Giải các phương trình:
a) \(\frac{{5x - 2}}{3} = \frac{{5 - 3x}}{2}\);
b) \(\frac{{10x + 3}}{{12}} = 1 + \frac{{6 + 8x}}{9}\);
c) \(\frac{{7x - 1}}{6} + 2x = \frac{{16 - x}}{5}\).
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HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
a) \(\sin \left( {2x - \frac{\pi }{3}} \right) = - \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}2x - \frac{\pi }{3} = - \frac{\pi }{3} + k2\pi \\2x - \frac{\pi }{3} = \pi + \frac{\pi }{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}2x = k2\pi \\2x = \frac{{5\pi }}{3} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = k\pi \\x = \frac{{5\pi }}{6} + k\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
Vậy phương trình có nghiệm là: \(x \in \left\{ {k\pi ;\frac{{5\pi }}{6} + k\pi } \right\}\)
b) \(\sin \left( {3x + \frac{\pi }{4}} \right) = - \frac{1}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}3x + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \\3x + \frac{\pi }{4} = \frac{{7\pi }}{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}3x = - \frac{{5\pi }}{{12}} + k2\pi \\3x = \frac{{11\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{{5\pi }}{{36}} + k\frac{{2\pi }}{3}\\x = \frac{{11\pi }}{{36}} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
HQ
Hà Quang Minh
Giáo viên
21 tháng 9 2023
c) \(\cos \left( {\frac{x}{2} + \frac{\pi }{4}} \right) = \frac{{\sqrt 3 }}{2}\)
\(\begin{array}{l} \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} + \frac{\pi }{4} = \frac{\pi }{6} + k2\pi \\\frac{x}{2} + \frac{\pi }{4} = - \frac{\pi }{6} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}\frac{x}{2} = - \frac{\pi }{{12}} + k2\pi \\\frac{x}{2} = - \frac{{5\pi }}{{12}} + k2\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = - \frac{\pi }{6} + k4\pi \\x = - \frac{{5\pi }}{6} + k4\pi \end{array} \right.\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
d) \(2\cos 3x + 5 = 3\)
\(\begin{array}{l} \Leftrightarrow \cos 3x = - 1\\ \Leftrightarrow \left[ \begin{array}{l}3x = \pi + k2\pi \\3x = - \pi + k2\pi \end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\\ \Leftrightarrow \left[ \begin{array}{l}x = \frac{\pi }{3} + k\frac{{2\pi }}{3}\\x = \frac{{ - \pi }}{3} + k\frac{{2\pi }}{3}\end{array} \right.\,\,\,\,\left( {k \in \mathbb{Z}} \right)\end{array}\)
24 tháng 3 2020
a) 7x - 35 = 0
<=> 7x = 0 + 35
<=> 7x = 35
<=> x = 5
b) 4x - x - 18 = 0
<=> 3x - 18 = 0
<=> 3x = 0 + 18
<=> 3x = 18
<=> x = 5
c) x - 6 = 8 - x
<=> x - 6 + x = 8
<=> 2x - 6 = 8
<=> 2x = 8 + 6
<=> 2x = 14
<=> x = 7
d) 48 - 5x = 39 - 2x
<=> 48 - 5x + 2x = 39
<=> 48 - 3x = 39
<=> -3x = 39 - 48
<=> -3x = -9
<=> x = 3
ND
4
19 tháng 4 2021
bạn tự kl nhé
a, \(\left|x+5\right|=3x+1\)
TH1 : \(x+5=3x+1\Leftrightarrow-2x=-4\Leftrightarrow x=2\)
TH2 : \(x+5=-3x-1\Leftrightarrow4x=-6\Leftrightarrow x=-\dfrac{3}{2}\)( ktm )
b, \(\left|-5x\right|=2x+21\)
TH1 : \(5x=2x+21\Leftrightarrow3x=21\Leftrightarrow x=7\)
TH2 : \(5x=-2x-21\Leftrightarrow7x=-21\Leftrightarrow x=-3\)
khó nhìn quá ko
3 tháng 1 2021
a) Ta có: \(\dfrac{5x+3}{2}+\dfrac{3x-8}{4}=4\)
\(\Leftrightarrow\dfrac{2\left(5x+3\right)}{4}+\dfrac{3x-8}{4}=4\)
\(\Leftrightarrow10x+6+3x-8=16\)
\(\Leftrightarrow13x-2=16\)
\(\Leftrightarrow13x=18\)
hay \(x=\dfrac{18}{13}\)
Vậy: \(x=\dfrac{18}{13}\)
b) Ta có: \(\dfrac{5x-6}{3}-\dfrac{5x+6}{12}=1\)
\(\Leftrightarrow\dfrac{4\left(5x-6\right)}{12}-\dfrac{5x+6}{12}=1\)
\(\Leftrightarrow20x-24-5x-6=12\)
\(\Leftrightarrow15x-30=12\)
\(\Leftrightarrow15x=42\)
hay \(x=\dfrac{14}{5}\)
Vậy: \(x=\dfrac{14}{5}\)
TG
14 tháng 1 2021
a) (x - 7)(2x + 8) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\2x=-8\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy: S = {7; -4}
b) Tương tự câu a
c) (x - 1)(2x + 7)(x2 + 2) = 0
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\\x^2+2=0\end{matrix}\right.\)
Mà: x2 + 2 > 0 với mọi x
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\2x=-7\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
Vậy: \(S=\left\{1;-\dfrac{7}{2}\right\}\)
d) (2x - 1)(x + 8)(x - 5) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=1\\x=-8\\x=5\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
Vậy \(S=\left\{\dfrac{1}{2};-8;5\right\}\)
TT
14 tháng 1 2021
a/ Pt \(\Leftrightarrow\left[{}\begin{matrix}x-7=0\\2x+8=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=7\\x=-4\end{matrix}\right.\)
Vậy \(S=\left\{7;-4\right\}\)
b/ pt \(\Leftrightarrow\left[{}\begin{matrix}3x+1=0\\5x-2=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=\dfrac{2}{5}\end{matrix}\right.\)
c/ pt \(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\2x+7=0\end{matrix}\right.\) (\(x^2+2>0\forall x\))\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-\dfrac{7}{2}\end{matrix}\right.\)
d/ pt \(\Leftrightarrow\left[{}\begin{matrix}2x-1=0\\x+8=0\\x-5=0\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=-8\\x=5\end{matrix}\right.\)
11 tháng 2 2020
\(\frac{5x-3}{6}-\frac{7x-1}{4}-\frac{4x+2}{7}+5=0\)
<=> \(\frac{14\left(5x-3\right)-21\left(7x-1\right)-12\left(4x+2\right)+420}{84}=0\)
<=> 70x - 42 - 147x + 21 - 48x -24 + 420 = 0
<=> -125x + 375 = 0
<=> -125x = -375
<=> x = 3
Vậy S = {3}
\(\frac{3\left(2x+1\right)}{4}-5-\frac{3x+2}{10}=\frac{2\left(3x-1\right)}{5}\)
<=> \(\frac{15\left(2x+1\right)-100-2\left(3x+2\right)}{20}=\frac{8\left(3x-1\right)}{20}\)
<=> 30x + 15 - 100 - 6x - 4 = 24x - 8
<=> 24x - 24x = -8 + 89
<=> 0x = 81
=> pt vô nghiệm
17 tháng 2 2019
\(\Leftrightarrow\frac{5\left(x+5\right)-3\left(x-3\right)}{15}=\frac{5\left(x+5\right)-3\left(x-3\right)}{\left(x-3\right)\left(x+5\right)}\)
\(\Leftrightarrow\frac{2x+34}{15}=\frac{2x+34}{x^2+2x-15}\Leftrightarrow\orbr{\begin{cases}2x+34=0\\x^2+2x-15=15\end{cases}}\)
\(\Leftrightarrow\orbr{\begin{cases}x=-17\\x^2+2x-30=0\end{cases}}\)
Từ đó tìm được \(S=\left\{-17;\sqrt{31}-1;-\sqrt{31}-1\right\}\)
AH
Akai Haruma
Giáo viên
27 tháng 4 2023
Bài 1:
a.
$(4x^2+4x+1)-x^2=0$
$\Leftrightarrow (2x+1)^2-x^2=0$
$\Leftrightarrow (2x+1-x)(2x+1+x)=0$
$\Leftrightarrow (x+1)(3x+1)=0$
$\Rightarrow x+1=0$ hoặc $3x+1=0$
$\Rightarrow x=-1$ hoặc $x=-\frac{1}{3}$
b.
$x^2-2x+1=4$
$\Leftrightarrow (x-1)^2=2^2$
$\Leftrightarrow (x-1)^2-2^2=0$
$\Leftrightarrow (x-1-2)(x-1+2)=0$
$\Leftrightarrow (x-3)(x+1)=0$
$\Leftrightarrow x-3=0$ hoặc $x+1=0$
$\Leftrightarrow x=3$ hoặc $x=-1$
c.
$x^2-5x+6=0$
$\Leftrightarrow (x^2-2x)-(3x-6)=0$
$\Leftrightarrow x(x-2)-3(x-2)=0$
$\Leftrightarrow (x-2)(x-3)=0$
$\Leftrightarrow x-2=0$ hoặc $x-3=0$
$\Leftrightarrow x=2$ hoặc $x=3$
AH
Akai Haruma
Giáo viên
27 tháng 4 2023
2c.
ĐKXĐ: $x\neq 0$
PT $\Leftrightarrow x-\frac{6}{x}=x+\frac{3}{2}$
$\Leftrightarrow -\frac{6}{x}=\frac{3}{2}$
$\Leftrightarrow x=-4$ (tm)
2d.
ĐKXĐ: $x\neq 2$
PT $\Leftrightarrow \frac{1+3(x-2)}{x-2}=\frac{3-x}{x-2}$
$\Leftrightarrow \frac{3x-5}{x-2}=\frac{3-x}{x-2}$
$\Rightarrow 3x-5=3-x$
$\Leftrightarrow 4x=8$
$\Leftrightarrow x=2$ (không tm)
Vậy pt vô nghiệm.
a)
\(\begin{array}{l}\,\,\,\,\,\,\,\frac{{5x - 2}}{3} = \frac{{5 - 3x}}{2}\\\frac{{2\left( {5x - 2} \right)}}{6} = \frac{{3\left( {5 - 3x} \right)}}{6}\\\,2\left( {5x - 2} \right) = 3\left( {5 - 3x} \right)\\\,\,\,\,\,\,10x - 4 = 15 - 9x\\\,\,\,10x + 9x = 15 + 4\\\,\,\,\,\,\,\,\,\,\,\,\,\,19x = 19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 19:19\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1.\end{array}\)
Vậy phương trình có nghiệm \(x = 1\).
b)
\(\begin{array}{l}\,\,\,\,\,\,\frac{{10x + 3}}{{12}} = 1 + \frac{{6 + 8x}}{9}\\\frac{{3\left( {10x + 3} \right)}}{{36}} = \frac{{36}}{{36}} + \frac{{4\left( {6 + 8x} \right)}}{{36}}\\\,3\left( {10x + 3} \right) = 36 + 4\left( {6 + 8x} \right)\\\,\,\,\,\,\,\,30x + 9 = 36 + 24 + 32x\\\,\,\,\,\,\,\,30x + 9 = 60 + 32x\\\,30x - 32x = 60 - 9\\\,\,\,\,\,\,\,\,\,\,\,\,\, - 2x = 51\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = - \frac{{51}}{2}.\end{array}\)
Vậy phương trình có nghiệm \(x = - \frac{{51}}{2}\).
c)
\(\begin{array}{l}\,\,\,\,\,\,\,\,\,\,\,\,\,\frac{{7x - 1}}{6} + 2x = \frac{{16 - x}}{5}\\\frac{{5\left( {7x - 1} \right)}}{{30}} + \frac{{30.2x}}{{30}} = \frac{{6\left( {16 - x} \right)}}{{30}}\\\,\,5\left( {7x - 1} \right) + 30.2x = 6\left( {16 - x} \right)\\\,\,\,\,\,\,\,\,\,\,35x - 5 + 60x = 96 - 6x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,95x - 5 = 96 - 6x\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,95x + 6x = 96 + 5\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,101x = 101\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 101:101\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,x = 1\end{array}\)
Vậy phương trình có nghiệm \(x = 1\).