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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
tớ ko bt lm abc , tớ lm d thôi nha , thứ lỗi
\(\frac{5}{2x-3}-\frac{1}{x+2}=\frac{5}{x-6}-\frac{7}{2x-1}\)
\(\frac{3x+13}{2x^2+x-6}=\frac{5}{x-6}+\frac{7}{1-2x}\)
\(\frac{3x+13}{\left(x+2\right)\left(2x-3\right)}=\frac{3x+37}{\left(x-6\right)\left(2x-1\right)}\)
\(\frac{10-9x}{-4x^3+32x^2-51x+18}=0\)
\(\Rightarrow\orbr{\begin{cases}x=-3\\x=\frac{10}{9}\end{cases}}\)
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) \(\frac{x-1}{2}+\frac{x-2}{3}+\frac{x-3}{4}=\frac{x-4}{5}+\frac{x-5}{6}\)
\(\left(\frac{x-1}{2}+1\right)+\left(\frac{x-2}{3}+3\right)+\left(\frac{x-3}{4}+1\right)=\left(\frac{x-4}{5}+1\right)+\left(\frac{x-5}{6}+1\right)\)
\(\frac{x-1}{2}+\frac{x-1}{3}+\frac{x-1}{4}=\frac{x-1}{5}+\frac{x-1}{6}\)
\(\left(x-1\right)\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}\right)\)=0
\(x-1=0\)
\(x=1\)
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
1. \(\frac{7x-1}{6}+2x=\frac{16-x}{5}\)
\(\Leftrightarrow5\left(7x-1\right)+60x=6\left(16-x\right)\)
\(\Leftrightarrow35x-5+60x=96-6x\)
\(\Leftrightarrow95x-5=96-6x\)
\(\Leftrightarrow95x+6x=96+5\)
\(\Leftrightarrow101x=101\)
\(\Leftrightarrow x=1\)
2. \(\frac{10x+3}{12}=1+\frac{6+8x}{9}\)
\(\Leftrightarrow3\left(10x+3\right)=36+4\left(6+8x\right)\)
\(\Leftrightarrow30x+9=36+24+32x\)
\(\Leftrightarrow30x+9=32x+60\)
\(\Leftrightarrow30x-32x=60-9\)
\(\Leftrightarrow-2x=51\)
\(\Leftrightarrow x=-\frac{51}{2}\)
3. \(\frac{8x-3}{4}-\frac{3x-2}{2}=\frac{2x-1}{2}+\frac{x+3}{4}\)
\(\Leftrightarrow8x-3-2\left(3x-2\right)=2\left(2x-1\right)+x+3\)
\(\Leftrightarrow8x-3-6x+4=4x-2+x+3\)
\(\Leftrightarrow2x+1=5x+1\)
\(\Leftrightarrow2x=5x\)
\(\Leftrightarrow x=0\)
4) \(\frac{3\left(3-x\right)}{8}+\frac{2\left(5-x\right)}{3}=\frac{1-x}{2}-2\)
=> \(\frac{9-3x}{8}+\frac{10-2x}{3}=\frac{1-x}{2}-\frac{2}{1}\)
=> \(\frac{3\left(9-3x\right)}{24}+\frac{8\left(10-2x\right)}{24}=\frac{12\left(1-x\right)}{24}-\frac{48}{24}\)
=> \(\frac{27-9x}{24}+\frac{80-16x}{24}=\frac{12-12x}{24}-\frac{48}{24}\)
=> \(\frac{27-9x+80-16x}{24}=\frac{12-12x-48}{24}\)
=> 27 - 9x + 80 - 16x = 12 - 12x - 48
=> 27 - 9x + 80 - 16x - 12 + 12x + 48 = 0
=> (27 + 80 - 12 + 48) + (-9x - 16x + 12x) = 0
=> 143 - 13x = 0
=> 13x = 143
=> x = 11
5) \(\frac{2\left(x-3\right)}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{2x-6}{7}+\frac{x-5}{3}-\frac{13x+4}{21}=0\)
=> \(\frac{3\left(2x-6\right)}{21}+\frac{7\left(x-5\right)}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18}{21}+\frac{7x-35}{21}-\frac{13x+4}{21}=0\)
=> \(\frac{6x-18+7x-35-13x-4}{21}=0\)
=> 6x - 18 + 7x - 35 - 13x - 4 = 0
=> (6x + 7x - 13x) + (-18 - 35 - 4) = 0
=> -57 = 0(vô nghiệm)
6) \(\frac{6x+5}{2}-\left(2x+\frac{2x+1}{2}\right)=\frac{10x+3}{4}\)
=> \(\frac{6x+5}{2}-\frac{10x+3}{4}=2x+\frac{2x+1}{2}\)
=> \(\frac{2\left(6x+5\right)}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{2\left(2x+1\right)}{4}\)
=> \(\frac{12x+10}{4}-\frac{10x+3}{4}=\frac{8x}{4}+\frac{4x+2}{4}\)
=> \(\frac{12x+10-\left(10x+3\right)}{4}=\frac{8x+4x+2}{4}\)
=> \(\frac{12x+10-10x-3}{4}=\frac{12x+2}{4}\)
=> \(12x+10-10x-3=12x+2\)
=> \(2x+10-3=12x+2\)
=> 2x + 10 - 3 - 12x - 2 = 0
=> (2x - 12x) + (10 - 3 - 2) = 0
=> -10x + 5 = 0
=> -10x = -5
=> x = 1/2
7) \(\frac{2x-1}{5}-\frac{x-2}{3}-\frac{x+7}{15}=0\)
=> \(\frac{3\left(2x-1\right)}{15}-\frac{5\left(x-2\right)}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3}{15}-\frac{5x-10}{15}-\frac{x+7}{15}=0\)
=> \(\frac{6x-3-\left(5x-10\right)-\left(x+7\right)}{15}=0\)
=> 6x - 3 - 5x + 10 - x - 7 = 0
=> (6x - 5x - x) + (-3 + 10 - 7) = 0
=> 0x + 0 = 0
=> 0x = 0
=> x tùy ý
Bài 8 tự làm nhé
\(\frac{x^2+5}{25-x^2}=\frac{3}{x+5}+\frac{x}{x-5}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3}{5+x}-\frac{x}{5-x}\)
\(\Leftrightarrow\frac{x^2+5}{\left(5-x\right)\left(5+x\right)}=\frac{3\left(5-x\right)-x\left(5+x\right)}{\left(5-x\right)\left(5+x\right)}\)
\(\Rightarrow x^2+5=3\left(5-x\right)-x\left(5+x\right)\)
\(\Leftrightarrow x^2+5=15-3x-5x-x^2\)
\(\Leftrightarrow15-3x-5x-x^2-x^2-5=0\)
\(\Leftrightarrow10-8x-2x^2=0\)
\(\Leftrightarrow2x^2+8x-10=0\)
\(\Leftrightarrow2\left(x^2+4x-5\right)=0\)
\(\Leftrightarrow2\left(x^2+5x-x-5\right)=0\)
\(\Leftrightarrow x^2-x+5x-5=0\)
\(\Leftrightarrow x\left(x-1\right)+5\left(x-1\right)=0\Leftrightarrow\left(x-1\right)\left(x+5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x+5=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=1\\x=-5\end{cases}}}\)
⇔ 5(5x + 2) – 10(8x – 1) = 6(4x + 2) – 150
⇔ 25x + 10 – 80x + 10 = 24x + 12 – 150
⇔ 20 – 55x = 24x – 138
⇔ -55x – 24x = -138 – 20
⇔ -79x = -158
⇔ x = 2.
Vậy phương trình có tập nghiệm S = {2}.