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17x + 3. ( -16x – 37) = 2x + 43 - 4x
<=>17x-48x-111=-2x+43
<=>-29x=154
<=> \(x=-\frac{154}{29}\)
-3. (2x + 5) -16 < -4. (3 – 2x)
\(\Leftrightarrow-6x-31< -12+8x.\)
\(\Leftrightarrow-14x< 19\Rightarrow x< -\frac{19}{14}\)
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Bài 1
2/5 - 3/5 : (0,75 + 30%)
= 2/5 - 3/5 : 21/20
= 2/5 - 4/7
= -6/35
‐--------------------
Bài 2
a) x/6 = -11/3
x = -11/3 . 6
x = -22
b) 16/5 của x là 48
x = 48 : 16/5
x = 15
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Câu 1:
a: =>-2x-x+17=34+x-25
=>-3x+17=x+9
=>-4x=-8
hay x=2
b: =>17x+16x+27=2x+43
=>33x+27=2x+43
=>31x=16
hay x=16/31
c: =>-2x-3x+51=34+2x-50
=>-5x+51=2x-16
=>-7x=-67
hay x=67/7
e: 3x-32>-5x+1
=>8x>33
hay x>33/8
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a) (−11)2 − 15(x − 2) = 134 − 16x
121 - 15x + 30 = 134 - 16x
16x - 15x = 134 - 121 - 30
x = -17
b) (4x + 1)(x2 − 16)=0
(4x + 1)(x - 4)(x + 4) = 0
\(\left[\begin{matrix}4x+1=0\\x-4=0\\x+4=0\end{matrix}\right.\)
\(\left[\begin{matrix}4x=-1\\x=4\\x=-4\end{matrix}\right.\)
\(\left[\begin{matrix}x=-\frac{1}{4}\\x=4\\x=-4\end{matrix}\right.\)
c) − 2(x − 3) + (− 2)2 = 4 − 3x
3x + 4 - 2x + 6 = 4
x = 4 - 4 - 6
x = - 6
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Khó hiểu chút, bạn cố gắng:
Ta thấy: 1+2+3= \(\frac{3}{2}\).(1+3) ; 1+2+3+4=\(\frac{4}{2}.\left(1+4\right)\)...... ; tương tự ta được 1+2+3..+16=\(\frac{16}{2}.\left(1+16\right)\)
Từ trên suy ra: \(1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16\left(1+2+...+16\right)}=1+\frac{1}{2}.3+\frac{1}{3}.\frac{3}{2}\left(1+2+3\right)+...+\frac{1}{16}.\frac{16}{2}.\left(1+2+...+16\right)\) \(=1+\frac{1}{2}.3+\frac{1}{2}.4+...+\frac{1}{2}.17=\frac{1}{2}.\left(3+4+...+17\right)\)
\(=1+\frac{1}{2}\left(3+4+...+17\right)=1+\frac{1}{2}.\frac{15}{2}.\left(3+17\right)\)
\(=1+75=76\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(\Leftrightarrow\left|x\cdot\dfrac{7}{3}-\dfrac{3}{4}\right|=1+\dfrac{1}{3}+\dfrac{2}{3}=2\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{7}{3}-\dfrac{3}{4}=2\\x\cdot\dfrac{7}{3}-\dfrac{3}{4}=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{33}{28}\\x=-\dfrac{15}{28}\end{matrix}\right.\)
b: \(\Leftrightarrow\left|x\cdot\dfrac{2}{3}-\dfrac{1}{3}\right|=\dfrac{6}{5}\)
\(\Leftrightarrow\left[{}\begin{matrix}x\cdot\dfrac{2}{3}-\dfrac{1}{3}=-\dfrac{6}{5}\\x\cdot\dfrac{2}{3}-\dfrac{1}{3}=\dfrac{6}{5}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-13}{10}\\x=\dfrac{23}{10}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(8-12x+6x^2-x^3\)
\(=\left(2-x\right)^3\)
\(125x^3-75x^2+15x-1\)
\(=\left(5x-1\right)^3\)
\(x^2-xz-9y^2+3yz\)
\(=\left(x-3y\right)\left(x+3y\right)-z\left(x-3y\right)\)
\(=\left(x-3y\right)\left(x+3y-z\right)\)
\(x^3-x^2-5x+125\)
\(=\left(x+5\right)\left(x^2-5x+25\right)-x\left(x+5\right)\)
\(=\left(x+5\right)\left(x^2-5x+25-x\right)\)
\(=\left(x+5\right)\left(x^2-6x+25\right)\)
\(x^3+2x^2-6x-27\)
\(=x^3+5x^2+9x-3x^2-15x-27\)
\(=x\left(x^2+5x+9\right)-3\left(x^2+5x+9\right)\)
\(=\left(x-3\right)\left(x^2+5x+9\right)\)
\(12x^3+4x^2-27x-9\)
\(=4x^2\left(3x+1\right)-9\left(3x+1\right)\)
\(=\left(3x+1\right)\left(4x^2-9\right)\)
\(=\left(3x+1\right)\left(2x-3\right)\left(2x+3\right)\)
\(4x^4+4x^3-x^2-x\)
\(=4x^3\left(x+1\right)-x\left(x+1\right)\)
\(=x\left(x+1\right)\left(4x^2-1\right)\)
\(=x\left(x+1\right)\left(2x-1\right)\left(2x+1\right)\)
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a; - \(\dfrac{1}{3}\).(15\(x-9\)) + \(\dfrac{2}{7}\).(- \(x-34\)) = 1 - \(\dfrac{3}{4}\).(-16\(x+4\))
- 5\(x\) + 3 - \(\dfrac{2}{7}\)\(x\) - \(\dfrac{68}{7}\) = 1 + 12\(x\) - 3
12\(x\) + 5\(x\) + \(\dfrac{2}{7}x\) = 3 - \(\dfrac{68}{7}\) - 1 + 3
17\(x\) + \(\dfrac{2}{7}x\) = (3 - 1 + 3) - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\)\(x\) = 5 - \(\dfrac{68}{7}\)
\(\dfrac{121}{7}\) \(x\) = - \(\dfrac{33}{7}\)
\(x\) = - \(\dfrac{33}{7}\): \(\dfrac{121}{7}\)
\(x\) = - \(\dfrac{3}{11}\)
Vậy \(x\) = - \(\dfrac{3}{11}\)
\(a,\frac{1}{2}x+\frac{2}{3}\left(x-2\right)=\frac{1}{3}\)
\(\frac{1}{2}x+\frac{2}{3}x-\frac{4}{3}=\frac{1}{3}\)
\(\frac{7}{6}x=\frac{5}{3}\)
\(x=\frac{10}{7}\)
\(b,16^{x-1}:16=4^4\)
\(16^{x-1}=4096\)
\(16^{x-1}=16^3\)
\(\Rightarrow x-1=3\)
\(x=4\)
=.= hk tốt!!
a) \(\frac{1}{2}x+\frac{2}{3}\left(x-2\right)=\frac{1}{3}\)
<=>\(\frac{x}{2}+\frac{2x}{3}-\frac{4}{3}-\frac{1}{3}=0\)<=>\(\frac{7x}{6}-\frac{5}{3}=0\)=>x=\(\frac{10}{7}\)
b)16x-1:16=256 => 16x-1=4096=163
T thấy x-1=3 =>x=2
Chúc bạn học tốt