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Bài 1: 

b: \(\dfrac{72-x}{7}=\dfrac{x-70}{9}\)

=>648-9x=7x-490

=>-16x=-1138

hay x=569/8

c: \(\Leftrightarrow x^2=\dfrac{36}{25}\)

hay \(x\in\left\{\dfrac{6}{5};-\dfrac{6}{5}\right\}\)

d: Đặt x/5=y/4=k

=>x=5k; y=4k

Ta có: xy=180

\(\Leftrightarrow20k^2=180\)

\(\Leftrightarrow k^2=9\)

Trường hợp 1: k=3

=>x=15; y=12

Trường hợp 2: k=-3

=>x=-15; y=-12

1: ĐKXĐ: \(x\ne-\dfrac{3}{2}\)

2: ĐKXĐ: \(x\ne\dfrac{1}{2}\)

1: \(y'=\dfrac{1}{4}\cdot2x-1=\dfrac{1}{2}x-1\)

2: \(y'=\left(sinx-1\right)'\cdot\left(2x-3\right)+\left(sinx-1\right)\cdot\left(2x-3\right)'\)

\(=\left(cosx\right)\cdot\left(2x-3\right)+\left(sinx-1\right)\cdot2\)

4: \(y'=\dfrac{\left(x-1\right)'\cdot\left(x+3\right)-\left(x-1\right)\cdot\left(x+3\right)'}{\left(x+3\right)^2}\)

\(=\dfrac{x+3-x+1}{\left(x+3\right)^2}=\dfrac{4}{\left(x+3\right)^2}\)

3 tháng 11 2023

\(\dfrac{8}{9}\) : ( 2 - 3 \(\times\) y) = \(\dfrac{5}{3}\) 

        2 - 3 \(\times\) y = \(\dfrac{8}{9}\) : \(\dfrac{5}{3}\)

        2 - 3 \(\times\) y = \(\dfrac{8}{15}\)

             3 \(\times\) y = 2 - \(\dfrac{8}{15}\)

             3 \(\times\) y = \(\dfrac{22}{15}\)

                   y  = \(\dfrac{22}{15}\) : 3 

                   y = \(\dfrac{22}{45}\)

             

10 tháng 8 2018

a)\(\dfrac{0,4}{x}=\dfrac{x}{0,9}\Rightarrow x^2=0,4.0,9=0,36\Rightarrow x=0,6;-0,6\)

\(b)\dfrac{0,2}{1\dfrac{1}{5}}=\dfrac{\dfrac{2}{3}}{6x+7}\Rightarrow6x+7=\dfrac{1\dfrac{1}{5}.\dfrac{2}{3}}{0,2}=4\Rightarrow6x=-3\Rightarrow x=-\dfrac{3}{6}=-\dfrac{1}{2}\)

c)\(\dfrac{13\dfrac{1}{3}}{1\dfrac{1}{3}}=\dfrac{26}{2x+1}\Rightarrow2x+1=\dfrac{1\dfrac{1}{3}.26}{13\dfrac{1}{3}}=2,6\Rightarrow2x=1,6\Rightarrow x=0,8\)

d) mk ko hiểu

e)\(\dfrac{-2,6}{x}=\dfrac{-12}{42}\Rightarrow x=\dfrac{-2,6.42}{-12}=9,1\)

f)\(\dfrac{x^2}{6}=\dfrac{24}{25}\Rightarrow x^2=\dfrac{6.24}{25}=5,76\Rightarrow x=-2,4;2,4\)

n)mk chịu thua

xin lỗi bạn nha

20 tháng 10 2021

\(x-y=-30\Rightarrow\dfrac{x}{-30}=\dfrac{1}{y}\\ y.z=-42\\ \Rightarrow\dfrac{z}{-42}=\dfrac{1}{y}\\ \Rightarrow\dfrac{x}{-30}=\dfrac{z}{-42}\)

Áp dụng TCDTSBN ta có:

\(\dfrac{x}{-30}=\dfrac{z}{-42}=\dfrac{z-x}{-42-\left(-30\right)}=\dfrac{-12}{-12}=1\)

\(\dfrac{x}{-30}=1\Rightarrow x=-30\\ \dfrac{z}{-42}=1\Rightarrow z=-42\)

\(x.y=-30\Rightarrow-30.y=-30\Rightarrow y=1\)

 

1: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-\left(x^3+54\right)\)

\(=x^3+27-x^3-54\)

=-27

2: Ta có: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)

\(=8x^3+y^3-8x^3+y^3\)

\(=2y^3\)

18 tháng 9 2021

\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)