}\)

*\(\frac{\left(1+2+3+...+100\right).\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(6,3.12-21.3,6\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{63}{10}.12-21.\frac{18}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).\left(\frac{378}{5}-\frac{378}{5}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}\)

=\(\frac{\left(1+2+3+...+100\right)\left(\frac{1}{3}-\frac{1}{5}-\frac{1}{7}-\frac{1}{9}\right).0}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{100}}=0\)

Giải:

a) \(2\dfrac{17}{20}-1\dfrac{15}{11}+6\dfrac{9}{20}:3\)

\(=\dfrac{57}{20}-\dfrac{26}{11}+\dfrac{129}{20}:3\) 

\(=\dfrac{107}{220}+\dfrac{43}{20}\)

\(=\dfrac{29}{11}\)

b) \(4\dfrac{3}{7}:\left(\dfrac{7}{5}.4\dfrac{3}{7}\right)\) 

\(=\dfrac{31}{7}:\left(\dfrac{7}{5}.\dfrac{31}{7}\right)\) 

\(=\dfrac{31}{7}:\dfrac{31}{5}\) 

\(=\dfrac{5}{7}\) 

c) \(\left(3\dfrac{2}{9}.\dfrac{15}{23}.1\dfrac{7}{29}\right):\dfrac{5}{23}\) 

\(=\left(\dfrac{29}{9}.\dfrac{15}{23}.\dfrac{36}{29}\right):\dfrac{5}{23}\) 

\(=\dfrac{60}{23}:\dfrac{5}{23}\) 

\(=12\)

a) \(6x^2-15x\)

b) \(x^2+5x+4\)

c) \(49-x^2\)

d) \(x^2+4x+4\)

e) \(9-12x+4x^2\)

f) \(x^3-8\)

\(a,=6x^2-15x\\ b,=x^2+5x+4\\ c,=49-x^2\\ d,=x^2+4x+4\\ e,=9-12x+4x^2\\ f,=x^3-8\)

khó quá bẹn gì đấy ơi

phá ngoặc ra mà tính đi bn ơi

a. \(\left(x^2-2x+1\right)-3x\left(x-1\right)=0\)

\(\Leftrightarrow x^2-2x+1-3x^2+3x=0\)

\(\Leftrightarrow-2x^2+x+1=0\)

\(\Leftrightarrow-2x^2+2x-x+1=0\)

\(\Leftrightarrow-2x\left(x-1\right)-\left(x-1\right)=0\)

\(\Leftrightarrow-\left(2x+1\right)\left(x-1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x+1=0\\x-1=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)

Vậy \(x\in\left\{-\frac{1}{2};1\right\}\)

b. \(4\left(7x-3\right)-\left(7x^2-3x\right)=0\)

\(\Leftrightarrow4\left(7x-3\right)-x\left(7x-3\right)=0\)

\(\Leftrightarrow\left(4-x\right)\left(7x-3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}4-x=0\\7x-3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=4\\x=\frac{3}{7}\end{cases}}\)

Vậy \(x\in\left\{4;\frac{3}{7}\right\}\)

c.\(\left(5-x\right)\left(2+3x\right)=4-9x^2\)

\(\Leftrightarrow\left(5-x\right)\left(2+3x\right)=\left(2-3x\right)\left(2+3x\right)\)

\(\Leftrightarrow\left(2+3x\right)\left(5-x-2+3x\right)=0\)

\(\Leftrightarrow\left(2+3x\right)\left(2x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2+3x=0\\2x+3=0\end{cases}}\)\(\Leftrightarrow\orbr{\begin{cases}x=-\frac{2}{3}\\x=-\frac{3}{2}\end{cases}}\)

Vậy \(x\in\left\{-\frac{2}{3};-\frac{3}{2}\right\}\)

d. \(7-\left(2x+4\right)=-\left(x+4\right)\)

\(\Leftrightarrow7-2x-4=-x-4\)

\(\Leftrightarrow7-4+4=-x+2x\)

\(\Leftrightarrow7=x\)

Vậy x = 7

e. \(\left(x-1\right)-\left(2x-1\right)=9\)

\(\Leftrightarrow x-1-2x+1=9\)

\(\Leftrightarrow-x=9\)

\(\Leftrightarrow x=-9\)

g. \(x^3+x^2+x+1=0\)

\(\Leftrightarrow x^2\left(x+1\right)+\left(x+1\right)=0\)

\(\Leftrightarrow\left(x^2+1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x^2+1=0\\x+1=0\end{cases}}\)Mà : \(x^2+1\ge1>0\)

\(\Leftrightarrow x+1=0\)

\(\Leftrightarrow x=-1\)

Vậy x = -1

Bài này đề bài phải là khai triển biểu thức, chứ không phải là tính em nhé.

Lời giải:

Ta áp dụng hằng đẳng thức đáng nhớ thôi.

a. $(3+2x)^3=3^3+3.3^2.2x+3.3.(2x)^2+(2x)^3$

$=8x^3+36x^2+54x+27$

b.

$(\frac{1}{2}-y)^3=(\frac{1}{2})^3-3.(\frac{1}{2})^2.y+3.\frac{1}{2}y^2-y^3$

$=-y^3+\frac{3}{2}y^2-\frac{3}{4}y+\frac{1}{8}$

c.

$(x-5)(x^2+5x+25)=(x-5)^2(x^2+5x+5^2)$

$=x^3-5^3=x^3-125$

d.

$(3x+\frac{1}{2})(9x^2-\frac{3}{2}x+\frac{1}{4})$

$=(3x+\frac{1}{2})[(3x)^2-3x.\frac{1}{2}+(\frac{1}{2})^2]$

$=(3x)^3+(\frac{1}{2})^3=27x^3+\frac{1}{8}$