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a) \(\frac{3}{4}x-\frac{1}{4}=2\left(x-3\right)+\frac{1}{4}x\)
\(\frac{3}{4}x-\frac{1}{4}=2x-6+\frac{1}{4}x\)
\(\frac{3}{4}x-2x-\frac{1}{4}x=\frac{1}{4}-6\)
\(x\left(\frac{3}{4}-2-\frac{1}{4}\right)=-\frac{23}{4}\)
\(-\frac{3}{2}x=-\frac{23}{4}\)
\(x=-\frac{23}{4}\div\left(-\frac{3}{2}\right)\)
\(x=\frac{23}{6}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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Chỗ cuối là (2x + 11) nha bạn
(2x + 1) + (2x + 3) + (2x + 5) +...+ (2x + 11) = 126
2x + 1 + 2x + 3 + 2x + 5 +...+ 2x + 11 = 126
(6 . 2x ) + (1 + 3 + 5 +...+ 11) = 126
Số số hạng của tổng (1 + 3 + 5 +...+ 11) là:
(11 - 1) : 2 + 1 = 6 (số)
Tổng của (1 + 3 + 5 +...+ 11) là:
(11 + 1) . 6 : 2 = 36
\(\Rightarrow\)12x + 36 = 126
12x = 126 - 36
12x = 90
\(\Rightarrow\) x = 90 : 12
x = 7,5
Vậy x = 7,5
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a: Bạn ghi lại đề nha bạn
b: \(30\left(x+2\right)-6\left(x-5\right)-24x=100\)
=>\(30x+60-6x+30-24x=100\)
=>\(\left(30x-6x-24x\right)+\left(60+30\right)=100\)
=>0x=100-90=10(vô lý)
c: \(\left(x-7\right)\left(x+3\right)< 0\)
TH1: \(\left\{{}\begin{matrix}x-7>0\\x+3< 0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x>7\\x< -3\end{matrix}\right.\)
=>\(x\in\varnothing\)
TH2: \(\left\{{}\begin{matrix}x-7< 0\\x+3>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}x< 7\\x>-3\end{matrix}\right.\)
=>-3<x<7
mà x nguyên
nên \(x\in\left\{-2;-1;0;1;2;3;4;5;6\right\}\)
d: -1<2x-1<4
=>\(-1+1< 2x< 4+1\)
=>0<2x<5
=>0<x<2,5
mà x nguyên
nên \(x\in\left\{1;2\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(\frac{4}{13}.\frac{6}{5}+\frac{4}{13}.\frac{2}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(\frac{4}{13}.\frac{8}{5}\right).\left(2x+1\right)^2=\frac{10}{13}\)
\(\frac{32}{65}.\left(2x+1\right)^2=\frac{10}{13}\)
\(\left(2x+1\right)^2=\frac{10}{13}\div\frac{32}{65}\)
\(\left(2x+1\right)^2=\frac{25}{16}\)
\(\Rightarrow2x+1\in\left\{\frac{5}{4};-\frac{5}{4}\right\}\)
\(\hept{\begin{cases}2x+1=\frac{5}{4}\\2x+1=-\frac{5}{4}\end{cases}\Rightarrow\hept{\begin{cases}2x=\frac{1}{4}\\2x=-\frac{9}{4}\end{cases}\Rightarrow}\hept{\begin{cases}x=\frac{1}{8}\\x=-\frac{9}{8}\end{cases}}}\)
Vậy \(x\in\left\{\frac{1}{8};-\frac{9}{8}\right\}\)
\(x^3-\frac{9}{16}.x=0\)
\(x\left(x^2-\frac{9}{16}\right)=0\)
\(\hept{\begin{cases}x=0\\x^2-\frac{9}{16}=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x^2=\frac{9}{16}\end{cases}\Rightarrow}\hept{\begin{cases}x=0\\x=\pm\frac{3}{4}\end{cases}}}\)
Vậy \(x\in\left\{0;\frac{3}{4};-\frac{3}{4}\right\}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1
a,\(\left|x-3\right|=13+7=20\)
\(TH1:x-3=20\Rightarrow x=23\)
\(TH2:x-3=-20\Rightarrow x=-17\)
b,
TH1:\(x-2=3\Rightarrow x=5\)
TH2:\(x-2=-3\Rightarrow x=-1\)
a,| x-3|-7=13
=>|x-3| =13+7
=>|x-3| =20
=>x-3=20;-20
* x-3=20 *x-3=-20
=>x=20+3 =>x=-20+3
=>x=23 thuộc Z =>x=-17 thuộc Z
Vậy x=23 hoặc x=-17
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\left|2x-3\right|=x-1\)
\(\Rightarrow\orbr{\begin{cases}2x-3=x-1\\2x-3=1-x\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}2x-x=-1+3\\2x+x=1+3\end{cases}}\)
\(\Rightarrow\orbr{\begin{cases}x=2\\x=\frac{4}{3}\end{cases}}\)
=>có 2 trg hp
trg hp 1:-2x-3=x-1
=>-2x-2=x
=>-3x-2=0
=>-3x=2
=>ko cs x
trg hp 2:2x-3=x-1
=>2x-2=x
=>x-2=0
=>x=2
vậy x=2
3x-5 . 32x = 81
3(x-5)+2x = 34
\(\Rightarrow\)( x - 5 ) + 2x = 4
\(\Rightarrow\)3x - 5 = 4
\(\Rightarrow\)3x = 4 + 5
\(\Rightarrow\)3x = 9
\(\Rightarrow\)x = 9 : 3
\(\Rightarrow\)x = 3