a, 1/3 < 6/x < 3/10
b, 1/3 < x/15 < 1/5
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a: \(\dfrac{1}{8}\cdot\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{1}{10}\cdot\dfrac{5}{3}=\dfrac{1}{2\cdot3}=\dfrac{1}{6}\)
b: \(=\dfrac{8-3}{12}\cdot\dfrac{6}{5}=\dfrac{6}{12}=\dfrac{1}{2}\)
c: \(=\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{3}{4}\)
d: =63/21+15/21-7/21=71/21
a \(\dfrac{1}{8}\times\dfrac{4}{5}\times\dfrac{10}{6}=\dfrac{1}{6}\)
b \(\left(\dfrac{8}{12}-\dfrac{3}{12}\right)\times\dfrac{6}{5}=\dfrac{5}{12}\times\dfrac{6}{5}=\dfrac{1}{2}\)
c \(\dfrac{24}{35}:\dfrac{32}{35}=\dfrac{24}{35}\times\dfrac{35}{32}=\dfrac{3}{4}\)
d \(\dfrac{63}{21}+\dfrac{15}{21}-\dfrac{7}{21}=\dfrac{71}{21}\)
a: Ta có: \(5\left(4x-1\right)+2\left(1-3x\right)-6\left(x+5\right)=10\)
\(\Leftrightarrow20x-5+2-6x-6x-30=10\)
\(\Leftrightarrow8x=43\)
hay \(x=\dfrac{43}{8}\)
b: ta có: \(2x\left(x+1\right)+3\left(x-1\right)\left(x+1\right)-5x\left(x+1\right)+6x^2=0\)
\(\Leftrightarrow2x^2+2x+3x^2-3-5x^2-5x+6x^2=0\)
\(\Leftrightarrow6x^2-3x-3=0\)
\(\Leftrightarrow2x^2-x-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{1}{2}\end{matrix}\right.\)
Bài 1:
a: 76-6(x-1)=10
\(\Leftrightarrow x-1=11\)
hay x=12
c: \(5x+15⋮x+2\)
\(\Leftrightarrow x+2=5\)
hay x=3
A/ -6/x = -5/10
B/ x - 3/2 = 1/4
Cho mình lời giải chi tiết ko tóm tắt lời giải hoặc ghi thẳng kết quả
\(\dfrac{-6}{x}=\dfrac{-5}{10}\\ x=\dfrac{10.\left(-6\right)}{5-}=12\\ x=\dfrac{1}{4}+\dfrac{3}{2}=\dfrac{7}{4}\)
\(a,PT\Leftrightarrow x^3-6x^2+12x-8-x^3+x+6x^2-18x-10=0\)
\(\Leftrightarrow-5x-18=0\)
\(\Leftrightarrow x=-\dfrac{18}{5}\)
Vậy ...
\(b,PT\Leftrightarrow x^3+3x^2+3x+1-x^3+3x^2-3x+1-6x^2+12x-6+10=0\)
\(\Leftrightarrow12x+6=0\)
\(\Leftrightarrow x=-\dfrac{1}{2}\)
Vậy ...
\(c,PT\Leftrightarrow\left(x+1\right)^3+3^3=0\)
\(\Leftrightarrow\left(x+1+3\right)\left(x^2+2x+1-3x-3+9\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x^2-x+7\right)=0\)
Thấy : \(x^2-\dfrac{2.x.1}{2}+\dfrac{1}{4}+\dfrac{27}{4}=\left(x-\dfrac{1}{2}\right)^2+\dfrac{27}{4}\ge\dfrac{27}{4}>0\)
\(\Rightarrow x+4=0\)
\(\Leftrightarrow x=-4\)
Vậy ...
\(d,PT\Leftrightarrow\left(x-2\right)^3+1^3=0\)
\(\Leftrightarrow\left(x-2+1\right)\left(x^2-4x+4-x+2+1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-5x+7\right)=0\)
Thấy : \(x^2-5x+7=x^2-\dfrac{5.x.2}{2}+\dfrac{25}{4}+\dfrac{3}{4}=\left(x-\dfrac{5}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\)
\(\Rightarrow x-1=0\)
\(\Leftrightarrow x=1\)
Vậy ...
a: =>4x^2-24x+36-4x^2+4x-1<10
=>-20x<10-35=-25
=>x>=5/4
b: =>x(x^2-25)-x^3-8<=3
=>x^3-25x-x^3-8<=3
=>-25x<=11
=>x>=-11/25
Hình như là đề sai nha bn
sửa đổi lại đề nha
tính lại xem nhé
tk nhé
a)\(\frac{1}{3}< \frac{6}{x}< \frac{3}{10}\)
\(\frac{6}{18}< \frac{6}{x}< \frac{6}{20}\)
=>x=19
b)\(\frac{1}{3}< \frac{x}{15}< \frac{1}{5}\)
\(\frac{5}{15}< \frac{x}{15}< \frac{3}{15}\)
=>x=4