\(\frac{x-1}{27}\)= \(\frac{-3}{1-x}\)
Tìm x
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Bài 1:
a) Ta có: \(\frac{7}{8}\cdot\frac{4}{9}+\frac{1}{14}:\frac{5}{14}\)
\(=\frac{28}{72}+\frac{1}{14}\cdot\frac{14}{5}\)
\(=\frac{28}{72}+\frac{1}{5}\)
\(=\frac{140}{360}+\frac{72}{360}\)
\(=\frac{212}{360}=\frac{53}{90}\)
Bài 2:
a) Ta có: \(\frac{2}{3}x+\frac{1}{4}x=\frac{-22}{27}\)
\(\Leftrightarrow x\cdot\left(\frac{2}{3}+\frac{1}{4}\right)=\frac{-22}{27}\)
\(\Leftrightarrow x\cdot\frac{11}{12}=\frac{-22}{27}\)
\(\Leftrightarrow x=\frac{-22}{27}:\frac{11}{12}=\frac{-22}{27}\cdot\frac{12}{11}=-\frac{8}{9}\)
Vậy: \(x=\frac{-8}{9}\)
(x - 2/3)3 = -1/27
=> (x - 2/3)3 = (-1/3)3
=> x - 2/3 = -1/3
=> x = -1/3 + 2/3
=> x = 1/3
Từ bài ra ta có \(\left(x-\frac{2}{3}\right)^3=\left(\frac{-1}{3}\right)^3\)
\(\Rightarrow x-\frac{2}{3}=\frac{-1}{3}\)
\(\Rightarrow x=\frac{-1}{3}+\frac{2}{3}\)
\(\Rightarrow x=\frac{1}{3}\)
Vậy ... nếu đúng thì k nha
\(\left(3x-1\right)^3=\left(\frac{2}{3}\right)^3\)
=> 3x -1 = 2/3
3x = 5/3
x = 5/9
học tốt ^^
\(\left(x^4\right)^2=x^{12-5}\)
\(x^8-x^7=0\)
\(x^7\cdot x-x^7=0\)
\(x^7\cdot\left(x-1\right)=0\)
+) x^7 = 0 => x = 0
+) x -1 = 0 => x = 1
Vậy,...........
học tốt ^^
a. \(A=\left[\frac{1}{3}+\frac{3}{x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(9-x^2\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x.\left(x-3\right)}{3.x.\left(x-3\right)}+\frac{3.3}{x\left(x-3\right).3}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{1}{x+3}\right]\)
\(=\left[\frac{x^2-3x+9}{3x.\left(x-3\right)}\right]:\left[\frac{x^2}{3.\left(3-x\right)\left(3+x\right)}+\frac{\left(3-x\right).3}{\left(x+3\right).\left(3-x\right).3}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}:\left[\frac{x^2+9-3x}{3.\left(3-x\right)\left(3+x\right)}\right]\)
\(=\frac{x^2-3x+9}{3x.\left(x-3\right)}.\frac{3.\left(3-x\right)\left(3+x\right)}{x^2-3x+9}\)
\(=\frac{-\left(x-3\right)\left(3+x\right)}{x-3}=-\left(3+x\right)\)
b. Để A < -1 thì:
-(3+x) < -1
=> -3 - x < -1
=> x < -3 - (-1) = -2
Vậy x < -2 thì A < -1.
a/ \(\Rightarrow\frac{\left(-3\right)^n}{81}=-27\Rightarrow\left(-3\right)^n=-2187\Rightarrow\left(-3\right)^n=\left(-3\right)^7\Rightarrow n=7\)
b/ \(\Rightarrow-\frac{3}{8}-x+\frac{5}{6}=\frac{4}{3}\Rightarrow\frac{11}{24}-x=\frac{4}{3}\Rightarrow x=-\frac{7}{8}\)
ở hàng thứ 3 tính cả đề, ở phân số thứ 2 trên tử là số 3 ak bn???
<=> \(\frac{-2x-1}{12}\)-\(\frac{2x+2}{3}=\frac{1-2x}{4}-\frac{3x-1}{12}\)
<=>\(\frac{-2x-1-8x-8-3+6x+3x-1}{12}=0\)
<=> -x-13=0=> x=-13
\(\frac{\frac{1}{2}-\frac{x+2}{3}}{2}-\frac{2}{3}\left(x+1\right)=\frac{1}{4}\left(1-2x\right)-\frac{\frac{1}{3}-\frac{1-x}{2}}{2}\)
<=>\(6.\left(\frac{1}{2}-\frac{x+2}{3}\right)-8.\left(x+1\right)=3\left(1-2x\right)-6.\left(\frac{1}{3}-\frac{1-x}{2}\right)\)
<=>3-2.(x+2)-8x-8=3-6x-2+3.(1-x)
<=>3-2x-4-8x-8=3-6x-2+3-3x
<=>-10x-9=-9x+4
<=>x=-13
\(\frac{x-1}{27}=-\frac{3}{1-x}\)
\(\Rightarrow\left(x-1\right).\left(1-x\right)=27.\left(-3\right)\)
\(\left(x-1\right).\left(-x+1\right)=-81\)
* x-1=-81 * -x+1=-81
x=-81+1 -x=-81-1
x=-80 -x=-82
x=82
vậy x=80 hoặc x=82
\(\frac{x-1}{27}=\frac{-3}{1-x}\)
\(\Rightarrow x-1.x-1=27.\left(-3\right)\)
\(\left(x-1\right)^2=-81\)
\(\left(x-1\right)^2=\left(-9\right)^2\)
\(x-1=-9hayx-1=9\)
\(x=-9+1hayx=9+1\)
\(x=-8hayx=10\)
Vay x=-8 hay x=10