1/x2 =x/27
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a: \(\Leftrightarrow\left(x-5\right)\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-1\\x=1\end{matrix}\right.\)
d: \(\Leftrightarrow\left(x+3\right)\left(x^2-4x+5\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left(x+1\right)\left(x+2\right)\left(x+5\right)-x^2\left(x+8\right)=27\)
\(\Leftrightarrow\left(x^2+3x+2\right)\left(x+5\right)-\left(x^3+8x^2\right)-27=0\)
\(\Leftrightarrow17x-17=0\\ \Leftrightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Điều kiện 1 ≤ x ≤ 7
Ta có: x + 2 7 − x = 2 x − 1 + − x 2 + 8 x − 7 + 1
⇔ 2 7 − x − x − 1 + x − 1 − x − 1 7 − x = 0 ⇔ 2 7 − x − x − 1 + x − 1 x − 1 − 7 − x = 0 ⇔ 7 − x − x − 1 2 − x − 1 = 0 ⇔ x − 1 = 2 x − 1 = 7 − x ⇔ x = 5 x = 4 ( t / m )
Vậy phương trình có hai nghiệm x= 4 và x= 5
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\(1,\Leftrightarrow x^2+10x+25=x^2-4x-21\\ \Leftrightarrow14x=-46\\ \Leftrightarrow x=-\dfrac{23}{7}\\ 2,\Leftrightarrow x^3+8=15+x^3+2x\\ \Leftrightarrow2x=-7\Leftrightarrow x=-\dfrac{7}{2}\\ 3,\Leftrightarrow\left(x+3\right)^2=0\\ \Leftrightarrow x=-3\\ 4,\Leftrightarrow x^3-9x^2+27x-27=0\\ \Leftrightarrow\left(x-3\right)^3=0\\ \Leftrightarrow x-3=0\Leftrightarrow x=3\\ 5,\Leftrightarrow4x^2+4x+1-4x^2-16x-16=9\\ \Leftrightarrow-12x=24\Leftrightarrow x=-2\\ 6,\Leftrightarrow x^2-3x+5x-15=0\\ \Leftrightarrow\left(x-3\right)\left(x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
1: \(\Leftrightarrow\left(x-3\right)\left(x+3\right)-\left(x-3\right)\left(5x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(-4x+1\right)=0\)
hay \(x\in\left\{3;\dfrac{1}{4}\right\}\)
2: \(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)-\left(x-1\right)\left(x^2-2x+16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1-x^2+2x-16\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(3x-15\right)=0\)
hay \(x\in\left\{1;5\right\}\)
3: \(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(2x+1\right)=0\)
hay \(x\in\left\{1;\dfrac{1}{2};-\dfrac{1}{2}\right\}\)
4: \(\Leftrightarrow x^2\left(x+4\right)-9\left(x+4\right)=0\)
\(\Leftrightarrow\left(x+4\right)\left(x-3\right)\left(x+3\right)=0\)
hay \(x\in\left\{-4;3;-3\right\}\)
5: \(\Leftrightarrow\left[{}\begin{matrix}3x+5=x-1\\3x+5=1-x\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=-6\\4x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=-1\end{matrix}\right.\)
6: \(\Leftrightarrow\left(6x+3\right)^2-\left(2x-10\right)^2=0\)
\(\Leftrightarrow\left(6x+3-2x+10\right)\left(6x+3+2x-10\right)=0\)
\(\Leftrightarrow\left(4x+13\right)\left(8x-7\right)=0\)
hay \(x\in\left\{-\dfrac{13}{4};\dfrac{7}{8}\right\}\)
1.
\(\Leftrightarrow\left(x-3\right)\left(x+3\right)=\left(x-3\right)\left(5x-2\right)\)
\(\Leftrightarrow x+3=5x-2\)
\(\Leftrightarrow4x=5\Leftrightarrow x=\dfrac{5}{4}\)
2.
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+1\right)=\left(x-1\right)\left(x^2-2x+16\right)\)
\(\Leftrightarrow x^2+x+1=x^2-2x+16\)
\(\Leftrightarrow3x=15\Leftrightarrow x=5\)
3.
\(\Leftrightarrow4x^2\left(x-1\right)-\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(4x^2-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{1}{2};x=-\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^3+27=-x^2+9\Leftrightarrow x^3+x^2+18=0\Leftrightarrow\left(x+3\right)\left(x^2-2x+6\right)=0\Leftrightarrow x+3=0\Leftrightarrow x=-3\)(do \(x^2-2x+6=\left(x-1\right)^2+5\ge5>0\))
Ta có: \(x^3+27=-x^2+9\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x-3\right)\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3
![](https://rs.olm.vn/images/avt/0.png?1311)
1).( 27,56 x 35 ) + ( 27,56 x 67 ) - ( 27,56 x 2)
= (964 + 1846,52) - 55,12
=2810,52 - 55,12
= 2755,4
2).( 4x 35 ) x ( 25 x 5 ) x 2
= ( 140 x 125 ) x2
= 17500 x 2
=35000
4). 3/10
5). 1188
6). 61/6
\(\dfrac{1}{x^2}=\dfrac{x}{27}\)
\(\Rightarrow x^2.x=1.27\)
\(\Rightarrow x^3=27\Rightarrow x=3\)
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