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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
ĐKXĐ: \(x\ne\left\{0;-1;-2;-3;-4;-5;-6;-7\right\}\)
\(\frac{1}{x}+\frac{1}{x+2}+\frac{1}{x+5}+\frac{1}{x+7}=\frac{1}{x+1}+\frac{1}{x+3}+\frac{1}{x+4}+\frac{1}{x+6}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{x+7}+\frac{1}{x+2}+\frac{1}{x+5}=\frac{1}{x+1}+\frac{1}{x+6}+\frac{1}{x+3}+\frac{1}{x+4}\)
\(\Rightarrow\frac{x+7+x}{x\left(x+7\right)}+\frac{x+5+x+2}{\left(x+2\right)\left(x+5\right)}=\frac{x+6+x+1}{\left(x+1\right)\left(x+6\right)}+\frac{x+4+x+3}{\left(x+3\right)\left(x+4\right)}\)
\(\Rightarrow\frac{2x+7}{x^2+7x}+\frac{2x+7}{x^2+7x+10}=\frac{2x+7}{x^2+7x+6}+\frac{2x+7}{x^2+7x+12}\)
\(\Rightarrow\left(2x+7\right)\left(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\right)=0\)
mà \(\frac{1}{x^2+7x}+\frac{1}{x^2+7x+10}-\frac{1}{x^2+7x+6}-\frac{1}{x^2+7x+12}\ne0\)
=> 2x + 7 = 0 => x = -7/2
Vậy x = -7/2
1) (2x - 3)2 = 4x2 - 8
<=> 4x2 - 12x + 9 = 4x2 - 8
<=> 12x + 9 = -8
<=> 12x = -17
<=> x = 17/12
1) (2x - 3)^2 = 4x^2 - 8
<=> 4x^2 - 12x + 9 = 4x^2 - 8
<=> 4x^2 - 12x + 9 - 4x^2 = -8
<=> -12x + 9 = -8
<=> -12x = -8 - 9
<=> -12x = -17
<=> x = 17/12
2) x - (x + 2)(x - 3) = 4 - x^2
<=> x - x^2 + 3x - 2x + 6 = 4 - x^2
<=> 2x - x^2 + 6 = 4 - x^2
<=> 2x - x^2 + 6 + x^2 = 4
<=> 2x + 6 = 4
<=> 2x = 4 + 6
<=> 2x = 10
<=> x = 5
3) 3x - (x - 3)(x + 1) = 6x - x^2
<=> 3x - x^2 - x + 3x + 3 = 6x - x^2
<=> 5x - x^2 + 3 = 6x - x^2
<=> 5x - x^2 + 3 + x^2 = 6x
<=> 5x + 3 = 6x
<=> 3 = 6x - 5x
<=> 3 = x
4) 3x/4 = 6
<=> 3x = 6.4
<=> 3x = 24
<=> x = 8
5) 7 + 5x/3 = x - 2
<=> 21 + 5x = 3x - 6
<=> 5x = 3x - 6 - 21
<=> 5x = 3x - 27
<=> 5x - 3x = -27
<=> 2x = -27
<=> x = -27/2
6) x + 4 = 2/5x - 3
<=> 5x + 20 = 2x - 15
<=> 5x + 20 - 2x = -15
<=> 3x + 20 = -15
<=> 3x = -15 - 20
<=> 3x = -35
<=> x = -35/3
7) 1 + x/9 = 4/3
<=> x/9 = 4/3 - 1
<=> x/9 = 1/3
<=> x = 3
Câu 1 :
a, \(\frac{3\left(2x+1\right)}{4}-\frac{5x+3}{6}=\frac{2x-1}{3}-\frac{3-x}{4}\)
\(\Leftrightarrow\frac{6x+3}{4}+\frac{3-x}{4}=\frac{2x-1}{3}+\frac{5x+3}{6}\)
\(\Leftrightarrow\frac{5x+6}{4}=\frac{9x+1}{6}\Leftrightarrow\frac{30x+36}{24}=\frac{36x+4}{24}\)
Khử mẫu : \(30x+36=36x+4\Leftrightarrow-6x=-32\Leftrightarrow x=\frac{32}{6}=\frac{16}{3}\)
tương tự
\(\frac{19}{4}-\frac{2\left(3x-5\right)}{5}=\frac{3-2x}{10}-\frac{3x-1}{4}\)
\(< =>\frac{19.5}{20}-\frac{8\left(3x-5\right)}{20}=\frac{2\left(3-2x\right)}{20}-\frac{5\left(3x-1\right)}{20}\)
\(< =>95-24x+40=6-4x-15x+5\)
\(< =>-24x+135=-19x+11\)
\(< =>5x=135-11=124\)
\(< =>x=\frac{124}{5}\)
1)
a) \(\frac{x+5}{3x-6}-\frac{1}{2}=\frac{2x-3}{2x-4}< =>\frac{2\left(x+5\right)}{2\left(3x-6\right)}-\frac{3x-6}{2\left(3x-6\right)}=\frac{3\left(2x-3\right)}{3\left(2x-4\right)}.\)
(đk:x khác \(\frac{1}{2}\))
\(\frac{2x+10}{6x-12}-\frac{3x-6}{6x-12}=\frac{6x-9}{6x-12}< =>2x+10-3x+6=6x-9< =>x=\frac{25}{7}\)
Vậy x=\(\frac{25}{7}\)
b) /7-2x/=x-3 \(x\ge\frac{7}{2}\)
(đk \(x\ge3,\frac{7}{2}< =>x\ge\frac{7}{2}\))
\(\Rightarrow\orbr{\begin{cases}7-2x=x-3\\7-2x=-\left(x-3\right)\end{cases}\Leftrightarrow\orbr{\begin{cases}x=\frac{10}{3}\left(< \frac{7}{2}\Rightarrow l\right)\\x=4\left(tm\right)\end{cases}}}\)
Vậy x=4
2)
\(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}>\frac{x-4}{5}+\frac{x-5}{6}\)
\(\Leftrightarrow\frac{30\left(x-1\right)}{60}+\frac{20\left(x-2\right)}{60}+\frac{15\left(x-3\right)}{60}-\frac{12\left(x-4\right)}{60}-\frac{10\left(x-5\right)}{60}>0\)
\(\Leftrightarrow30x-30+20x-40+15x-45-12x+48-10x+50>0\Leftrightarrow43x-17>0\Leftrightarrow x>\frac{17}{43}\)
a) 4 ( x + 5 )( x + 6 )( x + 10 )( x + 12 ) = 3x2
Do x = 0 không là nghiệm pt nên chia 2 vế pt cho \(x^2\ne0\), ta được :
\(\frac{4}{x^2}\left(x^2+60+17x\right)\left(x^2+60+16x\right)=3\)
\(\Leftrightarrow4\left(x+\frac{60}{x}+17\right)\left(x+\frac{60}{x}+16\right)=3\)
Đến đây ta đặt \(x+\frac{60}{x}+16=t\left(1\right)\)
Ta được :
\(4t\left(t+1\right)=3\Leftrightarrow4t^2+4t-3=0\Leftrightarrow\left(2t+3\right)\left(2t-1\right)=0\)
Từ đó ta lắp vào ( 1 ) tính được x
5 x - 1 + 2 6 - 7 x - 1 4 = 2 2 x + 1 7 - 5 ⇔ 5 x - 3 6 - 7 x - 1 4 = 4 x + 2 7 - 5
⇔ 14(5x – 3) – 21(7x – 1) = 12(4x + 2) – 5.84
⇔ 70x – 42 – 147x + 21 = 48x + 24 – 420
⇔ 70x – 147x – 48x = 24 – 420 + 42 – 21
⇔ -125x = -375
⇔ x = 3
Phương trình có nghiệm x = 3