Bài 1: Tìm x trong các tỉ lệ thức sau
a, (2x - 1): hỗn số1 3/7=hỗn số1 13/15 :hỗn số1 1/5
b, 72-x /7=x-70/9
c, x:0,16=9:x
d, x/5=y/4 và x.y=180
e,x/5=y/3=x2-y2=4(x>y>0)
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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: \(y'=\dfrac{\left(x-1\right)'\cdot\left(x+3\right)-\left(x-1\right)\cdot\left(x+3\right)'}{\left(x+3\right)^2}\)
\(=\dfrac{x+3-x+1}{\left(x+3\right)^2}=\dfrac{4}{\left(x+3\right)^2}\)
\(\dfrac{8}{9}\) : ( 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 = \(\dfrac{22}{45}\)
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
\(x-y=-30\Rightarrow\dfrac{x}{-30}=\dfrac{1}{y}\\ y.z=-42\\ \Rightarrow\dfrac{z}{-42}=\dfrac{1}{y}\\ \Rightarrow\dfrac{x}{-30}=\dfrac{z}{-42}\)
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: \(\left(2x+y\right)\left(4x^2-2xy+y^2\right)-\left(2x-y\right)\left(4x^2+2xy+y^2\right)\)
\(=8x^3+y^3-8x^3+y^3\)
\(=2y^3\)
\(1,=x^3+270-x^3-54=-27\\ 2,=8x^3+y^3-8x^3+y^3=2y^3\\ 3,=x^3-3x^2+3x-1-x^3-8+3x^2-48=3x-57\\ 4,=x^3-x-x^3-1=-x-1\\ 5,=8x^3-5\left(8x^3+1\right)=-32x^3-5\\ 6,=27+x^3-27=x^3\\ 7,làm.ở.câu.3\\ 8,=x^3-6x^2+12x-8+6x^2-12x+6-x^3-1+3x\\ =3x-3\)
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