( x - 2/3 ) x3=3/4
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TN
25 tháng 11 2021
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28 tháng 2 2021
a. (3x - 1)2 - (x + 3)2 = 0
\(\Leftrightarrow\left(3x-1+x+3\right)\left(3x-1-x-3\right)=0\)
\(\Leftrightarrow\left(4x+2\right)\left(2x-4\right)=0\)
\(\Leftrightarrow4x+2=0\) hoặc \(2x-4=0\)
1. \(4x+2=0\Leftrightarrow4x=-2\Leftrightarrow x=-\dfrac{1}{2}\)
2. \(2x-4=0\Leftrightarrow2x=4\Leftrightarrow x=2\)
S=\(\left\{-\dfrac{1}{2};2\right\}\)
28 tháng 2 2021
b. \(x^3=\dfrac{x}{49}\)
\(\Leftrightarrow49x^3=x\)
\(\Leftrightarrow49x^3-x=0\)
\(\Leftrightarrow x\left(49x^2-1\right)=0\)
\(\Leftrightarrow x\left(7x+1\right)\left(7x-1\right)=0\)
\(\Leftrightarrow x=0\) hoặc \(7x+1=0\) hoặc \(7x-1=0\)
1. x=0
2. \(7x+1=0\Leftrightarrow7x=-1\Leftrightarrow x=-\dfrac{1}{7}\)
3. \(7x-1=0\Leftrightarrow7x=1\Leftrightarrow x=\dfrac{1}{7}\)
TP
1
17 tháng 9 2021
1) \(\left(x^3-8\right):\left(x-2\right)=\left[\left(x-2\right)\left(x^2+2x+4\right)\right]:\left(x-2\right)=x^2+2x+4\)
2) \(\left(x^3-1\right):\left(x^2+x+1\right)=\left[\left(x-1\right)\left(x^2+x+1\right)\right]:\left(x^2+x+1\right)=x-1\)
3) \(\left(x^3+3x^2+3x+1\right):\left(x^2+2x+1\right)=\left(x+1\right)^3:\left(x+1\right)^2=x+1\)
4) \(\left(25x^2-4y^2\right):\left(5x-2y\right)=\left[\left(5x-2y\right)\left(5x+2y\right)\right]:\left(5x-2y\right)=5x+2y\)
1 tháng 3 2021
1) ĐKXĐ: \(x\notin\left\{-2;2\right\}\)
Ta có: \(\dfrac{x-1}{x+2}-\dfrac{9}{x^2-4}=\dfrac{-3}{x-2}\)
\(\Leftrightarrow\dfrac{\left(x-1\right)\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{9}{\left(x-2\right)\left(x+2\right)}=\dfrac{-3\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}\)
Suy ra: \(x^2-3x+2-9=-3x-6\)
\(\Leftrightarrow x^2-3x-7+3x+6=0\)
\(\Leftrightarrow x^2-1=0\)
\(\Leftrightarrow\left(x-1\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=-1\left(nhận\right)\end{matrix}\right.\)
Vậy: S={1;-1}
2)
Sửa đề: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)
ĐKXĐ: \(x\notin\left\{3;-3\right\}\)
Ta có: \(\dfrac{3x-3}{x^2-9}-\dfrac{1}{x-3}=\dfrac{x+1}{x+3}\)
\(\Leftrightarrow\dfrac{3x-3}{\left(x-3\right)\left(x+3\right)}-\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{\left(x+1\right)\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}\)
Suy ra: \(3x-3-x-3=x^2-3x+x-3\)
\(\Leftrightarrow x^2-2x-3=2x-6\)
\(\Leftrightarrow x^2-2x-3-2x+6=0\)
\(\Leftrightarrow x^2-4x+3=0\)
\(\Leftrightarrow x^2-x-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Vậy: S={1}
1 tháng 3 2021
`1)(x-1)/(x+2)-9/(x^2-4)=-3/(x-2)(x ne 2)`
`<=>x^2-3x+2-9=-3x-6`
`<=>x^2-1=0`
`<=>x=+-1`
30 tháng 9 2021
a)
=(x-2)3
b)\(\left(2-x\right)^3\)
c)\(\left(x+\dfrac{1}{3}\right)^3\)
d)\(\left(\dfrac{x}{2}+y\right)^3\)
e)
\(=\left(x-1\right)^2\left(x-1-15\right)+25\left[3\left(x-1\right)-5\right]\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-3-5\right)\)
\(=\left(x-1\right)^2\left(x-16\right)+25\left(3x-8\right)\)
Bài làm
\(\left(x-\frac{2}{3}\right)\times3=\frac{3}{4}\)
\(x-\frac{2}{3}=\frac{3}{4}\div3\)
\(x-\frac{2}{3}=\frac{3}{4}\times\frac{1}{3}\)
\(x-\frac{2}{3}=\frac{1}{4}\)
\(x=\frac{1}{4}+\frac{2}{3}\)
\(x=\frac{3}{12}+\frac{8}{12}\)
\(x=\frac{11}{12}\)
Vậy \(x=\frac{11}{12}\)
# Chúc các em học tốt #
(x-2/3)= 3/4:3
x-2/3 = 1/4
x=1/4+2/3
x=11/12
Nhớ k nhé